Apply Cantor's enumeration of the rational numbers q_n, n = 1, 2, 3,
... Cover each q_n by the interval
ε[q_n - sqrt(2)/2^n, q_n + sqrt(2)/2^n].
Let ε --> 0.
Then all intervals together have a measure m < 2ε*sqrt(2) --> 0.
By construction there are no rational numbers outside of the intervals. Further there are never two irrational numbers without a rational
number between them. This however would be the case if an irrational
number existed between two intervals with irrational ends.
(Even the existence of neighbouring intervals is problematic.)
Therefore there is nothing between the intervals, and the complete real
axis has measure 0.
This result is wrong but implied by the premise that Cantor's
enumeration is complete.
On 2024-11-03 08:38:01 +0000, WM said:
Apply Cantor's enumeration of the rational numbers q_n, n = 1, 2, 3,
... Cover each q_n by the interval
ε[q_n - sqrt(2)/2^n, q_n + sqrt(2)/2^n].
Let ε --> 0.
Then all intervals together have a measure m < 2ε*sqrt(2) --> 0.
By construction there are no rational numbers outside of the
intervals. Further there are never two irrational numbers without a
rational number between them. This however would be the case if an
irrational number existed between two intervals with irrational ends.
No, it would not. Between any two distinct numbers, whether rational or irrational, there are both rational and irrational numbers.
As long as ε > 0 the intervals overlap
Anyway, there are real numbers that are not in any interval.
Apply Cantor's enumeration of
the rational numbers q_n, n = 1, 2, 3, ...
Cover each q_n by the interval
ε[q_n - sqrt(2)/2^n, q_n + sqrt(2)/2^n].
Let ε --> 0.
Then all intervals together have
a measure m < 2ε*sqrt(2) --> 0.
By construction there are
no rational numbers outside of the intervals.
Further there are never
two irrational numbers without
a rational number between them.
This however would be the case
This however would [not(?)] be the case
if an irrational number existed between
two intervals with irrational ends.
(Even the existence of neighbouring intervals
is problematic.)
Therefore
there is nothing between the intervals,
and the complete real axis has measure 0.
This result is wrong
but implied by the premise that
Cantor's enumeration is complete.
There aren't any neighboring intervals.
Any two intervals have intervals between them.
On 2024-11-04 09:55:24 +0000, WM said:
On 03.11.2024 23:18, Jim Burns wrote:
There aren't any neighboring intervals.
Any two intervals have intervals between them.
That is wrong. The measure outside of the intervals is infinite. Hence
there exists a point outside. This point has two nearest intervals
No, it hasn't.
Between that point an an interval there are rational
numbers and therefore other intervals
Therefore the
point has no nearest interval.
On 03.11.2024 23:18, Jim Burns wrote:
There aren't any neighboring intervals.
Any two intervals have intervals between them.
That is wrong. The measure outside of the intervals is infinite. Hence
there exists a point outside. This point has two nearest intervals
On 04.11.2024 11:31, Mikko wrote:
On 2024-11-04 09:55:24 +0000, WM said:
On 03.11.2024 23:18, Jim Burns wrote:
There aren't any neighboring intervals.
Any two intervals have intervals between them.
That is wrong. The measure outside of the intervals is infinite.
Hence there exists a point outside. This point has two nearest intervals
No, it hasn't.
In geometry it has.
Between that point an an interval there are rational
numbers and therefore other intervals
I said the nearest one. There is no interval nearer than the nearest one.
Therefore the
point has no nearest interval.
That is an unfounded assertions and therefore not accepted.
Regards, WM
On 11/4/24 5:47 AM, WM wrote:
I said the nearest one. There is no interval nearer than the nearest one.
Unless "nearest" isn't a thing because things are dense.
Therefore the
point has no nearest interval.
That is an unfounded assertions and therefore not accepted.
No, it is a PROVEN statement, therefor true.
On 04.11.2024 11:31, Mikko wrote:
On 2024-11-04 09:55:24 +0000, WM said:
On 03.11.2024 23:18, Jim Burns wrote:
There aren't any neighboring intervals.
Any two intervals have intervals between them.
That is wrong. The measure outside of the intervals is infinite. Hence
there exists a point outside. This point has two nearest intervals
No, it hasn't.
In geometry it has.
Between that point an an interval there are rational
numbers and therefore other intervals
I said the nearest one. There is no interval nearer than the nearest one.
Therefore the
point has no nearest interval.
That is an unfounded assertions and therefore not accepted.
On 2024-11-04 10:47:19 +0000, WM said:
On 04.11.2024 11:31, Mikko wrote:
On 2024-11-04 09:55:24 +0000, WM said:
On 03.11.2024 23:18, Jim Burns wrote:
There aren't any neighboring intervals.
Any two intervals have intervals between them.
That is wrong. The measure outside of the intervals is infinite.
Hence there exists a point outside. This point has two nearest
intervals
No, it hasn't.
In geometry it has.
This discussion is about numbers, not geometry.
Between that point an an interval there are rational
numbers and therefore other intervals
I said the nearest one. There is no interval nearer than the nearest one.
There is no nearesst one. There is always a nearer one.
Therefore the
point has no nearest interval.
That is an unfounded assertions and therefore not accepted.
It is not unfounded.
foundation is not in anythhing relevant).
On 04.11.2024 13:28, Richard Damon wrote:
On 11/4/24 5:47 AM, WM wrote:
I said the nearest one. There is no interval nearer than the nearest
one.
Unless "nearest" isn't a thing because things are dense.
The intervals cannot be dense because they have a length of less than 3
in an infinite space. The rationals are dense. This proves that they are
not countable.>>
Therefore the
point has no nearest interval.
That is an unfounded assertions and therefore not accepted.
No, it is a PROVEN statement, therefor true.
It is proven in an inconsistent theory. I describe the true mathematics.
Regards, WM
On 11/4/24 11:55 AM, WM wrote:
Cantor showed how to count the Rationals in a countable infinity.
He showed that the Reals Could not be counted, not even a finite length
line of them.
Therefore the
point has no nearest interval.
That is an unfounded assertions and therefore not accepted.
No, it is a PROVEN statement, therefor true.
It is proven in an inconsistent theory. I describe the true mathematics.
No, you describe
On 04.11.2024 18:49, Mikko wrote:
On 2024-11-04 10:47:19 +0000, WM said:
On 04.11.2024 11:31, Mikko wrote:
On 2024-11-04 09:55:24 +0000, WM said:
On 03.11.2024 23:18, Jim Burns wrote:
There aren't any neighboring intervals.
Any two intervals have intervals between them.
That is wrong. The measure outside of the intervals is infinite. Hence >>>>> there exists a point outside. This point has two nearest intervals
No, it hasn't.
In geometry it has.
This discussion is about numbers, not geometry.
Geometry is only another language for the same thing.
There is no nearesst one. There is always a nearer one.Between that point an an interval there are rational
numbers and therefore other intervals
I said the nearest one. There is no interval nearer than the nearest one. >>
Nonsense.
Therefore the
point has no nearest interval.
That is an unfounded assertions and therefore not accepted.
It is not unfounded.
Of course it is. It is the purest nonsense.
On 2024-11-04 18:12:55 +0000, WM said:
There is no nearesst one. There is always a nearer one.
That you don't even try to support your clam to support your claim
indicates that you don't really believe it.
conclusions of proofs and you have not shown any error in the proofs.
You are free to deny one of more of the assumptions that constitue
the foudations of the results but you havn't.
On 05.11.2024 04:12, Richard Damon wrote:
On 11/4/24 11:55 AM, WM wrote:
Cantor showed how to count the Rationals in a countable infinity.
No. Then the real axis would have measure zero.
He showed that the Reals Could not be counted, not even a finite
length line of them.
That is the nonsense believed by matheologians. He assumes a fixed set
ℕ. But if Hilbert's hotel was real, then always a diagonal number could
be enumerated by inserting it into the list at line no. 1.
Therefore the
point has no nearest interval.
That is an unfounded assertions and therefore not accepted.
No, it is a PROVEN statement, therefor true.
It is proven in an inconsistent theory. I describe the true mathematics.
No, you describe
I describe that a point between two finite intervals has two finite
intervals around it. Even this simple conclusion must be denied by the believers in matheology.
Regards, WM
On 11/5/24 3:45 AM, WM wrote:
I describe that a point between two finite intervals has two finite
intervals around it. Even this simple conclusion must be denied by the
believers in matheology.
If you mean that between to finite length
intervals, which include there endpoints, and a point that is between
those two intervals, then YES there exist two other intervals between
the point and those two intervals.
On 05.11.2024 13:03, Richard Damon wrote:
On 11/5/24 3:45 AM, WM wrote:
I describe that a point between two finite intervals has two finite
intervals around it. Even this simple conclusion must be denied by
the believers in matheology.
If you mean that between to finite length intervals, which include
there endpoints, and a point that is between those two intervals, then
YES there exist two other intervals between the point and those two
intervals.
These other intervals also have irrational endpoints. Every point
outside of an interval is next to some endpoint which is irrational.
Regards, WM
On 05.11.2024 11:29, Mikko wrote:
On 2024-11-04 18:12:55 +0000, WM said:
There is no nearesst one. There is always a nearer one.
And always the endpoint is irrational.
That you don't even try to support your clam to support your claim
indicates that you don't really believe it.
My claim says that every point outside of intervals has an irrational interval end next to it. It does not matter how many intervals you
claim between the point and the nearest interval, because _all_
intervals have irrational endpoints.
Cantor's results are
conclusions of proofs and you have not shown any error in the proofs.
I have. This example for instance proves that he did not enumerate all rationals, because the rationals are dense, the intervals are not dense.
You are free to deny one of more of the assumptions that constitue
the foudations of the results but you havn't.
Cantor's bijections concern only potentially infinite sets, but are
assumed and claimed to concern the complete sets.
That is the grave mistake. His result says for all infinite "countable
sets" that they are infinite, nothing more.
On 2024-11-05 11:26:58 +0000, WM said:
Cantor's results are
conclusions of proofs and you have not shown any error in the proofs.
I have. This example for instance proves that he did not enumerate all
rationals, because the rationals are dense, the intervals are not dense.
You have not proven that. It is fairly easy to prove that there are
no positive rationals other than those enumerated by Cantor (if I
recall correctly he enumerated only positive rationals). To prove
that there are positive rationals that are not included in Cantor's enumeration it suffices to show one but you have not shown any.
You are free to deny one of more of the assumptions that constitue
the foudations of the results but you havn't.
Cantor's bijections concern only potentially infinite sets, but are
assumed and claimed to concern the complete sets.
Everything Cantor said was about complete sets. He did neither deny the possibility of potentially infinte sets nor said anything about them (as
far as I know and remember).
That is the grave mistake. His result says for all infinite "countable
sets" that they are infinite, nothing more.
He very clearly says and proves that all infinite sets are not
equinumerous.
On 11/5/24 7:34 AM, WM wrote:
On 05.11.2024 13:03, Richard Damon wrote:
On 11/5/24 3:45 AM, WM wrote:
I describe that a point between two finite intervals has two finite
intervals around it. Even this simple conclusion must be denied by
the believers in matheology.
If you mean that between to finite length intervals, which include
there endpoints, and a point that is between those two intervals,
then YES there exist two other intervals between the point and those
two intervals.
These other intervals also have irrational endpoints. Every point
outside of an interval is next to some endpoint which is irrational.
Or rational endpoints.
There is no "next to" on the dense line
On 06.11.2024 12:46, Richard Damon wrote:
On 11/5/24 7:34 AM, WM wrote:
On 05.11.2024 13:03, Richard Damon wrote:
On 11/5/24 3:45 AM, WM wrote:
I describe that a point between two finite intervals has two finite
intervals around it. Even this simple conclusion must be denied by
the believers in matheology.
If you mean that between to finite length intervals, which include
there endpoints, and a point that is between those two intervals,
then YES there exist two other intervals between the point and those
two intervals.
These other intervals also have irrational endpoints. Every point
outside of an interval is next to some endpoint which is irrational.
Or rational endpoints.
No, the rationals are centres of their intervals.
There is no "next to" on the dense line
Every positive point is nearer to zero than to any negative point.
Of -x and 0 the latter is next to any positive x.
Regards, WM
On 06.11.2024 15:59, Mikko wrote:
On 2024-11-05 11:26:58 +0000, WM said:
Cantor's results are
conclusions of proofs and you have not shown any error in the proofs.
I have. This example for instance proves that he did not enumerate all
rationals, because the rationals are dense, the intervals are not dense.
You have not proven that. It is fairly easy to prove that there are
no positive rationals other than those enumerated by Cantor (if I
recall correctly he enumerated only positive rationals). To prove
that there are positive rationals that are not included in Cantor's
enumeration it suffices to show one but you have not shown any.
I have shown that without rational numbers outside of the intervals
with irrational endpoints covering 3 of infinitely many units the real
axis has measure 3.
On 11/6/24 1:22 PM, WM wrote:
These other intervals also have irrational endpoints. Every point
outside of an interval is next to some endpoint which is irrational.
Or rational endpoints.
No, the rationals are centres of their intervals.
They can also be endpoints of intervals.
There is no "next to" on the dense line
Every positive point is nearer to zero than to any negative point.
Of -x and 0 the latter is next to any positive x.
But that isn't "next to".
On 07.11.2024 01:51, Richard Damon wrote:
On 11/6/24 1:22 PM, WM wrote:
These other intervals also have irrational endpoints. Every point
outside of an interval is next to some endpoint which is irrational.
Or rational endpoints.
No, the rationals are centres of their intervals.
They can also be endpoints of intervals.
Not of their own intervals.
There is no "next to" on the dense line
Every positive point is nearer to zero than to any negative point.
Of -x and 0 the latter is next to any positive x.
But that isn't "next to".
It is next to when between a point and the interval no further point
exists, like here:
Use the intervals J(n) = [n - 1/10, n + 1/10]. Without splitting or
modifying them they can be translated and reordered, to cover the whole positive axis and every rational as a midpoint - if Cantor was right.
Regards, WM
On 04.11.2024 11:31, Mikko wrote:
On 2024-11-04 09:55:24 +0000, WM said:
On 03.11.2024 23:18, Jim Burns wrote:
There aren't any neighboring intervals.
Any two intervals have intervals between them.
That is wrong. The measure outside of the intervals is infinite. Hence
there exists a point outside. This point has two nearest intervals
No, it hasn't.
In geometry it has.
On 2024-11-04 10:47:39 +0000, WM said:
That is wrong. The measure outside of the intervals is infinite.
Hence there exists a point outside. This point has two nearest
intervals
No, it hasn't.
In geometry it has.
Depends on the set of intervals.
On 21.11.2024 10:21, Mikko wrote:
On 2024-11-04 10:47:39 +0000, WM said:
That is wrong. The measure outside of the intervals is infinite. Hence >>>>> there exists a point outside. This point has two nearest intervals
No, it hasn't.
In geometry it has.
Depends on the set of intervals.
No. Every point in the complement is closer to the end of an interval
than to its contents of rationals.
On 2024-11-21 10:50:32 +0000, WM said:
On 21.11.2024 10:21, Mikko wrote:
On 2024-11-04 10:47:39 +0000, WM said:
That is wrong. The measure outside of the intervals is infinite.
Hence there exists a point outside. This point has two nearest
intervals
No, it hasn't.
In geometry it has.
Depends on the set of intervals.
No. Every point in the complement is closer to the end of an interval
than to its contents of rationals.
True but irrelevant because it may be even closer to the end of
another interval.
In particular with Cantor's set of intervals
where there is no nearest interval.
On 21.11.2024 12:01, Mikko wrote:
On 2024-11-21 10:50:32 +0000, WM said:
On 21.11.2024 10:21, Mikko wrote:
On 2024-11-04 10:47:39 +0000, WM said:
That is wrong. The measure outside of the intervals is infinite. Hence >>>>>>> there exists a point outside. This point has two nearest intervals >>>>>>No, it hasn't.
In geometry it has.
Depends on the set of intervals.
No. Every point in the complement is closer to the end of an interval
than to its contents of rationals.
True but irrelevant because it may be even closer to the end of
another interval.
Every end of any interval is irrational.
On 2024-11-21 19:22:39 +0000, WM said:
True but irrelevant because it may be even closer to the end of
another interval.
Every end of any interval is irrational.
Irrelevant to the claim that a point outside of all intervals has two
nearest intervals
and also to the claim that there is a point outside
of all intervals
and also to the topic specified on the subjec line.
All those claims are true for some sets of intervals and false for others.
On 22.11.2024 09:52, Mikko wrote:
On 2024-11-21 19:22:39 +0000, WM said:
True but irrelevant because it may be even closer to the end of
another interval.
Every end of any interval is irrational.
Irrelevant to the claim that a point outside of all intervals has two
nearest intervals
What else should it have?
and also to the claim that there is a point outside
of all intervals
That is proven by the measure of the intervals being less than 3.
and also to the topic specified on the subjec line.
That is concerned by the fact that in the complement of measure oo - 3 rational numbers must exist.
All those claims are true for some sets of intervals and false for others.
Let me know what else than an irrational number could be next to a
point in the complement, if there were no rational numbers.
On 2024-11-22 11:05:58 +0000, WM said:
On 22.11.2024 09:52, Mikko wrote:
On 2024-11-21 19:22:39 +0000, WM said:
True but irrelevant because it may be even closer to the end of
another interval.
Every end of any interval is irrational.
Irrelevant to the claim that a point outside of all intervals has two
nearest intervals
What else should it have?
For every point that is neither an endpoint nor interior point of any
of the intervals there is no nearest interval because between that point
and any interval there is another interval.
That is concerned by the fact that in the complement of measure oo - 3
rational numbers must exist.
As every rational number is the midpoint of one of the intervals it is obvious that no rational is outside of all intervals.
Let me know what else than an irrational number could be next to a
point in the complement, if there were no rational numbers.
No point is next to any other point because there are other points between the two.
On 24.11.2024 13:56, Mikko wrote:
On 2024-11-22 11:05:58 +0000, WM said:
On 22.11.2024 09:52, Mikko wrote:
On 2024-11-21 19:22:39 +0000, WM said:
True but irrelevant because it may be even closer to the end of
another interval.
Every end of any interval is irrational.
Irrelevant to the claim that a point outside of all intervals has two
nearest intervals
What else should it have?
For every point that is neither an endpoint nor interior point of any
of the intervals there is no nearest interval because between that point
and any interval there is another interval.
Wrong. Starting from that point a cursor, by simplest logic, will
either touch nothing or something. If it touches something, it can only
be an irrational endpoint.
On 2024-11-24 14:19:06 +0000, WM said:
For every point that is neither an endpoint nor interior point of any
of the intervals there is no nearest interval because between that point >>> and any interval there is another interval.
Wrong. Starting from that point a cursor, by simplest logic, will
either touch nothing or something. If it touches something, it can
only be an irrational endpoint.
Your "wrong" is wrong. It will either touch something or nothing. If it touches nothing it cannot touch an irrational endpoint.
On 25.11.2024 13:41, Mikko wrote:
On 2024-11-24 14:19:06 +0000, WM said:
For every point that is neither an endpoint nor interior point of any
of the intervals there is no nearest interval because between that point >>>> and any interval there is another interval.
Wrong. Starting from that point a cursor, by simplest logic, will
either touch nothing or something. If it touches something, it can only
be an irrational endpoint.
Your "wrong" is wrong. It will either touch something or nothing. If it
touches nothing it cannot touch an irrational endpoint.
But before touching a rational it will touch an irrational.
On 2024-11-25 13:55:57 +0000, WM said:
But before touching a rational it will touch an irrational.
Of course as the starting point is outside of all the intervals and
every rational is in some of the intervals and therefore must be
irrational. But when it has moved to another point it has already
moved over both infinitely many irrationals
On 26.11.2024 10:08, Mikko wrote:
On 2024-11-25 13:55:57 +0000, WM said:
But before touching a rational it will touch an irrational.
Of course as the starting point is outside of all the intervals and
every rational is in some of the intervals and therefore must be
irrational. But when it has moved to another point it has already
moved over both infinitely many irrationals
This is true in every case. The intermediate numbers cannot be
discerned. They are dark. This is so in fact between every pair of discernible real numbers: There are infinitely many dark numbers between them.
Regards, WM
On 11/26/24 6:05 AM, WM wrote:
On 26.11.2024 10:08, Mikko wrote:
On 2024-11-25 13:55:57 +0000, WM said:
But before touching a rational it will touch an irrational.
Of course as the starting point is outside of all the intervals and
every rational is in some of the intervals and therefore must be
irrational. But when it has moved to another point it has already
moved over both infinitely many irrationals
This is true in every case. The intermediate numbers cannot be
discerned. They are dark. This is so in fact between every pair of
discernible real numbers: There are infinitely many dark numbers
between them.
Name a pair of numbers that you can not see one of the numbers between
them.
All you have done is shown that to you *ALMOST* *ALL* numburs are "dark" because your logic can't handle them,
even though the mathematics that
defines them fully defines them so others CAN handle them.
Your "darkness" is just you closing your eyes to the contradictions that
you invalid use of finite logic on an infinite set
On 26.11.2024 14:11, Richard Damon wrote:
On 11/26/24 6:05 AM, WM wrote:
On 26.11.2024 10:08, Mikko wrote:
On 2024-11-25 13:55:57 +0000, WM said:
But before touching a rational it will touch an irrational.
Of course as the starting point is outside of all the intervals and
every rational is in some of the intervals and therefore must be
irrational. But when it has moved to another point it has already
moved over both infinitely many irrationals
This is true in every case. The intermediate numbers cannot be
discerned. They are dark. This is so in fact between every pair of
discernible real numbers: There are infinitely many dark numbers
between them.
Name a pair of numbers that you can not see one of the numbers between
them.
Between every pair of discernible real numbers potentially infinitely
many real numbers can be seen. But between seen numbers there are always actually infinitely many dark numbers which cannot be seen
All you have done is shown that to you *ALMOST* *ALL* numburs are
"dark" because your logic can't handle them,
Neither can you.
even though the mathematics that defines them fully defines them so
others CAN handle them.
It defines the set but not its individual elements.
Your "darkness" is just you closing your eyes to the contradictions
that you invalid use of finite logic on an infinite set
All logic is finite.
Regards, WM
It defines the set but not its individual elements.
Sure it does. What number exists that it can not define?
On 26.11.2024 15:50, Richard Damon wrote:
It defines the set but not its individual elements.
Sure it does. What number exists that it can not define?
It cannot define numbers of the second half of ℕ.
Regards, WM
On 26.11.2024 10:08, Mikko wrote:
On 2024-11-25 13:55:57 +0000, WM said:
But before touching a rational it will touch an irrational.
Of course as the starting point is outside of all the intervals and
every rational is in some of the intervals and therefore must be
irrational. But when it has moved to another point it has already
moved over both infinitely many irrationals
This is true in every case. The intermediate numbers cannot be
discerned. They are dark. This is so in fact between every pair of discernible real numbers: There are infinitely many dark numbers
between them.
On 2024-11-26 11:05:30 +0000, WM said:
On 26.11.2024 10:08, Mikko wrote:
On 2024-11-25 13:55:57 +0000, WM said:
But before touching a rational it will touch an irrational.
Of course as the starting point is outside of all the intervals and
every rational is in some of the intervals and therefore must be
irrational. But when it has moved to another point it has already
moved over both infinitely many irrationals
This is true in every case. The intermediate numbers cannot be
discerned. They are dark. This is so in fact between every pair of
discernible real numbers: There are infinitely many dark numbers
between them.
Some of the intermediate numbers can be expressed with a finite string.
In particular, every rational number can.
On 27.11.2024 10:52, Mikko wrote:
On 2024-11-26 11:05:30 +0000, WM said:
On 26.11.2024 10:08, Mikko wrote:
On 2024-11-25 13:55:57 +0000, WM said:
But before touching a rational it will touch an irrational.
Of course as the starting point is outside of all the intervals and
every rational is in some of the intervals and therefore must be
irrational. But when it has moved to another point it has already
moved over both infinitely many irrationals
This is true in every case. The intermediate numbers cannot be
discerned. They are dark. This is so in fact between every pair of
discernible real numbers: There are infinitely many dark numbers
between them.
Some of the intermediate numbers can be expressed with a finite string.
But most cannot.
In particular, every rational number can.
No. For every unit fraction there exist infinitely many smaller unit fractions, infinitely many of which cannot be expressed. They remain
simply to be smaller.
Regards, WM
Nope, EVERY unit fraction is expressible, by DEFINITION.
On 27.11.2024 15:19, Richard Damon wrote:
Nope, EVERY unit fraction is expressible, by DEFINITION.But not in fact. In fact most are not expressible because there are infinitely many but only finitely many are expressible. Infinitely many
minus finitely many are infinitely many.
Regards, WM
On 27.11.2024 10:52, Mikko wrote:
On 2024-11-26 11:05:30 +0000, WM said:
On 26.11.2024 10:08, Mikko wrote:
On 2024-11-25 13:55:57 +0000, WM said:
But before touching a rational it will touch an irrational.
Of course as the starting point is outside of all the intervals and
every rational is in some of the intervals and therefore must be
irrational. But when it has moved to another point it has already
moved over both infinitely many irrationals
This is true in every case. The intermediate numbers cannot be
discerned. They are dark. This is so in fact between every pair of
discernible real numbers: There are infinitely many dark numbers
between them.
Some of the intermediate numbers can be expressed with a finite string.
But most cannot.
In particular, every rational number can.
No. For every unit fraction there exist infinitely many smaller unit fractions, infinitely many of which cannot be expressed. They remain
simply to be smaller.
On 11/27/24 3:55 PM, WM wrote:
On 27.11.2024 15:19, Richard Damon wrote:
Nope, EVERY unit fraction is expressible, by DEFINITION.But not in fact. In fact most are not expressible because there are
infinitely many but only finitely many are expressible. Infinitely
many minus finitely many are infinitely many.
EVERY unit fraction is the recpircal of a natural number.
EVERY natural number is finitely expressable.
There are an infinite number of finitly expressible numbers, since there
is no fixed upper limit to the size of an expression to express it with.
On 2024-11-27 11:12:54 +0000, WM said:
On 27.11.2024 10:52, Mikko wrote:
On 2024-11-26 11:05:30 +0000, WM said:
On 26.11.2024 10:08, Mikko wrote:
On 2024-11-25 13:55:57 +0000, WM said:
But before touching a rational it will touch an irrational.
Of course as the starting point is outside of all the intervals and
every rational is in some of the intervals and therefore must be
irrational. But when it has moved to another point it has already
moved over both infinitely many irrationals
This is true in every case. The intermediate numbers cannot be
discerned. They are dark. This is so in fact between every pair of
discernible real numbers: There are infinitely many dark numbers
between them.
Some of the intermediate numbers can be expressed with a finite string.
But most cannot.
In particular, every rational number can.
No. For every unit fraction there exist infinitely many smaller unit
fractions, infinitely many of which cannot be expressed. They remain
simply to be smaller.
The number 0 can be expressed with a finite string. The successor of an expresssible number can be expressed. The ratio of two expressible numbers can be expressed. Nothing else is a rational number.
On 27.11.2024 22:24, Richard Damon wrote:
On 11/27/24 3:55 PM, WM wrote:
On 27.11.2024 15:19, Richard Damon wrote:
Nope, EVERY unit fraction is expressible, by DEFINITION.But not in fact. In fact most are not expressible because there are
infinitely many but only finitely many are expressible. Infinitely
many minus finitely many are infinitely many.
EVERY unit fraction is the recpircal of a natural number.
EVERY natural number is finitely expressable.
No. There are infinitely many, but only finitely many can be expressed.
There are an infinite number of finitly expressible numbers, since
there is no fixed upper limit to the size of an expression to express
it with.
That is potential infinity. Actual or completed infinity is a fixed
quantity larger than every natural number. The set of expressed numbers however remains finite in every instance.
Regards, WM
On 28.11.2024 12:46, Mikko wrote:
On 2024-11-27 11:12:54 +0000, WM said:
On 27.11.2024 10:52, Mikko wrote:
On 2024-11-26 11:05:30 +0000, WM said:But most cannot.
On 26.11.2024 10:08, Mikko wrote:
On 2024-11-25 13:55:57 +0000, WM said:
But before touching a rational it will touch an irrational.
Of course as the starting point is outside of all the intervals and >>>>>> every rational is in some of the intervals and therefore must be
irrational. But when it has moved to another point it has already
moved over both infinitely many irrationals
This is true in every case. The intermediate numbers cannot be
discerned. They are dark. This is so in fact between every pair of
discernible real numbers: There are infinitely many dark numbers
between them.
Some of the intermediate numbers can be expressed with a finite string. >>>
In particular, every rational number can.
No. For every unit fraction there exist infinitely many smaller unit
fractions, infinitely many of which cannot be expressed. They remain
simply to be smaller.
The number 0 can be expressed with a finite string. The successor of an
expresssible number can be expressed. The ratio of two expressible numbers >> can be expressed. Nothing else is a rational number.
That is potential infinity. Actual or completed infinity is a fixed
quantity larger than every natural number. The set of expressed numbers however remains finite in every instance.
On 2024-11-28 12:31:35 +0000, WM said:
No. For every unit fraction there exist infinitely many smaller unit
fractions, infinitely many of which cannot be expressed. They remain
simply to be smaller.
The number 0 can be expressed with a finite string. The successor of an
expresssible number can be expressed. The ratio of two expressible
numbers
can be expressed. Nothing else is a rational number.
That is potential infinity. Actual or completed infinity is a fixed
quantity larger than every natural number. The set of expressed
numbers however remains finite in every instance.
In mathematics everything is actual and nothing is potential.
There is no
concept of "expressed", only "expressible".
You can't change anything
about numbers. There are infinitely many finitely expressible numbers and that includes all rationals.
On 29.11.2024 11:12, Mikko wrote:
On 2024-11-28 12:31:35 +0000, WM said:
No. For every unit fraction there exist infinitely many smaller
unit fractions, infinitely many of which cannot be expressed. They
remain simply to be smaller.
The number 0 can be expressed with a finite string. The successor of an >>>> expresssible number can be expressed. The ratio of two expressible
numbers
can be expressed. Nothing else is a rational number.
That is potential infinity. Actual or completed infinity is a fixed
quantity larger than every natural number. The set of expressed
numbers however remains finite in every instance.
In mathematics everything is actual and nothing is potential.
Then you could avoid the ℵo unit fractions beneath every unit fraction.
But you cannot.
There is no
concept of "expressed", only "expressible".
But it is reality. Without expressing your ideas you could not do mathematics. If there is no concept, then it should be introduced.
You can't change anything
about numbers. There are infinitely many finitely expressible numbers and
that includes all rationals.
All expressed numbers are a finite set. It can be increased without
bound but it is never actually infinite like the remainder in ℕ.
According to Cantor actual infinity is a fixed quantity.
Regrads, WM
On 29.11.2024 11:12, Mikko wrote:
On 2024-11-28 12:31:35 +0000, WM said:
No. For every unit fraction there exist infinitely many smaller unit >>>>> fractions, infinitely many of which cannot be expressed. They remain >>>>> simply to be smaller.
The number 0 can be expressed with a finite string. The successor of an >>>> expresssible number can be expressed. The ratio of two expressible numbers >>>> can be expressed. Nothing else is a rational number.
That is potential infinity. Actual or completed infinity is a fixed
quantity larger than every natural number. The set of expressed numbers
however remains finite in every instance.
In mathematics everything is actual and nothing is potential.
Then you could avoid the ℵo unit fractions beneath every unit fraction.
But you cannot.
There is no
concept of "expressed", only "expressible".
But it is reality. Without expressing your ideas you could not do mathematics.
If there is no concept, then it should be introduced.
You can't change anything
about numbers. There are infinitely many finitely expressible numbers and
that includes all rationals.
All expressed numbers are a finite set. It can be increased without bound
but it is never actually infinite like the remainder in ℕ. According to Cantor actual infinity is a fixed quantity.
On 2024-11-29 13:53:39 +0000, WM said:
In mathematics everything is actual and nothing is potential.
Then you could avoid the ℵo unit fractions beneath every unit
fraction. But you cannot.
What makes you think so?
There is no
concept of "expressed", only "expressible".
But it is reality. Without expressing your ideas you could not do
mathematics.
But the mathematics would be the same anyway, independently of whether
anyone knows it.
All expressed numbers are a finite set. It can be increased without bound
but it is never actually infinite like the remainder in ℕ. According to
Cantor actual infinity is a fixed quantity.
The finite set of expressed numbers has no role in mathematics, only the infinite set of expressible numbers,
On 30.11.2024 10:54, Mikko wrote:
On 2024-11-29 13:53:39 +0000, WM said:
In mathematics everything is actual and nothing is potential.
Then you could avoid the ℵo unit fractions beneath every unit
fraction. But you cannot.
What makes you think so?
The fact that every unit fraction that can be defined has ℵo unit
fractions beneath it proves that you cannot avoid that fact.
There is no
concept of "expressed", only "expressible".
But it is reality. Without expressing your ideas you could not do
mathematics.
But the mathematics would be the same anyway, independently of whether
anyone knows it.
Anyone uses it. It is deplorable that most don't know what they need.
All expressed numbers are a finite set. It can be increased without
bound
but it is never actually infinite like the remainder in ℕ. According to >>> Cantor actual infinity is a fixed quantity.
The finite set of expressed numbers has no role in mathematics, only the
infinite set of expressible numbers,
That is always finite as proved by fact. Try to express a natural number having less than ℵo sussessors. Impossible. Since the set of natural numbers cannot be subdivided into two actually infinite consecutive
sets, the set of natural numbers that can be expressed is always finite whereupon almost all natural numbers are following.
Regards, WM
On 30.11.2024 10:54, Mikko wrote:
On 2024-11-29 13:53:39 +0000, WM said:
In mathematics everything is actual and nothing is potential.
Then you could avoid the ℵo unit fractions beneath every unit fraction. >>> But you cannot.
What makes you think so?
The fact that every unit fraction that can be defined has ℵo unit
fractions beneath it proves that you cannot avoid that fact.
There is no
concept of "expressed", only "expressible".
But it is reality. Without expressing your ideas you could not do mathematics.
But the mathematics would be the same anyway, independently of whether
anyone knows it.
Anyone uses it. It is deplorable that most don't know what they need.
All expressed numbers are a finite set. It can be increased without bound >>> but it is never actually infinite like the remainder in ℕ. According to >>> Cantor actual infinity is a fixed quantity.
The finite set of expressed numbers has no role in mathematics, only the
infinite set of expressible numbers,
That is always finite as proved by fact.
On 03.11.2024 14:57, Mikko wrote:
On 2024-11-03 08:38:01 +0000, WM said:
Apply Cantor's enumeration of the rational numbers q_n, n = 1, 2, 3,
... Cover each q_n by the interval
ε[q_n - sqrt(2)/2^n, q_n + sqrt(2)/2^n].
Let ε --> 0.
Then all intervals together have a measure m < 2ε*sqrt(2) --> 0.
By construction there are no rational numbers outside of the intervals.
Further there are never two irrational numbers without a rational
number between them. This however would be the case if an irrational
number existed between two intervals with irrational ends.
No, it would not. Between any two distinct numbers, whether rational or
irrational, there are both rational and irrational numbers.
Not between two adjacent intervals. Such intervals must exist because
space between intervals must exist. Choose a point of this space and go
in both directions, find the adjacent intervals.
As long as ε > 0 the intervals overlap
Let ε = 1. If all intervals overlap and there is no space "between",
then the measure of the real line is less than 2*sqrt(2). Therefore not
all intervals overlap.
Anyway, there are real numbers that are not in any interval.
That is not possible because between two adjacent intervals there is no rational number and hence no irrational number.
Between any two intevals there is space and that space contains other intervals.
On 13.12.2024 10:46, Mikko wrote:
Between any two intevals there is space and that space contains other
intervals.
No. Starting from a point in the complement the cursor will hit a first interval. This is true for all visible intervals.
On 2024-12-13 10:28:44 +0000, WM said:
On 13.12.2024 10:46, Mikko wrote:
Between any two intervals there is space and that space contains other
intervals.
No. Starting from a point in the complement the cursor will hit a
first interval. This is true for all visible intervals.
False. From a point that is not a part of an interval no interval is the nearest one because another interval is nearer.
On 14.12.2024 09:30, Mikko wrote:They are ALREADY there.
On 2024-12-13 10:28:44 +0000, WM said:IF ALL intervals and their endpoints are existing as invariable points
On 13.12.2024 10:46, Mikko wrote:False. From a point that is not a part of an interval no interval is
Between any two intervals there is space and that space containsNo. Starting from a point in the complement the cursor will hit a
other intervals.
first interval. This is true for all visible intervals.
the nearest one because another interval is nearer.
on the real line this cannot happen. In potential infinity however
between any two points new intervals come into being.
Am Sat, 14 Dec 2024 09:42:37 +0100 schrieb WM:
On 14.12.2024 09:30, Mikko wrote:They are ALREADY there.
On 2024-12-13 10:28:44 +0000, WM said:IF ALL intervals and their endpoints are existing as invariable points
On 13.12.2024 10:46, Mikko wrote:False. From a point that is not a part of an interval no interval is
Between any two intervals there is space and that space containsNo. Starting from a point in the complement the cursor will hit a
other intervals.
first interval. This is true for all visible intervals.
the nearest one because another interval is nearer.
on the real line this cannot happen. In potential infinity however
between any two points new intervals come into being.
On 14.12.2024 12:06, joes wrote:
Am Sat, 14 Dec 2024 09:42:37 +0100 schrieb WM:
On 14.12.2024 09:30, Mikko wrote:They are ALREADY there.
On 2024-12-13 10:28:44 +0000, WM said:IF ALL intervals and their endpoints are existing as invariable points
On 13.12.2024 10:46, Mikko wrote:False. From a point that is not a part of an interval no interval is
Between any two intervals there is space and that space containsNo. Starting from a point in the complement the cursor will hit a
other intervals.
first interval. This is true for all visible intervals.
the nearest one because another interval is nearer.
on the real line this cannot happen. In potential infinity however
between any two points new intervals come into being.
Therefore they cannot appear after the cursor has passed their
positions. Every interval and every end of an interval would be hit by
the cursor.
Regards, WM
On 12/14/24 10:46 AM, WM wrote:
On 14.12.2024 12:06, joes wrote:Where did the cursor come from in the first place?
Am Sat, 14 Dec 2024 09:42:37 +0100 schrieb WM:
On 14.12.2024 09:30, Mikko wrote:They are ALREADY there.
On 2024-12-13 10:28:44 +0000, WM said:IF ALL intervals and their endpoints are existing as invariable points >>>> on the real line this cannot happen. In potential infinity however
On 13.12.2024 10:46, Mikko wrote:False. From a point that is not a part of an interval no interval is >>>>> the nearest one because another interval is nearer.
Between any two intervals there is space and that space contains >>>>>>> other intervals.No. Starting from a point in the complement the cursor will hit a
first interval. This is true for all visible intervals.
between any two points new intervals come into being.
Therefore they cannot appear after the cursor has passed their
positions. Every interval and every end of an interval would be hit by
the cursor.
And why did it pass them when you tried to place t?
This is your old problem of there not being a "next" in a dense set.
On 14.12.2024 12:06, joes wrote:
Am Sat, 14 Dec 2024 09:42:37 +0100 schrieb WM:
On 14.12.2024 09:30, Mikko wrote:They are ALREADY there.
On 2024-12-13 10:28:44 +0000, WM said:IF ALL intervals and their endpoints are existing as invariable points
On 13.12.2024 10:46, Mikko wrote:False. From a point that is not a part of an interval no interval is
Between any two intervals there is space and that space containsNo. Starting from a point in the complement the cursor will hit a
other intervals.
first interval. This is true for all visible intervals.
the nearest one because another interval is nearer.
on the real line this cannot happen. In potential infinity however
between any two points new intervals come into being.
Therefore they cannot appear after the cursor has passed their
positions. Every interval and every end of an interval would be hit by
the cursor.
On 14.12.2024 09:30, Mikko wrote:
On 2024-12-13 10:28:44 +0000, WM said:
On 13.12.2024 10:46, Mikko wrote:
Between any two intervals there is space and that space contains other >>>> intervals.
No. Starting from a point in the complement the cursor will hit a first
interval. This is true for all visible intervals.
False. From a point that is not a part of an interval no interval is the
nearest one because another interval is nearer.
IF ALL intervals and their endpoints are existing as invariable points
on the real line this cannot happen.
On 14.12.2024 19:53, Richard Damon wrote:
On 12/14/24 10:46 AM, WM wrote:
On 14.12.2024 12:06, joes wrote:Where did the cursor come from in the first place?
Am Sat, 14 Dec 2024 09:42:37 +0100 schrieb WM:
On 14.12.2024 09:30, Mikko wrote:They are ALREADY there.
On 2024-12-13 10:28:44 +0000, WM said:IF ALL intervals and their endpoints are existing as invariable points >>>>> on the real line this cannot happen. In potential infinity however
On 13.12.2024 10:46, Mikko wrote:False. From a point that is not a part of an interval no interval is >>>>>> the nearest one because another interval is nearer.
Between any two intervals there is space and that space contains >>>>>>>> other intervals.No. Starting from a point in the complement the cursor will hit a >>>>>>> first interval. This is true for all visible intervals.
between any two points new intervals come into being.
Therefore they cannot appear after the cursor has passed their
positions. Every interval and every end of an interval would be hit by
the cursor.
It starts in the complement of the intervals of measure 3 covering
rational numbers. If the cursor is thrown by chance, the chance is 3/oo
= 0 that it hits an interval.
And why did it pass them when you tried to place t?
It passes an interval when it moves.
This is your old problem of there not being a "next" in a dense set.
In a geometry where all points exist, all points can be passed.
On 2024-12-14 15:46:04 +0000, WM said:
On 14.12.2024 12:06, joes wrote:
They are ALREADY there.
Therefore they cannot appear after the cursor has passed their
positions. Every interval and every end of an interval would be hit by
the cursor.
Yes, but not before another interval hits the cursor.
On 2024-12-14 08:42:37 +0000, WM said:
On 14.12.2024 09:30, Mikko wrote:
On 2024-12-13 10:28:44 +0000, WM said:
On 13.12.2024 10:46, Mikko wrote:
Between any two intervals there is space and that space contains other >>>>> intervals.
No. Starting from a point in the complement the cursor will hit a
first interval. This is true for all visible intervals.
False. From a point that is not a part of an interval no interval is the >>> nearest one because another interval is nearer.
IF ALL intervals and their endpoints are existing as invariable points
on the real line this cannot happen.
It can. Your { [q_n - sqrt(2)/2^n, q_n + sqrt(2)/2^n] | n = 1, 2, 3, ... }
is one such set.
On 2024-12-14 21:40:48 +0000, WM said:
In a geometry where all points exist, all points can be passed.
Yes but none of them can be passed before passing other points.
On 15.12.2024 11:51, Mikko wrote:
On 2024-12-14 21:40:48 +0000, WM said:
In a geometry where all points exist, all points can be passed.
Yes but none of them can be passed before passing other points.
That contradicts the actual existence of all. When other points a re
passed, the former has been passed before. Otherwise it would not be the former.
Regards, WM
So? The problem is the cursor can't move without immediately hitting segments, none of which are "next" because they are dense.
On 12/15/24 6:05 AM, WM wrote:
You believe that only afterwards the first interval comes into being?
That is not the infinity used in set theory.
There is no "next", only before or after in dense sets.
Next is a property of directly indexed sets
On 15.12.2024 13:52, Richard Damon wrote:
On 12/15/24 6:05 AM, WM wrote:
You believe that only afterwards the first interval comes into being?
That is not the infinity used in set theory.
There is no "next", only before or after in dense sets.
Next is a property of directly indexed sets
Next is a geometric property, in particular since the average distance
of intervals is infinitely larger than their sizes.
Regards, WM
On 12/15/24 2:29 PM, WM wrote:
Next is a geometric property, in particular since the average distance
of intervals is infinitely larger than their sizes.
Not sure where you get that the "average" distance of intervals is
infinitely larger than ther sizes.
On 15.12.2024 11:49, Mikko wrote:
On 2024-12-14 15:46:04 +0000, WM said:
On 14.12.2024 12:06, joes wrote:
They are ALREADY there.
Therefore they cannot appear after the cursor has passed their
positions. Every interval and every end of an interval would be hit by
the cursor.
Yes, but not before another interval hits the cursor.
You believe that only afterwards the first interval comes into being?
That is not the infinity used in set theory.
On 15.12.2024 11:51, Mikko wrote:
On 2024-12-14 21:40:48 +0000, WM said:
In a geometry where all points exist, all points can be passed.
Yes but none of them can be passed before passing other points.
That contradicts the actual existence of all. When other points a re
passed, the former has been passed before. Otherwise it would not be
the former.
On 15.12.2024 13:52, Richard Damon wrote:
On 12/15/24 6:05 AM, WM wrote:
You believe that only afterwards the first interval comes into being?
That is not the infinity used in set theory.
There is no "next", only before or after in dense sets.
Next is a property of directly indexed sets
Next is a geometric property, in particular since the average distance
of intervals is infinitely larger than their sizes.
On 15.12.2024 22:14, Richard Damon wrote:
On 12/15/24 2:29 PM, WM wrote:
Next is a geometric property, in particular since the average distance
of intervals is infinitely larger than their sizes.
Not sure where you get that the "average" distance of intervals is
infinitely larger than ther sizes.
The accumulated size of all intervals is less than 3 over the infinite length.
Hence
there is at least one location with a ratio oo between distance to the interval and length of the interval.
Start there with the cursor. It will hit one next interval. Crash.
On 15.12.2024 11:50, Mikko wrote:
On 2024-12-14 08:42:37 +0000, WM said:
On 14.12.2024 09:30, Mikko wrote:
On 2024-12-13 10:28:44 +0000, WM said:
On 13.12.2024 10:46, Mikko wrote:
Between any two intervals there is space and that space contains other >>>>>> intervals.
No. Starting from a point in the complement the cursor will hit a first >>>>> interval. This is true for all visible intervals.
False. From a point that is not a part of an interval no interval is the >>>> nearest one because another interval is nearer.
IF ALL intervals and their endpoints are existing as invariable points
on the real line this cannot happen.
It can. Your { [q_n - sqrt(2)/2^n, q_n + sqrt(2)/2^n] | n = 1, 2, 3, ... } >> is one such set.
That is nothing but an unfounded claim. In actual infinity of set
theory all intervals and their endpoints are existing as invariable
points from the beginning of the cursor's motion.
On 2024-12-15 11:05:48 +0000, WM said:
On 15.12.2024 11:49, Mikko wrote:
On 2024-12-14 15:46:04 +0000, WM said:
On 14.12.2024 12:06, joes wrote:
They are ALREADY there.
Therefore they cannot appear after the cursor has passed their
positions. Every interval and every end of an interval would be hit
by the cursor.
Yes, but not before another interval hits the cursor.
You believe that only afterwards the first interval comes into being?
That is not the infinity used in set theory.
There is no time in mathematics. Nothing happens. In particular, nothing comes into being.
On 2024-12-16 08:55:39 +0000, WM said:
On 15.12.2024 22:14, Richard Damon wrote:
On 12/15/24 2:29 PM, WM wrote:
Next is a geometric property, in particular since the average
distance of intervals is infinitely larger than their sizes.
Not sure where you get that the "average" distance of intervals is
infinitely larger than ther sizes.
The accumulated size of all intervals is less than 3 over the infinite
length.
True.
Hence
False.
there is at least one location with a ratio oo between distance to the
interval and length of the interval.
False. Regardless which interval is "the" interval the distance to that interval is finite and the length of the interval is non-zero so the
ratio is finite.
Start there with the cursor. It will hit one next interval. Crash.
No, it does not. It does not touch an interval before passing another interval.
An interval it touches after passing other intervals is not
the next interval.
On 15.12.2024 22:14, Richard Damon wrote:
On 12/15/24 2:29 PM, WM wrote:
Next is a geometric property, in particular since the average
distance of intervals is infinitely larger than their sizes.
Not sure where you get that the "average" distance of intervals is
infinitely larger than ther sizes.
The accumulated size of all intervals is less than 3 over the infinite length. Hence there is at least one location with a ratio oo between
distance to the interval and length of the interval. Start there with
the cursor. It will hit one next interval. Crash.
Regards, WM
On 12/16/24 3:55 AM, WM wrote:
On 15.12.2024 22:14, Richard Damon wrote:
On 12/15/24 2:29 PM, WM wrote:
Next is a geometric property, in particular since the average
distance of intervals is infinitely larger than their sizes.
Not sure where you get that the "average" distance of intervals is
infinitely larger than ther sizes.
The accumulated size of all intervals is less than 3 over the infinite
length. Hence there is at least one location with a ratio oo between
distance to the interval and length of the interval. Start there with
the cursor. It will hit one next interval. Crash.
Since none of the gaps are infinte, and none of the intervals are of 0
size, there is no "infinite" ratio of any gap to any interval.
On 17.12.2024 00:55, Richard Damon wrote:
On 12/16/24 3:55 AM, WM wrote:
On 15.12.2024 22:14, Richard Damon wrote:
On 12/15/24 2:29 PM, WM wrote:
Next is a geometric property, in particular since the average
distance of intervals is infinitely larger than their sizes.
Not sure where you get that the "average" distance of intervals is
infinitely larger than ther sizes.
The accumulated size of all intervals is less than 3 over the
infinite length. Hence there is at least one location with a ratio oo
between distance to the interval and length of the interval. Start
there with the cursor. It will hit one next interval. Crash.
Since none of the gaps are infinte, and none of the intervals are of 0
size, there is no "infinite" ratio of any gap to any interval.
There is no upper bound for the ratio between distance and size of
intervals. This excludes the density of intervals. This excludes
covering of all rationals by intervals. This excludes a bijection
between natural numbers and rational numbers.
Regards, WM
On 16.12.2024 10:51, Mikko wrote:
On 2024-12-16 08:55:39 +0000, WM said:
On 15.12.2024 22:14, Richard Damon wrote:
On 12/15/24 2:29 PM, WM wrote:
Next is a geometric property, in particular since the average distance >>>>> of intervals is infinitely larger than their sizes.
Not sure where you get that the "average" distance of intervals is
infinitely larger than ther sizes.
The accumulated size of all intervals is less than 3 over the infinite length.
True.
Hence
False.
there is at least one location with a ratio oo between distance to the
interval and length of the interval.
False. Regardless which interval is "the" interval the distance to that
interval is finite and the length of the interval is non-zero so the
ratio is finite.
Well, it is finite but huge. Much larger than the interval and
therefore the finite intervals are not dense.
On 16.12.2024 10:43, Mikko wrote:
On 2024-12-15 11:05:48 +0000, WM said:
On 15.12.2024 11:49, Mikko wrote:
On 2024-12-14 15:46:04 +0000, WM said:
On 14.12.2024 12:06, joes wrote:
They are ALREADY there.
Therefore they cannot appear after the cursor has passed their
positions. Every interval and every end of an interval would be hit by >>>>> the cursor.
Yes, but not before another interval hits the cursor.
You believe that only afterwards the first interval comes into being?
That is not the infinity used in set theory.
There is no time in mathematics. Nothing happens. In particular, nothing
comes into being.
Then your sentence "Yes, but not before another interval hits the
cursor." is false.
On 2024-12-16 10:50:22 +0000, WM said:
There is no time in mathematics. Nothing happens. In particular, nothing >>> comes into being.
Then your sentence "Yes, but not before another interval hits the
cursor." is false.
It is not false but it is not a mathematical statement as it refers to
your moving cursor that is not a mathematical object.
But your "cursor"
and its movment and therfore my "before" can be converted to mathematical presentation
On 2024-12-16 11:04:17 +0000, WM said:
False. Regardless which interval is "the" interval the distance to that
interval is finite and the length of the interval is non-zero so the
ratio is finite.
Well, it is finite but huge. Much larger than the interval and
therefore the finite intervals are not dense.
They are dense because there are other intervals between the point and the interval.
That's what "dense" means.
On 12/17/24 5:25 AM, WM wrote:
There is no upper bound for the ratio between distance and size of
intervals. This excludes the density of intervals. This excludes
covering of all rationals by intervals. This excludes a bijection
between natural numbers and rational numbers.
Nope,
On 17.12.2024 14:13, Mikko wrote:
On 2024-12-16 10:50:22 +0000, WM said:
There is no time in mathematics. Nothing happens. In particular, nothing >>>> comes into being.
Then your sentence "Yes, but not before another interval hits the
cursor." is false.
It is not false but it is not a mathematical statement as it refers to
your moving cursor that is not a mathematical object.
You are wrong. Even Cantor has given lectures on mechanics as a part of mathematics.
But your "cursor"
and its movment and therfore my "before" can be converted to mathematical
presentation
It is an object of applied mathematics.
On 17.12.2024 14:08, Mikko wrote:
On 2024-12-16 11:04:17 +0000, WM said:
False. Regardless which interval is "the" interval the distance to that >>>> interval is finite and the length of the interval is non-zero so the
ratio is finite.
Well, it is finite but huge. Much larger than the interval and
therefore the finite intervals are not dense.
They are dense because there are other intervals between the point and the >> interval.
The distance between intervals (in some location) is finite but much,
much larger than the finite length of the interval. This distance is
the distance between intervals which are next to each other. Therefore
there is nothing in between.
On 2024-12-17 19:32:51 +0000, WM said:
But your "cursor"
and its movment and therfore my "before" can be converted to
mathematical
presentation
It is an object of applied mathematics.
Matter of opinion, and so is whether applied mathematics is mahtematics.
On 2024-12-17 19:29:52 +0000, WM said:
On 17.12.2024 14:08, Mikko wrote:
On 2024-12-16 11:04:17 +0000, WM said:
False. Regardless which interval is "the" interval the distance to
that
interval is finite and the length of the interval is non-zero so the >>>>> ratio is finite.
Well, it is finite but huge. Much larger than the interval and
therefore the finite intervals are not dense.
They are dense because there are other intervals between the point
and the
interval.
The distance between intervals (in some location) is finite but much,
much larger than the finite length of the interval. This distance is
the distance between intervals which are next to each other. Therefore
there is nothing in between.
There is no next interval and therefore no distance to the next interval
as there are always other intervals nearer.
You haven't prove your claim and can't prove so it is just an unujustified opnion.
On 17.12.2024 13:34, Richard Damon wrote:
On 12/17/24 5:25 AM, WM wrote:
There is no upper bound for the ratio between distance and size of
intervals. This excludes the density of intervals. This excludes
covering of all rationals by intervals. This excludes a bijection
between natural numbers and rational numbers.
Nope,
It does. oo/3 is infinite. That means there are finite distances much
larger than the intervals of finite length between intervals which are
next to each other. This excludes that these intervals are dense.
Regards, WM
On 18.12.2024 11:16, Mikko wrote:
On 2024-12-17 19:29:52 +0000, WM said:
On 17.12.2024 14:08, Mikko wrote:
On 2024-12-16 11:04:17 +0000, WM said:
False. Regardless which interval is "the" interval the distance to that >>>>>> interval is finite and the length of the interval is non-zero so the >>>>>> ratio is finite.
Well, it is finite but huge. Much larger than the interval and
therefore the finite intervals are not dense.
They are dense because there are other intervals between the point and the >>>> interval.
The distance between intervals (in some location) is finite but much,
much larger than the finite length of the interval. This distance is
the distance between intervals which are next to each other. Therefore
there is nothing in between.
There is no next interval and therefore no distance to the next interval
as there are always other intervals nearer.
Mathematics says the covering by intervals is 3/oo. Therefore the ratio between not covered part and covered part of the positive real axis is
oo/3. That implies an average of oo/3 which in some locations must be realized.
That proves the existence of distances between next intervals much
larger than the finite lengths of the intervals.
It excludes density of intervals.
On 18.12.2024 11:23, Mikko wrote:
On 2024-12-17 19:32:51 +0000, WM said:
Anyhow my argument is correct.But your "cursor"
and its movment and therfore my "before" can be converted to mathematical >>>> presentation
It is an object of applied mathematics.
Matter of opinion, and so is whether applied mathematics is mahtematics.
On 2024-12-18 11:25:25 +0000, WM said:
Mathematics says the covering by intervals is 3/oo. Therefore the
ratio between not covered part and covered part of the positive real
axis is oo/3. That implies an average of oo/3 which in some locations
must be realized.
Yes.
That proves the existence of distances between next intervals much
larger than the finite lengths of the intervals.
No. There is no next interval because there are thore intervals nearer.
It excludes density of intervals.
No, it does not.
Not really. What is acceptable for applied mathematics depends on the application area, which you didn't specify.
On 19.12.2024 11:41, Mikko wrote:
Not really. What is acceptable for applied mathematics depends on the
application area, which you didn't specify.
It was obvious when the argument was discussed: The cursor moves from 0
to 1 on the real axis. For every unit fractions 1/n which it hits there
are smaller unit fractions which it had not hit before because they were
dark at the first time and came into being only later.
Regards, WM
On 12/19/24 10:47 AM, WM wrote:
On 19.12.2024 11:41, Mikko wrote:
Not really. What is acceptable for applied mathematics depends on the
application area, which you didn't specify.
It was obvious when the argument was discussed: The cursor moves from
0 to 1 on the real axis. For every unit fractions 1/n which it hits
there are smaller unit fractions which it had not hit before because
they were dark at the first time and came into being only later.
No, it means you missed them because you moved too far, because you
closed your eyes.
This shows that you can't move to the "first" (smallest valued) 1/n
because no such number actually exist,
On 20.12.2024 03:52, Richard Damon wrote:
On 12/19/24 10:47 AM, WM wrote:
On 19.12.2024 11:41, Mikko wrote:
Not really. What is acceptable for applied mathematics depends on the
application area, which you didn't specify.
It was obvious when the argument was discussed: The cursor moves from
0 to 1 on the real axis. For every unit fractions 1/n which it hits
there are smaller unit fractions which it had not hit before because
they were dark at the first time and came into being only later.
No, it means you missed them because you moved too far, because you
closed your eyes.
The cursor moves until it hits a unit fraction.
This shows that you can't move to the "first" (smallest valued) 1/n
because no such number actually exist,
But as soon the cursor has met a unit fraction, many smaller ones show
up. They had nor "actually" existed as visible unit fractions.
Regards, WM
On 12/20/24 9:50 AM, WM wrote:
On 20.12.2024 03:52, Richard Damon wrote:
On 12/19/24 10:47 AM, WM wrote:
On 19.12.2024 11:41, Mikko wrote:
Not really. What is acceptable for applied mathematics depends on the >>>>> application area, which you didn't specify.
It was obvious when the argument was discussed: The cursor moves
from 0 to 1 on the real axis. For every unit fractions 1/n which it
hits there are smaller unit fractions which it had not hit before
because they were dark at the first time and came into being only
later.
No, it means you missed them because you moved too far, because you
closed your eyes.
The cursor moves until it hits a unit fraction.
Then why did it it not stop till after it has passed one?
This shows that you can't move to the "first" (smallest valued) 1/n
because no such number actually exist,
But as soon the cursor has met a unit fraction, many smaller ones show
up. They had not "actually" existed as visible unit fractions.
No, they were always there, you just didn't look for them.
You find "dark numbers" because you seem to have a blind spot
On 20.12.2024 16:33, Richard Damon wrote:
On 12/20/24 9:50 AM, WM wrote:
On 20.12.2024 03:52, Richard Damon wrote:
On 12/19/24 10:47 AM, WM wrote:
On 19.12.2024 11:41, Mikko wrote:
Not really. What is acceptable for applied mathematics depends on the >>>>>> application area, which you didn't specify.
It was obvious when the argument was discussed: The cursor moves
from 0 to 1 on the real axis. For every unit fractions 1/n which it
hits there are smaller unit fractions which it had not hit before
because they were dark at the first time and came into being only
later.
No, it means you missed them because you moved too far, because you
closed your eyes.
The cursor moves until it hits a unit fraction.
Then why did it it not stop till after it has passed one?
Ask it! My answer is that it stops at the smallest unit fraction. But
you deny its existence. Then it can only stop where you allow it. But
then many smaller unit fractions show up. So your permission concerns
only visible unit fractions.
This shows that you can't move to the "first" (smallest valued) 1/n
because no such number actually exist,
But as soon the cursor has met a unit fraction, many smaller ones
show up. They had not "actually" existed as visible unit fractions.
No, they were always there, you just didn't look for them.
Use the function NUF(x). It shows the smallest unit fraction.
You find "dark numbers" because you seem to have a blind spot
The function NUF(x) has none. Like the function of endsegments:
If the complete sequence of indices indexing the Cantor list is
accepted, then it must be possible to construct it. When indexing the
entries by 1, 2, 3, ..., then always infinitely many natnumbers remain.
I call these sets endsegments
E(n) = {n+1, n+2, n+3, ...}
with
E(0) = ℕ
and
∀n ∈ ℕ : E(n+1) = E(n) \ {n+1}.
This means the sequence of endsegments can decrease only by one
natnumber per step. Therefore the sequence of endsegments cannot become
empty (i.e., not all natnumbers can be applied as indices) unless the
empty endsegment is reached, and before finite endsegments have been
passed. These however, if existing at all, cannot be seen. They are
dark. Therefore it is impossible to introduce the corresponding entries
in Cantor's list.
Regards, WM
On 12/21/24 6:00 AM, WM wrote:
The cursor moves until it hits a unit fraction.
Then why did it it not stop till after it has passed one?
Ask it! My answer is that it stops at the smallest unit fraction. But
you deny its existence. Then it can only stop where you allow it. But
then many smaller unit fractions show up. So your permission concerns
only visible unit fractions.
But you admit that it didn't because it ended up past it.
The numbers didn't just show up, they were always there and we didn't do anything that prohibited it from stopping at any of them.
The only thing that prevented it from stopping at the "first" unit
fraction is the fact that "the first unit fraction" (from the 0 end)
just doesn't exist.
Use the function NUF(x). It shows the smallest unit fraction.
You think circular definitions are acceptable.
Sure it does, as NUF(x) doesn't exist
E(n) = {n+1, n+2, n+3, ...}
with
E(0) = ℕ
and
∀n ∈ ℕ : E(n+1) = E(n) \ {n+1}.
This means the sequence of endsegments can decrease only by one
natnumber per step. Therefore the sequence of endsegments cannot
become empty (i.e., not all natnumbers can be applied as indices)
unless the empty endsegment is reached, and before finite endsegments
have been passed. These however, if existing at all, cannot be seen.
They are dark. Therefore it is impossible to introduce the
corresponding entries in Cantor's list.
What Natural Number can't be used as an index?
On 21.12.2024 14:19, Richard Damon wrote:
On 12/21/24 6:00 AM, WM wrote:
The cursor moves until it hits a unit fraction.
Then why did it it not stop till after it has passed one?
Ask it! My answer is that it stops at the smallest unit fraction. But
you deny its existence. Then it can only stop where you allow it. But
then many smaller unit fractions show up. So your permission concerns
only visible unit fractions.
But you admit that it didn't because it ended up past it.
The reason is that after every visible unit fraction there are more
visible unit fractions created. That is potential infinity. You can't understand that matter.
The numbers didn't just show up, they were always there and we didn't
do anything that prohibited it from stopping at any of them.
They were dark.
The only thing that prevented it from stopping at the "first" unit
fraction is the fact that "the first unit fraction" (from the 0 end)
just doesn't exist.
Before any unit fraction where it stops there show up many more - but
only afterwards.
Use the function NUF(x). It shows the smallest unit fraction.
You think circular definitions are acceptable.
It is not circular.
Sure it does, as NUF(x) doesn't exist
You can't grasp it. That's not my problem.
E(n) = {n+1, n+2, n+3, ...}
with
E(0) = ℕ
and
∀n ∈ ℕ : E(n+1) = E(n) \ {n+1}.
This means the sequence of endsegments can decrease only by one
natnumber per step. Therefore the sequence of endsegments cannot
become empty (i.e., not all natnumbers can be applied as indices)
unless the empty endsegment is reached, and before finite endsegments
have been passed. These however, if existing at all, cannot be seen.
They are dark. Therefore it is impossible to introduce the
corresponding entries in Cantor's list.
What Natural Number can't be used as an index?
The true infinite can be exhausted, but only collectively:
ℕ \ {1, 2, 3, ...} = { }. So it gets empty. The getting empty happens collectively but cannot be observed individually. We cannot construct a
list that assigns natural numbers to all natural numbers. Almost all
cannot be manipulated individually. But
∀n ∈ ℕ : E(n+1) = E(n) \ {n+1}
proves that the sequence of endsegments can decrease only by one
natnumber per step. Therefore the sequence of endsegments cannot become
empty (i.e., not all natnumbers can be applied as indices) unless the
empty endsegment is reached, and before finite endsegments, endsegments containing only 1, 2, 3, or n ∈ ℕ numbers, have been passed.
Regards, WM
On 12/21/24 4:40 PM, WM wrote:
The reason is that after every visible unit fraction there are more
visible unit fractions created. That is potential infinity. You can't
understand that matter.
No, they are not "created" they have always been there, they just
haven't been enumerated/discovered yet.
The numbers didn't just show up, they were always there and we didn't
do anything that prohibited it from stopping at any of them.
They were dark.
No, you just didn't see them because your logic closes its eyes and lies
to you.
On 22.12.2024 13:28, Richard Damon wrote:
On 12/21/24 4:40 PM, WM wrote:
The reason is that after every visible unit fraction there are more
visible unit fractions created. That is potential infinity. You can't
understand that matter.
No, they are not "created" they have always been there, they just
haven't been enumerated/discovered yet.
Why haven't they?
The numbers didn't just show up, they were always there and we
didn't do anything that prohibited it from stopping at any of them.
They were dark.
No, you just didn't see them because your logic closes its eyes and
lies to you.
Then show me a unit fraction that you see for the first time when going
from 0 to 1 which has no smaller unit fractions that you hadn't
discovered before.
Regards, WM
On 22.12.2024 13:28, Richard Damon wrote:
On 12/21/24 4:40 PM, WM wrote:
The reason is that after every visible unit fraction there are more
visible unit fractions created. That is potential infinity. You can't
understand that matter.
No, they are not "created" they have always been there, they just
haven't been enumerated/discovered yet.
Why haven't they?
The numbers didn't just show up, they were always there and we
didn't do anything that prohibited it from stopping at any of them.
They were dark.
No, you just didn't see them because your logic closes its eyes and
lies to you.
Then show me a unit fraction that you see for the first time when going
from 0 to 1 which has no smaller unit fractions that you hadn't
discovered before.
Regards, WM
On 12/22/24 10:04 AM, WM wrote:
No, they are not "created" they have always been there, they just
haven't been enumerated/discovered yet.
Why haven't they?
Because you just haven't had time to discover it,
Then show me a unit fraction that you see for the first time whenBut the question isn't "hadn't discovered befor" but which exists.
going from 0 to 1 which has no smaller unit fractions that you hadn't
discovered before.
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