Are you aware of the fact that in
{1}
{1, 2}
{1, 2, 3}
...
{1, 2, 3, ..., n}
...
up to every n infinitely many natural numbers of the whole set
{1, 2, 3, ...}
are missing? Infinitely many of them will never be mentioned
individually. They are dark.
On 2025-05-17 15:00:33 +0000, WM said:
Are you aware of the fact that in
{1}
{1, 2}
{1, 2, 3}
...
{1, 2, 3, ..., n}
...
up to every n infinitely many natural numbers of the whole set
{1, 2, 3, ...}
are missing? Infinitely many of them will never be mentioned
individually. They are dark.
For example, if we pick 5 for n we have
{1}
{1, 2}
{1, 2, 3}
{1, 2, 3, 4}
{1, 2, 3, 4, 5}
then 6 and infinitely many other numbers are missing. So numbers
6, and 7 are dark as are ingfinitely many other numbers.
On 05/18/2025 05:20 AM, WM wrote:
On 18.05.2025 12:30, Mikko wrote:Nein, das ist schlecht.
On 2025-05-17 15:00:33 +0000, WM said:Maybe for a 3-year old child. Doves can count to 7. Earthworms may fail
Are you aware of the fact that in
{1}
{1, 2}
{1, 2, 3}
...
{1, 2, 3, ..., n}
...
up to every n infinitely many natural numbers of the whole set
{1, 2, 3, ...}
are missing? Infinitely many of them will never be mentioned
individually. They are dark.
For example, if we pick 5 for n we have
{1}
{1, 2}
{1, 2, 3}
{1, 2, 3, 4}
{1, 2, 3, 4, 5}
then 6 and infinitely many other numbers are missing. So numbers
6, and 7 are dark as are ingfinitely many other numbers.
at 1 already. It depends on the system. But important is that no system
can get over the infinite gap of dark numbers.
Are you aware of the fact that in
{1}
{1, 2}
{1, 2, 3}
...
{1, 2, 3, ..., n}
...
up to every n infinitely many natural numbers of the whole set
{1, 2, 3, ...}
are missing? Infinitely many of them will never be mentioned
individually. They are dark.
On 18.05.2025 12:30, Mikko wrote:
On 2025-05-17 15:00:33 +0000, WM said:
Are you aware of the fact that in
{1}
{1, 2}
{1, 2, 3}
...
{1, 2, 3, ..., n}
...
up to every n infinitely many natural numbers of the whole set
{1, 2, 3, ...}
are missing? Infinitely many of them will never be mentioned
individually. They are dark.
For example, if we pick 5 for n we have
{1}
{1, 2}
{1, 2, 3}
{1, 2, 3, 4}
{1, 2, 3, 4, 5}
then 6 and infinitely many other numbers are missing. So numbers
6, and 7 are dark as are ingfinitely many other numbers.
Maybe for a 3-year old child. Doves can count to 7. Earthworms may
fail at 1 already.
It depends on the system. But important is that no system can get over
the infinite gap of dark numbers.
On 18.05.2025 16:36, Ross Finlayson wrote:
On 05/18/2025 05:20 AM, WM wrote:
On 18.05.2025 12:30, Mikko wrote:Nein, das ist schlecht.
On 2025-05-17 15:00:33 +0000, WM said:Maybe for a 3-year old child. Doves can count to 7. Earthworms may fail
Are you aware of the fact that in
{1}
{1, 2}
{1, 2, 3}
...
{1, 2, 3, ..., n}
...
up to every n infinitely many natural numbers of the whole set
{1, 2, 3, ...}
are missing? Infinitely many of them will never be mentioned
individually. They are dark.
For example, if we pick 5 for n we have
{1}
{1, 2}
{1, 2, 3}
{1, 2, 3, 4}
{1, 2, 3, 4, 5}
then 6 and infinitely many other numbers are missing. So numbers
6, and 7 are dark as are ingfinitely many other numbers.
at 1 already. It depends on the system. But important is that no system
can get over the infinite gap of dark numbers.
Aber nicht zu ändern.
WM <[email protected]> writes:
Are you aware of the fact that in
{1}
{1, 2}
{1, 2, 3}
...
{1, 2, 3, ..., n}
...
up to every n infinitely many natural numbers of the whole set
{1, 2, 3, ...}
are missing? Infinitely many of them will never be mentioned
individually. They are dark.
Presumably you are aware that for every n in ℕ, n will be mentioned in infinitely many such sets?
They are bathed in light.
Do they still let you teach this stuff?
On 5/18/2025 5:41 PM, Ben Bacarisse wrote:
WM <[email protected]> writes:
Are you aware of the fact that in
{1}
{1, 2}
{1, 2, 3}
...
{1, 2, 3, ..., n}
...
up to every n infinitely many natural numbers of the whole set
{1, 2, 3, ...}
are missing? Infinitely many of them will never be mentioned
individually. They are dark.
Presumably you are aware that for every n in ℕ, n will be mentioned in
infinitely many such sets? They are bathed in light.
Do they still let you teach this stuff?
Clearly you do not understand that any
action that takes an infinite amount of
time will never be completed. There is
always some n in ℕ that no one ever
got around to mentioning.
On 2025-05-18 15:12:32 +0000, WM said:
On 18.05.2025 16:36, Ross Finlayson wrote:
On 05/18/2025 05:20 AM, WM wrote:
On 18.05.2025 12:30, Mikko wrote:Nein, das ist schlecht.
On 2025-05-17 15:00:33 +0000, WM said:Maybe for a 3-year old child. Doves can count to 7. Earthworms may fail >>>> at 1 already. It depends on the system. But important is that no system >>>> can get over the infinite gap of dark numbers.
Are you aware of the fact that in
{1}
{1, 2}
{1, 2, 3}
...
{1, 2, 3, ..., n}
...
up to every n infinitely many natural numbers of the whole set
{1, 2, 3, ...}
are missing? Infinitely many of them will never be mentioned
individually. They are dark.
For example, if we pick 5 for n we have
{1}
{1, 2}
{1, 2, 3}
{1, 2, 3, 4}
{1, 2, 3, 4, 5}
then 6 and infinitely many other numbers are missing. So numbers
6, and 7 are dark as are ingfinitely many other numbers.
Aber nicht zu ändern.
If it is that bad then don't even try to chage it. Just drop it into
a garbage can.
On 2025-05-18 12:20:47 +0000, WM said:
On 18.05.2025 12:30, Mikko wrote:
On 2025-05-17 15:00:33 +0000, WM said:
Are you aware of the fact that in
{1}
{1, 2}
{1, 2, 3}
...
{1, 2, 3, ..., n}
...
up to every n infinitely many natural numbers of the whole set
{1, 2, 3, ...}
are missing? Infinitely many of them will never be mentioned
individually. They are dark.
For example, if we pick 5 for n we have
{1}
{1, 2}
{1, 2, 3}
{1, 2, 3, 4}
{1, 2, 3, 4, 5}
then 6 and infinitely many other numbers are missing. So numbers
6, and 7 are dark as are ingfinitely many other numbers.
Maybe for a 3-year old child. Doves can count to 7. Earthworms may
fail at 1 already.
Many animals can differentiate quantities up to about 7. As far as
we know most of them needn't and can't count. They just see the
difference. Accurate determination of larger quantities may require
counting.
None of which is relevant to may observation that if n = 5 then your definition makes 6 dark.
It depends on the system. But important is that no system can get over
the infinite gap of dark numbers.
Why not? Cantor quite obviously gets over quite large infinities.
On 19.05.2025 00:41, Ben Bacarisse wrote:...
Do they still let you teach this stuff?
I am one of the few Professors worldwide who do teach the correct view of infinity (if actual infinity exists at all).
On 19.05.2025 15:59, Mikko wrote:
On 2025-05-18 15:12:32 +0000, WM said:It is better to know the truth than to shut your eyes.
On 18.05.2025 16:36, Ross Finlayson wrote:
On 05/18/2025 05:20 AM, WM wrote:
On 18.05.2025 12:30, Mikko wrote:Nein, das ist schlecht.
On 2025-05-17 15:00:33 +0000, WM said:Maybe for a 3-year old child. Doves can count to 7. Earthworms may fail >>>>> at 1 already. It depends on the system. But important is that no system >>>>> can get over the infinite gap of dark numbers.
Are you aware of the fact that in
{1}
{1, 2}
{1, 2, 3}
...
{1, 2, 3, ..., n}
...
up to every n infinitely many natural numbers of the whole set
{1, 2, 3, ...}
are missing? Infinitely many of them will never be mentioned
individually. They are dark.
For example, if we pick 5 for n we have
{1}
{1, 2}
{1, 2, 3}
{1, 2, 3, 4}
{1, 2, 3, 4, 5}
then 6 and infinitely many other numbers are missing. So numbers
6, and 7 are dark as are ingfinitely many other numbers.
Aber nicht zu ändern.
If it is that bad then don't even try to chage it. Just drop it into
a garbage can.
On 19.05.2025 15:57, Mikko wrote:
On 2025-05-18 12:20:47 +0000, WM said:
On 18.05.2025 12:30, Mikko wrote:
On 2025-05-17 15:00:33 +0000, WM said:
Are you aware of the fact that in
{1}
{1, 2}
{1, 2, 3}
...
{1, 2, 3, ..., n}
...
up to every n infinitely many natural numbers of the whole set
{1, 2, 3, ...}
are missing? Infinitely many of them will never be mentioned
individually. They are dark.
For example, if we pick 5 for n we have
{1}
{1, 2}
{1, 2, 3}
{1, 2, 3, 4}
{1, 2, 3, 4, 5}
then 6 and infinitely many other numbers are missing. So numbers
6, and 7 are dark as are ingfinitely many other numbers.
Maybe for a 3-year old child. Doves can count to 7. Earthworms may
fail at 1 already.
Many animals can differentiate quantities up to about 7. As far as
we know most of them needn't and can't count. They just see the
difference. Accurate determination of larger quantities may require
counting.
None of which is relevant to may observation that if n = 5 then your
definition makes 6 dark.
If you have no idea of 6, it is dark for you. I you arbitrarily stop at
5 although you know 6, 5 is not dark for you.
WM <[email protected]> writes:
On 19.05.2025 00:41, Ben Bacarisse wrote:...
Do they still let you teach this stuff?
I am one of the few Professors worldwide who do teach the correct view of
infinity (if actual infinity exists at all).
Fortunately it's an optional course (at least it was) and your college
does not offer degrees in mathematics (or has that changed?). The harm
is very much limited.
WM <[email protected]> writes:
I am one of the few Professors worldwide who do teach the correct view of
infinity (if actual infinity exists at all).
Fortunately it's an optional course (at least it was) and your college
The harm
is very much limited.
On 2025-05-20 00:50:42 +0000, Ben Bacarisse said:
WM <[email protected]> writes:
On 19.05.2025 00:41, Ben Bacarisse wrote:...
Do they still let you teach this stuff?
I am one of the few Professors worldwide who do teach the correct
view of
infinity (if actual infinity exists at all).
Fortunately it's an optional course (at least it was) and your college
does not offer degrees in mathematics (or has that changed?). The harm
is very much limited.
And further limitations come from the truth being presented more often
than the falsehood.
On 2025-05-19 18:53:43 +0000, WM said:
On 19.05.2025 15:57, Mikko wrote:
On 2025-05-18 12:20:47 +0000, WM said:
On 18.05.2025 12:30, Mikko wrote:
On 2025-05-17 15:00:33 +0000, WM said:
Are you aware of the fact that in
{1}
{1, 2}
{1, 2, 3}
...
{1, 2, 3, ..., n}
...
up to every n infinitely many natural numbers of the whole set
{1, 2, 3, ...}
are missing? Infinitely many of them will never be mentioned
individually. They are dark.
For example, if we pick 5 for n we have
{1}
{1, 2}
{1, 2, 3}
{1, 2, 3, 4}
{1, 2, 3, 4, 5}
then 6 and infinitely many other numbers are missing. So numbers
6, and 7 are dark as are ingfinitely many other numbers.
Maybe for a 3-year old child. Doves can count to 7. Earthworms may
fail at 1 already.
Many animals can differentiate quantities up to about 7. As far as
we know most of them needn't and can't count. They just see the
difference. Accurate determination of larger quantities may require
counting.
None of which is relevant to may observation that if n = 5 then your
definition makes 6 dark.
If you have no idea of 6, it is dark for you. I you arbitrarily stop
at 5 although you know 6, 5 is not dark for you.
I do have an idea on numbers greated than n. But per OP they are dark
anyway.
On 2025-05-20 00:50:42 +0000, Ben Bacarisse said:
WM <[email protected]> writes:
On 19.05.2025 00:41, Ben Bacarisse wrote:...
Fortunately it's an optional course (at least it was) and your collegeDo they still let you teach this stuff?I am one of the few Professors worldwide who do teach the correct view
of
infinity (if actual infinity exists at all).
does not offer degrees in mathematics (or has that changed?). The harm
is very much limited.
And further limitations come from the truth being presented more often
than the falsehood.
On 20.05.2025 02:50, Ben Bacarisse wrote:
WM <[email protected]> writes:
I am one of the few Professors worldwide who do teach the correct view of >>> infinity (if actual infinity exists at all).Fortunately it's an optional course (at least it was) and your college
It is not a college but the Technische Hochschule Ausgsburg (THA).
The harm
is very much limited.
It is harm only for fanatic reactionaries who disregard mathematical
proofs.
WM <[email protected]> writes:
On 20.05.2025 02:50, Ben Bacarisse wrote:
WM <[email protected]> writes:
I am one of the few Professors worldwide who do teach the correct view of >>>> infinity (if actual infinity exists at all).Fortunately it's an optional course (at least it was) and your college
It is not a college but the Technische Hochschule Ausgsburg (THA).
Why is college (or maybe technical college) not a good translation of
that term?
The harm
is very much limited.
It is harm only for fanatic reactionaries who disregard mathematical
proofs.
That was indeed the harm I was referring to.
Do you disregard this mathematical proof https://www.whitman.edu/mathematics/higher_math_online/section04.10.htmlYes, there it is believed that all numbers can be used as individuals.
?
On 20.05.2025 09:18, Mikko wrote:
On 2025-05-19 18:53:43 +0000, WM said:
On 19.05.2025 15:57, Mikko wrote:
On 2025-05-18 12:20:47 +0000, WM said:
On 18.05.2025 12:30, Mikko wrote:
On 2025-05-17 15:00:33 +0000, WM said:
Are you aware of the fact that in
{1}
{1, 2}
{1, 2, 3}
...
{1, 2, 3, ..., n}
...
up to every n infinitely many natural numbers of the whole set
{1, 2, 3, ...}
are missing? Infinitely many of them will never be mentioned
individually. They are dark.
For example, if we pick 5 for n we have
{1}
{1, 2}
{1, 2, 3}
{1, 2, 3, 4}
{1, 2, 3, 4, 5}
then 6 and infinitely many other numbers are missing. So numbers
6, and 7 are dark as are ingfinitely many other numbers.
Maybe for a 3-year old child. Doves can count to 7. Earthworms may
fail at 1 already.
Many animals can differentiate quantities up to about 7. As far as
we know most of them needn't and can't count. They just see the
difference. Accurate determination of larger quantities may require
counting.
None of which is relevant to may observation that if n = 5 then your
definition makes 6 dark.
If you have no idea of 6, it is dark for you. I you arbitrarily stop at
5 although you know 6, 5 is not dark for you.
I do have an idea on numbers greated than n. But per OP they are dark
anyway.
If you can express them so that a reader can recognize them, then they
are not / no longer dark.
On 2025-05-20 11:17:31 +0000, WM said:
On 20.05.2025 09:18, Mikko wrote:
I do have an idea on numbers greated than n. But per OP they are dark
anyway.
If you can express them so that a reader can recognize them, then they
are not / no longer dark.
OP said otherwise.
On 22.05.2025 11:10, Mikko wrote:
On 2025-05-20 11:17:31 +0000, WM said:
On 20.05.2025 09:18, Mikko wrote:
I do have an idea on numbers greated than n. But per OP they are dark
anyway.
If you can express them so that a reader can recognize them, then they
are not / no longer dark.
OP said otherwise.
OP said: "many of them will never be mentioned individually."
Do you mean that every natural number is dark until
someone mentions it but no longer?
On 21.05.2025 03:17, Ben Bacarisse wrote:
WM <[email protected]> writes:According to my dictionaries Colleges are parts of universities, but also institutions not offering degrees attended by secretaries or
On 20.05.2025 02:50, Ben Bacarisse wrote:Why is college (or maybe technical college) not a good translation of
WM <[email protected]> writes:
I am one of the few Professors worldwide who do teach the correct view of >>>>> infinity (if actual infinity exists at all).Fortunately it's an optional course (at least it was) and your college
It is not a college but the Technische Hochschule Ausgsburg (THA).
that term?
hairdressers. According to this translation the Technische Hochschule Augsburg consists of several colleges. But the faculty of general studies covers the full university of applied sciences. All students can attend my courses.
...That was indeed the harm I was referring to.The harm
is very much limited.
It is harm only for fanatic reactionaries who disregard mathematical
proofs.
Do you disregard this mathematical proofYes, ...
https://www.whitman.edu/mathematics/higher_math_online/section04.10.html
?
On 23.05.2025 09:43, Mikko wrote:
Do you mean that every natural number is dark untilEvery natural number is dark in a system that cannot represent it in
someone mentions it but no longer?
any form like writing, thinking or whatever. The pocket calculator is
limited to decimal representations below 10^100, the universe is
limited to more or less sophisticated formulas requiring less than
10^80 bit.
In every system almost all natural numbers are and remain dark - if an
actual infinity of them exists.
WM <[email protected]> writes:
On 21.05.2025 03:17, Ben Bacarisse wrote:
WM <[email protected]> writes:According to my dictionaries Colleges are parts of universities, but also
On 20.05.2025 02:50, Ben Bacarisse wrote:Why is college (or maybe technical college) not a good translation of
WM <[email protected]> writes:It is not a college but the Technische Hochschule Ausgsburg (THA).
I am one of the few Professors worldwide who do teach the correct view ofFortunately it's an optional course (at least it was) and your college >>>>
infinity (if actual infinity exists at all).
that term?
institutions not offering degrees attended by secretaries or
hairdressers. According to this translation the Technische Hochschule
Augsburg consists of several colleges. But the faculty of general studies
covers the full university of applied sciences. All students can attend my >> courses.
Your dictionary is very odd. In common English usage some colleges are
parts of universities and some are not. Some award degrees and some
don't. Some offer PhD studies and some don't. I don't know what hairdressers have to so with it.
WM <[email protected]> writes:
On 21.05.2025 03:17, Ben Bacarisse wrote:
WM <[email protected]> writes:According to my dictionaries Colleges are parts of universities, but also
On 20.05.2025 02:50, Ben Bacarisse wrote:Why is college (or maybe technical college) not a good translation of
WM <[email protected]> writes:It is not a college but the Technische Hochschule Ausgsburg (THA).
I am one of the few Professors worldwide who do teach the correct view ofFortunately it's an optional course (at least it was) and your college >>>>
infinity (if actual infinity exists at all).
that term?
institutions not offering degrees attended by secretaries or
hairdressers. According to this translation the Technische Hochschule
Augsburg consists of several colleges. But the faculty of general studies
covers the full university of applied sciences. All students can attend my >> courses.
Your dictionary is very odd.
Do you disregard this mathematical proofYes, ...
https://www.whitman.edu/mathematics/higher_math_online/section04.10.html >>> ?
Good to know.
On 2025-05-23 08:31:27 +0000, WM said:
On 23.05.2025 09:43, Mikko wrote:
Do you mean that every natural number is dark untilEvery natural number is dark in a system that cannot represent it in
someone mentions it but no longer?
any form like writing, thinking or whatever. The pocket calculator is
limited to decimal representations below 10^100, the universe is
limited to more or less sophisticated formulas requiring less than
10^80 bit.
In every system almost all natural numbers are and remain dark - if an
actual infinity of them exists.
That is not a useful concept as it is not possible to know wich numbers are presentable in future sysems and which will be actually presented.
At the end of the web page https://mlevanto.github.io/lauseke.html there
is an arithmetic expression that evaluates to a 65600 digit number.
Although
the value of the expression is not written there I used that digit sequence (and several others, some even longer) when I wrote the page.
We don't know whether our universe is finite or infinite. or wheter it
can be fully described with finite information.
On 2025-05-23 13:21:21 +0000, Ben Bacarisse said:
WM <[email protected]> writes:
On 21.05.2025 03:17, Ben Bacarisse wrote:Your dictionary is very odd. In common English usage some colleges are
WM <[email protected]> writes:According to my dictionaries Colleges are parts of universities, but also >>> institutions not offering degrees attended by secretaries or
On 20.05.2025 02:50, Ben Bacarisse wrote:Why is college (or maybe technical college) not a good translation of
WM <[email protected]> writes:
I am one of the few Professors worldwide who do teach the correct view ofFortunately it's an optional course (at least it was) and your college >>>>> It is not a college but the Technische Hochschule Ausgsburg (THA).
infinity (if actual infinity exists at all).
that term?
hairdressers. According to this translation the Technische Hochschule
Augsburg consists of several colleges. But the faculty of general studies >>> covers the full university of applied sciences. All students can attend my >>> courses.
parts of universities and some are not. Some award degrees and some
don't. Some offer PhD studies and some don't. I don't know what
hairdressers have to so with it.
The point is that as a college can be significanlty less ambitious as
a superior school and in particular the Tehcnische Hochschule Augsburg
the word "college" is not a good translation of "hochschule".
On 23.05.2025 15:21, Ben Bacarisse wrote:...
WM <[email protected]> writes:
On 21.05.2025 03:17, Ben Bacarisse wrote:
Your dictionary is very odd.Why is college (or maybe technical college) not a good translation ofAccording to my dictionaries Colleges are parts of universities, but also >>> institutions not offering degrees attended by secretaries or
that term?
hairdressers. According to this translation the Technische Hochschule
Augsburg consists of several colleges. But the faculty of general studies >>> covers the full university of applied sciences. All students can attend my >>> courses.
Langenscheidt Collins e-Großwörterbuch Englisch 5.0
Good to know.Do you disregard this mathematical proofYes, ...
https://www.whitman.edu/mathematics/higher_math_online/section04.10.html >>>> ?
{1} has infinitely many (ℵo) successors.
If {1, 2, 3, ..., n} has infinitely many (ℵo) successors, then {1, 2, 3, ..., n, n+1} has infinitely many (ℵo) successors. For every n that can be defined.
Do you disregard this mathematical proof?
WM <[email protected]> writes:
On 23.05.2025 15:21, Ben Bacarisse wrote:...
WM <[email protected]> writes:
On 21.05.2025 03:17, Ben Bacarisse wrote:
Your dictionary is very odd.Why is college (or maybe technical college) not a good translation of >>>>> that term?According to my dictionaries Colleges are parts of universities, but also >>>> institutions not offering degrees attended by secretaries or
hairdressers. According to this translation the Technische Hochschule
Augsburg consists of several colleges. But the faculty of general studies >>>> covers the full university of applied sciences. All students can attend my >>>> courses.
Langenscheidt Collins e-Großwörterbuch Englisch 5.0
Can you quote the text for me? I'd like to see what it says about hairdressers and secretaries. I found that part very odd.
Good to know.Do you disregard this mathematical proofYes, ...
https://www.whitman.edu/mathematics/higher_math_online/section04.10.html >>>>> ?
{1} has infinitely many (ℵo) successors.
If {1, 2, 3, ..., n} has infinitely many (ℵo) successors, then {1, 2, 3, >> ..., n, n+1} has infinitely many (ℵo) successors. For every n that can be >> defined.
Do you disregard this mathematical proof?
I will consider it if you can write it correctly.
Mikko <[email protected]> writes:
On 2025-05-23 13:21:21 +0000, Ben Bacarisse said:
WM <[email protected]> writes:
On 21.05.2025 03:17, Ben Bacarisse wrote:Your dictionary is very odd. In common English usage some colleges are
WM <[email protected]> writes:According to my dictionaries Colleges are parts of universities, but also >>>> institutions not offering degrees attended by secretaries or
On 20.05.2025 02:50, Ben Bacarisse wrote:Why is college (or maybe technical college) not a good translation of >>>>> that term?
WM <[email protected]> writes:
I am one of the few Professors worldwide who do teach the correct view ofFortunately it's an optional course (at least it was) and your college >>>>>> It is not a college but the Technische Hochschule Ausgsburg (THA).
infinity (if actual infinity exists at all).
hairdressers. According to this translation the Technische Hochschule
Augsburg consists of several colleges. But the faculty of general studies >>>> covers the full university of applied sciences. All students can attend my >>>> courses.
parts of universities and some are not. Some award degrees and some
don't. Some offer PhD studies and some don't. I don't know what
hairdressers have to so with it.
The point is that as a college can be significanlty less ambitious as
a superior school and in particular the Tehcnische Hochschule Augsburg
the word "college" is not a good translation of "hochschule".
A college can be very ambitious.
On 24.05.2025 10:13, Mikko wrote:
On 2025-05-23 08:31:27 +0000, WM said:
On 23.05.2025 09:43, Mikko wrote:
Do you mean that every natural number is dark untilEvery natural number is dark in a system that cannot represent it in
someone mentions it but no longer?
any form like writing, thinking or whatever. The pocket calculator is
limited to decimal representations below 10^100, the universe is
limited to more or less sophisticated formulas requiring less than
10^80 bit.
In every system almost all natural numbers are and remain dark - if an
actual infinity of them exists.
That is not a useful concept as it is not possible to know wich numbers are >> presentable in future sysems and which will be actually presented.
But it is fact.
Further it need not be deteremined exactly what can be presented. It is sufficient, for many purposes, to know that most numbers cannot be
presented
At the end of the web page https://mlevanto.github.io/lauseke.html there
is an arithmetic expression that evaluates to a 65600 digit number. Although >> the value of the expression is not written there I used that digit sequence >> (and several others, some even longer) when I wrote the page.
The numbers that can be used belong to a potentially infinite set.
There may be much longer sequences. But most natural numbers remain
dark - if ℕ is actually infinite.
On 2025-05-24 11:29:53 +0000, WM said:
On 24.05.2025 10:13, Mikko wrote:
On 2025-05-23 08:31:27 +0000, WM said:
On 23.05.2025 09:43, Mikko wrote:
Do you mean that every natural number is dark untilEvery natural number is dark in a system that cannot represent it in
someone mentions it but no longer?
any form like writing, thinking or whatever. The pocket calculator
is limited to decimal representations below 10^100, the universe is
limited to more or less sophisticated formulas requiring less than
10^80 bit.
In every system almost all natural numbers are and remain dark - if
an actual infinity of them exists.
That is not a useful concept as it is not possible to know wich
numbers are
presentable in future sysems and which will be actually presented.
But it is fact.
But not a mathematical fact.
Further it need not be determined exactly what can be presented. It
is sufficient, for many purposes, to know that most numbers cannot be
presented
For many porposes it sufficient to know that most numbers needn't be presented.
At the end of the web page https://mlevanto.github.io/lauseke.html there >>> is an arithmetic expression that evaluates to a 65600 digit number.
Although
the value of the expression is not written there I used that digit
sequence
(and several others, some even longer) when I wrote the page.
The numbers that can be used belong to a potentially infinite set.
There may be much longer sequences. But most natural numbers remain
dark - if ℕ is actually infinite.
The set of natural numbers is actually infinite. There is nothing
potential in mathematics: what is is
and that is all that can be.
On 25.05.2025 03:27, Ben Bacarisse wrote:
WM <[email protected]> writes:
On 23.05.2025 15:21, Ben Bacarisse wrote:...
WM <[email protected]> writes:
On 21.05.2025 03:17, Ben Bacarisse wrote:
Can you quote the text for me? I'd like to see what it says aboutYour dictionary is very odd.Why is college (or maybe technical college) not a good translation of >>>>>> that term?According to my dictionaries Colleges are parts of universities, but also >>>>> institutions not offering degrees attended by secretaries or
hairdressers. According to this translation the Technische Hochschule >>>>> Augsburg consists of several colleges. But the faculty of general studies >>>>> covers the full university of applied sciences. All students can attend my
courses.
Langenscheidt Collins e-Großwörterbuch Englisch 5.0
hairdressers and secretaries. I found that part very odd.
With pleasure:
COLLEGE
College ist ein allgemeiner Oberbegriff für höhere Bildungsinstitute. In Großbritannien kann er sich auf Einrichtungen beziehen, in denen man in einzelnen Fachbereichen, wie Kunst oder Musik, einen Hochschulabschluss erwerben kann, aber ebenso auf Schulen ohne weiteren Abschluss, z. B. für Sekretärinnen oder Friseure. Einige britische Universitäten, darunter Oxford und Cambridge, setzen sich aus Colleges zusammen. In diesen
collegiate universities sorgen die Colleges für die Unterbringung und Ausbildung der Studenten, auch wenn die Universität dann die Abschlüsse verleiht. Zu den bekanntesten Colleges zählen wohl das Kings College in Cambridge und das Magdalen College in Oxford.
In den USA werden die Universitäten in Verwaltungseinheiten unterteilt, die als Colleges bezeichnet werden: zum Beispiel das College of Arts and
Sciences oder das College of Medicine. Graduate schools, die normalerweise Teil einer Universität sind, bieten auf dem bachelor aufbauende Studiengänge zur weiteren Spezialisierung an. Junior oder community
colleges sind Institute, an denen man nach zweijähriger Studienzeit einen berufsbezogenen Abschluss machen kann; sie bieten auch Weiterbildungen für Berufstätige an DEGREE, Oxbridge
© Langenscheidt KG, Berlin und München und HarperCollins Publishers Ltd
I will consider it if you can write it correctly.Good to know.Do you disregard this mathematical proofYes, ...
https://www.whitman.edu/mathematics/higher_math_online/section04.10.html >>>>>> ?
{1} has infinitely many (ℵo) successors.
If {1, 2, 3, ..., n} has infinitely many (ℵo) successors, then {1, 2, 3, >>> ..., n, n+1} has infinitely many (ℵo) successors. For every n that can be >>> defined.
Do you disregard this mathematical proof?
With pleasure:
For every n ∈ ℕ that can be defined, i.e., ∀n ∈ ℕ_def:
On 2025-05-25 01:09:59 +0000, Ben Bacarisse said:
Mikko <[email protected]> writes:
On 2025-05-23 13:21:21 +0000, Ben Bacarisse said:A college can be very ambitious.
WM <[email protected]> writes:The point is that as a college can be significanlty less ambitious as
On 21.05.2025 03:17, Ben Bacarisse wrote:Your dictionary is very odd. In common English usage some colleges are >>>> parts of universities and some are not. Some award degrees and some
WM <[email protected]> writes:According to my dictionaries Colleges are parts of universities, but also >>>>> institutions not offering degrees attended by secretaries or
On 20.05.2025 02:50, Ben Bacarisse wrote:
WM <[email protected]> writes:
I am one of the few Professors worldwide who do teach theFortunately it's an optional course (at least it was) and your college >>>>>>> It is not a college but the Technische Hochschule Ausgsburg (THA). >>>>>> Why is college (or maybe technical college) not a good translation of >>>>>> that term?
correct view of
infinity (if actual infinity exists at all).
hairdressers. According to this translation the Technische Hochschule >>>>> Augsburg consists of several colleges. But the faculty of general studies >>>>> covers the full university of applied sciences. All students can attend my
courses.
don't. Some offer PhD studies and some don't. I don't know what
hairdressers have to so with it.
a superior school and in particular the Tehcnische Hochschule Augsburg
the word "college" is not a good translation of "hochschule".
That does not mean that every college is.
On 25.05.2025 12:42, Mikko wrote:
On 2025-05-24 11:29:53 +0000, WM said:
On 24.05.2025 10:13, Mikko wrote:
On 2025-05-23 08:31:27 +0000, WM said:
On 23.05.2025 09:43, Mikko wrote:
Do you mean that every natural number is dark untilEvery natural number is dark in a system that cannot represent it in >>>>> any form like writing, thinking or whatever. The pocket calculator is >>>>> limited to decimal representations below 10^100, the universe is
someone mentions it but no longer?
limited to more or less sophisticated formulas requiring less than
10^80 bit.
In every system almost all natural numbers are and remain dark - if an >>>>> actual infinity of them exists.
That is not a useful concept as it is not possible to know wich numbers are
presentable in future sysems and which will be actually presented.
But it is fact.
But not a mathematical fact.
That depends on the definition of mathematics.
On 26.05.2025 02:52, Ben Bacarisse wrote:
WM <[email protected]> writes:
With pleasure:
For every n ∈ ℕ that can be defined, i.e., ∀n ∈ ℕ_def:
I can't comment on an argument that is based on a set you have not
defined.
Can you understand my proof by induction?
The resulting set is ℕ_def. (According to set theory however it is not
a set but a potentially infinity collection.)
WM <[email protected]> writes:
With pleasure:
For every n ∈ ℕ that can be defined, i.e., ∀n ∈ ℕ_def:
I can't comment on an argument that is based on a set you have not
defined.
Your textbook defies N
Mikko <[email protected]> writes:
On 2025-05-25 01:09:59 +0000, Ben Bacarisse said:
Mikko <[email protected]> writes:
On 2025-05-23 13:21:21 +0000, Ben Bacarisse said:A college can be very ambitious.
WM <[email protected]> writes:The point is that as a college can be significanlty less ambitious as
On 21.05.2025 03:17, Ben Bacarisse wrote:Your dictionary is very odd. In common English usage some colleges are >>>>> parts of universities and some are not. Some award degrees and some >>>>> don't. Some offer PhD studies and some don't. I don't know what
WM <[email protected]> writes:According to my dictionaries Colleges are parts of universities, but also
On 20.05.2025 02:50, Ben Bacarisse wrote:
WM <[email protected]> writes:It is not a college but the Technische Hochschule Ausgsburg (THA). >>>>>>> Why is college (or maybe technical college) not a good translation of >>>>>>> that term?
I am one of the few Professors worldwide who do teach theFortunately it's an optional course (at least it was) and your college
correct view of
infinity (if actual infinity exists at all).
institutions not offering degrees attended by secretaries or
hairdressers. According to this translation the Technische Hochschule >>>>>> Augsburg consists of several colleges. But the faculty of general studies
covers the full university of applied sciences. All students can attend my
courses.
hairdressers have to so with it.
a superior school and in particular the Tehcnische Hochschule Augsburg >>>> the word "college" is not a good translation of "hochschule".
That does not mean that every college is.
I don't know what point you are making anymore. For someone like me who
does not know exactly what a "hochschule" is, "college" is a good
translation as it is a general term. It may not be specific enough for
some purposes but it can't be wrong.
On 2025-05-25 11:38:23 +0000, WM said:
On 25.05.2025 12:42, Mikko wrote:
On 2025-05-24 11:29:53 +0000, WM said:
On 24.05.2025 10:13, Mikko wrote:
On 2025-05-23 08:31:27 +0000, WM said:
On 23.05.2025 09:43, Mikko wrote:
Do you mean that every natural number is dark untilEvery natural number is dark in a system that cannot represent it
someone mentions it but no longer?
in any form like writing, thinking or whatever. The pocket
calculator is limited to decimal representations below 10^100, the >>>>>> universe is limited to more or less sophisticated formulas
requiring less than 10^80 bit.
In every system almost all natural numbers are and remain dark -
if an actual infinity of them exists.
That is not a useful concept as it is not possible to know wich
numbers are
presentable in future sysems and which will be actually presented.
But it is fact.
But not a mathematical fact.
That depends on the definition of mathematics.
An exact definition of mathematics is not needed except by some
philosophers.
Even whithout a definition there understood meaning of the word varies only
a little. By the usual understand ot the word of a fact about people or
other
real world beings is not a mathematical fact.
On 2025-05-26 10:17:27 +0000, WM said:
On 26.05.2025 02:52, Ben Bacarisse wrote:
WM <[email protected]> writes:
With pleasure:
For every n ∈ ℕ that can be defined, i.e., ∀n ∈ ℕ_def:
I can't comment on an argument that is based on a set you have not
defined.
Can you understand my proof by induction?
The resulting set is ℕ_def. (According to set theory however it is not
a set but a potentially infinity collection.)
A proof by inductin does not define. It proves.
For example: if we assume that
0 ∈ ℕ_def
and that
∀n (n ∈ ℕ_def → (n + 1) ∈ ℕ_def)
then we can apply induction and prove that
ℕ ⊆ ℕ_def .
Your textbook defies N (incorrectly)
On 26.05.2025 02:52, Ben Bacarisse wrote:
WM <[email protected]> writes:
With pleasure:I can't comment on an argument that is based on a set you have not
For every n ∈ ℕ that can be defined, i.e., ∀n ∈ ℕ_def:
defined.
Can you understand my proof by induction?
The resulting set is ℕ_def. (According to set theory however it is not a set but a potentially infinity collection.)
Your textbook defies N
It defines ℕ_def.
Your textbook defies N (incorrectly)
My textbook defines the classical natural numbers, ℕ, meanwhile more precisely called ℕ_def, correctly.
1 ∈ M (4.1)
n ∈ M ⇒ (n + 1) ∈ M (4.2)
If M satisfies (4.1) and (4.2), then ℕ ⊆ M.
On 26.05.2025 02:52, Ben Bacarisse wrote:
Your textbook defies N (incorrectly)
My textbook defines...
On 2025-05-26 00:56:22 +0000, Ben Bacarisse said:
Mikko <[email protected]> writes:
On 2025-05-25 01:09:59 +0000, Ben Bacarisse said:I don't know what point you are making anymore. For someone like me who
Mikko <[email protected]> writes:That does not mean that every college is.
On 2025-05-23 13:21:21 +0000, Ben Bacarisse said:A college can be very ambitious.
WM <[email protected]> writes:The point is that as a college can be significanlty less ambitious as >>>>> a superior school and in particular the Tehcnische Hochschule Augsburg >>>>> the word "college" is not a good translation of "hochschule".
On 21.05.2025 03:17, Ben Bacarisse wrote:Your dictionary is very odd. In common English usage some colleges are >>>>>> parts of universities and some are not. Some award degrees and some >>>>>> don't. Some offer PhD studies and some don't. I don't know what
WM <[email protected]> writes:According to my dictionaries Colleges are parts of universities, but also
On 20.05.2025 02:50, Ben Bacarisse wrote:
WM <[email protected]> writes:It is not a college but the Technische Hochschule Ausgsburg (THA). >>>>>>>> Why is college (or maybe technical college) not a good translation of >>>>>>>> that term?
I am one of the few Professors worldwide who do teach the >>>>>>>>>>> correct view ofFortunately it's an optional course (at least it was) and your college
infinity (if actual infinity exists at all).
institutions not offering degrees attended by secretaries or
hairdressers. According to this translation the Technische Hochschule >>>>>>> Augsburg consists of several colleges. But the faculty of general studies
covers the full university of applied sciences. All students
can attend my
courses.
hairdressers have to so with it.
does not know exactly what a "hochschule" is, "college" is a good
translation as it is a general term. It may not be specific enough for
some purposes but it can't be wrong.
A translation that is good enough for you is not good enough for
everybody.
The term "college" does not imply a level of ambition any grater than that
of the least ambitious college.
The term "hochschule" does.
On 26.05.2025 12:26, Mikko wrote:
On 2025-05-25 11:38:23 +0000, WM said:
On 25.05.2025 12:42, Mikko wrote:
On 2025-05-24 11:29:53 +0000, WM said:
On 24.05.2025 10:13, Mikko wrote:
On 2025-05-23 08:31:27 +0000, WM said:
On 23.05.2025 09:43, Mikko wrote:
Do you mean that every natural number is dark untilEvery natural number is dark in a system that cannot represent it in >>>>>>> any form like writing, thinking or whatever. The pocket calculator is >>>>>>> limited to decimal representations below 10^100, the universe is >>>>>>> limited to more or less sophisticated formulas requiring less than >>>>>>> 10^80 bit.
someone mentions it but no longer?
In every system almost all natural numbers are and remain dark - if an >>>>>>> actual infinity of them exists.
That is not a useful concept as it is not possible to know wich numbers are
presentable in future sysems and which will be actually presented.
But it is fact.
But not a mathematical fact.
That depends on the definition of mathematics.
An exact definition of mathematics is not needed except by some philosophers.
Even whithout a definition there understood meaning of the word varies only >> a little. By the usual understand ot the word of a fact about people or other
real world beings is not a mathematical fact.
That is the common but mistaken view.
Without tools of the real world no mathematics is possible at all.
Therefore mathematics is limited by the power of these tools.
But even pure mathematics proves that most natural numbers will never
be definable:
{1} has infinitely many (ℵo) successors.
If {1, 2, 3, ..., n} has infinitely many (ℵo) successors, then {1, 2,
3, ..., n, n+1} has infinitely many (ℵo) successors, for every n that
can be defined.
On 26.05.2025 12:44, Mikko wrote:
On 2025-05-26 10:17:27 +0000, WM said:
On 26.05.2025 02:52, Ben Bacarisse wrote:
WM <[email protected]> writes:
With pleasure:
For every n ∈ ℕ that can be defined, i.e., ∀n ∈ ℕ_def:
I can't comment on an argument that is based on a set you have not
defined.
Can you understand my proof by induction?
The resulting set is ℕ_def. (According to set theory however it is not >>> a set but a potentially infinity collection.)
A proof by inductin does not define. It proves.
Here it proves that the natural numbers accessible by induction are not
all natural numbers.
For example: if we assume that
0 ∈ ℕ_def
and that
∀n (n ∈ ℕ_def → (n + 1) ∈ ℕ_def)
then we can apply induction and prove that
ℕ ⊆ ℕ_def .
Then you get a contradiction because by induction the natural numbers accessible by induction are not all natural numbers.
{1} has infinitely many (ℵo) successors.
If {1, 2, 3, ..., n} has infinitely many (ℵo) successors, then {1, 2,
3, ..., n, n+1} has infinitely many (ℵo) successors. For every n that
can be defined.
Do you accept this proof?
Mikko <[email protected]> writes:
On 2025-05-26 00:56:22 +0000, Ben Bacarisse said:
Mikko <[email protected]> writes:
On 2025-05-25 01:09:59 +0000, Ben Bacarisse said:I don't know what point you are making anymore. For someone like me who >>> does not know exactly what a "hochschule" is, "college" is a good
Mikko <[email protected]> writes:That does not mean that every college is.
On 2025-05-23 13:21:21 +0000, Ben Bacarisse said:A college can be very ambitious.
WM <[email protected]> writes:The point is that as a college can be significanlty less ambitious as >>>>>> a superior school and in particular the Tehcnische Hochschule Augsburg >>>>>> the word "college" is not a good translation of "hochschule".
On 21.05.2025 03:17, Ben Bacarisse wrote:Your dictionary is very odd. In common English usage some colleges are >>>>>>> parts of universities and some are not. Some award degrees and some >>>>>>> don't. Some offer PhD studies and some don't. I don't know what >>>>>>> hairdressers have to so with it.
WM <[email protected]> writes:According to my dictionaries Colleges are parts of universities, but also
On 20.05.2025 02:50, Ben Bacarisse wrote:
WM <[email protected]> writes:It is not a college but the Technische Hochschule Ausgsburg (THA). >>>>>>>>> Why is college (or maybe technical college) not a good translation of >>>>>>>>> that term?
I am one of the few Professors worldwide who do teach the >>>>>>>>>>>> correct view ofFortunately it's an optional course (at least it was) and your college
infinity (if actual infinity exists at all).
institutions not offering degrees attended by secretaries or
hairdressers. According to this translation the Technische Hochschule >>>>>>>> Augsburg consists of several colleges. But the faculty of general studies
covers the full university of applied sciences. All students
can attend my
courses.
translation as it is a general term. It may not be specific enough for
some purposes but it can't be wrong.
A translation that is good enough for you is not good enough for
everybody.
Absolutely agree.
The term "college" does not imply a level of ambition any grater than that >> of the least ambitious college.
That seems odd to me. I don't think that's how the term is interpreted
in many English speaking countries. Maybe the Colleges of Oxford and Cambridge have influenced how people here react to the term?
The term "hochschule" does.
But it's an almost unknown term in the UK so it conveys almost nothing
to most English speakers.
Mind you, all I know if the ambition of the institution in question is
that it allows nonsense to be taught, albeit as a entirely optional
course. That particular hochschule does not seem to be taking academic integrity very seriously.
WM <[email protected]> writes:
On 26.05.2025 02:52, Ben Bacarisse wrote:
WM <[email protected]> writes:
With pleasure:I can't comment on an argument that is based on a set you have not
For every n ∈ ℕ that can be defined, i.e., ∀n ∈ ℕ_def:
defined.
Can you understand my proof by induction?
Not without knowing what the set N_def is, since the argument starts
"For all n in N_def".
The resulting set is ℕ_def. (According to set theory however it is not a >> set but a potentially infinity collection.)
So you are not asking me to verify a proof at all but rather to accept a definition?
One that starts from claims about the thing being defined?
Your textbook defies N
It defines ℕ_def.
It claims to define N.
It's very poor form to tell students you are
defining N when you are not.
In another reply (please don't split threads -- you may have time to
discuss this stuff endlessly but I don't) you say:
Your textbook defies N (incorrectly)
My textbook defines the classical natural numbers, ℕ, meanwhile more
precisely called ℕ_def, correctly.
So when you write N and N_def you are referring to the same thing?
On 2025-05-26 13:38:00 +0000, WM said:
No alternative view is known to be
better.
But even pure mathematics proves that most natural numbers will never
be definable:
{1} has infinitely many (ℵo) successors.
(2) there are infinitely many (ℵo) possible definitions.
If {1, 2, 3, ..., n} has infinitely many (ℵo) successors, then {1, 2,
3, ..., n, n+1} has infinitely many (ℵo) successors, for every n that
can be defined.
You can't formulate that as a logically or mathematically valid proof.
On 2025-05-26 23:21:27 +0000, Ben Bacarisse said:
Mind you, all I know if the ambition of the institution in question is
that it allows nonsense to be taught, albeit as a entirely optional
course. That particular hochschule does not seem to be taking academic
integrity very seriously.
Maybe the idea is that an exposure to nonsense helps students to learn
to identify nonsense when they see it.
On 2025-05-26 13:44:06 +0000, WM said:
{1} has infinitely many (ℵo) successors.
If {1, 2, 3, ..., n} has infinitely many (ℵo) successors, then {1, 2,
3, ..., n, n+1} has infinitely many (ℵo) successors. For every n that
can be defined.
Do you accept this proof?
No. A proof should start with a clear presentation of the premises.
Then a sequence of sentences should follow, each with an indication
of how they follow from the previous sentence, and which earlier
sentences of the proof are also needed for the inference.
On 2025-05-26 23:21:27 +0000, Ben Bacarisse said:...
Mikko <[email protected]> writes:
The term "hochschule" does.
But it's an almost unknown term in the UK so it conveys almost nothing
to most English speakers.
Therefore some translation is desidered. The term "superior scool" at
least suggests a scool. ALthough the exact meaning of "superior" may
be unclear people probably have some idea of the intended meaning and
that may be sufficient for many purposes.
Mind you, all I know if the ambition of the institution in question is
that it allows nonsense to be taught, albeit as a entirely optional
course. That particular hochschule does not seem to be taking academic
integrity very seriously.
Maybe the idea is that an exposure to nonsense helps students to learn
to identify nonsense when they see it.
On 27.05.2025 01:57, Ben Bacarisse wrote:
WM <[email protected]> writes:
On 26.05.2025 02:52, Ben Bacarisse wrote:Not without knowing what the set N_def is, since the argument starts
WM <[email protected]> writes:
With pleasure:I can't comment on an argument that is based on a set you have not
For every n ∈ ℕ that can be defined, i.e., ∀n ∈ ℕ_def:
defined.
Can you understand my proof by induction?
"For all n in N_def".
It starts: For every n ∈ ℕ that can be defined.
Then it is proved that not every n ∈ ℕ can be defined.
The resulting set is ℕ_def. (According to set theory however it is not a >>> set but a potentially infinity collection.)So you are not asking me to verify a proof at all but rather to accept a
definition?
I am asking you to understand a proof by induction.
It claims to define N.Your textbook defies N
It defines ℕ_def.
Since this is the set used in applied mathematics.
It's very poor form to tell students you are
defining N when you are not.
It is the set defined by Peano and many others.
For every n ∈ ℕ:
{1} has infinitely many (ℵo) successors.
If {1, 2, 3, ..., n} has infinitely many (ℵo) successors, then {1, 2, 3, ..., n, n+1} has infinitely many (ℵo) successors.
On 27.05.2025 14:01, Mikko wrote:
On 2025-05-26 13:38:00 +0000, WM said:
No alternative view is known to be better.
That does not make this view good.
But even pure mathematics proves that most natural numbers will never
be definable:
{1} has infinitely many (ℵo) successors.
(2) there are infinitely many (ℵo) possible definitions.
No. All natural numbers can be manipulated collectively, for instance subtracted: ℕ \ {1, 2, 3, ...} = { }. Here all have disappeared.
Could all natural numbers be distinguished by individually defining
each one, then this subtraction could also happen but, caused by the well-order, a last number would disappear.
It is a valid proof by induction. Claim it for all natural numbers. GetIf {1, 2, 3, ..., n} has infinitely many (ℵo) successors, then {1, 2,
3, ..., n, n+1} has infinitely many (ℵo) successors, for every n that
can be defined.
You can't formulate that as a logically or mathematically valid proof.
a contradiction. But perhaps you prefer geometry?
The set of finite initial segments of natural numbers is potentially
infinite but not actually infinite.
(Actual infinity is a fixed number greater than all natural numbers.)
On 2025-05-27 15:09:30 +0000, WM said:
It is a valid proof by induction. Claim it for all natural numbers.
Get a contradiction. But perhaps you prefer geometry?
No, it is not. In order to use an inductive proof you must first specify
the theory you are using, and that theory must have an induction axiom.
There is no induction in plain logic.
An induction proof must prove P[0]
and P[n] -> P[n+1] before it can infer
that for all x P[x].
The set of finite initial segments of natural numbers is potentially
infinite but not actually infinite.
There is nothing potential in a set.
If there are infinitely many members
in a set then the set is infinite, otherwise it is finite.
(Actual infinity is a fixed number greater than all natural numbers.)
Infinity is not a number but a feature some sets have
WM <[email protected]> writes:
On 27.05.2025 01:57, Ben Bacarisse wrote:
WM <[email protected]> writes:
On 26.05.2025 02:52, Ben Bacarisse wrote:Not without knowing what the set N_def is, since the argument starts
WM <[email protected]> writes:
With pleasure:I can't comment on an argument that is based on a set you have not
For every n ∈ ℕ that can be defined, i.e., ∀n ∈ ℕ_def:
defined.
Can you understand my proof by induction?
"For all n in N_def".
It starts: For every n ∈ ℕ that can be defined.
"i.e. ∀n ∈ ℕ_def:".
Then it is proved that not every n ∈ ℕ can be defined.
The "proof" starts with an undefined collection.
We both know that you can't define N_def so you need to find some way of waffling about it that starts by assuming it is known.
It sounds as if you are saying that it (your book) defines N_def, and
that it (the set defined in your textbook) is the set defined by Peano
and many others.
That would make N_def and N the same. Really?
I see you cut the request to prove that 1 is in N (or it is N_def?)
using your junk "definition". Of course you cut it. You can't do it!
Can you even prove that 1 is in N using your definition?
How you prove that {1} "has ℵo" successors.
I'd like to see the base case proved.
Mikko <[email protected]> writes:
Maybe the idea is that an exposure to nonsense helps students to learn
to identify nonsense when they see it.
They have to regurgitate the nonsense to get the marks. I once asked WM
what would happen if a student presented real mathematics in the exam
and he said they would not get the marks.
On 28.05.2025 01:06, Ben Bacarisse wrote:
Mikko <[email protected]> writes:
Maybe the idea is that an exposure to nonsense helps students to learnThey have to regurgitate the nonsense to get the marks. I once asked WM
to identify nonsense when they see it.
what would happen if a student presented real mathematics in the exam
and he said they would not get the marks.
No, you are lying.
I would have asked him to explain his position
and would
have convinced him that your "real" mathematics is self-contradictory
On 28.05.2025 01:54, Ben Bacarisse wrote:
WM <[email protected]> writes:
On 27.05.2025 01:57, Ben Bacarisse wrote:"i.e. ∀n ∈ ℕ_def:".
WM <[email protected]> writes:
On 26.05.2025 02:52, Ben Bacarisse wrote:Not without knowing what the set N_def is, since the argument starts
WM <[email protected]> writes:
With pleasure:I can't comment on an argument that is based on a set you have not >>>>>> defined.
For every n ∈ ℕ that can be defined, i.e., ∀n ∈ ℕ_def:
Can you understand my proof by induction?
"For all n in N_def".
It starts: For every n ∈ ℕ that can be defined.
Then it is proved that not every n ∈ ℕ can be defined.The "proof" starts with an undefined collection.
Every n that can be expressed by digits should be known to you.
We both know that you can't define N_def so you need to find some way of
waffling about it that starts by assuming it is known.
Of course I can decide for every number whether it can be distinguished
from all other numbers. If so, it belongs to ℕ_def.
If you are unable to do so, simply assume that every natural number can be defined. Then you get the following contradiction:
All natural numbers can be manipulated collectively, for instance
subtracted: ℕ \ {1, 2, 3, ...} = { }. Here all numbers have disappeared.
Assume that all natural numbers can be defined/distinguished, then the
above subtraction could also happen but, caused by the well-order, a last number would disappear. Contradiction.
It sounds as if you are saying that it (your book) defines N_def, and
that it (the set defined in your textbook) is the set defined by Peano
and many others.
Yes.
That would make N_def and N the same. Really?
The above proof contradicts that statement.
I see you cut the request to prove that 1 is in N (or it is N_def?)
using your junk "definition". Of course you cut it. You can't do
it!
I have shown you the definition Below it is again.
Can you even prove that 1 is in N using your definition?
1 ∈ M (4.1)
n ∈ M ⇒ (n + 1) ∈ M (4.2)
If M satisfies (4.1) and (4.2), then ℕ ⊆ M.
Of course no intelligent reader need be told that this ℕ = ℕ_def also satisfies the axioms (4.1) and (4.2).
How you prove that {1} "has ℵo" successors.
I do not prove it
I'd like to see the base case proved.
It cannot be proved but only assumed.
On 28.05.2025 10:25, Mikko wrote:
On 2025-05-27 15:09:30 +0000, WM said:
It is a valid proof by induction. Claim it for all natural numbers. Get
a contradiction. But perhaps you prefer geometry?
No, it is not. In order to use an inductive proof you must first specify
the theory you are using, and that theory must have an induction axiom.
Why do you think has the induction axiom been devised at all? Right,
because the sequence of natural numbers has this property. When Pascal
and and Fermat first used induction, there was no axiom but the
property of natural numbers had been recognized.
There is no induction in plain logic.
But it is in the mathematics we apply.
An induction proof must prove P[0]
I have said: {1} has infinitely many (ℵo) successors.
and P[n] -> P[n+1] before it can infer
I did not expect that you need this explanation:
If {1, 2, 3, ..., n} has infinitely many (ℵo) successors, then {1, 2,
3, ..., n, n+1} has infinitely many (ℵo) successors because here the
number of successors has been reduced by 1, and ℵo - 1 = ℵo. There is
no way to avoid this conclusion if ℵo natural numbers are assumed to
exist. And that is the theory that I use.
that for all x P[x].
Just that is wrong because it is not true for all natural numbers but
only for definable ones.
The set of finite initial segments of natural numbers is potentially
infinite but not actually infinite.
There is nothing potential in a set.
Then call it a collection.
On 27.05.2025 14:27, Mikko wrote:
On 2025-05-26 13:44:06 +0000, WM said:
{1} has infinitely many (ℵo) successors.
If {1, 2, 3, ..., n} has infinitely many (ℵo) successors, then {1, 2,
3, ..., n, n+1} has infinitely many (ℵo) successors. For every n that
can be defined.
Do you accept this proof?
No. A proof should start with a clear presentation of the premises.
A proof as simple as this one should be understood as it stands. You
only try to suppress it.
Then a sequence of sentences should follow, each with an indication
of how they follow from the previous sentence, and which earlier
sentences of the proof are also needed for the inference.
{1} has infinitely many (ℵo) successors. If you don't understand, please ask.
For every number n that can be represented in decimals:
If {1, 2, 3, ..., n} has infinitely many (ℵo) successors, then {1, 2,
3, ..., n, n+1} has infinitely many (ℵo) successors. If you don't understand, please ask.
For every number n that can be represented in decimals:
If {1, 2, 3, ..., n} has infinitely many (ℵo) successors, then {1, 2,
3, ..., n, n+1} has infinitely many (ℵo) successors.
For every number n that can be represented in decimals:
If {1, 2, 3, ..., n} has infinitely many (ℵo) successors, then {1, 2,
3, ..., n, n+1} has infinitely many (ℵo) successors. If you don't understand, please ask.
Mikko <[email protected]> writes:
On 2025-05-26 23:21:27 +0000, Ben Bacarisse said:...
Mikko <[email protected]> writes:
The term "hochschule" does.
But it's an almost unknown term in the UK so it conveys almost nothing
to most English speakers.
Therefore some translation is desidered. The term "superior scool" at
least suggests a scool. ALthough the exact meaning of "superior" may
be unclear people probably have some idea of the intended meaning and
that may be sufficient for many purposes.
I think superior school has the same problem. It's unlikely to be
clear.
On 2025-05-27 23:06:14 +0000, Ben Bacarisse said:
Mikko <[email protected]> writes:
On 2025-05-26 23:21:27 +0000, Ben Bacarisse said:...
Mikko <[email protected]> writes:
I think superior school has the same problem. It's unlikely to beTherefore some translation is desidered. The term "superior scool" atThe term "hochschule" does.But it's an almost unknown term in the UK so it conveys almost nothing >>>> to most English speakers.
least suggests a scool. ALthough the exact meaning of "superior" may
be unclear people probably have some idea of the intended meaning and
that may be sufficient for many purposes.
clear.
Everything would be. Often the word "university" is used althogh it
hides the distiction between an "universit�t" and a "hochschule".
WM <[email protected]> writes:
Every n that can be expressed by digits should be known to you.
But the important fact, since it's /your/ proof, is what that means to
/you/ and I can not know that.
You are actually prepared to state that N (defined by Peano) and N_def (defined by your book) are the same and also that they are also not the
same?
Can you even prove that 1 is in N using your definition?
Nothing on this (of course).
1 ∈ M (4.1)
n ∈ M ⇒ (n + 1) ∈ M (4.2)
If M satisfies (4.1) and (4.2), then ℕ ⊆ M.
Of course no intelligent reader need be told that this ℕ = ℕ_def also
satisfies the axioms (4.1) and (4.2).
But it seems you can't prove that 1 is in N, can you?
It should be
easy, should it not?
It is simple using the correct definition, but
yours is junk.
How you prove that {1} "has ℵo" successors.
I do not prove it
But you need to. It's is the base case in the proof you asked everyone about. You can't make a proof by induction by simply asserting things.
On 2025-05-27 15:24:50 +0000, WM said:
{1} has infinitely many (ℵo) successors. If you don't understand,
please ask.
For every number n that can be represented in decimals:
If {1, 2, 3, ..., n} has infinitely many (ℵo) successors, then {1, 2,
3, ..., n, n+1} has infinitely many (ℵo) successors. If you don't
understand, please ask.
The indication that a sentence is a premise is still absent.
The indication that and how a sentence is a conseqence of earler sentences
is still absent.
WM <[email protected]> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte
des Unendlichen" and "Kleine Geschichte der Mathematik" at Technische Hochschule Augsburg.)
On 28.05.2025 01:06, Ben Bacarisse wrote:
Mikko <[email protected]> writes:
Maybe the idea is that an exposure to nonsense helps students to learn >>>> to identify nonsense when they see it.They have to regurgitate the nonsense to get the marks. I once asked WM >>> what would happen if a student presented real mathematics in the exam
and he said they would not get the marks.
No, you are lying.
That's harsh.
I may simply have misremembered what you said about this
before. If so I apologise. But I see you /don't/ in fact say they that would get the marks. You only say that they would need to be convinced
they were wrong. What if they were not convinced and stuck by the
answer they had written in the exam?
I would have asked him to explain his position
In the UK (at least at the universities I am familiar with), exam papers
must be marked according to a pre-written mark scheme. There is no
option to interview the student after they submit their paper.
Does
this really happen in Germany? And if so, does the interview have only
one outcome -- agree or else? Do you not have to write marking schemes
for your exams? And if in fact you do, what do yours say about
alternative answers?
On 2025-05-28 15:13:54 +0000, WM said:
There is no induction in plain logic.
But it is in the mathematics we apply.
It is in certain mathematical structures but not in all.
I have said: {1} has infinitely many (ℵo) successors.
But you navn't proven that this infinity is not begger than some other infinity.
and P[n] -> P[n+1] before it can infer
I did not expect that you need this explanation:
If {1, 2, 3, ..., n} has infinitely many (ℵo) successors, then {1, 2,
3, ..., n, n+1} has infinitely many (ℵo) successors because here the
number of successors has been reduced by 1, and ℵo - 1 = ℵo. There is
no way to avoid this conclusion if ℵo natural numbers are assumed to
exist. And that is the theory that I use.
To me this does not look like P[n] -> P[n+1].
Just that is wrong because it is not true for all natural numbers but
only for definable ones.
It is not wrong because you failed specify the theory you are using.
As I said the theory must be specified.
In Peano arithmetic the induction axiom is applicable to everything.
If you want something else you must specify some other theory, perhaps
some set theory.
The set of finite initial segments of natural numbers is potentially
infinite but not actually infinite.
There is nothing potential in a set.
Then call it a collection.
Things get soon complicated if we allow other than objects, first order functions and first order predicates.
a first order predicate so it does not complicate too much. But a
collection that is not a set would require another book and I don't
think I would read it.
On 29.05.2025 02:46, Ben Bacarisse wrote:
WM <[email protected]> writes:
On 28.05.2025 01:06, Ben Bacarisse wrote:That's harsh.
Mikko <[email protected]> writes:
Maybe the idea is that an exposure to nonsense helps students to learn >>>>> to identify nonsense when they see it.They have to regurgitate the nonsense to get the marks. I once asked WM >>>> what would happen if a student presented real mathematics in the exam
and he said they would not get the marks.
No, you are lying.
Your statement.
I may simply have misremembered what you said about this
before. If so I apologise. But I see you /don't/ in fact say they that
would get the marks. You only say that they would need to be convinced
they were wrong. What if they were not convinced and stuck by the
answer they had written in the exam?
That depends. What answer do you have in mind?
I would have asked him to explain his positionIn the UK (at least at the universities I am familiar with), exam papers
must be marked according to a pre-written mark scheme. There is no
option to interview the student after they submit their paper.
Above I assumed a personal discussion.
Does
this really happen in Germany? And if so, does the interview have only
one outcome -- agree or else? Do you not have to write marking schemes
for your exams? And if in fact you do, what do yours say about
In the exam there are questions like these:
- Beschreiben Sie, was man unter der Abz�hlbarkeit aller positiven Br�che
versteht und er�rtern Sie ein Gegenargument.
- Beschreiben Sie, was man unter der �berabz�hlbarkeit der reellen Zahlen
versteht, und er�rtern Sie ein Gegenargument.
- Beschreiben Sie das Spiel "Wir erobern den Bin�ren Baum" und die damit
verkn�pfte Aussage.
- Sehen Sie eine Parallele zwischen MacDuck und der Nummerierung aller
Br�che? Wenn ja, welche?
- Was halten Sie vom wissenschaftlichen Wert der Mengenlehre unter
Ber�cksichtigung von Banach-Tarski-Paradoxon und Verteilung der Br�che in
(0, 1) und (1, oo).
Try to answer. Then I will give you marks.
- Beschreiben Sie, was man unter der Abz�hlbarkeit aller positiven Br�che
versteht und er�rtern Sie ein Gegenargument.
- Beschreiben Sie, was man unter der �berabz�hlbarkeit der reellen Zahlen
versteht, und er�rtern Sie ein Gegenargument.
Do you not have to write marking schemes for your exams? And if in
fact you do, what do yours say about alternative answers?
On 29.05.2025 02:25, Ben Bacarisse wrote:
WM <[email protected]> writes:
Every n that can be expressed by digits should be known to you.But the important fact, since it's /your/ proof, is what that means to
/you/ and I can not know that.
I have often published it.
Definition: A natural number is "identified" or (individually) "defined" or "instantiated" if it can be communicated such that sender and receiver understand the same and can link it by a finite initial segment to the
origin 0. All other natural numbers are called dark natural numbers.
Communication can occur
- by direct description in the unary system like ||||||| or as many beeps,
flashes, or raps,
- by a finite initial segment of natural numbers (1, 2, 3, 4, 5, 6, 7)
called a FISON,
- as n-ary representation, for instance binary 111 or decimal 7,
- by indirect description like "the number of colours of the rainbow",
- by other words known to sender and receiver like "seven".
Only when a number n is identified we can use it in mathematical discourse and can determine the trichotomy properties of n and of every multiple kn
or power n^k with respect to every identified number k. ℕ_def is the set that contains all defined natural numbers as elements – and nothing else. ℕ_def is a potentially infinite set; therefore henceforth it will be called a collection.
You are actually prepared to state that N (defined by Peano) and N_def
(defined by your book) are the same and also that they are also not the
same?
You have not understood. They are the same. Both differ from Cantor's actually infinite set ℕ.
Nothing on this (of course).Can you even prove that 1 is in N using your definition?
The next lines show it. Aren't you ashamed?
1 ∈ M (4.1)But it seems you can't prove that 1 is in N, can you?
n ∈ M ⇒ (n + 1) ∈ M (4.2)
If M satisfies (4.1) and (4.2), then ℕ ⊆ M.
Of course no intelligent reader need be told that this ℕ = ℕ_def also >>> satisfies the axioms (4.1) and (4.2).
It requires a lot of stupidity or hate to put this question after seeing
the axiom that 1 is in ℕ.
How you prove that {1} "has ℵo" successors.
I do not prove it
But you need to. It's is the base case in the proof you asked everyone
about. You can't make a proof by induction by simply asserting things.
Of course. Based on the assumption that Cantor is right I can prove the existence of dark numbers.
I'd like to see the base case proved.
It cannot be proved but only assumed.
That is the usual way in mathematics and logic:
Given A it follows B. That is called an implication.
Mikko <[email protected]> writes:
On 2025-05-27 23:06:14 +0000, Ben Bacarisse said:
Mikko <[email protected]> writes:
On 2025-05-26 23:21:27 +0000, Ben Bacarisse said:...
Mikko <[email protected]> writes:
I think superior school has the same problem. It's unlikely to beTherefore some translation is desidered. The term "superior scool" atThe term "hochschule" does.But it's an almost unknown term in the UK so it conveys almost nothing >>>>> to most English speakers.
least suggests a scool. ALthough the exact meaning of "superior" may
be unclear people probably have some idea of the intended meaning and
that may be sufficient for many purposes.
clear.
Everything would be. Often the word "university" is used althogh it
hides the distiction between an "universität" and a "hochschule".
A university would offer a wide range of degree subjects (certainly mathematics) and would have PhD (and other post graduate degree)
programs. I don't know about all the hochschule, but I don't think the Technische Hochschule Augsburg has either.
On 29.05.2025 12:22, Mikko wrote:
On 2025-05-27 15:24:50 +0000, WM said:
{1} has infinitely many (ℵo) successors. If you don't understand, please ask.
For every number n that can be represented in decimals:
If {1, 2, 3, ..., n} has infinitely many (ℵo) successors, then {1, 2,
3, ..., n, n+1} has infinitely many (ℵo) successors. If you don't
understand, please ask.
The indication that a sentence is a premise is still absent.
This is {1} has infinitely many (ℵo) successors.
The indication that and how a sentence is a conseqence of earler sentences >> is still absent.
{1, 2, 3, ..., n, n+1} has infinitely many (ℵo) successors is a
consequence of {1, 2, 3, ..., n} has infinitely many (ℵo) successors,
and ℵo - 1 = ℵo.
On 29.05.2025 12:07, Mikko wrote:
On 2025-05-28 15:13:54 +0000, WM said:
There is no induction in plain logic.
But it is in the mathematics we apply.
It is in certain mathematical structures but not in all.
Anyhow a reader in sci.logic should understand it.
I have said: {1} has infinitely many (ℵo) successors.
But you navn't proven that this infinity is not begger than some other
infinity.
I am assuming Cantor's infinity. This is expressed by ℵo.
and P[n] -> P[n+1] before it can infer
I did not expect that you need this explanation:
If {1, 2, 3, ..., n} has infinitely many (ℵo) successors, then {1, 2,
3, ..., n, n+1} has infinitely many (ℵo) successors because here the
number of successors has been reduced by 1, and ℵo - 1 = ℵo. There is >>> no way to avoid this conclusion if ℵo natural numbers are assumed to
exist. And that is the theory that I use.
To me this does not look like P[n] -> P[n+1].
P[n]: {1, 2, 3, ..., n} has infinitely many (ℵo) successors.
P[n+1]: {1, 2, 3, ..., n, n+1} has infinitely many (ℵo) successors.
Do you doubt ℵo - 1 = ℵo?
It is basic mathematics as you learn it in the first semester.
As I said the theory must be specified.
In Peano arithmetic the induction axiom is applicable to everything.
If you want something else you must specify some other theory, perhaps
some set theory.
Induction is applied to every natural number of the Peano set. The
proof shows that it cannot be applied to every natural number of the
Cantor set.
The set of finite initial segments of natural numbers is potentially >>>>> infinite but not actually infinite.
There is nothing potential in a set.
Then call it a collection.
Things get soon complicated if we allow other than objects, first order
functions and first order predicates.
Here nothing gets complicated, but all remains very simple.
All Cantor's natural numbers can be manipulated collectively, for
instance subtracted: ℕ \ {1, 2, 3, ...} = { }. Here all have
disappeared.
Could all Cantor's natural numbers be distinguished, then this
subtraction could also happen but, caused by the well-order, a last one
would disappear. Contradiction.
On 29.05.2025 02:25, Ben Bacarisse wrote:
WM <[email protected]> writes:
Every n that can be expressed by digits should be known to you.
But the important fact, since it's /your/ proof, is what that means to
/you/ and I can not know that.
I have often published it.
Definition: A natural number is "identified" or (individually)
"defined" or "instantiated" if it can be communicated such that sender
and receiver understand the same and can link it by a finite initial
segment to the origin 0. All other natural numbers are called dark
natural numbers.
WM <[email protected]> writes:
Ah, so you can't do what you suggested and talk to the student, after
the exam, to try to persuade them that you are right! That did sound a
bit crazy.
In the exam there are questions like these:
- Beschreiben Sie, was man unter der Abzählbarkeit aller positiven Brüche >> versteht und erörtern Sie ein Gegenargument.
- Beschreiben Sie, was man unter der Überabzählbarkeit der reellen Zahlen >> versteht, und erörtern Sie ein Gegenargument.
- Beschreiben Sie das Spiel "Wir erobern den Binären Baum" und die damit
verknüpfte Aussage.
- Sehen Sie eine Parallele zwischen MacDuck und der Nummerierung aller
Brüche? Wenn ja, welche?
- Was halten Sie vom wissenschaftlichen Wert der Mengenlehre unter
Berücksichtigung von Banach-Tarski-Paradoxon und Verteilung der Brüche in
(0, 1) und (1, oo).
From this, it is not obvious that you want students to say anything I'd consider to be wrong in an exam. So maybe someone could indeed get full marks without having to deny mathematics.
Are there any claims in your
lectures that someone at the university down the road would object to?
Try to answer. Then I will give you marks.
I'll try a couple of questions...
- Beschreiben Sie, was man unter der Abzählbarkeit aller positiven Brüche >> versteht und erörtern Sie ein Gegenargument.
The positive fractions are said to be countable because the function
b(0) = 1
b(n+1) = s(b(n))
where s(q) = 1 / (2*floor(q) - q + 1)
is a bijection between the natural numbers and the positive fractions according to the definition in Prof. Mückenheim's textbook.
I am not aware of a valid counter argument since this is simply a
definition of what the term "countable" means.
- Beschreiben Sie, was man unter der Überabzählbarkeit der reellen Zahlen >> versteht, und erörtern Sie ein Gegenargument.
The real numbers are said to be uncountable because no bijective
function exists between N and R.
I am not aware of any valid counterargument because this theorem is well-established.
The way you word the questions does seem to allow for correct answers.
What does your mark scheme say for these questions?
Would you accept
any answers that I would consider to be wrong?
Do you not have to write marking schemes for your exams? And if in
fact you do, what do yours say about alternative answers?
WM <[email protected]> writes:
I thought it might be something cumbersome and vague like that. I can't
even tell if this is a inductive collection,
so I must decline any
request to review a proof by induction based on it.
You are actually prepared to state that N (defined by Peano) and N_def
(defined by your book) are the same and also that they are also not the
same?
You have not understood. They are the same. Both differ from Cantor's
actually infinite set ℕ.
Ah. That is just an assertion on your part. I will accept that you
believe it to be true.
Nothing on this (of course).Can you even prove that 1 is in N using your definition?
The next lines show it. Aren't you ashamed?
Of course not.
The axioms say that 1 is in M (I think you mean that it is in many
possible Ms) and that N is a subset of any M meeting the two axioms. At least that seems to be what you wrote.
Please prove that the subset you call N includes 1. There are lots of
sets that are subsets of every possible M, and many don't include 1.
[You might think that 4.1 and 4.2 uniquely define a set M
of which you
state N is a subset, but that does not help you show that 1 is in N.]
That is the usual way in mathematics and logic:
Given A it follows B. That is called an implication.
So write the proof correctly, stating the assumptions and the
consequences that follow. That way the reader can tell if, maybe, one
or more of the assumptions need to be rejected.
On 2025-05-29 14:52:45 +0000, WM said:
On 29.05.2025 12:22, Mikko wrote:
On 2025-05-27 15:24:50 +0000, WM said:
{1} has infinitely many (ℵo) successors. If you don't understand,
please ask.
For every number n that can be represented in decimals:
If {1, 2, 3, ..., n} has infinitely many (ℵo) successors, then {1,
2, 3, ..., n, n+1} has infinitely many (ℵo) successors. If you don't >>>> understand, please ask.
The indication that a sentence is a premise is still absent.
This is {1} has infinitely many (ℵo) successors.
The indication that and how a sentence is a conseqence of earler
sentences
is still absent.
{1, 2, 3, ..., n, n+1} has infinitely many (ℵo) successors is a
consequence of {1, 2, 3, ..., n} has infinitely many (ℵo) successors,
and ℵo - 1 = ℵo.
Still no proof.
On 2025-05-29 14:47:49 +0000, WM said:
It is in certain mathematical structures but not in all.
Anyhow a reader in sci.logic should understand it.
Everybody should understand at least arithmetic induction and its limitations. But everybody doesn't.
Cantor did not use ℵo for infinity in general but only for a particular kind of infinity.
P[n]: {1, 2, 3, ..., n} has infinitely many (ℵo) successors.
P[n+1]: {1, 2, 3, ..., n, n+1} has infinitely many (ℵo) successors.
But P[n] -> P[n+1] is not there.
Do you doubt ℵo - 1 = ℵo?
That can't be said in Peano arithmetic.
It is basic mathematics as you learn it in the first semester.
No, it is not that basic. There are no infinities there, and no
induction, either.
As I said the theory must be specified.
In Peano arithmetic the induction axiom is applicable to everything.
If you want something else you must specify some other theory, perhaps
some set theory.
Induction is applied to every natural number of the Peano set. The
proof shows that it cannot be applied to every natural number of the
Cantor set.
You have shown no proof that shows that.
It is not a contradiction that at least one disappears when all disappear.
On 30.05.2025 03:08, Ben Bacarisse wrote:
WM <[email protected]> writes:
I thought it might be something cumbersome and vague like that. I can't
even tell if this is a inductive collection,
It is obvious and clear. Do you know a case where a natural number can be
in it and cannot be in it? No. You can only curse. It is the same as
Peano's set. If you can't understand blame it on yourself.
so I must decline any
request to review a proof by induction based on it.
Of course. There is no counter argument. So you must decline.
Of course not.Nothing on this (of course).Can you even prove that 1 is in N using your definition?
The next lines show it. Aren't you ashamed?
...The axioms say that 1 is in M (I think you mean that it is in many
possible Ms) and that N is a subset of any M meeting the two axioms. At
least that seems to be what you wrote.
Please prove that the subset you call N includes 1. There are lots of
sets that are subsets of every possible M, and many don't include 1.
I told you already that I have written my book for intelligent
students. That means not to repeat the obvious. If ℕ should not obey the conditions put on M, then the two axioms would be ado about nothing. An intelligent reader understands that.
As I said that requires an intelligent reader recognizing that without ℕ obeying the axioms too the paragraph would be nonsense.
That is the usual way in mathematics and logic:So write the proof correctly, stating the assumptions and the
Given A it follows B. That is called an implication.
consequences that follow. That way the reader can tell if, maybe, one
or more of the assumptions need to be rejected.
All natural numbers of Cantor's set ℕ can be manipulated collectively, for instance subtracted: ℕ \ {1, 2, 3, ...} = { }. Here all have
disappeared.
On 30.05.2025 02:05, Ben Bacarisse wrote:
WM <[email protected]> writes:
Ah, so you can't do what you suggested and talk to the student, after
the exam, to try to persuade them that you are right! That did sound a
bit crazy.
Do you never have oral exams in England?
In the exam there are questions like these:From this, it is not obvious that you want students to say anything I'd
- Beschreiben Sie, was man unter der Abzählbarkeit aller positiven Brüche >>> versteht und erörtern Sie ein Gegenargument.
- Beschreiben Sie, was man unter der Überabzählbarkeit der reellen Zahlen >>> versteht, und erörtern Sie ein Gegenargument.
- Beschreiben Sie das Spiel "Wir erobern den Binären Baum" und die damit >>> verknüpfte Aussage.
- Sehen Sie eine Parallele zwischen MacDuck und der Nummerierung aller
Brüche? Wenn ja, welche?
- Was halten Sie vom wissenschaftlichen Wert der Mengenlehre unter
Berücksichtigung von Banach-Tarski-Paradoxon und Verteilung der Brüche in
(0, 1) und (1, oo).
consider to be wrong in an exam. So maybe someone could indeed get full
marks without having to deny mathematics.
Correct mathematics should not be denied.
Are there any claims in your
lectures that someone at the university down the road would object to?
Of course.
Try to answer. Then I will give you marks.I'll try a couple of questions...
- Beschreiben Sie, was man unter der Abzählbarkeit aller positiven Brüche >>> versteht und erörtern Sie ein Gegenargument.The positive fractions are said to be countable because the function
b(0) = 1
b(n+1) = s(b(n))
where s(q) = 1 / (2*floor(q) - q + 1)
is a bijection between the natural numbers and the positive fractions
according to the definition in Prof. Mückenheim's textbook.
I am not aware of a valid counter argument since this is simply a
definition of what the term "countable" means.
It has been shown to the student by many arguments that the bijection
fails.
Without giving one of these arguments (if desired also showing that and why it fails) you would get only half of the full score.
If a student had ever disproved my proofs he would have got additional points.Do you not have to write marking schemes for your exams? And if in
fact you do, what do yours say about alternative answers?
On 30.05.2025 11:36, Mikko wrote:
On 2025-05-29 14:47:49 +0000, WM said:
It is in certain mathematical structures but not in all.
Anyhow a reader in sci.logic should understand it.
Everybody should understand at least arithmetic induction and its
limitations. But everybody doesn't.
"Not everybody does" would be correct.
Cantor did not use ℵo for infinity in general but only for a particular
kind of infinity.
For all infinite sets of natural numbers he used it. That's what I
discuss here.
P[n]: {1, 2, 3, ..., n} has infinitely many (ℵo) successors.
P[n+1]: {1, 2, 3, ..., n, n+1} has infinitely many (ℵo) successors.
But P[n] -> P[n+1] is not there.
Do you doubt ℵo - 1 = ℵo?
That can't be said in Peano arithmetic.
Here I use induction in Cantor's set. That is allowed. Cantor did it too.
On 30.05.2025 11:51, Mikko wrote:
On 2025-05-29 14:52:45 +0000, WM said:
On 29.05.2025 12:22, Mikko wrote:
On 2025-05-27 15:24:50 +0000, WM said:
{1} has infinitely many (ℵo) successors. If you don't understand, please ask.
For every number n that can be represented in decimals:
If {1, 2, 3, ..., n} has infinitely many (ℵo) successors, then {1, 2, >>>>> 3, ..., n, n+1} has infinitely many (ℵo) successors. If you don't
understand, please ask.
The indication that a sentence is a premise is still absent.
This is {1} has infinitely many (ℵo) successors.
The indication that and how a sentence is a conseqence of earler sentences >>>> is still absent.
{1, 2, 3, ..., n, n+1} has infinitely many (ℵo) successors is a
consequence of {1, 2, 3, ..., n} has infinitely many (ℵo) successors,
and ℵo - 1 = ℵo.
Still no proof.
Proof that every definable natural number has more successors than predecessors.
On 2025-05-30 14:25:22 +0000, WM said:
Here I use induction in Cantor's set. That is allowed. Cantor did it too.
No, there is no artithmetic induction and no set induction there.
On 2025-05-30 14:46:55 +0000, WM said:
{1, 2, 3, ..., n, n+1} has infinitely many (ℵo) successors is a
consequence of {1, 2, 3, ..., n} has infinitely many (ℵo)
successors, and ℵo - 1 = ℵo.
Still no proof.
Proof that every definable natural number has more successors than
predecessors.
You have not shown that proof, either.
But because every natural number
has more successors than predessors it hardly is interesting, unless
you happen to have an interesting way to say 'more'.
WM <[email protected]> writes:
It has been shown to the student by many arguments that the bijection
fails.
Which of the conditions of being a bijection (as presented in your book)
does b fail to meet? (I'm betting you won't say.)
...
Without giving one of these arguments (if desired also showing that and why >> it fails) you would get only half of the full score.
You called me a liar for saying this. I /would/ actually have to accept
the nonsense you teach (or at least copy it out) to get full marks.
...
If a student had ever disproved my proofs he would have got additionalDo you not have to write marking schemes for your exams? And if in
fact you do, what do yours say about alternative answers?
points.
Do you always refuse to answer simple questions? Do you have to write marking schemes for your exams? I'm just trying to find out if there is documentary evidence of what a student at your college has to write to
get full marks.
WM <[email protected]> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte
des Unendlichen" and "Kleine Geschichte der Mathematik" at Technische Hochschule Augsburg.)
On 30.05.2025 03:08, Ben Bacarisse wrote:
WM <[email protected]> writes:
I thought it might be something cumbersome and vague like that. I can't >>> even tell if this is a inductive collection,
It is obvious and clear. Do you know a case where a natural number can be
in it and cannot be in it? No. You can only curse. It is the same as
Peano's set. If you can't understand blame it on yourself.
Can you prove it is an inductive set/collection?
so I must decline any
request to review a proof by induction based on it.
Of course. There is no counter argument. So you must decline.
No, I decline because I don't know if it is an inductive set. Do you?
(I note you deleted the cumbersome and vague definition.
If it really
were obvious and clear, I would have left it in to show the world how
wrong I was to call it cumbersome and vague.)
I see you've cut the incorrect definition and the claim that the axioms directly say that 1 is in N because, presumably, you now see that they
don't.
As I said that requires an intelligent reader recognizing that without ℕ >> obeying the axioms too the paragraph would be nonsense.
That's funny! Yes, an intelligent reader will see you've written a junk definition
All natural numbers of Cantor's set ℕ can be manipulated collectively, for >> instance subtracted: ℕ \ {1, 2, 3, ...} = { }. Here all have
disappeared.
What definition of N do you want your intelligent readers to assume?
On 31.05.2025 12:11, Mikko wrote:
On 2025-05-30 14:46:55 +0000, WM said:
{1, 2, 3, ..., n, n+1} has infinitely many (ℵo) successors is a
consequence of {1, 2, 3, ..., n} has infinitely many (ℵo) successors, >>>>> and ℵo - 1 = ℵo.
Still no proof.
Proof that every definable natural number has more successors than
predecessors.
You have not shown that proof, either.
I have above. You cannot understand it. That is a different thing.
On 31.05.2025 11:59, Mikko wrote:
On 2025-05-30 14:25:22 +0000, WM said:
Here I use induction in Cantor's set. That is allowed. Cantor did it too. >>No, there is no artithmetic induction and no set induction there.
"daß die Reihe
1, i2, i3, ..., i, ...
nur eine Permutation der Reihe
1, 2, 3, ..., , ...
ist. Dies beweisen wir durch vollständige Induktion,"
[Cantor, collected works, p. 305]
On 2025-05-31 13:40:12 +0000, WM said:
On 31.05.2025 11:59, Mikko wrote:
On 2025-05-30 14:25:22 +0000, WM said:
Here I use induction in Cantor's set. That is allowed. Cantor did it
too.
No, there is no artithmetic induction and no set induction there.
"daß die Reihe
1, i2, i3, ..., i, ...
nur eine Permutation der Reihe
1, 2, 3, ..., , ...
ist. Dies beweisen wir durch vollständige Induktion,"
[Cantor, collected works, p. 305]
Did Cantor acurally prove that with a complete induction? As far as
I have seen Cantor has proven what he promised to prove.
On 2025-05-31 13:47:51 +0000, WM said:
On 31.05.2025 12:11, Mikko wrote:
On 2025-05-30 14:46:55 +0000, WM said:
{1, 2, 3, ..., n, n+1} has infinitely many (ℵo) successors is a
consequence of {1, 2, 3, ..., n} has infinitely many (ℵo)
successors, and ℵo - 1 = ℵo.
Still no proof.
Proof that every definable natural number has more successors than
predecessors.
You have not shown that proof, either.
I have above. You cannot understand it. That is a different thing.
No proof of anything above. Besides, nothing that requires any
understanding beyond ordinary proof checking cannot be a proof.
On 31.05.2025 02:20, Ben Bacarisse wrote:
WM <[email protected]> writes:
It has been shown to the student by many arguments that the bijectionWhich of the conditions of being a bijection (as presented in your book)
fails.
does b fail to meet? (I'm betting you won't say.)
The condition to be definable.
...
Without giving one of these arguments (if desired also showing that and why >>> it fails) you would get only half of the full score.You called me a liar for saying this. I /would/ actually have to accept
the nonsense you teach (or at least copy it out) to get full marks.
You could also disprove it. But you carefully evade.
...
Do you always refuse to answer simple questions? Do you have to writeIf a student had ever disproved my proofs he would have got additionalDo you not have to write marking schemes for your exams? And if in >>>>>> fact you do, what do yours say about alternative answers?
points.
marking schemes for your exams? I'm just trying to find out if there is
documentary evidence of what a student at your college has to write to
get full marks.
Either understand and repeat what I teach or disprove it...
On 31.05.2025 02:02, Ben Bacarisse wrote:
WM <[email protected]> writes:
On 30.05.2025 03:08, Ben Bacarisse wrote:Can you prove it is an inductive set/collection?
WM <[email protected]> writes:
I thought it might be something cumbersome and vague like that. I can't >>>> even tell if this is a inductive collection,
It is obvious and clear. Do you know a case where a natural number can be >>> in it and cannot be in it? No. You can only curse. It is the same as
Peano's set. If you can't understand blame it on yourself.
See my book. The set is defined by induction.
Definition: A natural number is "identified" or (individually) "defined" or "instantiated" if it can be communicated such that sender and receiver understand the same and can link it by a finite initial segment to the
origin 0. All other natural numbers are called dark natural numbers.
Communication can occur
- by direct description in the unary system like ||||||| or as many beeps,
flashes, or raps,
- by a finite initial segment of natural numbers (1, 2, 3, 4, 5, 6, 7)
called a FISON,
- as n-ary representation, for instance binary 111 or decimal 7,
- by indirect description like "the number of colours of the rainbow",
- by other words known to sender and receiver like "seven".
If n is in it, then also n+1 is in it.
No, I decline because I don't know if it is an inductive set. Do you?so I must decline any
request to review a proof by induction based on it.
Of course. There is no counter argument. So you must decline.
Every mathematician knows that the definable natural numbers are an
inductive set.
(I note you deleted the cumbersome and vague definition.
It has been given to be understood. Now you have or have not understood. If not, the further presence would not help, I assume.
If it really
were obvious and clear, I would have left it in to show the world how
wrong I was to call it cumbersome and vague.)
I can give you a simpler and shorter definition: Every n that can be expressed by digits is definable.
I see you've cut the incorrect definition and the claim that the axioms
directly say that 1 is in N because, presumably, you now see that they
don't.
As I said that requires an intelligent reader recognizing that without ℕ >>> obeying the axioms too the paragraph would be nonsense.That's funny! Yes, an intelligent reader will see you've written a junk
definition
You are a dishonest liar.
But that is not relevant.
What definition of N do you want your intelligent readers to assume?
ℕ is Cantor's infinite set.
was /immediately/ followed by:What definition of N do you want your intelligent readers to assume?
so you could have just said "yes" but then you would not have been ablePresumably you don't want [them] to assume one Cantor uses
WM <[email protected]> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte
des Unendlichen" and "Kleine Geschichte der Mathematik" at Technische Hochschule Augsburg.)
On 31.05.2025 02:20, Ben Bacarisse wrote:
WM <[email protected]> writes:
It has been shown to the student by many arguments that the bijectionWhich of the conditions of being a bijection (as presented in your book) >>> does b fail to meet? (I'm betting you won't say.)
fails.
The condition to be definable.
There are two conditions (as you know perfectly well) and it meets both
(as you also know perfectly well). Here are the conditions:
bijektiv (oder eineindeutig), wenn f injektiv und surjektiv ist
Is b not injective? Is b not surjektiv? Here's b again so you can
check for yourself that it is both:
You called me a liar for saying this. I /would/ actually have to accept >>> the nonsense you teach (or at least copy it out) to get full marks.
You could also disprove it. But you carefully evade.
Eh? You just told me that what I wrote would not get full marks. I
would have to parrot at least some of your nonsense to get more. That's
all I was saying before when you said I was lying.
...
Do you always refuse to answer simple questions? Do you have to writeIf a student had ever disproved my proofs he would have got additional >>>> points.Do you not have to write marking schemes for your exams? And if in >>>>>>> fact you do, what do yours say about alternative answers?
marking schemes for your exams? I'm just trying to find out if there is >>> documentary evidence of what a student at your college has to write to
get full marks.
You really don't want to say, do you?
By the way, I can see why you don't want to show any marking scheme. It would have to include the junk you expect the students to say,
WM <[email protected]> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte
des Unendlichen" and "Kleine Geschichte der Mathematik" at Technische Hochschule Augsburg.)
On 31.05.2025 02:02, Ben Bacarisse wrote:
WM <[email protected]> writes:
On 30.05.2025 03:08, Ben Bacarisse wrote:Can you prove it is an inductive set/collection?
WM <[email protected]> writes:
I thought it might be something cumbersome and vague like that. I can't >>>>> even tell if this is a inductive collection,
It is obvious and clear. Do you know a case where a natural number can be >>>> in it and cannot be in it? No. You can only curse. It is the same as
Peano's set. If you can't understand blame it on yourself.
See my book. The set is defined by induction.
You are losing the plot. Here is the definition you offered for the
"obvious and clear" idea of N_def:
Definition: A natural number is "identified" or (individually) "defined" or >> "instantiated" if it can be communicated such that sender and receiver
understand the same and can link it by a finite initial segment to the
origin 0. All other natural numbers are called dark natural numbers.
Communication can occur
- by direct description in the unary system like ||||||| or as many beeps, >> flashes, or raps,
- by a finite initial segment of natural numbers (1, 2, 3, 4, 5, 6, 7)
called a FISON,
- as n-ary representation, for instance binary 111 or decimal 7,
- by indirect description like "the number of colours of the rainbow",
- by other words known to sender and receiver like "seven".
But if you really want to talk more about your junk definition of N...
If n is in it, then also n+1 is in it.
... then you fail. Because that's true of all the M but not of N which
is an unspecified subset of them.
I think it helps the reader to see your sleight of hand. I put it back
so they can see your switched from my talking about one definition --
the waffle that obviously can't be used to prove anything
So now we just have the problem that 1 is not provably in N as you
define it.
You are a dishonest liar.
Then just prove that 1 is in N as you define it.
ℕ is Cantor's infinite set.
Surely that can't be right. I thought your book is about potential
infinity, not actual infinity. You pretended to be happy with your
incorrect definition because your intelligent readers would assume the correct definition, but you don't want then to assume Cantor's infinite
set in your textbook, do you?
On 01.06.2025 13:53, Mikko wrote:
On 2025-05-31 13:40:12 +0000, WM said:
On 31.05.2025 11:59, Mikko wrote:
On 2025-05-30 14:25:22 +0000, WM said:
Here I use induction in Cantor's set. That is allowed. Cantor did it too. >>>>No, there is no artithmetic induction and no set induction there.
"daß die Reihe
1, i2, i3, ..., i, ...
nur eine Permutation der Reihe
1, 2, 3, ..., , ...
ist. Dies beweisen wir durch vollständige Induktion,"
[Cantor, collected works, p. 305]
Did Cantor acurally prove that with a complete induction? As far as
I have seen Cantor has proven what he promised to prove.
Cantor often used induction: vollständige Induktion. But that is irrelevant.
On 01.06.2025 13:58, Mikko wrote:
On 2025-05-31 13:47:51 +0000, WM said:
On 31.05.2025 12:11, Mikko wrote:
On 2025-05-30 14:46:55 +0000, WM said:
{1, 2, 3, ..., n, n+1} has infinitely many (ℵo) successors is a >>>>>>> consequence of {1, 2, 3, ..., n} has infinitely many (ℵo) successors, >>>>>>> and ℵo - 1 = ℵo.
Still no proof.
Proof that every definable natural number has more successors than
predecessors.
You have not shown that proof, either.
I have above. You cannot understand it. That is a different thing.
No proof of anything above. Besides, nothing that requires any
understanding beyond ordinary proof checking cannot be a proof.
There is no further understanding required.
On 2025-06-01 14:09:23 +0000, WM said:
On 01.06.2025 13:58, Mikko wrote:
On 2025-05-31 13:47:51 +0000, WM said:
On 31.05.2025 12:11, Mikko wrote:
On 2025-05-30 14:46:55 +0000, WM said:
{1, 2, 3, ..., n, n+1} has infinitely many (ℵo) successors is a >>>>>>>> consequence of {1, 2, 3, ..., n} has infinitely many (ℵo)
successors, and ℵo - 1 = ℵo.
Still no proof.
Proof that every definable natural number has more successors than >>>>>> predecessors.
You have not shown that proof, either.
I have above. You cannot understand it. That is a different thing.
No proof of anything above. Besides, nothing that requires any
understanding beyond ordinary proof checking cannot be a proof.
There is no further understanding required.
That contradicts your above "You cannot understand it".
On 2025-06-01 14:15:06 +0000, WM said:
On 01.06.2025 13:53, Mikko wrote:
On 2025-05-31 13:40:12 +0000, WM said:
On 31.05.2025 11:59, Mikko wrote:
On 2025-05-30 14:25:22 +0000, WM said:
Here I use induction in Cantor's set. That is allowed. Cantor did
it too.
No, there is no artithmetic induction and no set induction there.
"daß die Reihe
1, i2, i3, ..., i, ...
nur eine Permutation der Reihe
1, 2, 3, ..., , ...
ist. Dies beweisen wir durch vollständige Induktion,"
[Cantor, collected works, p. 305]
Did Cantor acurally prove that with a complete induction? As far as
I have seen Cantor has proven what he promised to prove.
Cantor often used induction: vollständige Induktion. But that is
irrelevant.
Indeed. An answer to my question would have been relevant but you didn't
give any. Apparently you don't know whther or how Cantor proved what he claimed in the sentence that is partially quoted.
As there is no complete
sentence in the quoted text fragemnt it is hard to say what exacly he was going to do but I can't imagine any completion of the sentence that does
not promise to use complete induction for something.
But if you really want to talk more about your junk definition of N...
If n is in it, then also n+1 is in it.... then you fail. Because that's true of all the M but not of N which
is an unspecified subset of them.
It is clear that ℕ has the same properties as all the M because
otherwise ℕ could be the empty set.
Then the axioms would be void. Every intelligent reader would
recognize that this cannot be the object of the paragraph.
So now we just have the problem that 1 is not provably in N as you
define it.
You are wrong.
You are a dishonest liar.
Yes, an intelligent reader will see you've written a junk definition
Then just prove that 1 is in N as you define it.
See above.
On 02.06.2025 03:56, Ben Bacarisse wrote:
WM <[email protected]> writes:
On 31.05.2025 02:20, Ben Bacarisse wrote:There are two conditions (as you know perfectly well) and it meets both
WM <[email protected]> writes:
It has been shown to the student by many arguments that the bijection >>>>> fails.Which of the conditions of being a bijection (as presented in your book) >>>> does b fail to meet? (I'm betting you won't say.)
The condition to be definable.
(as you also know perfectly well). Here are the conditions:
bijektiv (oder eineindeutig), wenn f injektiv und surjektiv ist
Is b not injective? Is b not surjektiv? Here's b again so you can
check for yourself that it is both:
Not all natural numbers of Cantor's set can be individually defined:
You really don't want to say, do you?Do you always refuse to answer simple questions? Do you have to write >>>> marking schemes for your exams? I'm just trying to find out if there is >>>> documentary evidence of what a student at your college has to write to >>>> get full marks.
Here are two:
https://www.hs-augsburg.de/~mueckenh/GU/Pruefung%20GU1001.pdf https://www.hs-augsburg.de/~mueckenh/GU/Pruefung%20GU1007.pdf
By the way, I can see why you don't want to show any marking scheme. It
would have to include the junk you expect the students to say,
Why do you call it junk?
You cannot contradict even one of many proofs.
If you could, you would not talk about nonsense like whether 1 is a
definable natural number.
On 03.06.2025 10:08, Mikko wrote:
On 2025-06-01 14:15:06 +0000, WM said:
On 01.06.2025 13:53, Mikko wrote:
On 2025-05-31 13:40:12 +0000, WM said:
On 31.05.2025 11:59, Mikko wrote:
On 2025-05-30 14:25:22 +0000, WM said:
Here I use induction in Cantor's set. That is allowed. Cantor did it too.
No, there is no artithmetic induction and no set induction there.
"daß die Reihe
1, i2, i3, ..., i, ...
nur eine Permutation der Reihe
1, 2, 3, ..., , ...
ist. Dies beweisen wir durch vollständige Induktion,"
[Cantor, collected works, p. 305]
Did Cantor acurally prove that with a complete induction? As far as
I have seen Cantor has proven what he promised to prove.
Cantor often used induction: vollständige Induktion. But that is irrelevant.
Indeed. An answer to my question would have been relevant but you didn't
give any. Apparently you don't know whther or how Cantor proved what he
claimed in the sentence that is partially quoted.
I know it, but it is rather tedious. If you are interested, you may
look it up yourself or refer to the shorter proof I attach below.
I had only to show that in Cantor's set theory proofs by arithmetic
induction are possible.
That confirms my proof:
ℵo - 1 = ℵo
P[1]: {1} has infinitely many (ℵo) successors.
P[n]: {1, 2, 3, ..., n} has infinitely many (ℵo) successors.
P[n+1]: {1, 2, 3, ..., n, n+1} has infinitely many (ℵo) successors.
As there is no complete
sentence in the quoted text fragemnt it is hard to say what exacly he was
going to do but I can't imagine any completion of the sentence that does
not promise to use complete induction for something.
Here Cantor shows a shorter application of induction:
Ich schicke folgenden allgemeinen, höchst einleuchtenden Hilfssatz
voraus: sind irgend zwei Mengen M und N äquivalent, so können sie (im allgemeinen auf viele Weisen) so in gegenseitig eindeutige und
vollständige Zuordnung gebracht werden, daß bei dieser Zurodnung einem beliebig vorgegebenen Elemente m von M ein ebenso beliebig gewähltes
Element n von N entspricht.
Und nun wird zum Beweise des in Rede stehenden Satzes ein vollständiges Induktionsverfahren eingeleitet.
Man setze eine Menge M voraus, welche keinem ihrer Bestandteile äquivalent ist; ich will zeigen, daß alsdann auch die aus M durch Hinzufügung eines neuen Elementes l hervorgehende Menge M + l dieselbe Eigenschaft hat, mit keinem ihrer Bestandteile äquivalent zu sein. Sei
N irgendein Bestandteil von M + l, so kann er zwei Fälle darbieten. 1)
Es gehört das Element l mit zu N, so daß N = N' + l. N' ist dann
offenbar auch Bestandteil von M. Wäre nun N ~ M + l, so könnte nach
obigem Hilfssatze zwischen den Mengen N und M + l eine solche
gegenseitig eindeutige und vollständige Korrespondenz hergestellt
werden, daß das Element l von N dem Element l von M + l entspricht;
durch diese Zuordnung würde auch eine Zuordnung zwischen N' nd M
hergestellt sein und es wäre M seinem Bestandteil N' äquivalent, gegen unsere Voraussetzung. 2) Es gehört l nicht mit zu N; dann ist N nicht
nur Bestandteil von M + l sondern auch von M. Wäre in diesem Falle N ~
M + l, so nehme man irgendeine gegenseitig eindeutige vollständige
Zuordnung der beiden Mengen M + l und N und es möge bei derselben dem Elemente l von M + l das Element n vonN entsprechen. Ist N =N' + n, so
wäre durch diese Zuordnung auch eine gegenseitig eindeutige und vollständige Korrespondenz zwischen N' und M hergestellt, was, da auch
hier N' Bestandteil von M ist, gegen die gemachte Voraussetzung
streitet, wonach M keinem ihrer Bestandteile äquivalent ist.
[Cantor's collected works p. 415]
On 2025-06-03 13:17:57 +0000, WM said:
I had only to show that in Cantor's set theory proofs by arithmetic
induction are possible.
Which you didn't show.
That confirms my proof:
ℵo - 1 = ℵo
P[1]: {1} has infinitely many (ℵo) successors.
P[n]: {1, 2, 3, ..., n} has infinitely many (ℵo) successors.
P[n+1]: {1, 2, 3, ..., n, n+1} has infinitely many (ℵo) successors.
So far good. But no P[n] -> P[n+1] and no induction.
As there is no complete
sentence in the quoted text fragemnt it is hard to say what exacly he
was
going to do but I can't imagine any completion of the sentence that does >>> not promise to use complete induction for something.
Here Cantor shows a shorter application of induction:
Ich schicke folgenden allgemeinen, höchst einleuchtenden Hilfssatz
voraus: sind irgend zwei Mengen M und N äquivalent, so können sie (im
allgemeinen auf viele Weisen) so in gegenseitig eindeutige und
vollständige Zuordnung gebracht werden, daß bei dieser Zurodnung einem
beliebig vorgegebenen Elemente m von M ein ebenso beliebig gewähltes
Element n von N entspricht.
Und nun wird zum Beweise des in Rede stehenden Satzes ein
vollständiges Induktionsverfahren eingeleitet.
Man setze eine Menge M voraus, welche keinem ihrer Bestandteile
äquivalent ist; ich will zeigen, daß alsdann auch die aus M durch
Hinzufügung eines neuen Elementes l hervorgehende Menge M + l dieselbe
Eigenschaft hat, mit keinem ihrer Bestandteile äquivalent zu sein.
Sei N irgendein Bestandteil von M + l, so kann er zwei Fälle
darbieten. 1) Es gehört das Element l mit zu N, so daß N = N' + l. N'
ist dann offenbar auch Bestandteil von M. Wäre nun N ~ M + l, so
könnte nach obigem Hilfssatze zwischen den Mengen N und M + l eine
solche gegenseitig eindeutige und vollständige Korrespondenz
hergestellt werden, daß das Element l von N dem Element l von M + l
entspricht; durch diese Zuordnung würde auch eine Zuordnung zwischen
N' nd M hergestellt sein und es wäre M seinem Bestandteil N'
äquivalent, gegen unsere Voraussetzung. 2) Es gehört l nicht mit zu N;
dann ist N nicht nur Bestandteil von M + l sondern auch von M. Wäre in
diesem Falle N ~ M + l, so nehme man irgendeine gegenseitig eindeutige
vollständige Zuordnung der beiden Mengen M + l und N und es möge bei
derselben dem Elemente l von M + l das Element n vonN entsprechen. Ist
N =N' + n, so wäre durch diese Zuordnung auch eine gegenseitig
eindeutige und vollständige Korrespondenz zwischen N' und M
hergestellt, was, da auch hier N' Bestandteil von M ist, gegen die
gemachte Voraussetzung streitet, wonach M keinem ihrer Bestandteile
äquivalent ist.
[Cantor's collected works p. 415]
That is an indirect proof.
You seem to prefer direct proofs.
WM <[email protected]> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte
des Unendlichen" and "Kleine Geschichte der Mathematik" at Technische Hochschule Augsburg.)
On 02.06.2025 03:56, Ben Bacarisse wrote:
WM <[email protected]> writes:
Not all natural numbers of Cantor's set can be individually defined:
Not an answer. Is b not injective? Is b not surjektiv?
...
You really don't want to say, do you?Do you always refuse to answer simple questions? Do you have to write >>>>> marking schemes for your exams? I'm just trying to find out if there is >>>>> documentary evidence of what a student at your college has to write to >>>>> get full marks.
Here are two:
https://www.hs-augsburg.de/~mueckenh/GU/Pruefung%20GU1001.pdf
https://www.hs-augsburg.de/~mueckenh/GU/Pruefung%20GU1007.pdf
Thank you. Why avoid saying just "yes" for so long?
Your students need to say thing like "Das Cantorsche Diagonalargument
ist falsch" to get full marks.
You cannot contradict even one of many proofs.
Not to your satisfaction, no.
Cantor, Church, Turing, heck, even my old professor Swinnerton-Dyer
could not get full marks on your silly questions.
Your proofs have a
lot of waffle, but sound proofs need good definitions
WM <[email protected]> writes:
It is clear that ℕ has the same properties as all the M because
otherwise ℕ could be the empty set.
It is clear that N /should/ have the same properties as all the M, but
your definition does not achieve that effect. N, as you define it, can indeed be empty.
Then the axioms would be void. Every intelligent reader would
recognize that this cannot be the object of the paragraph.
Yes. Every reader will know your definition of N is fundamentally wrong.
On 04.06.2025 08:43, Mikko wrote:
On 2025-06-03 13:17:57 +0000, WM said:
I had only to show that in Cantor's set theory proofs by arithmetic
induction are possible.
Which you didn't show.
Cantor shows it.
That confirms my proof:
ℵo - 1 = ℵo
P[1]: {1} has infinitely many (ℵo) successors.
P[n]: {1, 2, 3, ..., n} has infinitely many (ℵo) successors.
P[n+1]: {1, 2, 3, ..., n, n+1} has infinitely many (ℵo) successors.
So far good. But no P[n] -> P[n+1] and no induction.
ℵo - 1 = ℵo.
As there is no complete
sentence in the quoted text fragemnt it is hard to say what exacly he was >>>> going to do but I can't imagine any completion of the sentence that does >>>> not promise to use complete induction for something.
Here Cantor shows a shorter application of induction:
Ich schicke folgenden allgemeinen, höchst einleuchtenden Hilfssatz
voraus: sind irgend zwei Mengen M und N äquivalent, so können sie (im
allgemeinen auf viele Weisen) so in gegenseitig eindeutige und
vollständige Zuordnung gebracht werden, daß bei dieser Zurodnung einem >>> beliebig vorgegebenen Elemente m von M ein ebenso beliebig gewähltes
Element n von N entspricht.
Und nun wird zum Beweise des in Rede stehenden Satzes ein
vollständiges Induktionsverfahren eingeleitet.
Man setze eine Menge M voraus, welche keinem ihrer Bestandteile
äquivalent ist; ich will zeigen, daß alsdann auch die aus M durch
Hinzufügung eines neuen Elementes l hervorgehende Menge M + l dieselbe
Eigenschaft hat, mit keinem ihrer Bestandteile äquivalent zu sein. Sei >>> N irgendein Bestandteil von M + l, so kann er zwei Fälle darbieten. 1)
Es gehört das Element l mit zu N, so daß N = N' + l. N' ist dann
offenbar auch Bestandteil von M. Wäre nun N ~ M + l, so könnte nach
obigem Hilfssatze zwischen den Mengen N und M + l eine solche
gegenseitig eindeutige und vollständige Korrespondenz hergestellt
werden, daß das Element l von N dem Element l von M + l entspricht;
durch diese Zuordnung würde auch eine Zuordnung zwischen N' nd M
hergestellt sein und es wäre M seinem Bestandteil N' äquivalent, gegen >>> unsere Voraussetzung. 2) Es gehört l nicht mit zu N; dann ist N nicht
nur Bestandteil von M + l sondern auch von M. Wäre in diesem Falle N ~
M + l, so nehme man irgendeine gegenseitig eindeutige vollständige
Zuordnung der beiden Mengen M + l und N und es möge bei derselben dem
Elemente l von M + l das Element n vonN entsprechen. Ist N =N' + n, so
wäre durch diese Zuordnung auch eine gegenseitig eindeutige und
vollständige Korrespondenz zwischen N' und M hergestellt, was, da auch
hier N' Bestandteil von M ist, gegen die gemachte Voraussetzung
streitet, wonach M keinem ihrer Bestandteile äquivalent ist.
[Cantor's collected works p. 415]
That is an indirect proof.
It is applying induction in set theory.
You seem to prefer direct proofs.
That is irrelevant.
But in fact it supplies the shortest proof that not all natural numbers
of Cantor's set can be individually defined:
Since all natural numbers can be reduced to the empty set by
subtracting them collectively,
ℕ \ {1, 2, 3, ...} = { }
they could also be reduced to the empty set by subtracting them
individually - if this was possible. But then the well-order would
force the existence of a last one. Contradiction.
On 2025-06-04 17:32:28 +0000, WM said:
On 04.06.2025 08:43, Mikko wrote:
On 2025-06-03 13:17:57 +0000, WM said:
I had only to show that in Cantor's set theory proofs by arithmetic
induction are possible.
Which you didn't show.
Cantor shows it.
That confirms my proof:
ℵo - 1 = ℵo
P[1]: {1} has infinitely many (ℵo) successors.
P[n]: {1, 2, 3, ..., n} has infinitely many (ℵo) successors.
P[n+1]: {1, 2, 3, ..., n, n+1} has infinitely many (ℵo) successors.
So far good. But no P[n] -> P[n+1] and no induction.
ℵo - 1 = ℵo.
No -> there.
It is relevant to the extent that you cannot learn form it how a direct
proof should be presented.
The expression "subtracting them individually" should be represented mathematically, e.g. a sequence. Informal expressions tend to lead
to bad proofs.
On 04.06.2025 02:11, Ben Bacarisse wrote:
WM <[email protected]> writes:
It is clear that ℕ has the same properties as all the M becauseIt is clear that N /should/ have the same properties as all the M, but
otherwise ℕ could be the empty set.
your definition does not achieve that effect. N, as you define it, can
indeed be empty.
With that point I could sort out stupid students, if necessary.
Then the axioms would be void. Every intelligent reader wouldYes. Every reader will know your definition of N is fundamentally wrong.
recognize that this cannot be the object of the paragraph.
Ever stupid reader, unable to understand the meaning.
No reason to talk about that any longer. If you wanted to mimic an
idiot, I can congratulate you.
On 04.06.2025 02:35, Ben Bacarisse wrote:
WM <[email protected]> writes:
On 02.06.2025 03:56, Ben Bacarisse wrote:
WM <[email protected]> writes:
Not all natural numbers of Cantor's set can be individually defined:Not an answer. Is b not injective? Is b not surjektiv?
It is for the set of definable numbers, it is not for the dark
numbers.
Your students need to say thing like "Das Cantorsche Diagonalargument
ist falsch" to get full marks.
So it is. The reason is what you refuse to answer:
Not all natural numbers of Cantor's set can be individually defined:
All natural numbers can be thought as defining the diagonal but not individually. The well-order would force the existence of a last
one. Contradiction.
Therefore most indices of the diagonal elements are undefined, dark.
You cannot contradict even one of many proofs.Not to your satisfaction, no.
But I have shown my students how it goes.
On 05.06.2025 09:32, Mikko wrote:
On 2025-06-04 17:32:28 +0000, WM said:
On 04.06.2025 08:43, Mikko wrote:
On 2025-06-03 13:17:57 +0000, WM said:
I had only to show that in Cantor's set theory proofs by arithmetic
induction are possible.
Which you didn't show.
Cantor shows it.
That confirms my proof:So far good. But no P[n] -> P[n+1] and no induction.
ℵo - 1 = ℵo
P[1]: {1} has infinitely many (ℵo) successors.
P[n]: {1, 2, 3, ..., n} has infinitely many (ℵo) successors.
P[n+1]: {1, 2, 3, ..., n, n+1} has infinitely many (ℵo) successors. >>>>
ℵo - 1 = ℵo.
No -> there.
If {1, 2, 3, ..., n} has infinitely many (ℵo) successors, then {1, 2,
3, ..., n, n+1} has one successor less, that is infinitely many (ℵo) successors.
It is relevant to the extent that you cannot learn form it how a direct
proof should be presented.
I need not learn it, because I did it.
The expression "subtracting them individually" should be represented
mathematically, e.g. a sequence. Informal expressions tend to lead
to bad proofs.
This expression is represented by
((((ℕ \ {1}) \ {2}) \ {3}) ...) = { }
On 2025-06-05 20:36:45 +0000, WM said:
On 05.06.2025 09:32, Mikko wrote:
On 2025-06-04 17:32:28 +0000, WM said:
The expression "subtracting them individually" should be represented
mathematically, e.g. a sequence. Informal expressions tend to lead
to bad proofs.
This expression is represented by
((((ℕ \ {1}) \ {2}) \ {3}) ...) = { }
That may be good enough when you want to prove something that we already believe anyway. But for a sufficiently rigorous proof of something else
the "..." should be replaced with something more mathematicsl.
WM <[email protected]> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte
des Unendlichen" and "Kleine Geschichte der Mathematik" at Technische Hochschule Augsburg.)
On 04.06.2025 02:35, Ben Bacarisse wrote:
WM <[email protected]> writes:
On 02.06.2025 03:56, Ben Bacarisse wrote:
WM <[email protected]> writes:
Not all natural numbers of Cantor's set can be individually defined:Not an answer. Is b not injective? Is b not surjektiv?
It is for the set of definable numbers, it is not for the dark
numbers.
No, the topic is your exam papers and the nonsense
that a student must
accept to get full marks.
b is both injective and surjective.
(...(((ℕ \ {1}) \ {2}) \ {3}) ...) = { }Your students need to say thing like "Das Cantorsche Diagonalargument
ist falsch" to get full marks.
So it is. The reason is what you refuse to answer:
Not all natural numbers of Cantor's set can be individually defined:
All natural numbers can be thought as defining the diagonal but not
individually. The well-order would force the existence of a last
one. Contradiction.
Therefore most indices of the diagonal elements are undefined, dark.
You cannot contradict even one of many proofs.Not to your satisfaction, no.
But I have shown my students how it goes.
I feel for any student who knows how mathematics works. With luck they
know how German exams work as well and will just write stuff they know
to be wrong so they get the marks.
WM <[email protected]> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte
des Unendlichen" and "Kleine Geschichte der Mathematik" at Technische Hochschule Augsburg.)
On 04.06.2025 02:11, Ben Bacarisse wrote:
WM <[email protected]> writes:
It is clear that ℕ has the same properties as all the M becauseIt is clear that N /should/ have the same properties as all the M, but
otherwise ℕ could be the empty set.
your definition does not achieve that effect. N, as you define it, can
indeed be empty.
With that point I could sort out stupid students, if necessary.
With that definition of N I could sort out stupid teachers. Are you not
even a little ashamed to admit that your readers have to realise that
your definition of N is wrong and must simply assume you meant something else?
On 06.06.2025 09:37, Mikko wrote:
On 2025-06-05 20:36:45 +0000, WM said:
On 05.06.2025 09:32, Mikko wrote:
On 2025-06-04 17:32:28 +0000, WM said:
The expression "subtracting them individually" should be represented
mathematically, e.g. a sequence. Informal expressions tend to lead
to bad proofs.
This expression is represented by
((((ℕ \ {1}) \ {2}) \ {3}) ...) = { }
That may be good enough when you want to prove something that we already
believe anyway. But for a sufficiently rigorous proof of something else
the "..." should be replaced with something more mathematicsl.
The "..." can be replaced with the singletons of definable natural numbers
On 05.06.2025 23:43, Ben Bacarisse wrote:
WM <[email protected]> writes:
On 04.06.2025 02:11, Ben Bacarisse wrote:With that definition of N I could sort out stupid teachers. Are you not
WM <[email protected]> writes:
It is clear that ℕ has the same properties as all the M becauseIt is clear that N /should/ have the same properties as all the M, but >>>> your definition does not achieve that effect. N, as you define it, can >>>> indeed be empty.
otherwise ℕ could be the empty set.
With that point I could sort out stupid students, if necessary.
even a little ashamed to admit that your readers have to realise that
your definition of N is wrong and must simply assume you meant something
else?
It is not wrong. Text need to be interpreted.
Repeating the obvious
condition that also ℕ must adhere to the given axioms might be appropriate in a scientific paper but not in a textbook where long-winded sentences hinder the comprehension.
Pedagogical considerations have priority in my book for the first
semesters.
On 05.06.2025 23:51, Ben Bacarisse wrote:
WM <[email protected]> writes:
On 04.06.2025 02:35, Ben Bacarisse wrote:No, the topic is your exam papers and the nonsense
WM <[email protected]> writes:
On 02.06.2025 03:56, Ben Bacarisse wrote:
WM <[email protected]> writes:
Not all natural numbers of Cantor's set can be individually defined:Not an answer. Is b not injective? Is b not surjektiv?
It is for the set of definable numbers, it is not for the dark
numbers.
You cannot disprove any of my results.
that a student must
accept to get full marks.
Have you something substantial to say about dark numbers?
Q+ and you won't say whether it is not injective and/or surjective.
(...(((ℕ \ {1}) \ {2}) \ {3}) ...) = { }Your students need to say thing like "Das Cantorsche Diagonalargument
ist falsch" to get full marks.
So it is. The reason is what you refuse to answer:
is wrong for all singletons of definable numbers.
Not all natural numbers of Cantor's set can be individually defined:
All natural numbers can be thought as defining the diagonal but not
individually. The well-order would force the existence of a last
one. Contradiction.
Therefore most indices of the diagonal elements are undefined, dark.
You cannot contradict even one of many proofs.Not to your satisfaction, no.
Not at all. Then you would try it.
I feel for any student who knows how mathematics works. With luck they
But I have shown my students how it goes.
know how German exams work as well and will just write stuff they know
to be wrong so they get the marks.
You "know" what is wrong without being able to disprove it. That is
not the way to pass an exam.
WM <[email protected]> writes:
1 ∈ M (4.1)
n ∈ M ⇒ (n + 1) ∈ M (4.2)
If M satisfies (4.1) and (4.2), then ℕ ⊆ M.
So when I challenged you to prove that 1 is in N (as you define it),
your proof would now be:
Ignore where the definition says that ℕ ⊆ M
WM <[email protected]> writes:
You cannot disprove any of my results.
You have admitted that in WMaths you can't (yet?) define set membership,
Is this still the state of WMaths
Have you something substantial to say about dark numbers?
Have you nothing substantial to say about why the exam questions you
posed and challenged me to answer was wrong? I posted a function b : N
Q+ and you won't say whether it is not injective and/or surjective.
You "know" what is wrong without being able to disprove it.
I know how to prove that b is both injective and surjective.
On 2025-06-06 10:47:33 +0000, WM said:
On 06.06.2025 09:37, Mikko wrote:
On 2025-06-05 20:36:45 +0000, WM said:
On 05.06.2025 09:32, Mikko wrote:
On 2025-06-04 17:32:28 +0000, WM said:
The expression "subtracting them individually" should be represented >>>>> mathematically, e.g. a sequence. Informal expressions tend to lead
to bad proofs.
This expression is represented by
((((ℕ \ {1}) \ {2}) \ {3}) ...) = { }
That may be good enough when you want to prove something that we already >>> believe anyway. But for a sufficiently rigorous proof of something else
the "..." should be replaced with something more mathematicsl.
The "..." can be replaced with the singletons of definable natural
numbers
Then do so.
On 08.06.2025 07:10, Mikko wrote:
On 2025-06-06 10:47:33 +0000, WM said:
On 06.06.2025 09:37, Mikko wrote:Then do so.
On 2025-06-05 20:36:45 +0000, WM said:
On 05.06.2025 09:32, Mikko wrote:
On 2025-06-04 17:32:28 +0000, WM said:
The expression "subtracting them individually" should be represented >>>>>> mathematically, e.g. a sequence. Informal expressions tend to lead >>>>>> to bad proofs.
This expression is represented by
((((ℕ \ {1}) \ {2}) \ {3}) ...) = { }
That may be good enough when you want to prove something that we already >>>> believe anyway. But for a sufficiently rigorous proof of something else >>>> the "..." should be replaced with something more mathematicsl.
The "..." can be replaced with the singletons of definable natural numbers >>
See above. Three are already given. More can be inserted. But it is impossible to replace them with all natural numbers.
On 03.06.2025 10:11, Mikko wrote:
On 2025-06-01 14:09:23 +0000, WM said:
On 01.06.2025 13:58, Mikko wrote:
On 2025-05-31 13:47:51 +0000, WM said:
On 31.05.2025 12:11, Mikko wrote:
On 2025-05-30 14:46:55 +0000, WM said:
{1, 2, 3, ..., n, n+1} has infinitely many (ℵo) successors is a >>>>>>>>> consequence of {1, 2, 3, ..., n} has infinitely many (ℵo) successors,
and ℵo - 1 = ℵo.
Still no proof.
Proof that every definable natural number has more successors than >>>>>>> predecessors.
You have not shown that proof, either.
I have above. You cannot understand it. That is a different thing.
No proof of anything above. Besides, nothing that requires any
understanding beyond ordinary proof checking cannot be a proof.
There is no further understanding required.
That contradicts your above "You cannot understand it".
No, ordinary proof checking and your understanding are two different things.
Every proof checker can confirm that every definable natural number has finitely many predecessors.
If there are more than finitely many numbers following upon every
definable number as successors, then every definable number has more successors than predecessors.
All can be removed collectively. ℕ \ {1, 2, 3, ...} = { }.
If all could be removed as individuals, then a last one would be
removed. Contradiction.
This is a proof of dark natural numbers.
On 2025-06-03 13:26:11 +0000, WM said:
If there are more than finitely many numbers following upon every
definable number as successors, then every definable number has more
successors than predecessors.
Provable.
All can be removed collectively. ℕ \ {1, 2, 3, ...} = { }.
The word "collectively" is not a well defined mathematical term.
It can
be used in informal presentation but not in a rigorous proof.
If all could be removed as individuals, then a last one would be
removed. Contradiction.
Likewise, "as individuals" has no well defined mathematical meaning
so cannot be used in a rigorous proofs.
This is a proof of dark natural numbers.
It is not a proof of anything.
|ℕ \ {m ∈ ℕ | m < n+1}| = ℵo.shows that is impossible to extend definability to all natural numbers
On 2025-06-10 17:06:18 +0000, WM said:
This expression is represented by
((((ℕ \ {1}) \ {2}) \ {3}) ...) = { }
That may be good enough when you want to prove something that we
already
believe anyway. But for a sufficiently rigorous proof of something
else
the "..." should be replaced with something more mathematicsl.
The "..." can be replaced with the singletons of definable natural
numbers
Then do so.
See above. Three are already given. More can be inserted. But it is
impossible to replace them with all natural numbers.
I can see that you stll have no rigorous proofs,
and can't have as you
apparently can't even write your claim without "..." and other informal expressions.
|ℕ \ {m ∈ ℕ | m < n+1}| = ℵo.
On 11.06.2025 10:01, Mikko wrote:
On 2025-06-03 13:26:11 +0000, WM said:
If there are more than finitely many numbers following upon every
definable number as successors, then every definable number has more
successors than predecessors.
Provable.
All can be removed collectively. ℕ \ {1, 2, 3, ...} = { }.
The word "collectively" is not a well defined mathematical term.
It is well defined because collectively means the whole set.
It can
be used in informal presentation but not in a rigorous proof.
If all could be removed as individuals, then a last one would be
removed. Contradiction.
Likewise, "as individuals" has no well defined mathematical meaning
so cannot be used in a rigorous proofs.
"As an element" has a well defined meaning. But it is useless to be well-defined in a wrong theory like ZF where rigorous proofs fail to establish the existence of dark numbers.
Outside of ZF in correct mathematics this proof for all definable numbers n |ℕ \ {1}| = ℵo.
This is a proof of dark natural numbers.
It is not a proof of anything.
|ℕ \ {m ∈ ℕ | m < n}| = ℵo
|ℕ \ {m ∈ ℕ | m < n+1}| = ℵo.shows that is impossible to extend definability to all natural numbers
with none remaining undefined.
On 11.06.2025 09:50, Mikko wrote:
On 2025-06-10 17:06:18 +0000, WM said:
This expression is represented by
((((ℕ \ {1}) \ {2}) \ {3}) ...) = { }
That may be good enough when you want to prove something that we already >>>>>> believe anyway. But for a sufficiently rigorous proof of something else >>>>>> the "..." should be replaced with something more mathematicsl.
The "..." can be replaced with the singletons of definable natural numbers
Then do so.
See above. Three are already given. More can be inserted. But it is
impossible to replace them with all natural numbers.
I can see that you stll have no rigorous proofs,
I have facts.
and can't have as you
apparently can't even write your claim without "..." and other informal
expressions.
Of course I can:
For all natural numbers that can be chosen as individuals:
|ℕ \ {1, 2, 3, ..., n}| = ℵo.
The "..." can be removed by the inductive proof
for all definable n ∈ ℕ:
|ℕ \ {1}| = ℵo.
|ℕ \ {m ∈ ℕ | m < n}| = ℵo
|ℕ \ {m ∈ ℕ | m < n+1}| = ℵo.
Result: It is impossible to define all natural numbers with none
remaining undefined.
On 2025-06-11 11:38:52 +0000, WM said:
Outside of ZF in correct mathematics this proof for all definable
numbers n
|ℕ \ {1}| = ℵo.
|ℕ \ {m ∈ ℕ | m < n}| = ℵo
|ℕ \ {m ∈ ℕ | m < n+1}| = ℵo.shows that is impossible to extend definability to all natural numbers
with none remaining undefined.
It does not show that. The "proof" does not even mention definability.
The conclusion follows from the second sentence alone when n is understood
to be universally quantified, so the first sentence is not needed and
should
not be there.
On 2025-06-11 11:30:26 +0000, WM said:
For all natural numbers that can be chosen as individuals:
|ℕ \ {1, 2, 3, ..., n}| = ℵo.
The "..." can be removed by the inductive proof
for all definable n ∈ ℕ:
|ℕ \ {1}| = ℵo.
|ℕ \ {m ∈ ℕ | m < n}| = ℵo
|ℕ \ {m ∈ ℕ | m < n+1}| = ℵo.
That the set difference of an infinite set and a finite set is infinite
is well understood and therefore an uninteresting result.
Result: It is impossible to define all natural numbers with none
remaining undefined.
You have not derived that "result".
On 12.06.2025 10:00, Mikko wrote:
On 2025-06-11 11:38:52 +0000, WM said:
Outside of ZF in correct mathematics this proof for all definable numbers n >>> |ℕ \ {1}| = ℵo.
|ℕ \ {m ∈ ℕ | m < n}| = ℵo
|ℕ \ {m ∈ ℕ | m < n+1}| = ℵo.shows that is impossible to extend definability to all natural numbers
with none remaining undefined.
It does not show that. The "proof" does not even mention definability.
"this proof for all definable numbers n"
The conclusion follows from the second sentence alone when n is understood >> to be universally quantified, so the first sentence is not needed and should >> not be there.
It is there. Only definablenumbers can be quantified as individuals.
On 12.06.2025 09:56, Mikko wrote:
On 2025-06-11 11:30:26 +0000, WM said:
For all natural numbers that can be chosen as individuals:
|ℕ \ {1, 2, 3, ..., n}| = ℵo.
The "..." can be removed by the inductive proof
for all definable n ∈ ℕ:
|ℕ \ {1}| = ℵo.
|ℕ \ {m ∈ ℕ | m < n}| = ℵo
|ℕ \ {m ∈ ℕ | m < n+1}| = ℵo.
That the set difference of an infinite set and a finite set is infinite
is well understood and therefore an uninteresting result.
That is not the only result.
Interesting is that all defined numbers belong to a finite initial segment.
Therefore most numbers are dark.
On 2025-06-12 09:44:06 +0000, WM said:
On 12.06.2025 10:00, Mikko wrote:
On 2025-06-11 11:38:52 +0000, WM said:
Outside of ZF in correct mathematics this proof for all definable
numbers n
|ℕ \ {1}| = ℵo.
|ℕ \ {m ∈ ℕ | m < n}| = ℵo
|ℕ \ {m ∈ ℕ | m < n+1}| = ℵo.shows that is impossible to extend definability to all natural
numbers with none remaining undefined.
It does not show that. The "proof" does not even mention definability.
"this proof for all definable numbers n"
That is a part of the sentence that mentions the proof, not a part of
the proof itself.
The conclusion follows from the second sentence alone when n is
understood
to be universally quantified, so the first sentence is not needed and
should
not be there.
It is there. Only definablenumbers can be quantified as individuals.
Quantification is never "as individuals".
On 2025-06-12 09:41:51 +0000, WM said:
On 12.06.2025 09:56, Mikko wrote:
On 2025-06-11 11:30:26 +0000, WM said:
For all natural numbers that can be chosen as individuals:
|ℕ \ {1, 2, 3, ..., n}| = ℵo.
The "..." can be removed by the inductive proof
for all definable n ∈ ℕ:
|ℕ \ {1}| = ℵo.
|ℕ \ {m ∈ ℕ | m < n}| = ℵo
|ℕ \ {m ∈ ℕ | m < n+1}| = ℵo.
That the set difference of an infinite set and a finite set is infinite
is well understood and therefore an uninteresting result.
That is not the only result.
That is the result of your proof. Other results can be discussed in
other contexts.
Interesting is that all defined numbers belong to a finite initial
segment.
About natural numbers that is obvious. One interesting thing is that
rational numbers can be ordered so that every one of them belongs
to a finite initial segment.
Another intresting thing is that real
numbers cannot be ordered so.
Note that in order to dscuss infinities you must use second or higher
order logic. In first order logic you cannot say that an initial
segment is finite.
Therefore most numbers are dark.
That does not follow.
On 13.06.2025 11:53, Mikko wrote:
On 2025-06-12 09:41:51 +0000, WM said:
On 12.06.2025 09:56, Mikko wrote:
On 2025-06-11 11:30:26 +0000, WM said:
For all natural numbers that can be chosen as individuals:
|ℕ \ {1, 2, 3, ..., n}| = ℵo.
The "..." can be removed by the inductive proof
for all definable n ∈ ℕ:
|ℕ \ {1}| = ℵo.
|ℕ \ {m ∈ ℕ | m < n}| = ℵo
|ℕ \ {m ∈ ℕ | m < n+1}| = ℵo.
That the set difference of an infinite set and a finite set is infinite >>>> is well understood and therefore an uninteresting result.
That is not the only result.
That is the result of your proof. Other results can be discussed in
other contexts.
The following result cannot be circumvented: After mankind will have ceased,
there is a largest natural number ever named.
On 13.06.2025 12:02, Mikko wrote:
On 2025-06-12 09:44:06 +0000, WM said:
On 12.06.2025 10:00, Mikko wrote:
On 2025-06-11 11:38:52 +0000, WM said:
Outside of ZF in correct mathematics this proof for all definable numbers n
|ℕ \ {1}| = ℵo.
|ℕ \ {m ∈ ℕ | m < n}| = ℵo
|ℕ \ {m ∈ ℕ | m < n+1}| = ℵo.shows that is impossible to extend definability to all natural numbers >>>>> with none remaining undefined.
It does not show that. The "proof" does not even mention definability.
"this proof for all definable numbers n"
That is a part of the sentence that mentions the proof, not a part of
the proof itself.
It is part of the proof because it specifies the set which the proof is
based upon.
The conclusion follows from the second sentence alone when n is understood >>>> to be universally quantified, so the first sentence is not needed and should
not be there.
It is there. Only definablenumbers can be quantified as individuals.
Quantification is never "as individuals".
It is always concerning individuals only.
∀n ∈ ℕ: P(n) specifies that every natural number n as an individual has the property P.
On 2025-06-13 13:49:25 +0000, WM said:
On 13.06.2025 12:02, Mikko wrote:
On 2025-06-12 09:44:06 +0000, WM said:
On 12.06.2025 10:00, Mikko wrote:
On 2025-06-11 11:38:52 +0000, WM said:"this proof for all definable numbers n"
Outside of ZF in correct mathematics this proof for all definable
numbers n
|ℕ \ {1}| = ℵo.
|ℕ \ {m ∈ ℕ | m < n}| = ℵo
|ℕ \ {m ∈ ℕ | m < n+1}| = ℵo.shows that is impossible to extend definability to all natural
numbers with none remaining undefined.
It does not show that. The "proof" does not even mention definability. >>>>
That is a part of the sentence that mentions the proof, not a part of
the proof itself.
It is part of the proof because it specifies the set which the proof
is based upon.
Doesn't really matter because the "proof" does not end with an interesting concousion.
The conclusion follows from the second sentence alone when n is
understood
to be universally quantified, so the first sentence is not needed
and should
not be there.
It is there. Only definablenumbers can be quantified as individuals.
Quantification is never "as individuals".
It is always concerning individuals only.
Therefore "as individuals" is meaningless.
∀n ∈ ℕ: P(n) specifies that every natural number n as an individual
has the property P.
That is an ordinary quantification, not a quantification as individuals.
On 2025-06-13 13:45:20 +0000, WM said:
On 13.06.2025 11:53, Mikko wrote:
On 2025-06-12 09:41:51 +0000, WM said:
On 12.06.2025 09:56, Mikko wrote:
On 2025-06-11 11:30:26 +0000, WM said:
For all natural numbers that can be chosen as individuals:
|ℕ \ {1, 2, 3, ..., n}| = ℵo.
The "..." can be removed by the inductive proof
for all definable n ∈ ℕ:
|ℕ \ {1}| = ℵo.
|ℕ \ {m ∈ ℕ | m < n}| = ℵo
|ℕ \ {m ∈ ℕ | m < n+1}| = ℵo.
That the set difference of an infinite set and a finite set is
infinite
is well understood and therefore an uninteresting result.
That is not the only result.
That is the result of your proof. Other results can be discussed in
other contexts.
The following result cannot be circumvented: After mankind will have
ceased,
who will care?
there is a largest natural number ever named.
That has no mathematical significance. Whether a number is named or not
is a feature of mankind, not of the number.
On 14.06.2025 12:58, Mikko wrote:
On 2025-06-13 13:45:20 +0000, WM said:
On 13.06.2025 11:53, Mikko wrote:
On 2025-06-12 09:41:51 +0000, WM said:
On 12.06.2025 09:56, Mikko wrote:
On 2025-06-11 11:30:26 +0000, WM said:
For all natural numbers that can be chosen as individuals:
|ℕ \ {1, 2, 3, ..., n}| = ℵo.
The "..." can be removed by the inductive proof
for all definable n ∈ ℕ:
|ℕ \ {1}| = ℵo.
|ℕ \ {m ∈ ℕ | m < n}| = ℵo
|ℕ \ {m ∈ ℕ | m < n+1}| = ℵo.
That the set difference of an infinite set and a finite set is infinite >>>>>> is well understood and therefore an uninteresting result.
That is not the only result.
That is the result of your proof. Other results can be discussed in
other contexts.
The following result cannot be circumvented: After mankind will have ceased,
who will care?
People who live now and ponder about dark numbers.
there is a largest natural number ever named.
That has no mathematical significance. Whether a number is named or not
is a feature of mankind, not of the number.
Of course the border between visible and dark numbers depends on time
and system.
But the mathematically interesting fact is that never all numbers can
become visible because completeness, well-order and infinity are
mutually incompatible.
ℕ \ {1, 2, 3, ...} = { } means that all natural numbers can be
subtracted - completely such that none remains. If all were visible
with a known well-order, then a last one would be subtracted.
Therefore not all can have a known well-order.
On 14.06.2025 13:02, Mikko wrote:
On 2025-06-13 13:49:25 +0000, WM said:
On 13.06.2025 12:02, Mikko wrote:
On 2025-06-12 09:44:06 +0000, WM said:
On 12.06.2025 10:00, Mikko wrote:
On 2025-06-11 11:38:52 +0000, WM said:"this proof for all definable numbers n"
Outside of ZF in correct mathematics this proof for all definable numbers n
|ℕ \ {1}| = ℵo.
|ℕ \ {m ∈ ℕ | m < n}| = ℵo
|ℕ \ {m ∈ ℕ | m < n+1}| = ℵo.shows that is impossible to extend definability to all natural numbers >>>>>>> with none remaining undefined.
It does not show that. The "proof" does not even mention definability. >>>>>
That is a part of the sentence that mentions the proof, not a part of
the proof itself.
It is part of the proof because it specifies the set which the proof is
based upon.
Doesn't really matter because the "proof" does not end with an interesting >> concousion.
Shows however that your objection was wrong.
The conclusion follows from the second sentence alone when n is understood
to be universally quantified, so the first sentence is not needed and should
not be there.
It is there. Only definablenumbers can be quantified as individuals.
Quantification is never "as individuals".
It is always concerning individuals only.
Therefore "as individuals" is meaningless.
It is not meaningless because it reminds mathematicians of this meaning.
∀n ∈ ℕ: P(n) specifies that every natural number n as an individual has
the property P.
That is an ordinary quantification, not a quantification as individuals.
Ordinary quantification is always concerning individuals only, That
should remind you of the facts.
On 2025-06-14 14:00:36 +0000, WM said:
Of course the border between visible and dark numbers depends on time
and system.
And therefore has no mathenatical meaning.
But the mathematically interesting fact is that never all numbers can
become visible because completeness, well-order and infinity are
mutually incompatible.
What kind of comleteness do you mean?
Therefore not all can have a known well-order.
Then can as they do. The arithmetic order of natural numbers is a known well-order. But a well-order does not imply that there is a last one.
only that there is a first one.
On 15.06.2025 12:32, Mikko wrote:
On 2025-06-14 14:00:36 +0000, WM said:
Of course the border between visible and dark numbers depends on time
and system.
And therefore has no mathenatical meaning.
That is only true if mathematics is restricted to definable numbers.
But that is not the usual approach. The undefinable prime numbers for instance belong to present mathematics.
But the mathematically interesting fact is that never all numbers can
become visible because completeness, well-order and infinity are
mutually incompatible.
What kind of comleteness do you mean?
The complete set of numbers / prime numbers is infinite. If each number
were known and subtracted, a last one would appear.
Therefore not all can have a known well-order.
Then can as they do. The arithmetic order of natural numbers is a known
well-order. But a well-order does not imply that there is a last one.
only that there is a first one.
But this does: ℕ \ {1, 2, 3, ...} = { } .
On 2025-06-15 16:15:39 +0000, WM said:
On 15.06.2025 12:32, Mikko wrote:
On 2025-06-14 14:00:36 +0000, WM said:
Of course the border between visible and dark numbers depends on
time and system.
And therefore has no mathenatical meaning.
That is only true if mathematics is restricted to definable numbers.
No, it is true anyway.
Mathematics is restricted to mathematical things
even when the inspiration comes from something non-mathematical, as it
often comes. Things that depend on time or otherwise on the real world
are not mathematical.
But that is not the usual approach. The undefinable prime numbers for
instance belong to present mathematics.
Definability is a mathematical property. Every natural number and in particular every prime number is definable.
Then can as they do. The arithmetic order of natural numbers is a known
well-order. But a well-order does not imply that there is a last one.
only that there is a first one.
But this does: ℕ \ {1, 2, 3, ...} = { } .
No, it does not. There is no last one in { }.
On 16.06.2025 09:41, Mikko wrote:
On 2025-06-15 16:15:39 +0000, WM said:
On 15.06.2025 12:32, Mikko wrote:
On 2025-06-14 14:00:36 +0000, WM said:
Of course the border between visible and dark numbers depends on time >>>>> and system.
And therefore has no mathenatical meaning.
That is only true if mathematics is restricted to definable numbers.
No, it is true anyway.
Please don't claim but argue by arguments:
The complete set of prime numbers is infinite. Only a finite set is
known. If each number were known and subtracted from its set, then a
last one would be subtracted.
On 2025-06-16 10:50:58 +0000, WM said:
The error above is that the expression "the last one" is used without
proving that there is a last one.
Every natural number and in
particular every prime number is definable.
On 17.06.2025 12:16, Mikko wrote:
On 2025-06-16 10:50:58 +0000, WM said:
The error above is that the expression "the last one" is used without
proving that there is a last one.
If all natnumbers are subtracted, none remains.
If all natnumbers are subtracted in their order, one after the other,
none remains.
On 2025-06-17 10:57:01 +0000, WM said:
On 17.06.2025 12:16, Mikko wrote:
On 2025-06-16 10:50:58 +0000, WM said:
The error above is that the expression "the last one" is used without
proving that there is a last one.
If all natnumbers are subtracted, none remains.
True.
If all natnumbers are subtracted in their order, one after the other,
none remains.
The order does not matter,
the result is the same anyway.
But that does not prove that your "the last one" denotes anything.
On 18.06.2025 11:39, Mikko wrote:
On 2025-06-17 10:57:01 +0000, WM said:
On 17.06.2025 12:16, Mikko wrote:
On 2025-06-16 10:50:58 +0000, WM said:
The error above is that the expression "the last one" is used without
proving that there is a last one.
If all natnumbers are subtracted, none remains.
True.
If all natnumbers are subtracted in their order, one after the other,
none remains.
The order does not matter,
But the order exists for all defined natural numbers. We know of each
one its predecessor and its successor, if those are existing.
the result is the same anyway.
That means that none remains. I case of known order we know the last
one subtracted.
But that does not prove that your "the last one" denotes anything.
How can an ordered set be completely subtracted in its given order
without a last one?
On 2025-06-18 13:48:13 +0000, WM said:
On 18.06.2025 11:39, Mikko wrote:
On 2025-06-17 10:57:01 +0000, WM said:
On 17.06.2025 12:16, Mikko wrote:
On 2025-06-16 10:50:58 +0000, WM said:
The error above is that the expression "the last one" is used without >>>>> proving that there is a last one.
If all natnumbers are subtracted, none remains.
True.
If all natnumbers are subtracted in their order, one after the
other, none remains.
The order does not matter,
But the order exists for all defined natural numbers. We know of each
one its predecessor and its successor, if those are existing.
It is also possible to order them differently. Removing all leaves
nothing, whther you remove in some order or another or all at the
same time.
the result is the same anyway.
That means that none remains. I case of known order we know the last
one subtracted.
But that does not prove that your "the last one" denotes anything.
How can an ordered set be completely subtracted in its given order
without a last one?
Euclid did not specify how to draw a circle. He merely postulated that
for any given centre and radius it can be done.
Likewise a set theory
does not specify how one set can subtracted from another.
It merely postulates that it can be done.
On 19.06.2025 09:16, Mikko wrote:
On 2025-06-18 13:48:13 +0000, WM said:
On 18.06.2025 11:39, Mikko wrote:
On 2025-06-17 10:57:01 +0000, WM said:
On 17.06.2025 12:16, Mikko wrote:
On 2025-06-16 10:50:58 +0000, WM said:
The error above is that the expression "the last one" is used without >>>>>> proving that there is a last one.
If all natnumbers are subtracted, none remains.
True.
If all natnumbers are subtracted in their order, one after the other, >>>>> none remains.
The order does not matter,
But the order exists for all defined natural numbers. We know of each
one its predecessor and its successor, if those are existing.
It is also possible to order them differently. Removing all leaves
nothing, whther you remove in some order or another or all at the
same time.
But removing all in their natural or any given order (that does not
allow two or more to take the same place) implies a last removed one.
the result is the same anyway.
That means that none remains. I case of known order we know the last
one subtracted.
But that does not prove that your "the last one" denotes anything.
How can an ordered set be completely subtracted in its given order
without a last one?
Euclid did not specify how to draw a circle. He merely postulated that
for any given centre and radius it can be done.
The radius is a length that allows to construct every point of the circumference. Further compasses are in use for constructions.
Likewise a set theory
does not specify how one set can subtracted from another.
But it postulates or accepts that an order exists.
It merely postulates that it can be done.
I ask how it can be done. Am I the first? Is it forbidden?
On 2025-06-19 14:13:49 +0000, WM said:
But removing all in their natural or any given order (that does not
allow two or more to take the same place) implies a last removed one.
No, it does not. That implication is not acceptable without a proof.
the result is the same anyway.
That means that none remains. I case of known order we know the last
one subtracted.
But that does not prove that your "the last one" denotes anything.
How can an ordered set be completely subtracted in its given order
without a last one?
Euclid did not specify how to draw a circle. He merely postulated that
for any given centre and radius it can be done.
The radius is a length that allows to construct every point of the
circumference. Further compasses are in use for constructions.
Euclid's postulates say nothing about that.
Likewise a set theory
does not specify how one set can subtracted from another.
But it postulates or accepts that an order exists.
Cantor originally did. The ZF, which is the most commonly used formal
set theory, doesn't. The nearest it has is the subset relation, which
is a partial order.
It merely postulates that it can be done.
I ask how it can be done. Am I the first? Is it forbidden?
It is outside of the scope of the theory. In a particular interpretation
or application of the theory you can and possibly need to answer that question.
On 20.06.2025 11:00, Mikko wrote:
On 2025-06-19 14:13:49 +0000, WM said:
But removing all in their natural or any given order (that does not
allow two or more to take the same place) implies a last removed one.
No, it does not. That implication is not acceptable without a proof.
Removing all means that none remains.
On 2025-06-20 12:40:41 +0000, WM said:
On 20.06.2025 11:00, Mikko wrote:
On 2025-06-19 14:13:49 +0000, WM said:
But removing all in their natural or any given order (that does not
allow two or more to take the same place) implies a last removed one.
No, it does not. That implication is not acceptable without a proof.
Removing all means that none remains.
True. But only those are removed that were there. What never was there
is not removed, and the last natural number was not.
On 21.06.2025 12:11, Mikko wrote:
On 2025-06-20 12:40:41 +0000, WM said:
On 20.06.2025 11:00, Mikko wrote:
On 2025-06-19 14:13:49 +0000, WM said:
But removing all in their natural or any given order (that does notNo, it does not. That implication is not acceptable without a proof.
allow two or more to take the same place) implies a last removed one. >>>>
Removing all means that none remains.
True. But only those are removed that were there. What never was there
is not removed, and the last natural number was not.
Consider only what has been removed. If all have been be removed, then
a last one has been removed.
As I already said, only what was there was removed.
On 22.06.2025 11:37, Mikko wrote:
As I already said, only what was there was removed.
There is a first removed element. If the removal is complete, then
there is a last removed element.
On 2025-06-22 15:41:23 +0000, WM said:
On 22.06.2025 11:37, Mikko wrote:
As I already said, only what was there was removed.
There is a first removed element. If the removal is complete, then
there is a last removed element.
In general there needs not be a first or a last. For example, the set difference of real numbers and irrational numbers is rational numbers.
But there is no first or last in irrational numbers.
On 23.06.2025 09:52, Mikko wrote:
On 2025-06-22 15:41:23 +0000, WM said:The natural numbers are well-ordered, from the first to the last which
On 22.06.2025 11:37, Mikko wrote:
As I already said, only what was there was removed.
There is a first removed element. If the removal is complete, then
there is a last removed element.
In general there needs not be a first or a last. For example, the set
difference of real numbers and irrational numbers is rational numbers.
But there is no first or last in irrational numbers.
is subtracted. Does this change if all can be subtracted?
On 17.06.2025 12:16, Mikko wrote:They are already "defined". Otherwise see the GIMPS.
On 2025-06-16 10:50:58 +0000, WM said:
Forgotten to answer:
Every natural number and in particular every prime number isThen define the prime numbers such that none remains undefined. Or
definable.
define a hitherto undefined prime number.
On 18.06.2025 11:39, Mikko wrote:Only if you "subtract" infinitely many.
On 2025-06-17 10:57:01 +0000, WM said:
On 17.06.2025 12:16, Mikko wrote:
On 2025-06-16 10:50:58 +0000, WM said:
The error above is that the expression "the last one" is used without
proving that there is a last one.
If all natnumbers are subtracted in their order, one after the other,
none remains.
Trust me, they do (except for the predecessor of 0).The order does not matter,But the order exists for all defined natural numbers. We know of each
one its predecessor and its successor, if those are existing.
No, if there were a last one, its successors would remain.the result is the same anyway.That means that none remains. I case of known order we know the last one subtracted.
By an infinite "process".But that does not prove that your "the last one" denotes anything.How can an ordered set be completely subtracted in its given order
without a last one?
On 14.06.2025 12:58, Mikko wrote:*person
On 2025-06-13 13:45:20 +0000, WM said:
People who live now and ponder about dark numbers.The following result cannot be circumvented: After mankind will havewho will care?
ceased,
But the mathematically interesting fact is that never all numbers canThat's not a mathematical but physical fact, if time has an end. If not,
become visible because completeness, well-order and infinity are
mutually incompatible.
ℕ \ {1, 2, 3, ...} = { } means that all natural numbers can beOr not all can be subtracted. I think that is closer to your intuition.
subtracted - completely such that none remains. If all were visible with
a known well-order, then a last one would be subtracted. Therefore not
all can have a known well-order.
On 16.06.2025 09:41, Mikko wrote:(There are also infinite subsets of the primes...)
On 2025-06-15 16:15:39 +0000, WM said:
On 15.06.2025 12:32, Mikko wrote:
On 2025-06-14 14:00:36 +0000, WM said:
The complete set of prime numbers is infinite.
Geometry doesn't depend on time; perfect lines do not exist. DoesMathematics is restricted to mathematical things even when theThat is a wrong opinion, as best can be seen by geometry. But also the restrictions of reality do limit what can be expressed in mathematics.
inspiration comes from something non-mathematical, as it often comes.
Things that depend on time or otherwise on the real world are not
mathematical.
We're working on it. Do you think the largest known prime is undefinedThen define the prime numbers such that none remains undefined. OrBut that is not the usual approach. The undefinable prime numbers forDefinability is a mathematical property. Every natural number and in
instance belong to present mathematics.
particular every prime number is definable.
define a hitherto undefined prime number.
No, order does not force finity. Infinite sets can be well-ordered.When the natural numbers are subtracted one by one and when finally none remains, then a last one has been subtracted. This is forced by theirNo, it does not. There is no last one in { }.Then can as they do. The arithmetic order of natural numbers is aBut this does: ℕ \ {1, 2, 3, ...} = { } .
known well-order. But a well-order does not imply that there is a
last one. only that there is a first one.
order.
On 2025-06-23 10:59:08 +0000, WM said:
The natural numbers are well-ordered, from the first to the last which
is subtracted. Does this change if all can be subtracted?
There is no last natural number. Every subset of natural numbers
has a first member (in the arithmetic order) but infinite subsets
don't have a last one.
On 29.05.2025 02:25, Ben Bacarisse wrote:Did you mean less than 10^80?
WM <[email protected]> writes:
Every n that can be expressed by digits should be known to you.
But the important fact, since it's /your/ proof, is what that means to
/you/ and I can not know that.
Definition: A natural number is "identified" or (individually) "defined"
or "instantiated" if it can be communicated such that sender and
receiver understand the same and can link it by a finite initial segment
to the origin 0. All other natural numbers are called dark natural
numbers.
Only when a number n is identified we can use it in mathematicalSo not actually a fixed set.
discourse and can determine the trichotomy properties of n and of every multiple kn or power n^k with respect to every identified number k.
ℕ_def is the set that contains all defined natural numbers as elements – and nothing else. ℕ_def is a potentially infinite set; therefore henceforth it will be called a collection.
All supersets of N satisfy the axioms.It requires a lot of stupidity or hate to put this question after seeing1 ∈ M (4.1)But it seems you can't prove that 1 is in N, can you?
n ∈ M ⇒ (n + 1) ∈ M (4.2)
If M satisfies (4.1) and (4.2), then ℕ ⊆ M.
Of course no intelligent reader need be told that this ℕ = ℕ_def also >>> satisfies the axioms (4.1) and (4.2).
the axiom that 1 is in ℕ.
I bet Cantor proved there are no "dark numbers".Of course. Based on the assumption that Cantor is right I can prove the existence of dark numbers.But you need to. It's is the base case in the proof you asked everyoneHow you prove that {1} "has ℵo" successors.I do not prove it
about. You can't make a proof by induction by simply asserting things.
On 29.05.2025 12:07, Mikko wrote:You have doubted that.
On 2025-05-28 15:13:54 +0000, WM said:
and P[n] -> P[n+1] before it can infer
If {1, 2, 3, ..., n} has infinitely many (ℵo) successors, then {1, 2,
3, ..., n, n+1} has infinitely many (ℵo) successors because here the
number of successors has been reduced by 1, and ℵo - 1 = ℵo. There is >>> no way to avoid this conclusion if ℵo natural numbers are assumed to
exist. And that is the theory that I use.
To me this does not look like P[n] -> P[n+1].
P[n]: {1, 2, 3, ..., n} has infinitely many (ℵo) successors.
P[n+1]: {1, 2, 3, ..., n, n+1} has infinitely many (ℵo) successors.
Do you doubt ℵo - 1 = ℵo?
Lol. Is the successor of every "definable" number always "definable"?As I said the theory must be specified.Induction is applied to every natural number of the Peano set. The proof shows that it cannot be applied to every natural number of the Cantor
In Peano arithmetic the induction axiom is applicable to everything. If
you want something else you must specify some other theory, perhaps
some set theory.
set.
Cool. Sets don't change.Here nothing gets complicated, but all remains very simple.Things get soon complicated if we allow other than objects, first orderThen call it a collection.The set of finite initial segments of natural numbers is potentially >>>>> infinite but not actually infinite.There is nothing potential in a set.
functions and first order predicates.
On 24.06.2025 11:15, Mikko wrote:
On 2025-06-23 10:59:08 +0000, WM said:
The natural numbers are well-ordered, from the first to the last which
is subtracted. Does this change if all can be subtracted?
There is no last natural number. Every subset of natural numbers
has a first member (in the arithmetic order) but infinite subsets
don't have a last one.
Then it is impossible to remove all of them in their order.
On 2025-06-24 17:33:20 +0000, WM said:
On 24.06.2025 11:15, Mikko wrote:
On 2025-06-23 10:59:08 +0000, WM said:
The natural numbers are well-ordered, from the first to the last
which is subtracted. Does this change if all can be subtracted?
There is no last natural number. Every subset of natural numbers
has a first member (in the arithmetic order) but infinite subsets
don't have a last one.
Then it is impossible to remove all of them in their order.
Maybe so. There expression "to remove all of them in their order" is
an informal expression that might refer to something that can be
expressed mathematically but maybe it does not.
On 25.06.2025 09:53, Mikko wrote:
On 2025-06-24 17:33:20 +0000, WM said:
On 24.06.2025 11:15, Mikko wrote:
On 2025-06-23 10:59:08 +0000, WM said:
The natural numbers are well-ordered, from the first to the last which >>>>> is subtracted. Does this change if all can be subtracted?
There is no last natural number. Every subset of natural numbers
has a first member (in the arithmetic order) but infinite subsets
don't have a last one.
Then it is impossible to remove all of them in their order.
Maybe so. There expression "to remove all of them in their order" is
an informal expression that might refer to something that can be
expressed mathematically but maybe it does not.
You confuse mathematics with formalism. The greatest mathematicians
have lived before formalism had been invented. Formalism is only a
crutch for those who cannot think without crutches.
If all naturals can be subtracted then this can be done in their order.
For enumerating another set the order is even a precondition.
On 2025-06-25 20:08:06 +0000, WM said:
On 25.06.2025 09:53, Mikko wrote:
On 2025-06-24 17:33:20 +0000, WM said:
On 24.06.2025 11:15, Mikko wrote:
On 2025-06-23 10:59:08 +0000, WM said:
The natural numbers are well-ordered, from the first to the last
which is subtracted. Does this change if all can be subtracted?
There is no last natural number. Every subset of natural numbers
has a first member (in the arithmetic order) but infinite subsets
don't have a last one.
Then it is impossible to remove all of them in their order.
Maybe so. There expression "to remove all of them in their order" is
an informal expression that might refer to something that can be
expressed mathematically but maybe it does not.
You confuse mathematics with formalism. The greatest mathematicians
have lived before formalism had been invented. Formalism is only a
crutch for those who cannot think without crutches.
Formalism was invented to avoid or correct the errrors the greatest mathematicians made before formalism was invented. The invoention
of formalism was not a sudden event but a gradual process that took
thousands of years.
If all naturals can be subtracted then this can be done in their
order. For enumerating another set the order is even a precondition.
For enumeration an order is necessary but not sufficient. But for
subtraction it is not.
For example, irrational numbers have their
arithmetic order but cannot be enumerated.
from the set of real numbers and the result is the set of rational
numbers.
On 26.06.2025 12:36, Mikko wrote:
On 2025-06-25 20:08:06 +0000, WM said:
On 25.06.2025 09:53, Mikko wrote:
On 2025-06-24 17:33:20 +0000, WM said:
On 24.06.2025 11:15, Mikko wrote:
On 2025-06-23 10:59:08 +0000, WM said:
The natural numbers are well-ordered, from the first to the last which >>>>>>> is subtracted. Does this change if all can be subtracted?
There is no last natural number. Every subset of natural numbers
has a first member (in the arithmetic order) but infinite subsets
don't have a last one.
Then it is impossible to remove all of them in their order.
Maybe so. There expression "to remove all of them in their order" is
an informal expression that might refer to something that can be
expressed mathematically but maybe it does not.
You confuse mathematics with formalism. The greatest mathematicians
have lived before formalism had been invented. Formalism is only a
crutch for those who cannot think without crutches.
Formalism was invented to avoid or correct the errrors the greatest
mathematicians made before formalism was invented. The invoention
of formalism was not a sudden event but a gradual process that took
thousands of years.
Say about 100 years.
If all naturals can be subtracted then this can be done in their order.
For enumerating another set the order is even a precondition.
For enumeration an order is necessary but not sufficient. But for
subtraction it is not.
If we subtract in the order that is used for enumerating then a last
one is necessary.
For example, irrational numbers have their
arithmetic order but cannot be enumerated.
The reason is that infinite sets cannot be enumerated. Also there
completion would necessitate a last one.
< But they can be sutracted
from the set of real numbers and the result is the set of rational
numbers.
That can be done collectively only.
On 2025-06-26 13:09:32 +0000, WM said:
If we subtract in the order that is used for enumerating then a last
one is necessary.
No, there is no last one in an infinete enumeration.
The word "infinite"
originally meant "having no end".
We can remove all odd numbers from the
natural numbers, leaving the even numbers, but there is no last number removed.
For example, irrational numbers have their
arithmetic order but cannot be enumerated.
The reason is that infinite sets cannot be enumerated. Also there
completion would necessitate a last one.
Infinite enumerable sets can. Then the enumeration is an infinte sequence.
< But they can be sutracted
from the set of real numbers and the result is the set of rational
numbers.
That can be done collectively only.
Doesn't matter.
On 27.06.2025 09:33, Mikko wrote:
On 2025-06-26 13:09:32 +0000, WM said:
If we subtract in the order that is used for enumerating then a last
one is necessary.
No, there is no last one in an infinete enumeration.
Then it is not finished or completed.
The word "infinite"
originally meant "having no end".
That is true. But if infinite sets are complete, then they are infinite
only because the end is not visible.
We can remove all odd numbers from the
natural numbers, leaving the even numbers, but there is no last number
removed.
Then not all are removed. All completely, never ending and in order
implies a contradiction.
For example, irrational numbers have their
arithmetic order but cannot be enumerated.
The reason is that infinite sets cannot be enumerated. Also there
completion would necessitate a last one.
Infinite enumerable sets can. Then the enumeration is an infinte sequence.
Then it is never completed. Going on and on is not a proof of completeness.
< But they can be sutracted
from the set of real numbers and the result is the set of rational
numbers.
That can be done collectively only.
Doesn't matter.
That is very important because these things are confused very often.
On 2025-06-27 19:36:41 +0000, WM said:
On 27.06.2025 09:33, Mikko wrote:
On 2025-06-26 13:09:32 +0000, WM said:
If we subtract in the order that is used for enumerating then a last
one is necessary.
No, there is no last one in an infinete enumeration.
Then it is not finished or completed.
No, but it can be continued.
The word "infinite"
originally meant "having no end".
That is true. But if infinite sets are complete, then they are
infinite only because the end is not visible.
There is no mathematical meaning of "complete" that could be applied
to a set.
We can remove all odd numbers from the
natural numbers, leaving the even numbers, but there is no last number
removed.
Then not all are removed. All completely, never ending and in order
implies a contradiction.
All are removed when all are removed.
Being completed is not a mathematical concept. An infinite sequence just
is infinite.
< But they can be sutracted
from the set of real numbers and the result is the set of rational
numbers.
That can be done collectively only.
Doesn't matter.
That is very important because these things are confused very often.
Confusions are best avoided by not using non-mathematical words and using mathematical words only in their mathematical meanings.
On 28.06.2025 11:56, Mikko wrote:
On 2025-06-27 19:36:41 +0000, WM said:
On 27.06.2025 09:33, Mikko wrote:
On 2025-06-26 13:09:32 +0000, WM said:
If we subtract in the order that is used for enumerating then a last >>>>> one is necessary.
No, there is no last one in an infinete enumeration.
Then it is not finished or completed.
No, but it can be continued.
That is potential infinity. But Cantor claimed complete enumeration.
The notion set can only be applied to complete sets. i.e., sets which
cannot be continued.
We can remove all odd numbers from the
natural numbers, leaving the even numbers, but there is no last number >>>> removed.
Then not all are removed. All completely, never ending and in order
implies a contradiction.
All are removed when all are removed.
When done in natural order, then a last one is to be removed before all
are removed. ℕ \ {1, 2, 3, ...} = { }.
This cannot be accomplished
by any definable natural number because
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| =/= 0.
Being completed is not a mathematical concept. An infinite sequence just
is infinite.
1.1 Cantor's original German terminology on infinite sets
The reader fluent in German may be interested in the subtleties of
Cantor's terminology on actual infinity the finer distinctions of which
are not easy to express in English. While Cantor early used
"vollständig" and "vollendet" to express "complete" and "finished", the
term "fertig", expressing "finished" too but being also somewhat
reminiscent of "ready", for the first time appeared in a letter to
Hilbert of 26 Sep 1897, where all its appearances had later been added
to the letter.
But Cantor already knew that there are incomplete, i.e., potentially infinite sets like the set of all cardinal numbers. He called them "absolutely infinite". The details of this enigmatic notion are
explained in section 1.2 (see also section 4.1. – Unfortunately it has turned out impossible to strictly separate Cantor's mathematical and religious arguments.)
1.1.1 Vollständig
"Wenn zwei wohldefinierte Mannigfaltigkeiten M und N sich eindeutig und vollständig, Element für Element, einander zuordnen lassen (was, wenn
es auf eine Art möglich ist, immer auch noch auf viele andere Weisen geschehen kann), so möge für das Folgende die Ausdrucksweise gestattet sein, daß diese Mannigfaltigkeiten gleiche Mächtigkeit haben, oder
auch, daß sie äquivalent sind." [Cantor, p. 119]
"gegenseitig eindeutige und vollständige Korrespondenz" [Cantor, p. 238]
"Die sämtlichen Punkte l unsrer Menge L sind also in gegenseitig
eindeutige und vollständige Beziehung zu sämtlichen Punkten f der Menge
F gebracht," [Cantor, p. 241]
"Zwei wohlgeordnete Mengen M und N heissen von gleichem Typus oder auch
von gleicher Anzahl, wenn sie sich gegenseitig eindeutig und
vollständig unter beidseitiger Wahrung der Rangfolge ihrer Elemente auf einander beziehen, abbilden lassen;" [G. Cantor, letter to R. Lipschitz
(19 Nov 1883)]
"Zwei bestimmte Mengen M und M1 nennen wir äquivalent (in Zeichen: M ~
M1), wenn es möglich ist, dieselben gesetzmäßig, gegenseitig eindeutig
und vollständig, Element für Element, einander zuzuordnen." [Cantor, p. 412]
"doch gibt es immer viele, im allgemeinen sogar unzählig viele Zuordnungsgesetze, durch welche zwei äquivalente Mengen in gegenseitig eindeutige und vollständige Beziehung zueinander gebracht werden
können." [Cantor, p. 413]
"eine solche gegenseitig eindeutige und vollständige Korrespondenz hergestellt [...] irgendeine gegenseitig eindeutige und vollständige Zuordnung der beiden Mengen [...] auch eine gegenseitig eindeutige und vollständige Korrespondenz" [Cantor, p. 415]
"Zwei n-fach geordnete Mengen M und N werden 'ähnlich' genannt, wenn es möglich ist, sie gegenseitig eindeutig und vollständig, Element für Element, einander so zuzuordnen," [Cantor, p. 424]
1.1.2 Vollendet
"Zu dem Gedanken, das Unendlichgroße [...] auch in der bestimmten Form
des Vollendet-unendlichen mathematisch durch Zahlen zu fixieren, bin
ich fast wider meinen Willen, weil im Gegensatz zu mir wertgewordenen Traditionen, durch den Verlauf vieljähriger wissenschaftlicher
Bemühungen und Versuche logisch gezwungen worden," [Cantor, p. 175]
"In den 'Grundlagen' formulire ich denselben Protest, indem ich an verschiedenen Stellen mich gegen die Verwechslung des Uneigentlich-unendlichen (so nenne ich das veränderliche Endliche) mit
dem Eigentlich-unendlichen (so nenne ich das bestimmte, das vollendete Unendliche, oder auch das Transfinite, Überendliche) ausspreche. Das Irrthümliche in jener Gauss'schen Stelle besteht darin, dass er sagt,
das Vollendetunendliche könne nicht Gegenstand mathematischer
Betrachtungen werden; dieser Irrthum hängt mit dem andern Irrthum
zusammen, dass er [...] das Vollendetunendliche mit dem Absoluten, Göttlichen identificirt, [...] Das Vollendetunendliche findet sich allerdings in gewissem Sinne in den Zahlen , + 1, ..., , ...; sie
sind Zeichen für gewisse Modi des Vollendetunendlichen und weil das Vollendetunendliche in verschiedenen, von einander mit der äussersten Schärfe durch den sogenannten 'endlichen, menschlichen Verstand' unterscheidbaren Modificationen auftreten kann, so sieht man hieraus
deutlich wie weit man vom Absoluten entfernt ist, obgleich man das Vollendetunendliche sehr wohl fassen und sogar mathematisch auffassen
kann." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]
"da nun jeder Typus auch im letzteren Falle etwas in sich Bestimmtes, vollendetes ist, so gilt ein gleiches von der zu ihm gehörigen Zahl.
[...] 'Eigentlichunendlichem = Transfinitum = Vollendetunendlichem = Unendlichseiendem = kategorematice infinitum' [...] dieser Unterschied ändert aber nichts daran, daß als ebenso bestimmt und vollendet anzusehen ist, wie 2," [G. Cantor, letter to K. Laßwitz (15 Feb 1884). Cantor, p. 395]
"Wir wollen nun zu einer genaueren Untersuchung der perfekten Mengen übergehen. Da jede solche Punktmenge gewissermaßen in sich begrenzt, abgeschlossen und vollendet ist, so zeichnen sich die perfekten Mengen
vor allen anderen Gebilden durch besondere Eigenschaften aus." [Cantor,
p. 236]
1.1.3 Fertig
"Die Totalität aller Alefs ist nämlich eine solche, welche nicht als
eine bestimmte, wohldefinirte fertige Menge aufgefaßt werden kann.
[...] 'Wenn eine bestimmte wohldefinirte fertige Menge eine
Cardinalzahl haben würde, die mit keinem der Alefs zusammenfiele, so
müßte sie Theilmengen enthalten, deren Cardinalzahl irgend ein Alef
ist, oder mit anderen Worten, die Menge müßte die Totalität aller Alefs
in sich tragen.' Daraus ist leicht zu folgern, daß unter der eben
genannten Voraussetzung (einer best. Menge, deren Cardinalzahl kein
Alef wäre) auch die Totalität aller Alefs als eine best. wohldefinirte fertige Menge aufgefaßt werden könnte." [G. Cantor, letter to D.
Hilbert (26 Sep 1897)]
"In meinen Untersuchungen habe ich, allgemein gesprochen, 'fertige
Mengen' im Auge und verstehe darunter solche, bei denen die
Zusammenfassung aller Elemente zu einem Ganzen, zu einem Ding für sich möglich ist, so daß eine 'fertige M.' eventuell selbst als Element
einer andern Menge gedacht werden kann. [...] Derartige Mengen, die die Bedingung 'fertig' nicht erfüllen, nenne ich 'absolut unendliche'
Mengen.
Nehmen wir einmal an, es könnten alle Alefs coexistieren, so führt uns
dies zu einem Widerspruch. Denn alsdann würden alle Alefs, wenn wir sie
nach ihrer Größe geordnet denken, eine wohlgeordnete, fertige Menge M bilden. Mit jeder wohlgeordneten fertigen Menge M von Alefs ist aber
nach dem Bildungsgesetz der Alefs ein bestimmtes Alef gegeben, welches
der Größe nach auf alle Individuen von M nächstfolgt.
Hier hätten wir also den Widerspruch eines Alefs, das größer wäre als
alle Alefs, folglich auch größer als es selbst. Ich schließe also, daß alle Alefs nicht coexistent sind, nicht zu einem 'Ding für sich' zusammengefasst werden können, daß sie mit anderen Worten keine
'fertige Menge' bilden.
Der Widerspruch erscheint mir so, als wenn wir von einer 'endlichen Zahl' sprechen wollten, die größer wäre als 'alle endlichen Zahlen'.
Nur ist hier der Unterschied, daß alle endlichen Zahlen eine fertige
Menge bilden, die nach oben von der kleinsten transfiniten Cardinalzahl
0 gewissermaßen begrenzt wird. Die absolute Grenzenlosigkeit der Menge aller Alefs erscheint als Grund der Unmöglichkeit, sie zu einem Ding
für sich zusammenzufassen.
In dem von Ihnen vorgetragenen Beispiele wird aber die Menge aller Alefs als eine 'fertige M.' vorausgesetzt und damit löst und erklärt
sich der Widerspruch, auf den Sie durch Anwendung von Sätzen geführt werden, die nur für fertige Mengen bewiesen und gültig sind." [G.
Cantor, letter to D. Hilbert (6 Oct 1898)]
"Aus der Definition: 'Unter einer fertigen Menge verstehe man jede
Vielheit, bei welcher alle Elemente ohne Widerspruch als zusammenseiend
und daher als ein Ding für sich gedacht werden können.' ergeben sich mancherlei Sätze, unter Anderm diese:
I 'Ist M eine fert. Menge, so ist auch jede Theilmenge von M eine fert. Menge.'
II 'Substituirt man in einer fert. M. an Stelle der Elemente fertige
Mengen, so ist die hieraus resultirende Vielheit eine fertige M.'
III 'Ist von zwei aequivalenten Vielheiten die eine eine fert. M., so
ist es auch die andere.'
IV 'Die Vielheit aller Theilmengen einer fertigen Menge M ist eine
fertige Menge.' Denn alle Theilmengen von M sind 'zusammen' in M
enthalten; der Umstand, daß sie sich theilweise decken, schadet hieran nichts.
Daß die 'abzählbaren' Vielheiten {} fertige Mengen sind, scheint mir
ein axiomatisch sicherer Satz zu sein, auf welchem die ganze Functionentheorie beruht. Dagegen scheint mir der Satz 'Das
Linearcontinuum ist eine fertige Menge' ein beweisbarer Satz zu sein
und zwar so: Das Linearcont. ist aequivalent der Menge S = {f()} wo
f() die Werthe 0 oder 1 haben kann. [...] Ich behaupte also S ist eine 'fertige Menge'. [...] Nach Satz IV ist aber die Vielheit aller
Theilmengen von {} eine fertige Menge; dasselbe gilt also nach Satz
III auch für S und für das Linearcontinuum.
Ebenso dürfte das Prädicat 'fertig' für die Mengen 1, 2, ... beweisbar sein." [G. Cantor, letter to D. Hilbert (10 Oct 1898)]
"Unter Bezugnahme auf mein Schreiben v. 10ten, stellt sich bei
genauerer Erwägung heraus, daß der Beweis des Satzes IV keineswegs so leicht geht. Der Umstand, daß die Elemente der 'Vielheit aller
Theilmengen einer fertigen Menge' sich theilweise decken, macht ihn illusorisch. In die Definition der fert. Menge wird die Voraussetzung
des Getrenntseins resp. Unabhängigseins der Elemente als wesentlich aufzunehmen sein." [G. Cantor, letter to D. Hilbert (12 Oct 1898)]
"Ich habe mich jetzt daran gewöhnt, das was ich früher 'fertig'
genannt, durch den Ausdruck 'consistent' zu ersetzen;" [G. Cantor,
letter to D. Hilbert (9 May 1899)]
"Die Totalität der Alefs lässt sich nicht als eine bestimmte fertige
Menge auffassen." [G. Cantor, letter to A. Schönflies via D. Hilbert
(28 Jun 1899)]
"Eine Vielheit kann nämlich so beschaffen sein, daß die Annahme eines 'Zusammenseins' aller ihrer Elemente auf einen Widerspruch führt, so
daß es unmöglich ist, die Vielheit als eine Einheit, als 'ein fertiges Ding' aufzufassen. Solche Vielheiten nenne ich absolut unendliche oder inkonsistente Vielheiten." [G. Cantor, letter to R. Dedekind (3 Aug
1899)]
"Zu Elementen einer Vielheit, können nur fertige Dinge genommen werden,
nur Mengen, nicht aber inconsistente Vielheiten, in deren Wesen es
liegt, daß sie nie als fertig und actuell existirend gedacht werden
kann." [G. Cantor, letter to P. Jourdain (9 Jul 1904)]
"'Unter einer Menge verstehen wir jede Zusammenfassung von ... zu einem Ganzen', worin doch liegt, daß Vielheiten, denen das Gepräge des
fertigen Ganzen oder der Dinglichkeit nicht nachgesagt werden kann,
nicht als 'Mengen' im eigentlichen Sinne des Wortes anzusehen sind."
[G. Cantor, letter to G. Chisholm-Young (9 Mar 1907)]
ZFC has one word for the meaning of completeness: set.< But they can be sutracted
from the set of real numbers and the result is the set of rational >>>>>> numbers.
That can be done collectively only.
Doesn't matter.
That is very important because these things are confused very often.
Confusions are best avoided by not using non-mathematical words and using
mathematical words only in their mathematical meanings.
On 2025-06-28 13:56:57 +0000, WM said:
On 28.06.2025 11:56, Mikko wrote:
On 2025-06-27 19:36:41 +0000, WM said:
On 27.06.2025 09:33, Mikko wrote:
On 2025-06-26 13:09:32 +0000, WM said:
If we subtract in the order that is used for enumerating then a
last one is necessary.
No, there is no last one in an infinite enumeration.
Then it is not finished or completed.
No, but it can be continued.
That is potential infinity. But Cantor claimed complete enumeration.
There is no mathematical definiton of "complete enumeration"
so it is
possible that Cantor's enumeartion is "complete" is one sense and "incomplete" in another.
The notion set can only be applied to complete sets. i.e., sets which
cannot be continued.
Saying that every set is "complete" does not mean anything,
All are removed when all are removed.
When done in natural order, then a last one is to be removed before
all are removed. ℕ \ {1, 2, 3, ...} = { }.
No. You said that every set is complete, so {1, 2, 3, ...}, which must
be a set in order to be valid for the context is complete and so is
ℕ \ {1, 2, 3, ...}, which is just another way to say { }-
This cannot be accomplished
There is nothing to accomplish. What is is, that's all.
Being completed is not a mathematical concept. An infinite sequence just >>> is infinite.
1.1 Cantor's original German terminology on infinite sets
The reader fluent in German may be interested in the subtleties of
Cantor's terminology on actual infinity the finer distinctions of
which are not easy to express in English. While Cantor early used
"vollständig" and "vollendet" to express "complete" and "finished",
the term "fertig", expressing "finished" too but being also somewhat
reminiscent of "ready", for the first time appeared in a letter to
Hilbert of 26 Sep 1897, where all its appearances had later been added
to the letter.
But Cantor already knew that there are incomplete, i.e.,
potentially infinite sets like the set of all cardinal numbers. He
called them "absolutely infinite". The details of this enigmatic
notion are explained in section 1.2 (see also section 4.1. –
Unfortunately it has turned out impossible to strictly separate
Cantor's mathematical and religious arguments.)
There is nothing religious in Cantor's arguments. The only traces of
his religious motivations are in the choice of his symbols, in paricular aleph and omega.
1.1.1 Vollständig
"Wenn zwei wohldefinierte Mannigfaltigkeiten M und N sich eindeutig
und vollständig, Element für Element, einander zuordnen lassen (was,
wenn es auf eine Art möglich ist, immer auch noch auf viele andere
Weisen geschehen kann), so möge für das Folgende die Ausdrucksweise
gestattet sein, daß diese Mannigfaltigkeiten gleiche Mächtigkeit
haben, oder auch, daß sie äquivalent sind." [Cantor, p. 119]
Above "vollständig" qualifies the verb "zuordnen" so the meaning may
differe from what it would mean when qualifying the word "set" or
any word that refers to all or some sets. It could be traslated as
"fully" or "completely", meaning that no member of either set is
unpaired.
"gegenseitig eindeutige und vollständige Korrespondenz" [Cantor, p. 238]
Usually "eindeutige und vollständige" is expressed in English with "one-to-one", or "gegenseitig eindeutige und vollständige Korrespondenz"
is expressed as "bijection". The word "gegenseitig" is not really
necessaty but at the time the idea was new and therefore greater clarity
was needed.
"Die sämtlichen Punkte l unsrer Menge L sind also in gegenseitig
eindeutige und vollständige Beziehung zu sämtlichen Punkten f der
Menge F gebracht," [Cantor, p. 241]
The same meaning and translation ("one-to-one" or "bijection") applies
here, too.
In all these example "eindeutig and vollständig" is an feature of
the correspondence, not of any set.
A correspondence can be expressed
with a set but was not in the above exmples (because "corresspondence"
was well understood at the time but "set" was not).
So no example of a set of being "complete".
1.1.2 Vollendet
"Zu dem Gedanken, das Unendlichgroße [...] auch in der bestimmten Form
des Vollendet-unendlichen mathematisch durch Zahlen zu fixieren, bin
ich fast wider meinen Willen, weil im Gegensatz zu mir wertgewordenen
Traditionen, durch den Verlauf vieljähriger wissenschaftlicher
Bemühungen und Versuche logisch gezwungen worden," [Cantor, p. 175]
This basically says that there is no real difference between actual
and potential infinity.
"da nun jeder Typus auch im letzteren Falle etwas in sich Bestimmtes,
vollendetes ist, so gilt ein gleiches von der zu ihm gehörigen Zahl.
[...] 'Eigentlichunendlichem = Transfinitum = Vollendetunendlichem =
Unendlichseiendem = kategorematice infinitum' [...] dieser Unterschied
ändert aber nichts daran, daß als ebenso bestimmt und vollendet
anzusehen ist, wie 2," [G. Cantor, letter to K. Laßwitz (15 Feb
1884). Cantor, p. 395]
"Wir wollen nun zu einer genaueren Untersuchung der perfekten Mengen
übergehen. Da jede solche Punktmenge gewissermaßen in sich begrenzt,
abgeschlossen und vollendet ist, so zeichnen sich die perfekten Mengen
vor allen anderen Gebilden durch besondere Eigenschaften aus."
[Cantor, p. 236]
Thie says that sets are always complete, so what has been said about uncompleted infinities either aplies to completed infinities as well
or does not apply to sets.
1.1.3 Fertig
"Die Totalität aller Alefs ist nämlich eine solche, welche nicht als
eine bestimmte, wohldefinirte fertige Menge aufgefaßt werden kann.
[...] 'Wenn eine bestimmte wohldefinirte fertige Menge eine
Cardinalzahl haben würde, die mit keinem der Alefs zusammenfiele, so
müßte sie Theilmengen enthalten, deren Cardinalzahl irgend ein Alef
ist, oder mit anderen Worten, die Menge müßte die Totalität aller
Alefs in sich tragen.' Daraus ist leicht zu folgern, daß unter der
eben genannten Voraussetzung (einer best. Menge, deren Cardinalzahl
kein Alef wäre) auch die Totalität aller Alefs als eine best.
wohldefinirte fertige Menge aufgefaßt werden könnte." [G. Cantor,
letter to D. Hilbert (26 Sep 1897)]
This says that if there were a set of all cardinals that would create
a contradiction.
Here "fertig" can be translated as "completed".
These are various ways to point out that sets are immutable, not
in the middle of a construction process.
ZFC has one word for the meaning of completeness: set.
ZFC (or plain ZF) does not specify any meaning for "set".
On 29.06.2025 12:25, Mikko wrote:Yes it is actually. That is why it's called infinite. You have to take
On 2025-06-28 13:56:57 +0000, WM said:
On 28.06.2025 11:56, Mikko wrote:
On 2025-06-27 19:36:41 +0000, WM said:
On 27.06.2025 09:33, Mikko wrote:No, but it can be continued.
On 2025-06-26 13:09:32 +0000, WM said:Then it is not finished or completed.
If we subtract in the order that is used for enumerating then aNo, there is no last one in an infinite enumeration.
last one is necessary.
Elements aren't "found", they either are or are not. You seem to meanIt means that no further element can be found later on.The notion set can only be applied to complete sets. i.e., sets whichSaying that every set is "complete" does not mean anything,
cannot be continued.
For the infinitieth time: no, that does not follow. There is no "last"Then it cannot be. If it is that all natural numbers are subtracted inNo. You said that every set is complete, so {1, 2, 3, ...}, which mustAll are removed when all are removed.When done in natural order, then a last one is to be removed before
all are removed. ℕ \ {1, 2, 3, ...} = { }.
be a set in order to be valid for the context is complete and so is ℕ \
{1, 2, 3, ...}, which is just another way to say { }-
This cannot be accomplishedThere is nothing to accomplish. What is is, that's all.
their order, then it is that a last one is subtracted.
Nah. It just has a higher cardinality.Being completed is not a mathematical concept. An infinite sequenceBut Cantor already knew that there are incomplete, i.e.,
just is infinite.
potentially infinite sets like the set of all cardinal numbers. He
called them "absolutely infinite".
How do you mean?"gegenseitig eindeutige und vollständige Korrespondenz" [Cantor, p.
238]
Usually "eindeutige und vollständige" is expressed in English with
"one-to-one", or "gegenseitig eindeutige und vollständige
Korrespondenz"
is expressed as "bijection". The word "gegenseitig" is not really
necessaty but at the time the idea was new and therefore greater
clarity was needed.
Yes. But "vollständig" is important. Otherwise "countable" would have no meaning.
Do you think N \ {0} and N \ {3} can not be bijected?Yes. both bijection and one-to-one imply completeness."Die sämtlichen Punkte l unsrer Menge L sind also in gegenseitigThe same meaning and translation ("one-to-one" or "bijection") applies
eindeutige und vollständige Beziehung zu sämtlichen Punkten f der
Menge F gebracht," [Cantor, p. 241]
here, too.
A set of pairs. An "incomplete bijection" would be a bijection betweenIn all these example "eindeutig and vollständig" is an feature of theA bijection is a set too.
correspondence, not of any set.
How so?So no example of a set of being "complete".Without completeness countability and uncountability both would be meaningless
Exactly. "... _also_ in the form of the 'completed'..." Nothing aboutNo. Here he talks about the "Form des Vollendet-unendlichen". Vollendet"Zu dem Gedanken, das Unendlichgroße [...] auch in der bestimmten FormThis basically says that there is no real difference between actual and
des Vollendet-unendlichen mathematisch durch Zahlen zu fixieren, bin
ich fast wider meinen Willen, weil im Gegensatz zu mir wertgewordenen
Traditionen, durch den Verlauf vieljähriger wissenschaftlicher
Bemühungen und Versuche logisch gezwungen worden," [Cantor, p. 175]
potential infinity.
means completed.
Too bad nobody else does. Does Cantor? With context, please.Thie says that sets are always complete, so what has been said aboutUncompleted does not apply to sets. Therefore I use the notion
uncompleted infinities either aplies to completed infinities as well or
does not apply to sets.
collection for the potentially infinite.
Primes are not infinite themselves.This says that if there were a set of all cardinals that would create aYes. If all prime numbers could be known, the same contradiction would
contradiction.
arise.
I'm pretty sure you can download that somwhere.These are various ways to point out that sets are immutable, not in theYes. But the collection of known prime numbers, for instance, is never completed.
middle of a construction process.
Am Mon, 30 Jun 2025 20:21:09 +0200 schrieb WM:
On 29.06.2025 12:25, Mikko wrote:Yes it is actually. That is why it's called infinite. You have to take
On 2025-06-28 13:56:57 +0000, WM said:
On 28.06.2025 11:56, Mikko wrote:
On 2025-06-27 19:36:41 +0000, WM said:
On 27.06.2025 09:33, Mikko wrote:No, but it can be continued.
On 2025-06-26 13:09:32 +0000, WM said:Then it is not finished or completed.
If we subtract in the order that is used for enumerating then a >>>>>>>> last one is necessary.No, there is no last one in an infinite enumeration.
it, like Cantor, as it's own "finished" thing. That is where you fail.
Elements aren't "found", they either are or are not.It means that no further element can be found later on.The notion set can only be applied to complete sets. i.e., sets whichSaying that every set is "complete" does not mean anything,
cannot be continued.
Then it cannot be. If it is that all natural numbers are subtracted inFor the infinitieth time: no, that does not follow.
their order, then it is that a last one is subtracted.
There is no "last"
number, and yet you can "subtract" an infinity of elements - in an
infinity of "steps", naturally.
Nah. It just has a higher cardinality.But Cantor already knew that there are incomplete, i.e.,
potentially infinite sets like the set of all cardinal numbers. He
called them "absolutely infinite".
Yes. both bijection and one-to-one imply completeness.Do you think N \ {0} and N \ {3} can not be bijected?
A set of pairs.In all these example "eindeutig and vollständig" is an feature of theA bijection is a set too.
correspondence, not of any set.
How so?So no example of a set of being "complete".Without completeness countability and uncountability both would be
meaningless
Uncompleted does not apply to sets. Therefore I use the notionToo bad nobody else does.
collection for the potentially infinite.
Yes. If all prime numbers could be known, the same contradiction wouldPrimes are not infinite themselves.
arise.
On 29.06.2025 12:25, Mikko wrote:
On 2025-06-28 13:56:57 +0000, WM said:
On 28.06.2025 11:56, Mikko wrote:
On 2025-06-27 19:36:41 +0000, WM said:
On 27.06.2025 09:33, Mikko wrote:
On 2025-06-26 13:09:32 +0000, WM said:
If we subtract in the order that is used for enumerating then a last >>>>>>> one is necessary.
No, there is no last one in an infinite enumeration.
Then it is not finished or completed.
No, but it can be continued.
That is potential infinity. But Cantor claimed complete enumeration.
There is no mathematical definiton of "complete enumeration"
The definition of bijection requires completeness.
so it is
possible that Cantor's enumeartion is "complete" is one sense and
"incomplete" in another.
No. Cantor claims all, every and complete:
"The infinite sequence thus defined has the peculiar property to
contain the positive rational numbers completely, and each of them only
once at a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]
"thus we get the epitome (ω) of all real algebraic numbers [...] and
with respect to this order we can talk about the th algebraic number
where not a single one of this epitome () has been forgotten." [E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und philosophischen Inhalts", Springer, Berlin (1932) p. 116]
"such that every element of the set stands at a definite position of
this sequence"
The notion set can only be applied to complete sets. i.e., sets which
cannot be continued.
Saying that every set is "complete" does not mean anything,
It means that no further element can be found later on.
All are removed when all are removed.
When done in natural order, then a last one is to be removed before all
are removed. ℕ \ {1, 2, 3, ...} = { }.
No. You said that every set is complete, so {1, 2, 3, ...}, which must
be a set in order to be valid for the context is complete and so is
ℕ \ {1, 2, 3, ...}, which is just another way to say { }-
This cannot be accomplished
There is nothing to accomplish. What is is, that's all.
Then it cannot be. If it is that all natural numbers are subtracted in
their order, then it is that a last one is subtracted.
Being completed is not a mathematical concept. An infinite sequence just >>>> is infinite.
1.1 Cantor's original German terminology on infinite sets
The reader fluent in German may be interested in the subtleties of
Cantor's terminology on actual infinity the finer distinctions of which
are not easy to express in English. While Cantor early used
"vollständig" and "vollendet" to express "complete" and "finished", the >>> term "fertig", expressing "finished" too but being also somewhat
reminiscent of "ready", for the first time appeared in a letter to
Hilbert of 26 Sep 1897, where all its appearances had later been added
to the letter.
But Cantor already knew that there are incomplete, i.e.,
potentially infinite sets like the set of all cardinal numbers. He
called them "absolutely infinite". The details of this enigmatic notion
are explained in section 1.2 (see also section 4.1. – Unfortunately it >>> has turned out impossible to strictly separate Cantor's mathematical
and religious arguments.)
There is nothing religious in Cantor's arguments. The only traces of
his religious motivations are in the choice of his symbols, in paricular
aleph and omega.
You are wrong. Here are only few pages of my Book Transfinity:
4.1 Cantor on theology
On 2025-06-30 18:21:09 +0000, WM said:
On 29.06.2025 12:25, Mikko wrote:
That is potential infinity. But Cantor claimed complete enumeration.
There is no mathematical definiton of "complete enumeration"
The definition of bijection requires completeness.
No, it doesn't.
However, that doesn't really matter as the distinction between complete
and incomplete is not mathematical.
It means that no further element can be found later on.
Whether an element is "found" has no mathematical meaning and in particular does not affect its being or not a member of some set.
Then it cannot be. If it is that all natural numbers are subtracted in
their order, then it is that a last one is subtracted.
Given two sets there is a set that is their difference. There is no
opeartion of subtraction in order.
You are wrong. Here are only few pages of my Book Transfinity:
4.1 Cantor on theology
Theology is not mathematics.
On 02.07.2025 09:45, Mikko wrote:Yeah, nothing about "finding" in there.
On 2025-06-30 18:21:09 +0000, WM said:
On 29.06.2025 12:25, Mikko wrote:
"Numerals constitute a potential infinity. Given any numeral, we can construct a new numeral by prefixing it with S." [E. Nelson: "Hilbert's mistake" (2007) p. 3]It means that no further element can be found later on.Whether an element is "found" has no mathematical meaning and in
particular does not affect its being or not a member of some set.
Bijections don't require order. Set difference has no order.The set ℕ has an intrinsic order which can be used at any time.Then it cannot be. If it is that all natural numbers are subtracted inGiven two sets there is a set that is their difference. There is no
their order, then it is that a last one is subtracted.
opeartion of subtraction in order.
Bijecting sets presupposes and requires order. Further the difference of
sets depends strongly on the order assumed.
On 02.07.2025 21:05, joes wrote:Nothing about "finding" in there either.
Am Wed, 02 Jul 2025 15:51:01 +0200 schrieb WM:"Should we briefly characterize the new view of the infinite introduced
On 02.07.2025 09:45, Mikko wrote:Yeah, nothing about "finding" in there.
On 2025-06-30 18:21:09 +0000, WM said:
On 29.06.2025 12:25, Mikko wrote:
"Numerals constitute a potential infinity. Given any numeral, we canIt means that no further element can be found later on.Whether an element is "found" has no mathematical meaning and in
particular does not affect its being or not a member of some set.
construct a new numeral by prefixing it with S." [E. Nelson:
"Hilbert's mistake" (2007) p. 3]
by Cantor, we could certainly say: In analysis we have to deal only with
the infinitely small and the infinitely large as a limit-notion, as
something becoming, emerging, produced, i.e., as we put it, with the potential infinite. But this is not the proper infinite. That we have
for instance when we consider the entirety of the numbers 1, 2, 3, 4,
... itself as a completed unit, or the points of a line as an entirety
of things which is completely available. That sort of infinity is named actual infinite." [D. Hilbert: "Über das Unendliche", Mathematische
Annalen 95 (1925) p. 167]
Do you also have your own words to miss the topic?"thus we get the epitome (ω) of all real algebraic numbers [...] andBijections don't require order. Set difference has no order.The set ℕ has an intrinsic order which can be used at any time.Then it cannot be. If it is that all natural numbers are subtractedGiven two sets there is a set that is their difference. There is no
in their order, then it is that a last one is subtracted.
opeartion of subtraction in order.
Bijecting sets presupposes and requires order. Further the difference
of sets depends strongly on the order assumed.
with respect to this order we can talk about the th algebraic number
where not a single one of this epitome () has been forgotten." [E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und philosophischen Inhalts", Springer, Berlin (1932) p. 116]
Am Wed, 02 Jul 2025 15:51:01 +0200 schrieb WM:
On 02.07.2025 09:45, Mikko wrote:
On 2025-06-30 18:21:09 +0000, WM said:
On 29.06.2025 12:25, Mikko wrote:
Yeah, nothing about "finding" in there."Numerals constitute a potential infinity. Given any numeral, we canIt means that no further element can be found later on.Whether an element is "found" has no mathematical meaning and in
particular does not affect its being or not a member of some set.
construct a new numeral by prefixing it with S." [E. Nelson: "Hilbert's
mistake" (2007) p. 3]
Bijections don't require order. Set difference has no order.The set ℕ has an intrinsic order which can be used at any time.Then it cannot be. If it is that all natural numbers are subtracted in >>>> their order, then it is that a last one is subtracted.Given two sets there is a set that is their difference. There is no
opeartion of subtraction in order.
Bijecting sets presupposes and requires order. Further the difference of
sets depends strongly on the order assumed.
On 02.07.2025 09:45, Mikko wrote:
On 2025-06-30 18:21:09 +0000, WM said:
On 29.06.2025 12:25, Mikko wrote:
That is potential infinity. But Cantor claimed complete enumeration.
There is no mathematical definiton of "complete enumeration"
Obviously you don't know much of mathematics.
The definition of bijection requires completeness.
No, it doesn't.
The function is injective, or one-to-one, if each element of the
codomain is mapped to by at most one element of the domain,
The function is surjective, or onto, if each element of the codomain is mapped to by at least one element of the domain; Wikipedia
Bijection = injection and surjection.
Note that no element must be missing. That means completeness.
However, that doesn't really matter as the distinction between complete
and incomplete is not mathematical.
Obviously you don't know the most important parts of mathematics.
"Cantor's belief in the actual existence of the infinite of Set Theory
still predominates in the mathematical world today." [A. Robinson: "The metaphysics of the calculus", in I. Lakatos (ed.): "Problems in the philosophy of mathematics", North Holland, Amsterdam (1967) p. 39]
Note belief and predominate.
"The arguments using infinity, including the Differential Calculus of
Newton and Leibniz, do not require the use of infinite sets." [T. Jech:
"Set theory", Stanford Encyclopedia of Philosophy (2002)]
"Should we briefly characterize the new view of the infinite introduced
by Cantor, we could certainly say: In analysis we have to deal only
with the infinitely small and the infinitely large as a limit-notion,
as something becoming, emerging, produced, i.e., as we put it, with the potential infinite. But this is not the proper infinite. That we have
for instance when we consider the entirety of the numbers 1, 2, 3, 4,
... itself as a completed unit, or the points of a line as an entirety
of things which is completely available. That sort of infinity is named actual infinite." [D. Hilbert: "Über das Unendliche", Mathematische
Annalen 95 (1925) p. 167]
It means that no further element can be found later on.
Whether an element is "found" has no mathematical meaning and in particular >> does not affect its being or not a member of some set.
"Numerals constitute a potential infinity. Given any numeral, we can construct a new numeral by prefixing it with S." [E. Nelson: "Hilbert's mistake" (2007) p. 3]
Then it cannot be. If it is that all natural numbers are subtracted in
their order, then it is that a last one is subtracted.
Given two sets there is a set that is their difference. There is no
opeartion of subtraction in order.
The set ℕ has an intrinsic order which can be used at any time.
Bijecting sets presupposes and requires order. Further the difference
of sets depends strongly on the order assumed.
You are wrong. Here are only few pages of my Book Transfinity:
4.1 Cantor on theology
Theology is not mathematics.
Set theory is theology. You are right, set theory is not mathematics.
Am Wed, 02 Jul 2025 21:23:22 +0200 schrieb WM:
On 02.07.2025 21:05, joes wrote:Nothing about "finding" in there either.
Am Wed, 02 Jul 2025 15:51:01 +0200 schrieb WM:"Should we briefly characterize the new view of the infinite introduced
On 02.07.2025 09:45, Mikko wrote:Yeah, nothing about "finding" in there.
On 2025-06-30 18:21:09 +0000, WM said:
On 29.06.2025 12:25, Mikko wrote:
"Numerals constitute a potential infinity. Given any numeral, we canIt means that no further element can be found later on.Whether an element is "found" has no mathematical meaning and in
particular does not affect its being or not a member of some set.
construct a new numeral by prefixing it with S." [E. Nelson:
"Hilbert's mistake" (2007) p. 3]
by Cantor, we could certainly say: In analysis we have to deal only with
the infinitely small and the infinitely large as a limit-notion, as
something becoming, emerging, produced, i.e., as we put it, with the
potential infinite. But this is not the proper infinite. That we have
for instance when we consider the entirety of the numbers 1, 2, 3, 4,
... itself as a completed unit, or the points of a line as an entirety
of things which is completely available. That sort of infinity is named
actual infinite." [D. Hilbert: "Über das Unendliche", Mathematische
Annalen 95 (1925) p. 167]
Do you also have your own words to miss the topic?"thus we get the epitome (ω) of all real algebraic numbers [...] andBijections don't require order. Set difference has no order.The set ℕ has an intrinsic order which can be used at any time.Then it cannot be. If it is that all natural numbers are subtracted >>>>>> in their order, then it is that a last one is subtracted.Given two sets there is a set that is their difference. There is no
opeartion of subtraction in order.
Bijecting sets presupposes and requires order. Further the difference
of sets depends strongly on the order assumed.
with respect to this order we can talk about the th algebraic number
where not a single one of this epitome () has been forgotten." [E.
Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und
philosophischen Inhalts", Springer, Berlin (1932) p. 116]
On 2025-07-02 13:51:01 +0000, WM said:
The definition of bijection requires completeness.
No, it doesn't.
The function is injective, or one-to-one, if each element of the
codomain is mapped to by at most one element of the domain,
The function is surjective, or onto, if each element of the codomain
is mapped to by at least one element of the domain; Wikipedia
Bijection = injection and surjection.
Note that no element must be missing. That means completeness.
It does not mean that the bijection is completely known.
"Cantor's belief in the actual existence of the infinite of Set Theory
still predominates in the mathematical world today." [A. Robinson:
"The metaphysics of the calculus", in I. Lakatos (ed.): "Problems in
the philosophy of mathematics", North Holland, Amsterdam (1967) p. 39]
Note belief and predominate.
Mathematics is about definitions and theorems, not beliefs. Peaple may
have beliefs about open problems or other things but those beliefs have
no mathematical significance.
Mathematical existence of many kinds of infinities has a firm mathematical basis.
"The arguments using infinity, including the Differential Calculus of
Newton and Leibniz, do not require the use of infinite sets." [T.
Jech: "Set theory", Stanford Encyclopedia of Philosophy (2002)]
Differential calculus does not require sets at all.
"Numerals constitute a potential infinity. Given any numeral, we can
construct a new numeral by prefixing it with S." [E. Nelson:
"Hilbert's mistake" (2007) p. 3]
That is a possible way to view them.
But a different view does not lead
to different mathematical conclusion as they are irrelevant to inferences from axioms and postulates.
That N has an order and can be given other orders is irrelevant.
One of the first things
Cantor specified in the introduction of the concept of set was that sets
have no order, i.e., the order is not a part of a set.
On 03.07.2025 11:35, Mikko wrote:Which are the case for Cantor's function.
On 2025-07-02 13:51:01 +0000, WM said:
The function is injective, or one-to-one, if each element of the
codomain is mapped to by at most one element of the domain,
The function is surjective, or onto, if each element of the codomain
is mapped to by at least one element of the domain; Wikipedia
Bijection = injection and surjection.
Note that no element must be missing. That means completeness.
It does not mean that the bijection is completely known.
It means that every element of the domain and of the codomain is
involved.
The domain must be complete by the definition of mapping, and the
codomain must be complete by the definition of surjectivity
The rule of subset proves that every proper subset has fewer elementsNo such rule for infinite sets.
than its superset. So there are more natural numbers than prime numbers,No, you can number the primes.
The rule of construction yields the number of integers |Z| = 2|N| + 1Those numbers are equal.
and the number of fractions |Q| = 2|N|^2 + 1.
It doesn't need "actual infinities".But it needs potential infinity. Therefore your "the distinction between complete and incomplete is not mathematical." is wrong."The arguments using infinity, including the Differential Calculus ofDifferential calculus does not require sets at all.
Newton and Leibniz, do not require the use of infinite sets." [T.
Jech: "Set theory", Stanford Encyclopedia of Philosophy (2002)]
Uh, no?"Numerals constitute a potential infinity. Given any numeral, we can
construct a new numeral by prefixing it with S." [E. Nelson:
"Hilbert's mistake" (2007) p. 3]
That is a possible way to view them.
But a different view does not lead to different mathematical conclusion
as they are irrelevant to inferences from axioms and postulates.
Potential infinity is based upon other axioms than actual infinity and
has other results.
That's a different function.That N has an order and can be given other orders is irrelevant.Not for bijections. The enumeration of the rational numbers is
impossible in the natural order by size for instance.
Am Thu, 03 Jul 2025 15:08:25 +0200 schrieb WM:
The rule of subset proves that every proper subset has fewer elementsNo such rule for infinite sets.
than its superset. So there are more natural numbers than prime numbers,No, you can number the primes.
The rule of construction yields the number of integers |Z| = 2|N| + 1Those numbers are equal.
and the number of fractions |Q| = 2|N|^2 + 1.
On 03.07.2025 11:35, Mikko wrote:
On 2025-07-02 13:51:01 +0000, WM said:
The definition of bijection requires completeness.
No, it doesn't.
The function is injective, or one-to-one, if each element of the
codomain is mapped to by at most one element of the domain,
The function is surjective, or onto, if each element of the codomain is
mapped to by at least one element of the domain; Wikipedia
Bijection = injection and surjection.
Note that no element must be missing. That means completeness.
It does not mean that the bijection is completely known.
It means that every element of the domain and of the codomain is involved.
The domain must be complete by the definition of mapping, and the
codomain must be complete by the definition of surjectivity
"Cantor's belief in the actual existence of the infinite of Set Theory
still predominates in the mathematical world today." [A. Robinson: "The
metaphysics of the calculus", in I. Lakatos (ed.): "Problems in the
philosophy of mathematics", North Holland, Amsterdam (1967) p. 39]
Note belief and predominate.
Mathematics is about definitions and theorems, not beliefs. Peaple may
have beliefs about open problems or other things but those beliefs have
no mathematical significance.
Cantor's beliefs have induced a large filed of mathematics.
On 03.07.2025 16:12, joes wrote:No, this is not an accepted theorem of mainstream mathematics.
Am Thu, 03 Jul 2025 15:08:25 +0200 schrieb WM:
For all sets.The rule of subset proves that every proper subset has fewer elementsNo such rule for infinite sets.
...you can enumerate them in the sense of a bijection to N:Yes, there are only few known primes.than its superset. So there are more natural numbers than primeNo, you can number the primes.
numbers,
And half of infinity is still infinite.Only for cranks. In mathematics we use the limit. For every large enough interval the even numbers are half as many as the natural numbers. ThisThe rule of construction yields the number of integers |Z| = 2|N| + 1Those numbers are equal.
and the number of fractions |Q| = 2|N|^2 + 1.
does never change. Consequently it holds in the limit for infinite sets.
Am Thu, 03 Jul 2025 21:10:39 +0200 schrieb WM:
On 03.07.2025 16:12, joes wrote:No, this is not an accepted theorem of mainstream mathematics.
Am Thu, 03 Jul 2025 15:08:25 +0200 schrieb WM:For all sets.
The rule of subset proves that every proper subset has fewer elementsNo such rule for infinite sets.
...you can enumerate them in the sense of a bijection to N:Yes, there are only few known primes.than its superset. So there are more natural numbers than primeNo, you can number the primes.
numbers,
there is a first prime, a second and so on for every natural,
which is infinitely many.
And half of infinity is still infinite.Only for cranks. In mathematics we use the limit. For every large enoughThe rule of construction yields the number of integers |Z| = 2|N| + 1Those numbers are equal.
and the number of fractions |Q| = 2|N|^2 + 1.
interval the even numbers are half as many as the natural numbers. This
does never change. Consequently it holds in the limit for infinite sets.
On 04.07.2025 10:38, joes wrote:It is incorrect: infinite sets have subsets of the same cardinality.
Am Thu, 03 Jul 2025 21:10:39 +0200 schrieb WM:But it is correct. See the last paragraph. Can you contradict it?
On 03.07.2025 16:12, joes wrote:No, this is not an accepted theorem of mainstream mathematics.
Am Thu, 03 Jul 2025 15:08:25 +0200 schrieb WM:For all sets.
The rule of subset proves that every proper subset has fewerNo such rule for infinite sets.
elements
I didn't, and you didn't the first time. I am aware that we don't know all primes, but you can't mean that. There aren't more naturals than primes,I mentioned the *known* primes....you can enumerate them in the sense of a bijection to N: there is aYes, there are only few known primes.than its superset. So there are more natural numbers than primeNo, you can number the primes.
numbers,
first prime, a second and so on for every natural, which is infinitely
many.
That is what I said.That does not change the correct mathematics: For every interval [0, 2n] there are half as many even numbers as natural numbers. That is trueAnd half of infinity is still infinite.Only for cranks. In mathematics we use the limit. For every largeThe rule of construction yields the number of integers |Z| = 2|N| +Those numbers are equal.
1 and the number of fractions |Q| = 2|N|^2 + 1.
enough interval the even numbers are half as many as the natural
numbers. This does never change. Consequently it holds in the limit
for infinite sets.
also in the limit.
On 04.07.2025 13:18, joes wrote:Cardinality IS mathematics.
Am Fri, 04 Jul 2025 12:15:16 +0200 schrieb WM:But cardinality is nonsense because mathematics contradicts it.
On 04.07.2025 10:38, joes wrote:It is incorrect: infinite sets have subsets of the same cardinality.
Am Thu, 03 Jul 2025 21:10:39 +0200 schrieb WM:But it is correct. See the last paragraph. Can you contradict it?
On 03.07.2025 16:12, joes wrote:No, this is not an accepted theorem of mainstream mathematics.
Am Thu, 03 Jul 2025 15:08:25 +0200 schrieb WM:For all sets.
The rule of subset proves that every proper subset has fewerNo such rule for infinite sets.
elements
All primes can be known and there is no contradiction.I said: Yes. If all prime numbers could be known, the same contradiction would arise.I didn't, and you didn't the first time.I mentioned the *known* primes....you can enumerate them in the sense of a bijection to N: there isNo, you can number the primes.Yes, there are only few known primes.
a first prime, a second and so on for every natural, which is
infinitely many.
You: Primes are not infinite themselves.Yeah, that we know only finitely many primes has nothing to do with that.
Me: But the known primes are a potentially infinite collection.
Huh? There is a prime to every natural.I am aware that we don't know all primes, but you can't mean that.That shows that this "bijection" does not capture all natural numbers.
There aren't more naturals than primes,
one doesn't run out of primes when counting them.
Yes, Aleph_0 of both.No. You said above "There aren't more naturals than primes".That is what I said.And half of infinity is still infinite.That does not change the correct mathematics: For every interval
[0, 2n] there are half as many even numbers as natural numbers.
That is true also in the limit.
Am Fri, 04 Jul 2025 12:15:16 +0200 schrieb WM:
On 04.07.2025 10:38, joes wrote:It is incorrect: infinite sets have subsets of the same cardinality.
Am Thu, 03 Jul 2025 21:10:39 +0200 schrieb WM:But it is correct. See the last paragraph. Can you contradict it?
On 03.07.2025 16:12, joes wrote:No, this is not an accepted theorem of mainstream mathematics.
Am Thu, 03 Jul 2025 15:08:25 +0200 schrieb WM:For all sets.
The rule of subset proves that every proper subset has fewerNo such rule for infinite sets.
elements
I didn't, and you didn't the first time.I mentioned the *known* primes....you can enumerate them in the sense of a bijection to N: there is aYes, there are only few known primes.than its superset. So there are more natural numbers than primeNo, you can number the primes.
numbers,
first prime, a second and so on for every natural, which is infinitely
many.
I am aware that we don't know all
primes, but you can't mean that. There aren't more naturals than primes,
one doesn't run out of primes when counting them.
That is what I said.That does not change the correct mathematics: For every interval [0, 2n]And half of infinity is still infinite.Only for cranks. In mathematics we use the limit. For every largeThe rule of construction yields the number of integers |Z| = 2|N| + >>>>>> 1 and the number of fractions |Q| = 2|N|^2 + 1.Those numbers are equal.
enough interval the even numbers are half as many as the natural
numbers. This does never change. Consequently it holds in the limit
for infinite sets.
there are half as many even numbers as natural numbers. That is true
also in the limit.
On 2025-07-03 13:08:25 +0000, WM said:
On 03.07.2025 11:35, Mikko wrote:
On 2025-07-02 13:51:01 +0000, WM said:
The definition of bijection requires completeness.
No, it doesn't.
The function is injective, or one-to-one, if each element of the
codomain is mapped to by at most one element of the domain,
The function is surjective, or onto, if each element of the codomain
is mapped to by at least one element of the domain; Wikipedia
Bijection = injection and surjection.
Note that no element must be missing. That means completeness.
It does not mean that the bijection is completely known.
It means that every element of the domain and of the codomain is
involved.
Being involved is not the same as being known.
The domain must be complete by the definition of mapping, and the
codomain must be complete by the definition of surjectivity
Cantor's opinion was that everything is complete, and at lest every set is.
Cantor's beliefs have induced a large filed of mathematics.
Only becase he could support his beliefs with proofs.
Mathematicians
don't care about beliefs but they do care about proofs.
On 04.07.2025 09:51, Mikko wrote:Dunno, what do you want?
On 2025-07-03 13:08:25 +0000, WM said:
There is no proof of an actual infinity. How should that work?Cantor's beliefs have induced a large filed of mathematics.Only becase he could support his beliefs with proofs.
And it does! Twice infinite is just infinite again. No contradiction.Mathematicians don't care about beliefs but they do care about proofs.His "proofs" contradict mathematics, according to which in every
interval (0, n] and i the limit there are twice as many natural nu8mbers
as eve numbers. Every correct counting method must confirm this result.
Am Fri, 04 Jul 2025 14:32:55 +0200 schrieb WM:
On 04.07.2025 09:51, Mikko wrote:
On 2025-07-03 13:08:25 +0000, WM said:
Dunno, what do you want?There is no proof of an actual infinity. How should that work?Cantor's beliefs have induced a large filed of mathematics.Only becase he could support his beliefs with proofs.
And it does! Twice infinite is just infinite again. No contradiction.Mathematicians don't care about beliefs but they do care about proofs.His "proofs" contradict mathematics, according to which in every
interval (0, n] and i the limit there are twice as many natural nu8mbers
as eve numbers. Every correct counting method must confirm this result.
Am Fri, 04 Jul 2025 14:23:57 +0200 schrieb WM:
On 04.07.2025 13:18, joes wrote:Cardinality IS mathematics.
Am Fri, 04 Jul 2025 12:15:16 +0200 schrieb WM:But cardinality is nonsense because mathematics contradicts it.
On 04.07.2025 10:38, joes wrote:It is incorrect: infinite sets have subsets of the same cardinality.
Am Thu, 03 Jul 2025 21:10:39 +0200 schrieb WM:But it is correct. See the last paragraph. Can you contradict it?
On 03.07.2025 16:12, joes wrote:No, this is not an accepted theorem of mainstream mathematics.
Am Thu, 03 Jul 2025 15:08:25 +0200 schrieb WM:For all sets.
The rule of subset proves that every proper subset has fewerNo such rule for infinite sets.
elements
All primes can be known and there is no contradiction.
You: Primes are not infinite themselves.Yeah, that we know only finitely many primes has nothing to do with that.
Me: But the known primes are a potentially infinite collection.
On 04.07.2025 09:51, Mikko wrote:
On 2025-07-03 13:08:25 +0000, WM said:
On 03.07.2025 11:35, Mikko wrote:Being involved is not the same as being known.
On 2025-07-02 13:51:01 +0000, WM said:
The definition of bijection requires completeness.
No, it doesn't.
The function is injective, or one-to-one, if each element of the
codomain is mapped to by at most one element of the domain,
The function is surjective, or onto, if each element of the codomain is >>>>> mapped to by at least one element of the domain; Wikipedia
Bijection = injection and surjection.
Note that no element must be missing. That means completeness.
It does not mean that the bijection is completely known.
It means that every element of the domain and of the codomain is involved. >>
I only said: The definition of bijection requires completeness.
You: No, it doesn't.
On 2025-07-04 12:32:55 +0000, WM said:
I only said: The definition of bijection requires completeness.
;You: No, it doesn't.
I also said what is worng in your claim: bijection only requires that
there is one and only one element of co-domain for each element of
domain, regardless of completeness.
On 05.07.2025 10:37, Mikko wrote:
On 2025-07-04 12:32:55 +0000, WM said:
I only said: The definition of bijection requires completeness.
You: No, it doesn't.
I also said what is worng in your claim: bijection only requires that
there is one and only one element of co-domain for each element of
domain, regardless of completeness.
Bijection requires completeness of domain and codomain.
On 05.07.2025 10:37, Mikko wrote:
On 2025-07-04 12:32:55 +0000, WM said:
I only said: The definition of bijection requires completeness.
;You: No, it doesn't.
I also said what is worng in your claim: bijection only requires that
there is one and only one element of co-domain for each element of
domain, regardless of completeness.
Bijection requires completeness of domain and codomain.
On 2025-07-05 13:15:11 +0000, WM said:
On 05.07.2025 10:37, Mikko wrote:
On 2025-07-04 12:32:55 +0000, WM said:
I only said: The definition of bijection requires completeness.
;You: No, it doesn't.
I also said what is worng in your claim: bijection only requires that
there is one and only one element of co-domain for each element of
domain, regardless of completeness.
Bijection requires completeness of domain and codomain.
So you say but cannot prove.
On 06.07.2025 10:34, Mikko wrote:
On 2025-07-05 13:15:11 +0000, WM said:
On 05.07.2025 10:37, Mikko wrote:
On 2025-07-04 12:32:55 +0000, WM said:
I only said: The definition of bijection requires completeness.
;You: No, it doesn't.
I also said what is worng in your claim: bijection only requires that
there is one and only one element of co-domain for each element of
domain, regardless of completeness.
Bijection requires completeness of domain and codomain.
So you say but cannot prove.
It is so by definition. See e.g. W. Mückenheim: "Mathematik für die
ersten Semester", 4th ed., De Gruyter, Berlin (2015).
Bijection requires completeness of domain and codomain.
So you say but cannot prove.
It is so by definition. See e.g. W. Mückenheim: "Mathematik für die
ersten Semester", 4th ed., De Gruyter, Berlin (2015).
Can you refer to some better author?
On 07.07.2025 10:29, Mikko wrote:
Bijection requires completeness of domain and codomain.
So you say but cannot prove.
It is so by definition. See e.g. W. Mückenheim: "Mathematik für die
ersten Semester", 4th ed., De Gruyter, Berlin (2015).
Can you refer to some better author?
That is hardly feasible. But you can look up the definition in every
textbook of your choice. You will find the same result. Even Wikipedia
will be sufficient: a bijection is a relation between two sets such
that each element of either set is paired with exactly one element of
the other set.
On 08.07.2025 09:46, Mikko wrote:None is *unpaired*. Nothing about "completeness" of (co)domain, whatever
On 2025-07-07 15:37:08 +0000, WM said:"Each element" means that none is missing.
On 07.07.2025 10:29, Mikko wrote:
That is hardly feasible. But you can look up the definition in everyCan you refer to some better author?It is so by definition. See e.g. W. Mückenheim: "Mathematik für die >>>>> ersten Semester", 4th ed., De Gruyter, Berlin (2015).Bijection requires completeness of domain and codomain.So you say but cannot prove.
textbook of your choice. You will find the same result. Even Wikipedia
will be sufficient: a bijection is a relation between two sets such
that each element of either set is paired with exactly one element of
the other set.
So no requirement of completeness.
On 2025-07-07 15:37:08 +0000, WM said:
On 07.07.2025 10:29, Mikko wrote:
Bijection requires completeness of domain and codomain.
So you say but cannot prove.
It is so by definition. See e.g. W. Mückenheim: "Mathematik für die
ersten Semester", 4th ed., De Gruyter, Berlin (2015).
Can you refer to some better author?
That is hardly feasible. But you can look up the definition in every
textbook of your choice. You will find the same result. Even Wikipedia
will be sufficient: a bijection is a relation between two sets such
that each element of either set is paired with exactly one element of
the other set.
So no requirement of completeness.
On 08.07.2025 09:46, Mikko wrote:
On 2025-07-07 15:37:08 +0000, WM said:
On 07.07.2025 10:29, Mikko wrote:
Bijection requires completeness of domain and codomain.
So you say but cannot prove.
It is so by definition. See e.g. W. Mückenheim: "Mathematik für die >>>>> ersten Semester", 4th ed., De Gruyter, Berlin (2015).
Can you refer to some better author?
That is hardly feasible. But you can look up the definition in every
textbook of your choice. You will find the same result. Even Wikipedia
will be sufficient: a bijection is a relation between two sets such
that each element of either set is paired with exactly one element of
the other set.
So no requirement of completeness.
"Each element" means that none is missing.
Am Tue, 08 Jul 2025 17:47:12 +0200 schrieb WM:
On 08.07.2025 09:46, Mikko wrote:None is *unpaired*.
On 2025-07-07 15:37:08 +0000, WM said:"Each element" means that none is missing.
On 07.07.2025 10:29, Mikko wrote:
That is hardly feasible. But you can look up the definition in everyCan you refer to some better author?It is so by definition. See e.g. W. Mückenheim: "Mathematik für die >>>>>> ersten Semester", 4th ed., De Gruyter, Berlin (2015).Bijection requires completeness of domain and codomain.So you say but cannot prove.
textbook of your choice. You will find the same result. Even Wikipedia >>>> will be sufficient: a bijection is a relation between two sets such
that each element of either set is paired with exactly one element of
the other set.
So no requirement of completeness.
Nothing about "completeness" of (co)domain, whatever
that may look like.
On 2025-07-08 15:47:12 +0000, WM said:
On 08.07.2025 09:46, Mikko wrote:
On 2025-07-07 15:37:08 +0000, WM said:
On 07.07.2025 10:29, Mikko wrote:
Bijection requires completeness of domain and codomain.
So you say but cannot prove.
It is so by definition. See e.g. W. Mückenheim: "Mathematik für
die ersten Semester", 4th ed., De Gruyter, Berlin (2015).
Can you refer to some better author?
That is hardly feasible. But you can look up the definition in every
textbook of your choice. You will find the same result. Even
Wikipedia will be sufficient: a bijection is a relation between two
sets such that each element of either set is paired with exactly one
element of the other set.
So no requirement of completeness.
"Each element" means that none is missing.
No, it does not. What is said about each element applies to missing
elements, too.
On 09.07.2025 10:34, Mikko wrote:
On 2025-07-08 15:47:12 +0000, WM said:In mathematics, a surjective function (also known as surjection, or
On 08.07.2025 09:46, Mikko wrote:
On 2025-07-07 15:37:08 +0000, WM said:
On 07.07.2025 10:29, Mikko wrote:
Bijection requires completeness of domain and codomain.
So you say but cannot prove.
It is so by definition. See e.g. W. Mückenheim: "Mathematik für die >>>>>>> ersten Semester", 4th ed., De Gruyter, Berlin (2015).
Can you refer to some better author?
That is hardly feasible. But you can look up the definition in every >>>>> textbook of your choice. You will find the same result. Even Wikipedia >>>>> will be sufficient: a bijection is a relation between two sets such
that each element of either set is paired with exactly one element of >>>>> the other set.
So no requirement of completeness.
"Each element" means that none is missing.
No, it does not. What is said about each element applies to missing
elements, too.
onto function is a function f such that, for every element y of the function's codomain, ... [Wiki]
There is none missing.
On 08.07.2025 17:59, joes wrote:So which element of a bijection isn't?
Am Tue, 08 Jul 2025 17:47:12 +0200 schrieb WM:
On 08.07.2025 09:46, Mikko wrote:
On 2025-07-07 15:37:08 +0000, WM said:
On 07.07.2025 10:29, Mikko wrote:
Bijection requires completeness of domain and codomain.So you say but cannot prove.
Each element of either set is in the bijection.None is *unpaired*."Each element" means that none is missing.Wikipedia will be sufficient: a bijection is a relation between twoSo no requirement of completeness.
sets such that each element of either set is paired with exactly one >>>>> element of the other set.
Your concept of "completeness" seems to differ from that.Nothing about "completeness" of (co)domain, whatever that may lookLook up surjection, for instance in Wikipedia: ...
like.
On 2025-07-09 16:35:02 +0000, WM said:
On 09.07.2025 10:34, Mikko wrote:
On 2025-07-08 15:47:12 +0000, WM said:In mathematics, a surjective function (also known as surjection, or
On 08.07.2025 09:46, Mikko wrote:
On 2025-07-07 15:37:08 +0000, WM said:
On 07.07.2025 10:29, Mikko wrote:
Bijection requires completeness of domain and codomain.
So you say but cannot prove.
It is so by definition. See e.g. W. Mückenheim: "Mathematik für >>>>>>>> die ersten Semester", 4th ed., De Gruyter, Berlin (2015).
Can you refer to some better author?
That is hardly feasible. But you can look up the definition in
every textbook of your choice. You will find the same result. Even >>>>>> Wikipedia will be sufficient: a bijection is a relation between
two sets such that each element of either set is paired with
exactly one element of the other set.
So no requirement of completeness.
"Each element" means that none is missing.
No, it does not. What is said about each element applies to missing
elements, too.
onto function is a function f such that, for every element y of the
function's codomain, ... [Wiki]
That text does not say that
There is none missing.
On 10.07.2025 11:47, Mikko wrote:
On 2025-07-09 16:35:02 +0000, WM said:
On 09.07.2025 10:34, Mikko wrote:
On 2025-07-08 15:47:12 +0000, WM said:In mathematics, a surjective function (also known as surjection, or
On 08.07.2025 09:46, Mikko wrote:
On 2025-07-07 15:37:08 +0000, WM said:
On 07.07.2025 10:29, Mikko wrote:
Bijection requires completeness of domain and codomain.
So you say but cannot prove.
It is so by definition. See e.g. W. Mückenheim: "Mathematik für die >>>>>>>>> ersten Semester", 4th ed., De Gruyter, Berlin (2015).
Can you refer to some better author?
That is hardly feasible. But you can look up the definition in every >>>>>>> textbook of your choice. You will find the same result. Even Wikipedia >>>>>>> will be sufficient: a bijection is a relation between two sets such >>>>>>> that each element of either set is paired with exactly one element of >>>>>>> the other set.
So no requirement of completeness.
"Each element" means that none is missing.
No, it does not. What is said about each element applies to missing
elements, too.
onto function is a function f such that, for every element y of the
function's codomain, ... [Wiki]
That text does not say that
There is none missing.
Try to learn to understand mathematical texts. Every means none is
missing, formally abbreviated by the universal quantifier ∀. Until you
got it: EOD.
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