On 10/26/2024 12:22 PM, Ross Finlayson wrote:

On 10/25/2024 12:44 PM, Jim Burns wrote:

On 10/23/2024 1:38 PM, Ross Finlayson wrote:

[...] that the most direct mapping between
discrete domain and continuous range is
this totally simple continuum limit of n/d
for natural integers as only d is not finite
and furthermore
is constant monotone strictly increasing
with a bounded range in [a,b], an infinite domain.

The continuum limit is not the continuum.
I know:
it sounds like it should be, but it isn't.

The continuum limit is
the spacing of a lattice approaching 0.

If we are _already_ working in the continuum,
the lattice points _in the limit_
are sufficient to
uniquely determine a _continuous_ function.
For many purposes,
uniquely determining a continuous function
is sufficient for that purpose.

But that isn't the continuum.
In a continuum,
each split has a point at the split,
either one which ends the foresplit
or one which begins the hindsplit
_which is different_

Do you yet recall that these properties:
extent density completeness measure,
would establish that ran(f) that being ran(EF)
is a continuous domain?

I still recall
you claiming that
EF(ℕ) is Dedekind.complete [0,1]ᴿ
You establishing that, not so much.

Do you recall that the continuum limit
is not the continuum?

The continuum limit is
letting the spacing of a lattice approach 0.

Then that completeness is as simply trivial
that it's defined that
the least-upper-bound of the set is
an element of the set, that
for f(...m) that f(m+1) is this?

Consider your
n/d n->d d->oo

Is that complete real interval [0,1]ᴿ ?

If
n/d n->d d->oo
means
limᵈ⁻ᐣⁱⁿᶠlimⁿ⁻ᐣᵈn/d
then no.

limᵈ⁻ᐣⁱⁿᶠlimⁿ⁻ᐣᵈn/d = limᵈ⁻ᐣⁱⁿᶠd/d = 1


If
n/d n->d d->oo
means
limᵈ⁻ᐣⁱⁿᶠ[0,d]ᴺ/d
(integer.interval [0,d]ᴺ ᵉᵃᶜʰ/d)
then also no.

⟨ [0,d]ᴺ/d ⟩ᵈ⁼¹ᐧᐧᐧⁱⁿᶠ is
the infinite sequence of sets [0,d]ᴺ/d

E([0,c]ᴺ/c) is an end.segment of ⟨ [0,d]ᴺ/d ⟩ᵈ⁼¹ᐧᐧᐧⁱⁿᶠ E([0,c]ᴺ/c) = { [0,c]ᴺ/c [0,c+1]ᴺ/(c+1) [0,c+2]ᴺ/(c+2) ... }

⋃E([0,c]ᴺ/c) is the supremum of end.segment E([0,c]ᴺ/c)

Each end.segment.supremum ⋃E([0,c]ᴺ/c) is
a superset of any set.limit of E([0,c]ᴺ/c)
-- if that set.limit exists.

⟨ ⋃E([0,c]ᴺ/c) ⟩ᶜ⁼¹ᐧᐧᐧⁱⁿᶠ is
an infinite sequence of supersets of
any set.limit of ⟨ [0,d]ᴺ/d ⟩ᵈ⁼¹ᐧᐧᐧⁱⁿᶠ
-- if that set.limit exists.

⋂⁰ᑉᶜ⋃E([0,c]ᴺ/c) is
also a superset of
any set.limit of ⟨ [0,d]ᴺ/d ⟩ᵈ⁼¹ᐧᐧᐧⁱⁿᶠ
-- if that set.limit exists.

However,
⋂⁰ᑉᶜ⋃E([0,c]ᴺ/c) = rational interval [0,1]ꟴ
[0,1]ꟴ is not Dedekind.complete.
Each subset of [0,1]ꟴ is not Dedekind.complete.
Any set.limit of ⟨ [0,d]ᴺ/d ⟩ᵈ⁼¹ᐧᐧᐧⁱⁿᶠ
is not Dedekind.complete
-- if that set.limit exists.

Either
limᵈ⁻ᐣⁱⁿᶠ[0,d]ᴺ/d ≠ [0,1]ᴿ
because complete [0,1]ᴿ ⊈ rational [0,1]ꟴ
or
limᵈ⁻ᐣⁱⁿᶠ[0,d]ᴺ/d ≠ [0,1]ᴿ
because limᵈ⁻ᐣⁱⁿᶠ[0,d]ᴺ/d isn't anything.


If
n/d n->d d->oo
means
[0,1]ᴿ
_by definition_
then who cares?

You have drawn a conclusion
no more sure.footed than
whatever that intuition was which
led you to make that definition.
And, anyway, a bare intuition is not shareable.

That's why we make proofs.

Then, about the "anti" and "only", and there being
this way that this ultimately tenuous continuum
limit (I'm glad at least we've arrived at that
being a word, "continuum-limit"),

[...]

makes for that
its range is a "continuous domain" itself

No.
That's not what the continuum limit is. https://en.wikipedia.org/wiki/Continuum_limit

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 10/28/2024 5:45 PM, Ross Finlayson wrote:

On 10/28/2024 01:15 PM, Jim Burns wrote:

[...]

Of course
you can keep in mind that "continuum limit"

...AKA letting lattice spacing approach 0...

was not admitted to mathematics
since modern mathematics
since at least for the past few decades,
since when for example
"everybody's favorite non-standard
not-a-real-function with standard analytical character,
function,
Dirac's delta the unit impulse function",
has of course that EF is a
non-standard and not-a-real-function,
in the usual sense,
yet has real analytical character itself.

I thought that I at least knew what it was
which you are talking about. Not so much, now.

Note, on the Dirac delta not.a.function:
It does not have a single non.0 value
such that it integrates to 1

That is an informal gloss of the not.a.function.
A useful intuition.builder, but
the existence of the Dirac delta doesn't prove
that a thick point like that exists.

----

Well, the property that
"a set is complete
if it contains each of its least-upper-bounds",

...if it contains,
for each of its bounded nonempty subsets,
a least upper bound...

is completeness,
is the one ascribed to line-reals,
ran(EF).

It just clicked with me that
you (RF) have been saying "range" and
I (JB) have been thinking "image".
My mistake.

Perhaps you should be saying "image".

It is the image EF(N) which is countable,
assuming a single.valued ("Cartesian"?) function.

The range is a superset of the image.
It might or might not also be countable.

You can't use the range to argue against
Cantor's uncountability proof.
Having a countable subset doesn't prove
countability,

It's an upper-bound, it's least, rather trivially
as either a finite set contains its upper-bound,
or, a finite set has a next-greater upper-bound,
one or the other of those is discernible and
one or the other of those exists and
one or the other of those is an upper-bound and
one or the other of those is least,
thusly the least-upper-bound "LUB" property holds,
completeness.

You have completely lost me.

as either a finite set contains its upper-bound,
or, a finite set has a next-greater upper-bound,

A finite nonempty set contains its upper bound.
So, yes,
a finite set has the least.upper.bound property.
Was it finite sets you've been talking about?

A finite set has the LUB property, yes,
but a finite set isn't a superset of Q
The smallest superset of Q with the LUB property
is R, which is what I've been talking about.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 10/30/2024 1:27 AM, Ross Finlayson wrote:

On 10/29/2024 07:38 AM, Jim Burns wrote:

A finite set has the LUB property, yes,
but a finite set isn't a superset of Q
The smallest superset of Q with the LUB property
is R, which is what I've been talking about.

That ran(EF)

Do you still define ran(EF) as n/d n->d d->oo ?

That ran(EF) is complete is rather simple,
i.e. it's trivial, in as to where,
as you for whatever reason chose to simply ignore:
the LUB can be simply either max(n), or, max(n)+1,
being or making a critical point.

I think that the reason that it seems that
I'm ignoring you is that, for some reason,
you can't see me telling you that

you are describing ran(EF) to be finite.

Whatever you mean by n/d n->d d->oo
if what you mean by it is finite,
you aren't saying what you think you're saying.

⎛ Necessary and sufficient conditions for finiteness
⎜
⎜ 3. (Paul Stäckel) S can be given a total ordering
⎜ which is well-ordered both forwards and backwards.
⎜ That is, every non-empty subset of S has
⎜ both a least and a greatest element in the subset.
⎝
https://en.wikipedia.org/wiki/Finite_set

Consider a set ran(EF) such that min.ran(EF) = 0 and,
for each non-empty subset B of ran(EF),
max.B and 1+max.B are in ran(EF)

Both a least and a greatest element is in each non-empty B
That makes ran(EF) finite.

max.B is the greatest element in B
B ᵉᵃᶜʰ≤ max.B

If 0 ∈ B, 0 is the least element in B
If 0 ∉ B
consider the non-empty set L of
strict.lower.bounds of B
L = {x ∈ ran(EF): x <ᵉᵃᶜʰ B}

max.L is a strict.lower.bound of B
1+max.L isn't a strict.lower.bound of B
max.L <ᵉᵃᶜʰ B
¬(1+max.L <ᵉᵃᶜʰ B)

∃x ∈ B: max.L < x ∧ ¬(1+max.L < x)
∃x ∈ B: 1+max.L = x
1+max.L ∈ B
1+max.L ≤ᵉᵃᶜʰ B
1+max.L is the least element in B

Both a least and a greatest element is in B
Both a least and a greatest element is in each non-empty B
That makes ran(EF) finite.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 10/30/2024 2:34 PM, Ross Finlayson wrote:

On 10/30/2024 09:18 AM, Jim Burns wrote:

On 10/30/2024 1:27 AM, Ross Finlayson wrote:

That ran(EF) is complete is rather simple,
i.e. it's trivial, in as to where,
as you for whatever reason chose to simply ignore:
the LUB can be simply either max(n), or, max(n)+1,
being or making a critical point.

⎛ Necessary and sufficient conditions for finiteness
⎜
⎜ 3. (Paul Stäckel) S can be given a total ordering
⎜ which is well-ordered both forwards and backwards.
⎜ That is, every non-empty subset of S has
⎜ both a least and a greatest element in the subset.
⎝
https://en.wikipedia.org/wiki/Finite_set

That that there's
"0/d, ..., { ... infinitely-many ... }, ..., d/d",
has that the "infinitely-many" is a usual sort of word
that is among the few words that end up in "math mode",
like "lim".

Contradiction.
If there are infinitely.many between 0/d and d/d
then there isn't max.B and 1+max.B
for each non.empty subset B

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)