On 5/21/2024 7:22 AM, Ross Finlayson wrote:

On 05/20/2024 02:28 AM, FromTheRafters wrote:

[...]

Continuum mechanics, has that there are
at least three definitions of a continuous domain,
line-reals, field-reals, and signal-reals.

What makes the real numbers important is that
they insert points in the rational numbers so that,
if a function is continuous at each point,
then that function does not jump.

One way to "insert points" is to regard
each equivalence class of all Cauchy sequences which
converge to each other
as a single point.

Another way to "insert points" is to regard
each open foresplit of the rational numbers
as a single point.

⎛ These are called "constructions" but
⎜ nothing is built in any non.mathematical sense,
⎝ They have been recognized as existing.

Your "continuous domain" is probably
our "connected domain".

Your line-reals field-reals and signal-reals
can't all be our Dedekind.complete-reals.
It is the Dedekind.complete-reals which
physicists and their ilk know and love.

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On 5/21/2024 4:06 PM, Ross Finlayson wrote:

On 05/21/2024 10:36 AM, Jim Burns wrote:

On 5/21/2024 7:22 AM, Ross Finlayson wrote:

Continuum mechanics, has that there are
at least three definitions of a continuous domain,
line-reals, field-reals, and signal-reals.

Your "continuous domain" is probably
our "connected domain".

Your line-reals field-reals and signal-reals
can't all be our Dedekind.complete-reals.
It is the Dedekind.complete-reals which
physicists and their ilk know and love.

The real numbers "exist", mathematically,
as a model of a linear continuum and that a
linear continuum is a model of real numbers.

The Dedekind.complete.real.number axioms
answer the question:
| This so.called "continuum" of which you.all speak,
| what is that?

Now, when somebody like Hilbert says,
"you know, Euclid is great and all, yet there's
always been an implicit postulate of continuity",

Add to the list:
"You know, Newton was great and all, yet
he completely screwed up the bending of starlight
near the Sun and the perihelion advance of Mercury."

I admit that
there are days when it sounds like a dark joke;
nonetheless, it is so: _humanity advances_
Especially, seen over thousands of years.

Now, when somebody like Hilbert says,
"you know, Euclid is great and all, yet there's
always been an implicit postulate of continuity",
you wonder, is the drawing of a line, as of putting
pencil to paper, drawing a line, and lifting the pencil,
underdefined, or is it really primitive itself?

By considerable effort, we have learned that
the essential part of the description of "continuum"
is that, when lines cross, they intersect.

crossed lines intersecting
Dedekind completeness
intermediate value theorem
least upper bound property
are all
different ways of saying the same thing,
which is: _what a continuum is_

Subsets of rationals in rationals are rationals.

They definitely are not.

Broadly speaking, sets are not their elements.

There is the occasional exception.
Frex, some sets of von Neumann ordinals
are themselves von Neumann ordinals.
Mostly not.

One problem with sets of rationals being rationals
is that there are more sets than numbers.

For any F: ℚ → 𝒫(ℚ)
F is not.onto 𝒫(ℚ)

In particular,
any f such that
F(f) = {q ∈ ℚ: q ∉ F(q) ∈ 𝒫(Q)}
is contradictory, so non.existent.

Furthermore,
the partitions of rationals are countable.

They definitely are not.

Consider
a countably.infinite sequence ⟨F₁ F₂ F₃ ...⟩ of
open.foresplits Fⱼ of ℚ

Each Fⱼ ∈ ⟨F₁ F₂ F₃ ...⟩ is finitely preceded
in ⟨F₁ F₂ F₃ ...⟩

There is a countably.infinite sequence of
nested in.line intervals such that
any sequenced foresplit
between in.line endpoints of an interval
is after in.sequence the endpoints.

A sequenced foresplit _in all intervals_
would need to be NOT.finitely preceded,
which is unlike any sequenced foresplit.
Therefore,
no _sequenced_ foresplit is in all intervals.

However,
some foresplit IS in all nested intervals.

That foresplit is the union of
all sequenced foresplits which stand
less.than in.line any nested interval.
That union (foresplit) is
-- an open foresplit of ℚ
-- in all nested intervals

That union is an open foresplit of ℚ
which is NOT sequenced by ⟨F₁ F₂ F₃ ...⟩

Generalizing,
NO countably.infinite sequence of open.foresplits of ℚ
holds all open.foresplits of ℚ

The open foresplits of ℚ are NOT countable.

Your attachment to Dedekind cuts as a putative
model of reals is well-known, yet that doesn't make
it any different, than compoundedly countable.

Not that they're countable, but
their non.countability status is a sideshow.

The least.upper.bound of
a bounded.nonempty.set of open.foresplits of ℚ
is the _union_ of that bounded.nonempty.set of
open.foresplits.

Therefore,
the open.foresplits of ℚ have the least.upper.bound
property.

Therefore,
lines (composed of open.foresplits of ℚ)
which cross intersect.

Therefore,
the open.foresplits of ℚ are what we mean by
"continuum".

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On 5/21/2024 7:05 PM, Ross Finlayson wrote:

On 05/21/2024 03:53 PM, Jim Burns wrote:

On 5/21/2024 4:06 PM, Ross Finlayson wrote:

Subsets of rationals in rationals are rationals.

The open foresplits of ℚ are NOT countable.

Partitions of rationals, contain a rational,
that all lesser partitions, in the natural
and total order of them, don't contain,
to be distinct.

I want to say that you and I have already done this.
I can't lay my hands on those posts at the moment,
though.

Did you, Ross, once upon a time have a paper that
made that argument?

However,
between each irrational xₖ and each rational qᵢ < xₖ
there is an irrational xⱼ: qᵢ < xⱼ < xₖ
(Proof upon request.)

For each open.foresplit ℚᑉˣᵏ with LUB xₖ
there does not exist rational qᵢ ∈ ℚᑉˣᵏ
such that
for each open.foresplit ℚᑉˣʲ ≠⊂ ℚᑉˣᵏ
qᵢ ∉ ℚᑉˣʲ

because qᵢ < xⱼ < xₖ

Therefore,
no,
no open.foresplit of rationals contains
a rational that
all lesser open.foresplits don't contain.

Each, ....

So, how many are there again?

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On 5/21/2024 10:45 PM, Ross Finlayson wrote:

On 05/21/2024 05:50 PM, Jim Burns wrote:

On 5/21/2024 7:05 PM, Ross Finlayson wrote:

On 05/21/2024 03:53 PM, Jim Burns wrote:

On 5/21/2024 4:06 PM, Ross Finlayson wrote:

Subsets of rationals in rationals are rationals.

The open foresplits of ℚ are NOT countable.

Partitions of rationals, contain a rational,
that all lesser partitions, in the natural
and total order of them, don't contain,
to be distinct.

between each irrational xₖ and each rational qᵢ < xₖ
there is an irrational xⱼ:  qᵢ < xⱼ < xₖ
(Proof upon request.)

For each open.foresplit ℚᑉˣᵏ with LUB xₖ
there does not exist rational qᵢ ∈ ℚᑉˣᵏ
such that
for each open.foresplit ℚᑉˣʲ ≠⊂ ℚᑉˣᵏ
qᵢ ∉ ℚᑉˣʲ

because  qᵢ < xⱼ < xₖ

Therefore,
no,
no open.foresplit of rationals contains
a rational that
all lesser open.foresplits don't contain.

The other day I'm reading a brief biography
by Dauben about Robinso(h)n, who was sort of
in the orbit of Tarski or Teitelbaum and was
mostly known for "Non-Standard Analysis", which
after the language of set theory is sort of
about non-standard analysis, except
it's only a "conservative" extension of sorts,
about a "halo of hyper-reals" in
a little contiguous cloud scattered about
each point on the real line.

Something important to keep in mind about
non.standard analysis is that
it's not standard analysis.

An essential aspect of
the kind of conclusions which we here draw
is: what.it.is about which we draw them.
Obvious? Oh, HELL, yes, it's obvious.
But obvious does not mean unimportant.

Real numbers are a certain thing.
Natural numbers are a certain thing.
Right triangles are a certain thing.

A line such that crossed curves must intersect
does not have a little contiguous cloud
scattered about each point.
(Proof upon request.)

That isn't a problem for standard analysis
and it isn't a problem for non.standard analysis.
Different conclusions about different things.

It's a real thing,
and helps a lot to be conscientious formalists 24/7,
and Sundays are just off.

(That's a cultural reference to
the guy who joked about
mathematicians being week-day formalists,
yet who really believed in a sense in platonism,
yet didn't want to be teased at school
for not being fictionalists.
Or, you know,
when their self-respect wasn't
a matter of others' humiliation,
that it was their real working theory.)

Another moral of the story of
the guy who joked about Sunday.Platonism
is that
_it doesn't affect our results_

Maybe you think there's
an immaterial unchangeable realm of ideals.
Maybe I think only formulas really exist
in chalk, in pixels, or anywhere.

You and I reach the same conclusions
about the same things.
Or, if you like,
Sunday.I and Wednesday.I reach
the same unchanging conclusions.

So, where Aristotle already has
at least two definitions of continuity, and
furthermore sort of puts "potential and actual",
I think that's great.
Then that most people don't know the half of it,
I figure is they, you know, don't know,
thusly I'm sort of like, "don't care".

I won't speak for others, but
what I see,
when you introduce considerations which
have no effect on the question on the table,
is that you are no longer interested in
the question on the table.
Which is your right.
Just as it is my right to continue
to be interested in that question, and
my right to not.want to
hare off in another direction.

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On 5/21/2024 11:28 PM, Ross Finlayson wrote:

On 05/21/2024 07:45 PM, Ross Finlayson wrote:

On 05/21/2024 05:50 PM, Jim Burns wrote:

Anyways, each partition, of the rationals, in
their normal ordering,
contains a rational, not contained, in
any partition that's lesser,

That's still incorrect,
Platonically and formalistically incorrect.

Not:
[anyways,] each partition of the rationals in
their normal ordering
contains a rational not contained in
any partition that's lesser.

Not:
each partition ℚᑉˣᵏ of the rationals in
their normal ordering '⊆'
contains a rational qᵢ not contained in
any partition ℚᑉˣʲ that's lesser.

¬∀ℚᑉˣᵏ: ∃qᵢ ∈ ℚᑉˣᵏ:
∀ℚᑉˣʲ: ℚᑉˣʲ < ℚᑉˣᵏ ⇒ qᵢ ∉ cℚᑉˣʲ

¬∀xₖ ∈ ℝ\ℚ: ∃qᵢ ∈ ℚ: qᵢ < xₖ ∧
∀xⱼ ∈ ℝ\ℚ: xⱼ < xₖ ⇒ qᵢ ≮ xⱼ

¬∀ᴿᐠꟴxₖ: ∃ꟴqi < xₖ:
∀ᴿᐠꟴxⱼ < xₖ: qi ≮ xⱼ

∃ᴿᐠꟴxₖ: ∀ꟴqi < xₖ:
∃ᴿᐠꟴxⱼ: qi < xⱼ < xₖ

Upon request, I'll prove
∀ᴿᐠꟴxₖ: ∀ꟴqi < xₖ:
∃ᴿᐠꟴxⱼ: qi < xⱼ < xₖ

which, for non.∅ ℝ\ℚ, contradicts
| Anyways, each partition of the rationals in
| their normal ordering
| contains a rational not contained in
| any partition that's lesser.

furthermore,
each partitions contains
a rational not contained
in any partition that's greater,
where a partition is by any real number, say,
yet here mostly about the rationals by themselves.

The problem with that claim is that
whether partitionsᴿꟳ (AKA open.foresplits) are
greater or lesser
is defined by the subset relation '⊆'
A partitionᴿꟳ holding rationals not.in a superset
doesn't begin to make sense.

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On 5/21/2024 11:28 PM, Ross Finlayson wrote:

On 05/21/2024 07:45 PM, Ross Finlayson wrote:

On 05/21/2024 05:50 PM, Jim Burns wrote:

[...]

[...]

One time there was a thread here about
"Well-Ordering the Reals".
This of course is where I came up with,
"So, well-order the reals", or,
"well-order the reals, while you're at it".

It sounds as though we are not communicating here.
I expect that you heard the same back in Ye Olde Thredde.

A well.order exists or not.exists.
We can determine which not.first.false.ly
from the description of whatever we're discussing.
Well.ordering is not an activity we engage in.
"Just do it" is a category error here.

So anyways it sort of gets into that
there's not an uncountable well-ordering,
that's in the real's normal ordering,

If there is a 1.to.1 map from the reals to
an uncountable ordinal, then there is
a well.ordering of the reals.

That well.ordering will not be
the normal ordering of the reals,
which is not a well.ordering.

The set of all countable ordinals is
an uncountable ordinal (a limit ordinal),
reminiscent of how
the set of all finite ordinals is
an infinite ordinal (a limit ordinal).

because, and, you know, be-cause, so-caused,
there are rationals between each those, because
rationals are dense in the real numbers,
so-caused as fore-stalled.

Yet, then somehow
all these members of a well-ordering of the reals,
are, not in the normal ordering,
so, ...,
they're in the reverse ordering, and
it's kind of a matter of peek-a-boo as it were,
which works great to astonish infants.

Each real number is in the normal ordering and
each real number is in the well.order.
They are total orders, so
they would not be total otherwise.

It's sort of less nouveau to us more mature sorts,
you know, with already
a universe full of everything and all.

I think that the less mature sorts have it right.
An infinite universe is packed with astonishments.

One of the finite ordinals
⎛ ∀S⅏: ℕ⅏ ⊆ S
⎝ S⅏ :⇔ S ∋ 0 ∧ ∀k ∈ S ∋ k⁺¹
encodes _all_ the published media, past and future,
of the human race and any descendant races.
Indulge yourself with an astonishment.
I swear, it's not illegal, immoral, or fattening.

So, when you flip the labels of
rationals and irrationals in your little decider there,
which is an opinion,

No.
Opinions and theorems are different.

and only justified insofar as
otherwise you'd be staring at the precipice of
the crevasse of the impasse
that your induction guarantees, the inductive impasse,
when you're provided the same example except where
the irrationals and rationals are only given as
properties their density in the reals,
and not their cardinal, which
in a briefer theory the density is
the only thing that applies, then,
there's that the rationals are, ..., "huge".

There is a very short proof of
the theorem.not.opinion that,
if there is any infinity,
then there are multiple infinities.

The set of rationals is huge.
The power set of the rationals is huger.
The power set of the power set of the rationals
is huger still.

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On 5/23/2024 9:50 AM, Ross Finlayson wrote:

On 05/22/2024 11:33 AM, Jim Burns wrote:

[...]

The partitions of the rationals,

{P⊆ℚ:∅≠Pᣔ<ᘁPᣔ<ᣔℚ\P≠∅}

have a smaller cardinal, than
the powerset of the rationals, where

You're accepting the Continuum Hypothesis.
CH is independent of ZFC. [Gödel,Cohen]

More to the point,
{P⊆ℚ:∅≠Pᣔ<ᘁPᣔ<ᣔℚ\P≠∅} has
a larger cardinal than ℚ

Upthread:

Subsets of rationals in rationals are rationals.
Furthermore, the partitions of rationals are countable.

</WM>
Date: Tue, 21 May 2024 13:06:23 -0700

The partitionsᴿꟳ aren't countable.

⎛ {P⊆ℚ:∅≠Pᣔ<ᘁPᣔ<ᣔℚ\P≠∅} is the set of
⎜ partitionsᴿꟳ == open.foresplits of ℚ
⎜ P⊆ℚ subset of rationals P
⎜ ∅≠P nonempty P
⎜ Pᣔ<ᘁP ⇔ ∀r ∈ P: ∃s ∈ P: r<s
⎜ Pᣔ<ᣔℚ\P ⇔ ∀r ∈ P: ∀s ∈ ℚ\P: r<s
⎝ ℚ\P≠∅ nonempty ℚ\P

according to the laws of arithmetic,
a partition of the set of rationals into two,
where the partitioning's partitions are
those above a value

That's not {P⊆ℚ:∅≠Pᣔ<ᘁPᣔ<ᣔℚ\P≠∅}

Pᣔ<ᣔℚ\P ⇔ ∀r ∈ P: ∀s ∈ ℚ\P: r<s
Each of P < each of ℚ\P
There is no requirement for
a rational point between P and ℚ\P
and examples of P without it are easy to find.

according to
the laws of trichotomy in arithmetic,
and below, respectively,
there aren't more partitionings than
there are rationals.

Pᣔ<ᣔℚ\P

Of course,
each partitioning is only distinct,
according to trichotomy,
by have at least one value unique for itself
not shared with all less than it,

No.
Pᣔ<ᘁP ⇔ ∀r ∈ P: ∃s ∈ P: r<s

There are _at least_ partitionsᴿꟳ {q ∈ ℚ: q<r}
for each rational r

For each rational qᵢ in partitionᴿꟳ Pₖ
Pₖᣔ<ᘁPₖ
qᵢ is not max Pₖ
qᵢ < rⱼ ∈ Pₖ

Pⱼ = {q ∈ Q: q < rⱼ}
qᵢ ∈ Pⱼ
qᵢ is shared between Pₖ and Pⱼ among others.

There is no unshared element in any partitionᴿꟳ

That's not the same as saying that
☠ there are different partitionsᴿꟳ which
☠ have all the same elements.

and at least one value unique for itself
not shared all all greater than it.

The partitionsᴿꟳ are totally.ordered by
the subset relation ⊆
Pⱼ ≤ Pₖ ⇔ Pⱼ ⊆ Pₖ

There are no elements of Pⱼ not shared with
greater Pₖ because subset.

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On 5/27/2024 4:08 PM, Ross Finlayson wrote:

On 05/23/2024 10:45 AM, Jim Burns wrote:

[...]

It seems you've axiomatized least-upper-bound for
partitionings of rationals then made a loop.

No.

I'm going to try something else.

x refers to both
a single infinite set ​͆x of rationals and
a single point ​̇x in the real line

x has two roles, and,
when x changes roles, it changes hats,
​͆x to ​̇x or ​̇x to ​͆x

Consider the set {​͆x⊆ℚ:∅≠​͆xᴬ<ᴱ​͆xᴬ<ᴬℚ\​͆x≠∅}
of sets ​͆x of rationals
and the set ℝ of points ​̇x

⎛ xᴬ<ᴱ​͆x ⟺ ∀ꟴr ∈ ​͆x: ∃ꟴs ∈ ​͆x: r<s
​⎝ xᴬ<ᴬℚ\​͆x ⟺ ∀ꟴr ∈ ​͆x: ∀ꟴs ∈ ℚ\​͆x: r<s

​̇x ∈ ℝ ⟺ ​͆x ∈ {​͆x⊆ℚ:∅≠​͆xᴬ<ᴱ​͆xᴬ<ᴬℚ\​͆x≠∅}

Define the order '≤' on ℝ by
referring to {​͆x⊆ℚ:∅≠​͆xᴬ<ᴱ​͆xᴬ<ᴬℚ\​͆x≠∅} ​̇y ≤ ​̇z ⟺ ​͆y ⊆ ​͆z

Prove that ​̇y ≤ ​̇z is total, ie,
​̇y ≤ ​̇z ∨ ​̇y ≥ ​̇z
using what we know about ​͆y and ​͆z as sets which
are elements of {​͆x⊆ℚ:∅≠​͆xᴬ<ᴱ​͆xᴬ<ᴬℚ\​͆x≠∅}, eg,
​͆y ᴬ<ᴬ ℚ\​͆y
​͆z ᴬ<ᴬ ℚ\​͆z

Proof.
​͆y ⊇ ​͆z ∨ ​͆y ⊉ ​͆z

1.
| Assume ​͆y ⊇ ​͆z
| ​͆y ⊇ ​͆z ∨ ​͆y ⊆ ​͆z

2.
| Assume ​͆y ⊉ ​͆z
|
| r ∈ ​͆z
| r ∉ ​͆y
|
| r ∈ ℚ\​͆y
| ​͆y ᴬ< r
| ​͆y ⊆ (∞,r)
|
| r ∈ ​͆z
| (∞,r) ᴬ∈ ​͆z
| (∞,r) ⊆ ​͆z
|
| ​͆y ⊆ (∞,r) ⊆ ​͆z
| ​͆y ⊆ ​͆z
| ​͆y ⊇ ​͆z ∨ ​͆y ⊆ ​͆z

Therefore,
​͆y ⊇ ​͆z ∨ ​͆y ⊆ ​͆z
Therefore,
​̇y ≥ ​̇z ∨ ​̇y ≤ ​̇z

It is theorem, not an axiom, that
the order '≤' on ℝ is total.
It follows from
meaning {​͆x⊆ℚ:∅≠​͆xᴬ<ᴱ​͆xᴬ<ᴬℚ\​͆x≠∅} by ℝ and meaning '⊆' by '≤'

It seems you've axiomatized least-upper-bound for
partitionings of rationals then made a loop.

No.

We can read the least.upper.bound property for ℝ
as a related claim about {​͆x⊆ℚ:∅≠​͆xᴬ<ᴱ​͆xᴬ<ᴬℚ\​͆x≠∅}
and prove the related claim, thereby proving
the least.upper.bound property for that version of ℝ

Let B be a bounded, nonempty subset of {​͆x⊆ℚ:∅≠​͆xᴬ<ᴱ​͆xᴬ<ᴬℚ\​͆x≠∅}

We can prove that
∅ ≠ ⋃B
∀ꟴr ∈ ⋃B: ∃ꟴs ∈ ⋃B: r<s
∀ꟴr ∈ ⋃B: ∀ꟴs ∈ ℚ\⋃B: r<s
ℚ\⋃B ≠ ∅
∀​͆y ∈ B: ​͆y ⊆ ⋃B (Bᴬ⊆⋃B)
∀​͆z ∈ {​͆x⊆ℚ:∅≠​͆xᴬ<ᴱ​͆xᴬ<ᴬℚ\​͆x≠∅}: Bᴬ⊆​͆z ⟹ Bᴬ⊆⋃B⊆​͆z

and, by proving all of the above,
we prove that,
for each bounded, nonempty subset B of ℝ
there exists a least.upper.bound ​̇b (=⋃B) in ℝ

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On 5/28/2024 3:56 PM, Ross Finlayson wrote:

On 05/28/2024 12:03 PM, Jim Burns wrote:

x refers to  both
a single infinite set ​͆x of rationals  and
a single point ​̇x in the real line

x has two roles, and,
when x changes roles, it changes hats,
​͆x to ​̇x or ​̇x to ​͆x

Consider the set {​͆x⊆ℚ:∅≠​͆xᴬ<ᴱ​͆xᴬ<ᴬℚ\​͆x≠∅}
of sets ​͆x of rationals
and the set ℝ of points ​̇x

⎛ xᴬ<ᴱ​͆x ⟺ ∀ꟴr ∈ ​͆x: ∃ꟴs ∈ ​͆x: r<s
​⎝ xᴬ<ᴬℚ\​͆x ⟺ ∀ꟴr ∈ ​͆x: ∀ꟴs ∈ ℚ\​͆x: r<s

​̇x ∈ ℝ ⟺ ​͆x ∈ {​͆x⊆ℚ:∅≠​͆xᴬ<ᴱ​͆xᴬ<ᴬℚ\​͆x≠∅}

Well, that's R, reals, not Q, rationals.

ℝ = {​͆x⊆ℚ:∅≠​͆xᴬ<ᴱ​͆xᴬ<ᴬℚ\​͆x≠∅} is
a set of sets of rationals,
not a set of rationals.

ℝ = {​͆x⊆ℚ:∅≠​͆xᴬ<ᴱ​͆xᴬ<ᴬℚ\​͆x≠∅} ⊆ 𝒫(ℚ)

ℝ = {​͆x⊆ℚ:∅≠​͆xᴬ<ᴱ​͆xᴬ<ᴬℚ\​͆x≠∅} provably has
all the properties that the complete ordered field has,
which means those _sets_ of rationals are real numbers
in the sense relevant here.

Yes,
ℚ does not have the least.upper.bound property.
But
ℝ = {​͆x⊆ℚ:∅≠​͆xᴬ<ᴱ​͆xᴬ<ᴬℚ\​͆x≠∅} provably has
the least.upper.bound property, and
ℝ = {​͆x⊆ℚ:∅≠​͆xᴬ<ᴱ​͆xᴬ<ᴬℚ\​͆x≠∅} is
the model of the real numbers.
ℚ isn't the model.

The argument via induction that
infinite sets are inexhaustible, [...]

What is it to be finite?
In linear order,
each nonempty.subset is two.ended.

To be infinite is the opposite of that,
to be such that
not all nonempty.subsets are two.ended.

Pick a non.empty.subset with fewer.than.two ends.
Delete an element, and create
another non.empty.subset with fewer.than.two ends.
Repeat ad infinitum.


Conclusions by cisfinite induction follow from
the description of what's.being.induced.
Cisfinitely,
all the things which can be counted.to
are reached _boy our description_ of them,
NOT by any kind of supertask.

It works the same way that
all the triangles with right angles are reached:
_by the description_ of a right triangle,
NOT by some supertask operation.

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On 5/29/2024 5:59 AM, Ross Finlayson wrote:

On 05/28/2024 02:23 PM, Jim Burns wrote:

NOT by any kind of supertask.

NOT by some supertask operation.

Yet, then you claim you can't be bothered, ....

Deduction arrives at that _something_
arrives at that, "supertask" I guess
you call it - really it's a great idea.

I (JB) think that
the large (to me) efforts I have made
in aid of bringing home certain ideas
have been mistaken by you (RF) as
not bothering about other ideas.

To clarify:
I don't want to move on from these ideas
while you are still incorrect about them.

There was a time when,
if you and I were rhetorically wrestling over X
and you changed the subject to Y,
I would have assumed that
you had understood my point and
you no longer disagreed, but, possibly.because
"I was wrong" sticks in some throats,
I shouldn't look for an explicit acknowledgement.
And I would move on.

Since that time, I have learned better
what you (RF) mean by changing the subject.
It isn't agreement.

You have said that the real numbers are countable.
They aren't.
Until I hear otherwise from you,
I will think that you think that
the real numbers are countable.

That, right there, is more than enough for me.
Until that's been resolved, I don't see any need
for me to hunt down more issues to hold forth on.

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On 5/29/2024 3:50 PM, Ross Finlayson wrote:

On 05/29/2024 10:07 AM, Jim Burns wrote:

So, here I have three different models of sets
that implement continuous domains, i.e., that
see arrived at the properties of extent, density,
completeness, and measure, in the values of the
real numbers, thus making for the Intermediate
Value Theorem and the Fundamental Theorems of
Calculus.

So, of these three, line-reals and field-reals
and signal-reals, the line-reals are countable.

The line.reals ℝ⁞ are countable.
∃index⁻¹: ℕ → ℝ⁞: f(ℕ) = ℝ⁞

index⁻¹: ℕ → ℝ⁞ determines a sequence of line.reals
λ₀ λ₁ λ₂ λ₃ ...

λ₀ λ₁ λ₂ λ₃ ... determines a sequence of
nested intervals
(κ₀,μ₀) (κ₁,μ₁) (κ₂,μ₂) (κ₃,μ₃) ...
κⱼ < κⱼ₊₁ < μⱼ₊₁ < μⱼ
such that,
for each interval (κⱼ,μⱼ)
endpoints are indexed before interior points
∀λ ∈ ℝ⁞: κⱼ < λ < μⱼ ⇒
index(κⱼ) < index(λ)
index(μⱼ) < index(λ)

Each index(λᵢ) ∈ ℕ and is finite.
Each λᵢ is preceded by finitely.many line.reals
in λ₀ λ₁ λ₂ λ₃ ...

There is NO line.real in all infinitely.many (κⱼ,μⱼ)
because
such a line.real would have infinitely.many
line.reals preceding it in λ₀ λ₁ λ₂ λ₃ ...
and
there are NO infinitely.preceded line.reals.

Thus,
the sequence of intervals
(κ₀,μ₀) (κ₁,μ₁) (κ₂,μ₂) (κ₃,μ₃) ...
_splits_ the line.reals into F H
points before one of the intervals
points after one of the intervals
F = {λ ∈ ℝ⁞: ∃(κⱼ,μⱼ): λ ≤ κⱼ}
H = {λ ∈ ℝ⁞: ∃(κⱼ,μⱼ): μⱼ ≤ λ}

and there is NO line.real between F and H


Consider the line.real function
f : ℝ⁞ → {0,1}
f(λ) =
⎡ 0 for λ ∈ F
⎣ 1 for λ ∈ H

For each line.real λ
f() is continuous in an open interval holding λ

However,
f() jumps over all the points between 0 and 1

That isn't what we want from a "continuous" function.
That isn't what we want from flying.rainbow.reals
or line.reals or you.name.it.reals.

It might be that Aristotle thought otherwise.
I haven't read up on that, myself.
However, I know that, after considerable work,
we decided that Dedekind had better thoughts.

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On 5/29/2024 8:31 PM, Ross Finlayson wrote:

On 05/29/2024 03:26 PM, Jim Burns wrote:

For each line.real λ
f() is continuous in an open interval holding λ

However,
f() jumps over all the points between 0 and 1

That isn't what we want from a "continuous" function.
That isn't what we want from flying.rainbow.reals
or line.reals or you.name.it.reals.

It might be that Aristotle thought otherwise.
I haven't read up on that, myself.
However, I know that, after considerable work,
we decided that Dedekind had better thoughts.

Once I wrote some results just establishing that
in each neighborhood of [0,1]
there's an element of ran(f), where
f = n/d, 0 <= n <= d, d -> oo.

fᵈ(n) = n/d
⟦0,d⟧ = {n ∈ ℕ: 0 ≤ n ≤ d}
fᵈ⟦0,d⟧ = {fᵈ(n): n ∈ ⟦0,d⟧}
fᵈ⟦0,d⟧ = {n/d ∈ ℚ: 0 ≤ n ≤ d ∧ n ∈ ℕ}

I think that you're saying that
the limit of fᵈ⟦0,d⟧ as d → ∞ is
the interval [0,1] in the line.reals ℝ⁞
lim(d → ∞) fᵈ⟦0,d⟧ = [0,1]∩ℝ⁞

That limit is
the interval [0,1] in the rationals ℚ
lim(d → ∞) fᵈ⟦0,d⟧ = [0,1]∩ℚ

Consider the sequence
f¹⟦0,1⟧ f²⟦0,2⟧ f³⟦0,3⟧ ... f¹⁰⁰⁰⟦0,1000⟧ ...

Each rational n/d in [0,1]∩ℚ
appears somewhere in the sequence
f¹⟦0,1⟧ f²⟦0,2⟧ f³⟦0,3⟧ ... f¹⁰⁰⁰⟦0,1000⟧ ...

but ONLY rationals in [0,1]∩ℚ
appear anywhere in the sequence
f¹⟦0,1⟧ f²⟦0,2⟧ f³⟦0,3⟧ ... f¹⁰⁰⁰⟦0,1000⟧ ...

The _limit_ might be a little murky because
rationals keep appearing and disappearing.
For example f¹⁰⁰⁰⟦0,1000⟧ and f¹⁰⁰¹⟦0,1001⟧
only hold 0 and 1 in common.
But all of f¹⁰⁰⁰⟦0,1000⟧ and all of f¹⁰⁰¹⟦0,1001⟧
appear later in f¹⁰⁰¹⁰⁰⁰⟦0,1001000⟧

It's clearer with the sub.sequence
f¹⟦0,1⟧ f²⟦0,2⟧ f⁶⟦0,6⟧ f²⁴⟦0,24⟧ f¹²⁰⟦0,120⟧ ...
which is
f¹ꜝ⟦0,1!⟧ f²ꜝ⟦0,2!⟧ f³ꜝ⟦0,3!⟧ ... f¹⁰⁰⁰ꜝ⟦0,1000!⟧ ...

Rationals appear and do NOT disappear in
f¹ꜝ⟦0,1!⟧ f²ꜝ⟦0,2!⟧ f³ꜝ⟦0,3!⟧ ... f¹⁰⁰⁰ꜝ⟦0,1000!⟧ ...
Each rational appears and none disappears.
Only rationals appear.

For a nested set.sequence
lim(d → ∞) fᵈꜝ⟦0,d!⟧ =
⋃{fᵈꜝ⟦0,d!⟧: 0 < d < ∞} =
[0,1]∩ℚ

But
f¹ꜝ⟦0,1!⟧ f²ꜝ⟦0,2!⟧ f³ꜝ⟦0,3!⟧ ... f¹⁰⁰⁰ꜝ⟦0,1000!⟧ ...
is a sub.sequence of
f¹⟦0,1⟧ f²⟦0,2⟧ f³⟦0,3⟧ ... f¹⁰⁰⁰⟦0,1000⟧ ...

f¹⟦0,1⟧ f²⟦0,2⟧ f³⟦0,3⟧ ... f¹⁰⁰⁰⟦0,1000⟧ ...
cannot converge to any set to which
f¹ꜝ⟦0,1!⟧ f²ꜝ⟦0,2!⟧ f³ꜝ⟦0,3!⟧ ... f¹⁰⁰⁰ꜝ⟦0,1000!⟧ ...
does not converge.

Thus,
lim(d → ∞) fᵈ⟦0,d⟧ = [0,1]∩ℚ = [0,1]∩ℝ⁞
ℚ = ℝ⁞
The line.reals are the rationals.

Once I wrote some results just establishing that
in each neighborhood of [0,1]
there's an element of ran(f), where
f = n/d, 0 <= n <= d, d -> oo.

In each neighborhood of (our real) [0,1]
there's an element of ℚ

Then,
it doesn't say which integer n has it
that f^-1(0.5) = n,
only that it exists.

For even d,
(fᵈ)⁻¹(0.5) = 0.5⋅d

In the limit,
there is no infinite 0.5⋅d ∈ ℕ

(This is courtesy that there are
infinitely-many integers.)

I.e. the properties "extent", "density",
"completeness", and "measure", are each established.

In each neighborhood of (our real) [0,1]
there's an element of ℝ⁞ = ℚ
However,
that's not completeness.

√½ = 0.70710678118... does not appear in fᵈ⟦0,d⟧
for any d

Is √½ in [0,1]∩ℝ⁞ ⟵ fᵈ⟦0,d⟧ ?
What definition of 'limit' says that?

Is √½ not.in [0,1]∩ℝ⁞ ⟵ fᵈ⟦0,d⟧ ?
Then an "all.continuous" function might jump
from [0,√½) to (√½,1]
and [0,1]∩ℝ⁞ is not complete.

I was able to convince the Gemini bot
this was so, ....

Well, it's only human...

You know,
line-reals are well-ordered in the sense that
their normal ordering is a well-ordering, i.e.,
in the usual sense.

I know that
the nonempty.subset (0,1]∩ℝ⁞ does not contain
a least element, which means
the usual order of the line.reals
is not a well.order.

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On 5/30/2024 10:52 PM, Ross Finlayson wrote:

On 05/30/2024 09:44 AM, Jim Burns wrote:

[...]

First, one might aver
"it's not a real function, doesn't exist",
yet these days
the "continuum limit" is rather a thing.

A function fₓ not.exists from
cisfinite.inductive ℕ _onto_
Dedekind.complete ℝ, fₓ(ℕ) = ℝ

...because
|fₓ(ℕ)| cannot be bigger than |ℕ|
|ℝ| is bigger than |ℕ|
thus fₓ(ℕ) cannot be ℝ
so such an fₓ is not a thing.


Yes,
the Dedekind.complete.continuum [0,1]∩ℝ is a thing.
∀B ∈ {A⊆ℝ: ℝᴱ≥ᴬA≠∅}: ∃β ∈ {b∈ℝ: b≥ᴬB} ᴬ≥ β

Yes,
the set of line.reals [0,1]∩ℝ⁞ is a thing.
[0,1]∩ℝ⁞ = lim(d → ∞) ⟦0/d!,1/d!,…,d!/d!⟧

But
they are not the same thing.

Then, one might aver
"it's merely countable", yet,
it's analytical and real character is
established for itself.

It itself has done a poor job of establishing.
Perhaps you (RF) should be doing the establishing.

About least-upper-bound,
these iota-values are iota-cuts, and
least-upper-bound for {f( < m}
is quite simply f(m).

Therefore,
each value has a bounded.nonempty.set of which
it is least.upper.bound.

However, what is needed is that
each bounded.non.empty.set has a value which
is least.upper.bound of it.

First, one might aver
"it's not a real function, doesn't exist",
yet these days
the "continuum limit" is rather a thing.

Your use of 'aver',
while very cool and mathematical.in.tone,
suggests to me agreement,
stating that "one" has said a thing _true_
The rest of your sentence suggests disagreement.
I find that confusing.

Perhaps that is a better place for 'assert',
of which my impression is it does not get into
what you think of what I say.

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On 5/31/2024 9:37 PM, Ross Finlayson wrote:

On 05/31/2024 11:32 AM, Jim Burns wrote:

[...]

This putative f called
EF the equivalency function,
f(n) = n/d, 0 <= n <= d,
d -> oo in the continuum limit,
_can not see its elements exchanged_,
it _can not be re-ordered_,
due the contrivance of how
it's standardly modeled as
a limit of real functions, and that
the continuum limit is not-a-real-function,
due the contrivance
it cannot be found its disorder,
it's "not Cartesian".

The Dirac delta function δ(x) is
not.a.real.function.

Whatever it's _called_
δ(x) is _a certain kind_ of limit of
a sequence of increasingly peaky functions.
There are many good choices for peaky functions.
For example, dₙ(x) = n⋅exp(-½⋅n²x²)/√​̅2​̅π

for x≠0: limₙ dₙ(x) = 0
for β>0: limₙ ∫₋ᵦ⁺ᵝdₙ(x)⋅dx = 1

A typical use of δ(x) is
∫ f(x)⋅δ(x-x₀)⋅dx = f(x₀)

_What that means_ is
limₙ ∫ f(x)⋅dₙ(x-x₀)⋅dx = f(x₀)

The peakiness limit n ⟶ ∞ comes last.

δ(x) is not.a.real.function.
δ(x) is not a limit of functions dₙ(x).
δ(x) is a placeholder for a peaky function,
for a limit.to.be.determined.later.

δ(x) is not.a.real.function.
δ(x) is not a function, full stop.
No value δ(0) exists such that
for β>0: ∫₋ᵦ⁺ᵝδ(0)⋅dx = 1


Consider your EF equivalency function,
| f(n) = n/d, 0 <= n <= d,
| d -> oo in the continuum limit,

EF is not a function, full stop.

If EF is a function,
then EF can be composed with a function which
transposes y1 and y2
Since it can't, it isn't.

I don't know that EF _needs_ to be a function,
but its non.function.hood makes
the lack of sense of EF⁻¹(0.5) make sense,
just as the lack of sense of δ(0) makes sense.

----
What is "the continuum limit"?
You spend considerable electrons detailing
all the usual ideas you are using.
But "the continuum limit" isn't a usual limit.

The usual limit(s) all give
the rationals and only the rationals.
lim(d → ∞) ⟦0/d!,1/d!,…,d!/d!⟧ = [0,1]∩ℚ
and so on.

I'd be interested in seeing something that
shows me I'm wrong, but
your "continuum limit" looks to me like
you (RF) have assumed that the limit is the continuum.

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On 6/2/2024 4:43 PM, Ross Finlayson wrote:

On 06/02/2024 12:23 PM, Jim Burns wrote:

What is "the continuum limit"?
You spend considerable electrons detailing
all the usual ideas you are using.
But "the continuum limit" isn't a usual limit.

The usual limit(s) all give
the rationals and only the rationals.
lim(d → ∞) ⟦0/d!,1/d!,…,d!/d!⟧ = [0,1]∩ℚ
and so on.

I'd be interested in seeing something that
shows me I'm wrong, but
your "continuum limit" looks to me like
you (RF) have assumed that the limit is
the continuum.

Nope,
"continuum limit" is just that in the limit,
that it's _named_ continuum limit
because its limit _is_ the continuum,
not that its limit _is_ the continuum
because it's _named_ that.

| In mathematical physics and mathematics,
| the continuum limit or
| scaling limit of a lattice model
| characterizes its behaviour in the limit
| as the lattice spacing goes to zero.
| It is often useful to use lattice models
| to approximate real-world processes,
| such as Brownian motion.
| Indeed, according to Donsker's theorem,
| the discrete random walk would,
| in the scaling limit,
| approach the true Brownian motion.
|
https://en.wikipedia.org/wiki/Continuum_limit

Okay,
the lattice spacing goes to zero, here,
so this is the continuum limit.
lim(d → ∞) ⟦0/d!,1/d!,…,d!/d!⟧

However,
the continuum limit
(the lattice spacing going to zero)
isn't the continuum
(everywhere.continuous functions never jumping)
Continuum.limited ℚ is not Dedekind.complete.

_For continuous functions_
values at lattice spacing going to zero
determines a unique answer.
So, for continuous physical processes,
that might well be a distinction without a difference.

However,
lim(d → ∞) ⟦0/d!,1/d!,…,d!/d!⟧ is
is the.continuum.limit but
is.not the.continuum.

Consider a sequence of
geometrically.shrinking intervals which
cover each point in ⟦0/d!,1/d!,…,d!/d!⟧
⎛ (-β/2²+0/d!,0/d!+β/2²)
⎜ (-β/2³+1/d!,1/d!+β/2³)
⎜ (-β/2⁴+2/d!,2/d!+β/2⁴)
⎜ ...
⎝ (-β/2ᵈꜝ⁺²+d!/d!,d!/d!+β/2ᵈꜝ⁺²)

with lengths which add up to
β/2¹ + β/2² + β/2³ + ... + β/2ᵈꜝ⁺¹ =
β⋅(1-⅟2ᵈꜝ⁺¹)

Even in the limit,
there is _positive_ number which
must be less than those covering intervals.
That isn't the continuum.

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On 6/3/2024 12:32 AM, Ross Finlayson wrote:

On 06/02/2024 08:51 PM, Jim Burns wrote:

[...]

Well, the idea is that
it goes with the same notion as
"ubiquitous ordinals".

What is it going with ubiquitous ordinals?
What are ubiquitous ordinals when they're at home?

That is to say,
the domain and range are natural integers,
and the quotients only defined on those.

Is this the same as the rationals?

I.e., you can notice there isn't any other
model of real numbers built yet or available,
that's all there is to it at that point.

Are you (RF) intending quotients of integers
to be a model of the continuum?

There isn't the field or rationals
yet defined there,

Defined where?

only those integer quotients representing parts,
there's nothing between those integers on the lattice,
except ran(f) this non-integer part between each.

Is this the same as the rationals?
Are you (RF) intending quotients of integers
to be a model of the continuum?

The "ubiquitous ordinals"
is for set theory and
a contradistinction

:
I'll probably regret asking, but YOLO

A _distinction_ makes things _distinct_ different.
Is a _contradistinction_ the opposite of that,
a similarity?

a contradistinction between numbering and counting,
where it's so that
successor is order type, and also, is powerset,

Be right back. Head exploding.

Please, please say what you mean another way..
Colorless green ideas sleep furiously.

so, that there's no missing set for
otherwise the usual what's called "anti-diagonal"
or for the powerset result properly,

The powerset result which seems
closest to relevant here
has the set {x ∈ A: x ∉ F(x)} missing
from the image F(A) of F: A → 𝒫(A)

Is that the result you refer to?

If that is the result you refer to,
how is that result affected by being
about numbering and/or counting?

that's also arithmetized and
called the diagonal method, or here
the anti-diagonal.
This reflects that for an integer continuum,

Are there any distinctions you (RF) draw
between "integer continuum" and
"four.cornered triangle"?

This reflects that for an integer continuum,
there's a sort of "only-increment"
to complement this "only-diagonal" construction,
as I've written since about twenty-five years.

See, I always knew I was right, so, ....

Do you have any theories about why
no one seems to have noticed
you always being right?

I'm glad you've found
"continuum limit" on the Wiki these days.

Do you agree with the Wiki that
continuum limit does not imply continuum
?

In the old days
we just proved it for ourselves.

In the old days
did you say what you meant by "continuum limit"?

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On 6/4/2024 3:23 AM, Ross Finlayson wrote:

On 06/03/2024 03:52 AM, Jim Burns wrote:

[...]

Establishing completeness or gaplessness rather
for this "natural/unit equivalency function"

EF(ℕ) = lim(d → ∞) {0/d!,1/d!,…,d!/d!}

is rather simple,
the lim sup or least-upper-bound for {f(< m)}
is f(m).

Even in gappy ℚ
bounded.nonempty.sets exist ⊆ ℚ
with least.upper.bounds ∈ ℚ
That's not gaplessness.

In hypothetically.gapless EF′(ℕ)
NO bounded.non.empty.sets exist ⊆ EF′(ℕ)
WITHOUT least.upper.bounds ∈ EF′(ℕ)

That
the lim sup or least-upper-bound for {f(< m)}
is f(m)
shows the first, and does not show the second.

Compare to
the ℝ.model 𝒫ᶜᵘᵗ(ℚ) the Dedekind.cuts of ℚ
𝒫ᶜᵘᵗ(ℚ) = {C⊆ℚ:∅≠Cᴬ<ᴱCᴬ<ᴬℚ\C≠∅}

𝒫ᶜᵘᵗ(ℚ) is a set of sets of rationals.

Bounded nonempty subset ℬ ⊆ 𝒫ᶜᵘᵗ(ℚ) is
a set of sets of rationals.

B = ⋃ℬ is a set of rationals for which
we can prove B⊆ℚ:∅≠Bᴬ<ᴱBᴬ<ᴬℚ\B≠∅ and which
is the smallest (⊆) set including each of ℬ

Because ⋃ℬ exists
least.upper.bound B exists.

⎛ For each bounded nonempty ℬ ⊆ 𝒫ᶜᵘᵗ(ℚ)
⎝ least.upper.bound B ∈ 𝒫ᶜᵘᵗ(ℚ) exists.

𝒫ᶜᵘᵗ(ℚ) has the least.upper.bound property.

EF(ℕ) does not have the least.upper.bound property.
EF(ℕ) =
lim(d → ∞) {0/d!,1/d!,…,d!/d!} =
⋃(d ∈ ℕ⁺) {0/d!,1/d!,…,d!/d!}

For example,
∀d ∈ ℕ⁺: √½ ∉ {0/d!,1/d!,…,d!/d!}
√½ ∉ ⋃(d ∈ ℕ⁺) {0/d!,1/d!,…,d!/d!}
√½ ∉ lim(d → ∞) {0/d!,1/d!,…,d!/d!}
√½ ∉ EF(ℕ)

The set {n/d ∈ EF(ℕ): (n/d)² < ½}
is a bounded nonempty subset ⊆ EF(ℕ)
without a least.upper.bound ∈ EF(ℕ)

lim(d → ∞) {0/d!,1/d!,…,d!/d!} is the continuum limit.
lim(d → ∞) {0/d!,1/d!,…,d!/d!} is not the continuum.

You know, Dirac's delta is only defined non-zero
at one point, ....

Therefore,
not.exists function δ:ℝ→ℝ
∀β>0 ∫₋ᵦ⁺ᵝδ(x)⋅dx = 1

exists family.of.functions δᵧ:ℝ→ℝ
∀β>0
lim(γ→0) ∫₊ᵦ⁺ᵒᵒδᵧ(x)⋅dx = 0
lim(γ→0) ∫₋ᵦ⁺ᵝδᵧ(x)⋅dx = 1
lim(γ→0) ∫₋ₒₒ⁻ᵝδᵧ(x)⋅dx = 0

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On 6/4/2024 3:23 AM, Ross Finlayson wrote:

On 06/03/2024 03:52 AM, Jim Burns wrote:

On 6/3/2024 12:32 AM, Ross Finlayson wrote:

This reflects that for an integer continuum,

Are there any distinctions you (RF) draw
between "integer continuum" and
"four.cornered triangle"?

The concept of the Integer Continuum is much like
that of the Linear Continuum, the idea of giving a
number to each thing, though, the infinite integers
instead of the infinite points of the lnes.

(That infinitely-many and infinitely-grand get conflated
is just a natural fact of the terms involved.)

So, the Integer Continuum is a concept often ascribed
to Duns Scotus and Spinoza, while, of course it's also
yet more antique and antiquarian.

It's kind of like the notion of "limit", when there's
introduced Weierstrass and Cauchy, it's already
arrived at in Aristotle and for Eudoxus and Archimedes,
and including that it's called "limit" then with regards
to "infinite limit".

The Integer Continuum then is a mathematical concept
that goes along with the Linear Continuum.

These things are what they are.

Then
the integer continuum is not the continuum
and also the continuum limit is not the continuum,
where,
by "continuum", I refer to domain and range in which
continuous curves which cross must intersect.

'Continuum' is a concept we have struggled with
at least back to Zeno of Elea.
It looks to me as though Dedekind ended the struggle.
I don't see the point of returning to
pre.Dedekind thinking on this question,
other than for its historical interest.

Still, you don't need to be bound by my opinion.
I'm only explaining why I'm not impressed by name.checking
the admittedly.very.impressive Duns Scotus and Spinoza.

I stand on the shoulders of giants.
From up there, from time to time,
I can see farther than earlier giants.
I accept the gift.

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On 6/4/2024 6:22 PM, Ross Finlayson wrote:

On 06/04/2024 12:04 PM, Jim Burns wrote:

[...]

About your contentedness with "Dedekind completeness",
one of the great things about the Integer Continuum
is that it reflects the language of definability.

One of the great things about the Dedekind.complete
continuum is that _it is equivalent to_
continuous curves which cross must intersect.

I admit it.
I am content with
continuous curves which cross never intersecting,
_also known as_ Dedekind completeness (of ℚ)

Now, each of the members of the complete ordered field
has a representation, an infinite expression as it is,
in positional notation its digits. So, all the
representations that ever be thusly named, are countable.

No, that's not how it works.
If representations can be infinite expressions,
then there can be uncountably.many of them.

It's if representations only are finite.length, that
they must all be encodable in finitely.preceded numbers,
also known as, they are countable.

Considering you must know about Skolem,
for example,
there's a countable model of them,
all definable.

I am confused about how you (RF) are using 'definable'
Must definitions be finite.length, or
can definitions be infinite.length?

What I know is that,
if a theory has an infinite model,
then it has a model of all infinite cardinalities.
So,
there's a countable model of the theory, and
there's an inaccessible.cardinal.sized model https://en.wikipedia.org/wiki/Inaccessible_cardinal
and so on, ad infinitum and then some.

So, the Integer Continuum, is just
giving a natural number to every thing,
in terms of an arithmetization,
of any given definable domain of discourse.

This sounds more like Gödel than
Duns Scotus and Spinoza.

Is it possible that that concept to which you refer
as "integer continuum" has, uhm, _evolved_ while
writing this post?

It makes it difficult to discern what thing
you're talking about,
if that thing isn't a thing, but several things.

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On 6/4/2024 8:43 PM, Jim Burns wrote:

On 6/4/2024 6:22 PM, Ross Finlayson wrote:

On 06/04/2024 12:04 PM, Jim Burns wrote:

[...]

About your contentedness with "Dedekind completeness",
one of the great things about the Integer Continuum
is that it reflects the language of definability.

One of the great things about the Dedekind.complete
continuum is that _it is equivalent to_
continuous curves which cross must intersect.

I admit it.
I am content with
continuous curves which cross never intersecting,

s/never/always

_also known as_ Dedekind completeness (of ℚ)

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On 6/5/2024 8:26 PM, Ross Finlayson wrote:

On 06/04/2024 05:43 PM, Jim Burns wrote:

[...]

So anyways,
you are rejecting or balking at that
line-reals, or the iota-values, the iota-cuts,
are complete, yet
you are looking at
features of numbers that don't relate to them,
which are only
integers infinite-ly divided while all together.

By 'ℭ' I mean 'continuum'
whatever that turns out to be.

Consider a curve in the plane ℭ×ℭ which
can be represented by a well.defined single.valued
function f: ℭ → ℭ
from the continuum to the continuum.

⎛ Not.All.Curves. Not x²+y²=1, for example.
⎝ But enough curves to be interesting and useful.

Describe f

⎛ That is the essential first step for
⎜ finite beings exploring the infinite:
⎜⎜ Make a claim true of each one of
⎜⎜ infinitely.many to.be.explored.
⎜
⎜ Then,
⎜⎜ augment the description with
⎜⎜ finitely.many not.first.false claims.
⎜ We'll have more true claims which we'll _know_
⎜ are true of each one of the ones described,
⎝ infinite.many though the described ones are.

Describe f:
f is continuous at each x

f is continuous at x ⇔
f(x₂) can be kept near f(x)
by keeping x₂ near enough x

However,
that isn't enough description.

For a too.small ℭᑉꟲ "continuum"
f can be continuous at each x in ℭᑉꟲ
without being a connected curve.

Describe ℭ
so that ℭ is not too.small.

This is where
my devotion to Dedekind.completeness has its source.
Dedekind.completed ℚ is not too.small.

The problem with an integer "continuum" or
a {0/d!,1/d!,.,d!/d!}.limit "continuum"
is that,
_even if_ f is continuous at
each integer."continuum".point or at
each {0/d!,1/d!,.,d!/d!}.limit."continuum".point
they are insufficient to restrict the discussion
to connected curves.

That it to say, and it makes itself clear,
the lim sup and the least-upper-bound of
{ f( < m ) } _is_ f(m), and, exists in ran(f).

Some bounded.nonempty.sets with least.upper.bounds
is insufficient.

No bounded.non.empty.sets without least.upper.bounds
is sufficient.

You (RF) keep taking about the first
as though it is the second.

Then the idea of associating and relating to it
the properties of the complete ordered field

Complete.ordered.field, enough points.

or really just the rationals,

Rationals, not enough points.

has that these iota-cuts

{0/d!,1/d!,.,d!/d!}.limit ?
Not enough points.

have all the properties requisite of
your Dedekind-cuts

Dedekind cuts of ℚ enough points.

and don't even have it that
non-integer integers exist when
the iota-values are seen to be made, while of
course Dedekind's sort of at a loss
explaining where ir-rational numbers exist
in the rational numbers

Between any two distinct points x₁ x₂ in ℭ
there are ℵ₀ rational points.
(Infinity is different from finity.)

Each point x in ℭ has an open.foresplit Fₓ of ℚ
Fₓ is different from any other open.foresplit of ℚ
by ℵ₀.many rationals.

Each open.foresplit of ℚ has a least.upper.bound in ℭ
though mostly not.in ℚ ⊆ ℭ

Therefore,
the open.foresplits of ℚ _biject with_ points of ℭ

Open.foresplits of ℚ and points of ℚ are different.

without having distinct partitions,

They have distinct partitionsᴿꟳ/open.foresplitsᴶᴮ

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On 6/6/2024 8:19 PM, Ross Finlayson wrote:

On 06/06/2024 10:10 AM, Jim Burns wrote:

[...]

Each of these iota-values

elements of
{0/d!,1/d!,…,d!/d!}.limit = [0,1]⁞ line.reals ?

in their neighborhoods

usual open sets in
{0/d!,1/d!,…,d!/d!}.limit = [0,1]⁞ line.reals ?

is a critical point,
you've agreed, they're "dense",
where critical point is
the usual definition from topology where
each neighborhood contains a neighborhood and
each neighborhood contains points,

I agree that
{0/d!,1/d!,…,d!/d!}.limit is dense in
Dedekind.complete [0,1]

I agree that, with the usual open sets,
each neighborhood contains a neighborhood and
each neighborhood contains points.

I've tried looking up 'critical point' in
the context of topology, and I've gotten no joy.

There is the following, but it doesn't sound like
what you (RF) are calling a "critical point". https://en.wikipedia.org/wiki/Critical_point_(mathematics)#Application_to_topology
Critical point (mathematics)
Application to topology
|
| It follows that
| the number of connected components of V
| is bounded above by the number of critical points.

then that what you don't agree is that
ran(f) is a continuous domain

Yes,
I don't agree that
{0/d!,1/d!,…,d!/d!}.limit = [0,1]⁞
can't be partitioned into more.than.1 open set --
which is what I think you mean by
"ran(f) is a continuous domain"

I don't agree because
[0,1]⁞ can be partitioned into open sets
{p e [0,1]⁞ p² < ½} and {p e [0,1]⁞ ½ < p²}

which partition [0,1]⁞ because
√½ isn't in [0,1]⁞ =
lim(d → ∞) {0/d!,1/d!,…,d!/d!} -
⋃(d ∈ ℕ⁺) {0/d!,1/d!,…,d!/d!}

I.e., it's crank-ish to ignore this result framed
in the same way as it is as other relevant results,
the very notion of un-countability itself after
the anti-diagonal of Cartesian functions, this
only-diagonal of not-Cartesian functions, in
this case a very special function that is the
limit, the infinite limit of dividing a segment
of the linear continuum into discrete points,
and vice-versa, these iota-multiples, these
iota-values, that are iota-cuts, complete as
they are

_exactly courtesy the definition of LUB_

that you choose to ignore.

The method by which I choose to ignore
the least.upper.bound property is by
pointing out _several times_ by now that
you (RF) have got wrong
what the least.upper.bound property is.

Not everyone would consider that
ignoring on my part.
Ignoring on your part, though...

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On 6/7/2024 4:24 PM, Ross Finlayson wrote:

On 06/06/2024 09:41 PM, Jim Burns wrote:

On 6/6/2024 8:19 PM, Ross Finlayson wrote:

On 06/06/2024 10:10 AM, Jim Burns wrote:

[...]

Each of these iota-values

elements of
{0/d!,1/d!,…,d!/d!}.limit = [0,1]⁞ line.reals ?

in their neighborhoods

usual open sets in
{0/d!,1/d!,…,d!/d!}.limit = [0,1]⁞ line.reals ?

is a critical point,
you've agreed, they're "dense",
where critical point is
the usual definition from topology where
each neighborhood contains a neighborhood and
each neighborhood contains points,

I agree that
{0/d!,1/d!,…,d!/d!}.limit is dense in
Dedekind.complete [0,1]

I agree that, with the usual open sets,
each neighborhood contains a neighborhood and
each neighborhood contains points.

I've tried looking up 'critical point' in
the context of topology, and I've gotten no joy.

There is the following, but it doesn't sound like
what you (RF) are calling a "critical point".
https://en.wikipedia.org/wiki/Critical_point_(mathematics)#Application_to_topology

Critical point (mathematics)
Application to topology
|
| It follows that
| the number of connected components of V
| is bounded above by the number of critical points.

then that what you don't agree is that
ran(f) is a continuous domain

Yes,
I don't agree that
{0/d!,1/d!,…,d!/d!}.limit = [0,1]⁞
can't be partitioned into more.than.1 open set --
which is what I think you mean by
"ran(f) is a continuous domain"

I don't agree because
[0,1]⁞ can be partitioned into open sets
{p e [0,1]⁞  p² < ½} and {p e [0,1]⁞ ½ < p²}

which partition [0,1]⁞ because
√½ isn't in [0,1]⁞  =
lim(d → ∞) {0/d!,1/d!,…,d!/d!}  -
⋃(d ∈ ℕ⁺)  {0/d!,1/d!,…,d!/d!}

I.e., it's crank-ish to ignore this result framed
in the same way as it is as other relevant results,
the very notion of un-countability itself after
the anti-diagonal of Cartesian functions, this
only-diagonal of not-Cartesian functions, in
this case a very special function that is the
limit, the infinite limit of dividing a segment
of the linear continuum into discrete points,
and vice-versa, these iota-multiples, these
iota-values, that are iota-cuts, complete as
they are

_exactly courtesy the definition of LUB_

that you choose to ignore.

The method by which I choose to ignore
the least.upper.bound property is by
pointing out _several times_ by now that
you (RF) have got wrong
what the least.upper.bound property is.

Not everyone would consider that
ignoring on my part.
Ignoring on your part, though...

Ha, I put them both in one hand,
you either get both or nothing.

And no, I didn't "spend a long time
developing a tolerance to iocane powder",
which is a great sketch from a fantastic movie.

A reference to Vizzini seems more on point. https://www.youtube.com/watch?v=BUg2cp23rGE
|
| Let me put it this way:
| Have you ever heard of Plato, Aristotle, Socrates?
| Morons.

https://www.imdb.com/title/tt0093779/quotes/?item=qt0482733&ref_=ext_shr_lnk

Now, I may not be able to convince the cool kids that
the Village People have some sick beats and there's
something positive in the YMCA and the Navy,
yet, you know, can't make a horse drink.

https://www.youtube.com/watch?v=XLYqTZKEpvs

In the past, you've expressed what seemed to be concern
about my "balking and clamming up".

Are you (RF) no longer concerned about eliciting
responses, or
do you have something there to respond to?

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On 6/7/2024 8:29 PM, Ross Finlayson wrote:

On 06/07/2024 02:15 PM, Jim Burns wrote:

[...]

The least-upper-bound property is exactly that
for an increasing sequence, that represents a
series in the usual setting that represents a
value, for a topology, any old topology that
results "connected" will do, not necessarily
"the usual open topology", here particularly
some sort of "continuous topology" if you recall
that brief conversation we had a few months ago,
the least-upper-bound property is exactly that
the limit, as it were, of such a "critical" "denseness",
to use the words of topology correctly and appropriately,
that it exists as an element of the same set, of the
infinite expression as it may be, or whatever else
results the "critical denseness", the _completion_
what results, how it is.

| In mathematics, the least-upper-bound property
| (sometimes called completeness, supremum property or
| l.u.b. property) is a fundamental property of
| the real numbers. More generally,
| a partially ordered set X has
| the least-upper-bound property if
| every non-empty subset of X with an upper bound has
| a least upper bound (supremum) in X.
| Not every (partially) ordered set has
| the least upper bound property. For example,
| the set ℚ of all rational numbers with its natural order
| does not have the least upper bound property.
|
https://en.wikipedia.org/wiki/Least-upper-bound_property

Least.Upper.Bound.Property(X) ⟺
∀B ⊆ X: B ≠ ∅ ≠ {b∈X|Bᴬ≤b} ⟹
{b∈X|Bᴬ≤b} ᴱ≤ᴬ {b∈X|Bᴬ≤b}

Bᴬ≤b ⟺ ∀x∈B: x≤b
C ᴱ≤ᴬ C ⟺ ∃x∈C:∀y∈C: x≤y

So, as noted it's simply that {f( < m) } for
elements in ran(f), has a least-upper-bound f(m),
which is in ran(f), because, m, is a primitive
sort of natural integer.

Yes,
for the natural numbers and
for sets with similar order types,
each bounded nonempty set is finite,
and, so, it has a least.upper.bound,
a maximum, since it's finite.

However,
the set of
elements which
begin somewhere in one of the sequence.sets
{0/1,1/1}
{0/2,1/2,2/2}
{0/6,1/6,.,6/6}
{0/24,1/24,.,24/24}
...
{0/d!,1/d!,…,d!/d!}
...
and never leave the sequence.sets after
is
a set which does not have the least.upper.bound
property.

I refer to {0/d!,1/d!,…,d!/d!}.limit
{0/d!,1/d!,…,d!/d!}.limit = ℚ∩[0,1]

It looks to me like
what you define the line.reals to be is
{0/d!,1/d!,…,d!/d!}.limit = ℚ∩[0,1]

Yes,
{0/d!,1/d!,…,d!/d!}.limit is the continuum limit.

"Continuum limit" means the distance between
nearest neighbors approaches 0, as it does in ℚ

"Continuum limit" does not mean "continuum".

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On 6/8/2024 11:28 AM, Ross Finlayson wrote:

On 06/07/2024 10:35 PM, Jim Burns wrote:

On 6/7/2024 8:29 PM, Ross Finlayson wrote:

The least-upper-bound property is exactly that
for an increasing sequence, that represents a
series in the usual setting that represents a
value, for a topology, any old topology that
results "connected" will do, not necessarily
"the usual open topology", here particularly
some sort of "continuous topology" if you recall
that brief conversation we had a few months ago,
the least-upper-bound property is exactly that
the limit, as it were, of such a "critical" "denseness",
to use the words of topology correctly and appropriately,
that it exists as an element of the same set, of the
infinite expression as it may be, or whatever else
results the "critical denseness", the _completion_
what results, how it is.

| In mathematics, the least-upper-bound property
| (sometimes called completeness, supremum property or
| l.u.b. property) is a fundamental property of
| the real numbers. More generally,
| a partially ordered set X has
| the least-upper-bound property if
| every non-empty subset of X with an upper bound has
| a least upper bound (supremum) in X.
| Not every (partially) ordered set has
| the least upper bound property. For example,
| the set ℚ of all rational numbers with its natural order
| does not have the least upper bound property.
|
https://en.wikipedia.org/wiki/Least-upper-bound_property

Least.Upper.Bound.Property(X)  ⟺
∀B ⊆ X:  B ≠ ∅ ≠ {b∈X|Bᴬ≤b}  ⟹
{b∈X|Bᴬ≤b} ᴱ≤ᴬ {b∈X|Bᴬ≤b}

Bᴬ≤b  ⟺  ∀x∈B: x≤b
C ᴱ≤ᴬ C  ⟺  ∃x∈C:∀y∈C: x≤y

So, as noted it's simply that {f( < m) } for
elements in ran(f), has a least-upper-bound f(m),
which is in ran(f), because, m, is a primitive
sort of natural integer.

Yes,
for the natural numbers and
for sets with similar order types,
each bounded nonempty set is finite,
and, so, it has a least.upper.bound,
a maximum, since it's finite.

However,
the set of
elements which
begin somewhere in one of the sequence.sets
{0/1,1/1}
{0/2,1/2,2/2}
{0/6,1/6,.,6/6}
{0/24,1/24,.,24/24}
...
{0/d!,1/d!,…,d!/d!}
...
and never leave the sequence.sets after
is
a set which does not have the least.upper.bound
property.

I refer to {0/d!,1/d!,…,d!/d!}.limit
{0/d!,1/d!,…,d!/d!}.limit  =  ℚ∩[0,1]

It looks to me like
what you define the line.reals to be is
{0/d!,1/d!,…,d!/d!}.limit  =  ℚ∩[0,1]

Yes,
{0/d!,1/d!,…,d!/d!}.limit is the continuum limit.

"Continuum limit" means the distance between
nearest neighbors approaches 0, as it does in ℚ

"Continuum limit" does not mean "continuum".

I'm delighted that you note that the set ran(f)
by its values, "doesn't not" meet the definition
of least-upper-bound, then insofar as that it does.

Whether ran(f) meets the definition depends upon
what ran(f) is.

ℚ∩[0,1] my (JB's) best guess at your (RF's) ran(f)
_does not_ meet the definition.

Is √½ = 0.70710678118... in ran(f) ?
Please explain.

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On 6/9/2024 8:11 PM, Ross Finlayson wrote:

On 06/08/2024 10:47 AM, Jim Burns wrote:

On 6/8/2024 11:28 AM, Ross Finlayson wrote:

On 06/07/2024 10:35 PM, Jim Burns wrote:

It looks to me like
what you define the line.reals to be is
{0/d!,1/d!,…,d!/d!}.limit  =  ℚ∩[0,1]

Yes,
{0/d!,1/d!,…,d!/d!}.limit is the continuum limit.

"Continuum limit" means the distance between
nearest neighbors approaches 0, as it does in ℚ

"Continuum limit" does not mean "continuum".

I'm delighted that you note that the set ran(f)
by its values, "doesn't not" meet the definition
of least-upper-bound, then insofar as that it does.

Whether ran(f) meets the definition depends upon
what ran(f) is.

ℚ∩[0,1]  my (JB's) best guess at your (RF's) ran(f)
_does not_ meet the definition.

Is √½ = 0.70710678118... in ran(f) ?
Please explain.

What ratios?

I refer to the ratios in my best guess at ran(f)

| This putative f
| called EF the equivalency function,
| f(n) = n/d, 0 <= n <= d, d -> oo
| in the continuum limit,
|
Date: Fri, 31 May 2024 18:37:34 -0700

I guess
| 0 <= n <= d
means
{0,1,…,d}

I guess
| f(n) = n/d, 0 <= n <= d
means
f{0,1,…,d} =
{0/d,1/d,…,d/d}

I guess
| f(n) = n/d, 0 <= n <= d, d -> oo
means
lim(d → ∞) f(n) = n/d, 0 <= n <= d =
lim(d → ∞) {0/d,1/d,…,d/d}

My guess is that
ran(f) = lim(d → ∞) {0/d,1/d,…,d/d}

Ross Finlayson,
is ran(f) = lim(d → ∞) {0/d,1/d,…,d/d} ?


While you (RF) ponder whether to perform
a simple, easy task which would greatly reduce
the uncertainty around what you are talking about,
I will carry on as though you had answered "yes".


I know what lim(d → ∞) {0/d,1/d,…,d/d} is,
forc reasonable values of "set limit":
lim(d → ∞) {0/d,1/d,…,d/d} =
⋂(0<dᵢ<∞) U(dᵢ<dᵤ<∞) {0/dᵤ,1/dᵤ,…,dᵤ/dᵤ) =
ℚ∩[0,1]

It is the same as the limit of
the sub.sequence of factorial.denominators,
which is a nested.set sequence, so
lim(d → ∞) {0/d!,1/d!,…,d!/d!} =
U(0<dᵤ<∞) {0/dᵤ!,1/dᵤ!,…,dᵤ!/dᵤ!) =
ℚ∩[0,1]

√½ ∉ ℚ∩[0,1]

Is √½ = 0.70710678118... in ran(f) ?
Please explain.

I'm delighted that you note that
the set ran(f) by its values, "doesn't not"
meet the definition of least-upper-bound,
then insofar as that it does.

If ran(f) = ℚ∩[0,1]
⎛ which, if it isn't, I'd like you to say it isn't,
⎝ and, if it is, I'd like you to say it is,
then ran(f) "doesn't not not" meet the definition
of having the least upper bound property.

These are only integer fractions
in the continuum limit, so the ordered field
doesn't even exist yet.

| With the order defined above,
| ℚ is an ordered field that has no subfield
| other than itself,
| and is the smallest ordered field,
| in the sense that every ordered field contains
| a unique subfield isomorphic to ℚ
|
https://en.wikipedia.org/wiki/Rational_number

ℚ isn't the compete ordered field.
ℚ is the smallest ordered field.

ℚ is the continuum limit, but
the continuum limit is not the continuum.

So, that
elements of the complete ordered field in [0,1],
like root two over two,
have values that are real values that
happen to equate to a value in ran(EF) in [0,1],
a unique value, and that,
there is no real value in [0,1] that
is not an element of ran(EF),
just has an existence result that
of the infinitely many distinct integers, and
the infinitely many distinct reals in [0,1],
they're 1 to 1.

Please explain.

Does
| f(n) = n/d, 0 <= n <= , d -> oo
mean something else other than
| lim(d → ∞) {0/d,1/d,…,d/d}
?

Is your use of "limit" in
| lim(d → ∞) {0/d,1/d,…,d/d}
something else other than
| ⋂(0<dᵢ<∞) U(dᵢ<dᵤ<∞) {0/dᵤ,1/dᵤ,…,dᵤ/dᵤ)
?

If something else, then what else?

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On 6/10/2024 4:48 PM, Ross Finlayson wrote:

On 06/09/2024 09:46 PM, Jim Burns wrote:

[...]

I like where you're going with this.

Please help me by telling me,
when I have guessed at what you mean,
whether I've guessed correctly or not.

| This putative f
| called EF the equivalency function,
| f(n) = n/d, 0 <= n <= d, d -> oo
| in the continuum limit,
|
Date: Fri, 31 May 2024 18:37:34 -0700

I guess
| 0 <= n <= d
means
{0,1,…,d}

I guess
| f(n) = n/d, 0 <= n <= d
means
f{0,1,…,d} =
{0/d,1/d,…,d/d}

I guess
| f(n) = n/d, 0 <= n <= d, d -> oo
means
lim(d → ∞) f(n) = n/d, 0 <= n <= d =
lim(d → ∞) {0/d,1/d,…,d/d}

My guess is that
ran(f) = lim(d → ∞) {0/d,1/d,…,d/d}

Ross Finlayson,
is ran(f) = lim(d → ∞) {0/d,1/d,…,d/d} ?

Thank you in advance.

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On 6/10/2024 11:46 PM, Ross Finlayson wrote:

On 06/10/2024 08:36 PM, Jim Burns wrote:

On 6/10/2024 4:48 PM, Ross Finlayson wrote:

On 06/09/2024 09:46 PM, Jim Burns wrote:

[...]

I like where you're going with this.

Please help me by telling me,
when I have guessed at what you mean,
whether I've guessed correctly or not.

Please help me by telling me,
when I have guessed at what you mean,
whether I've guessed correctly or not.
Surely, you can talk about whatever you like
AND do that?

| This putative f
| called EF the equivalency function,
| f(n) = n/d, 0 <= n <= d, d -> oo
| in the continuum limit,
|
Date: Fri, 31 May 2024 18:37:34 -0700

I guess
| 0 <= n <= d
means
{0,1,…,d}

I guess
| f(n) = n/d, 0 <= n <= d
means
f{0,1,…,d}  =
{0/d,1/d,…,d/d}

I guess
| f(n) = n/d, 0 <= n <= d, d -> oo
means
lim(d → ∞) f(n) = n/d, 0 <= n <= d  =
lim(d → ∞) {0/d,1/d,…,d/d}

My guess is that
ran(f)  =  lim(d → ∞) {0/d,1/d,…,d/d}

Ross Finlayson,
is  ran(f)  =  lim(d → ∞) {0/d,1/d,…,d/d}  ?

Thank you in advance.

[...]

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On 6/10/2024 11:46 PM, Ross Finlayson wrote:

On 06/10/2024 08:36 PM, Jim Burns wrote:

[...]

There's a deductive argument that for particular
infinite series, that are convergent, that
the limit, the infinite limit: _is_ the sum.

The items aₖ of Cauchy sequence ⟨a₁ a₂ a₃ …⟩
draw closer to each other
without a positive lower bound (¬∃ε>0)

The items bₖ of convergent sequence ⟨b₁ b₂ b₃ …⟩
draw closer to a single point
without a positive lower bound (¬∃ε>0)
-
There are Cauchy sequences of _rationals_
for which, in the _rationals_
not.exists a point drawn.closer.to
without a positive lower bound (¬∃ε>0)
Those Cauchy sequences don't converge
in the _rationals_

The _reals_ complete the _rationals_ with
points where limit points must be
in order that
there is NO Cauchy sequence of rationals or reals
which does NOT converge in the reals,
which is WITHOUT a single point drawn.closer.to
in the reals.

There's a deductive argument that for particular
infinite series, that are convergent, that
the limit, the infinite limit: _is_ the sum.

For a sequence of finite initial.sums of a series,
if the limit of that initial.sum.sequence exists,
then that limit is _defined_ to be the infinite.sum.

For the reals,
if the initial.sum.sequence is Cauchy
(initial.sum.sequence isn't summand.sequence),
then there is a single point drawn.closer.to
and its infinite.sum (that point) exists.

For the rationals,
that point might not exist in the _rationals_
If so, then the infinite.sum doesn't exist
in the _rationals_

I.e., the deductive argument is that
it can not not be, the sum.
For, were it not, then in the "double reductio",
it would never amount to anything, at all.

It's so that
real analysis is only correct, in the infinite limit,
and as so usually after Riemann sums,
that the measure is as of the continuum limit of those.

Broadly speaking,
the continuum limit allows Cauchy sequences.

_If we are working in the continuum_ then
the continuum limit allows Cauchy sequences
and Cauchy sequences converge.

If we are constructing a domain, whether
| This putative f
| called EF the equivalency function,
| f(n) = n/d, 0 <= n <= d, d -> oo
| in the continuum limit,
or
| ⋃(0<d<∞) {0/d!,1/d!,…,d!/d!}
or
| {​̅x⊆ℚ:∅≠​̅xᴬ<ᴱ​̅xᴬ<ᴬℚ\​̅x≠∅}
|
then we need to know if
the domain we've constructed
is the continuum or it isn't.

You (RF) appear to have assumed that
set.limit(d → ∞) {0/d!,1/d!,…,d!/d!}
is the continuum and then
used that assumption to derive
contradictions of well.known theorems.

Please explain the first part,
why it's the continuum.

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Am 11.06.2024 um 22:49 schrieb Chris M. Thomasson:

On 6/10/2024 8:46 PM, Ross Finlayson wrote:

There's a deductive argument that for particular
infinite series, [namely those] that are convergent,
[...] the limit [...] _is_ the sum.

Well, that "deductive argument" is simply a _definition_.

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Am 12.06.2024 um 01:13 schrieb Ross Finlayson:

There's a difference between "any precision" and "no difference".

Indeed!

In other words, "any precision" and "no difference" is not the same. :-)

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On 6/13/2024 3:04 PM, Ross Finlayson wrote:

On 06/11/2024 11:37 AM, Jim Burns wrote:

[...]

Then, for analysis, and a continuous domain,
which is a domain that has the properties
of extent, density, completeness, and measure,

https://en.wikipedia.org/wiki/Linear_continuum

| According to Raymond Wilder (1965),
| there are four axioms that make a set C and
| the relation < into a linear continuum:
|
| • C is simply [totally] ordered with respect to <.
| • If [A,B] is a cut of C, then
| either A has a last element
| or B has a first element.
| (compare Dedekind cut)
| • There exists a non-empty, countable subset S of C
| such that, if x,y ∈ C such that x < y,
| then there exists z ∈ S such that x < z < y.
| (separability axiom)
| • C has no first element and no last element.
| (Unboundedness axiom)
|
| These axioms characterize the order type of
| the real number line.

I can match Extent, Density, and Completeness
to Wilder's axioms, and I doubt that you object
to Line.ish.ness.

However,
Measure looks odd in this context because
I usually see the real numbers first, and
then they are used to define 'measure'.
Your use looks other.way.around.

of partitions of values among the integers,
or their ratios, is that
for this putative function f,
which is a function this way,
each of "extent density completeness measure"
is observed via inspection

How do you get completeness via inspection?

I understand that
_if that's the real line_
the points that you describe
n/d: 0≤n≤d: d → ∞
are sufficient to hold
sequences which converge to
any point of the _complete_ [0,1]

I can't tell whether
(1)
you're saying n/d: 0≤n≤d: d → ∞ includes
all its limit points, or
(2)
you're saying, poof! add the limit points to
n/d: 0≤n≤d: d → ∞

How would it be if you told me if you mean 1 or 2?
Do you think you might enjoy doing that?

after the
"constant monotone strictly increasing"
and the bounds, 0 and 1.

So, it's continuous.

Completeness and
⎛ constant monotone strictly.increasing
⎝ and the bounds, 0 and 1,
and infinitely.many
is not a thing.

Dedekind.completeness conflicts with
nonzero infinitesimals.

Complete, so
⟨⅟1 ⅟2 ⅟3 …⟩ has greatest.lower.bound β

If β = 0
then
NO x > 0 is without x > ⅟n > 0
(x isn't a greater.than.0 lower bound)
NO x is infinitesimal

If β > 0
then
2⋅β > β > ½⋅β > 0
exists ⅟n < 2⋅β > β
not.exists ⅟m < ½⋅β < β

However
⅟n < 2⋅β
(⅟n)/4 < (2⋅β)/4
⅟(4⋅n) < ½⋅β
Contradiction

Therefore,
NO x is infinitesimal

"The" continuum, or "the Continuum" or
"The Continuum", is a bit overloaded as a term,

I.am.not.a.historian, but perhaps
'continuum' was overloaded once.upon.a.time
because we had not yet worked out
a satisfactory description of a continuum.
(See also Zeno of Alea)

However, that was then.
"Lines which cross must intersect"
is necessary and sufficient for our purposes.
And "Lines which cross must intersect"
describes our Dedekind.complete continuum.

So, thanks for your request, it's not un-noticed,

Why not answer?

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On 6/14/2024 4:12 PM, Ross Finlayson wrote:

On 06/13/2024 08:40 PM, Jim Burns wrote:

[...]

So, I'll address further basically that measure is
axiomatized in the standard way and that it arrives
from length assignment of line-reals is the thing,

Are your line.reals Dedekind.complete?

| • If [A,B] is a cut of C, then
| either A has a last element
| or B has a first element.

Are the line.reals in [0,1]
n/d: 0≤n≤d: d → ∞
?

Is n/d: 0≤n≤d: d → ∞
lim(d → ∞) {0/d,1/d,…,d/d}
?

Standardly,
lim(d → ∞) {0/d,1/d,…,d/d} =
⋂(0<dᵢ<∞) ⋃(dᵢ<dᵤ<∞) {0/dᵤ,1/dᵤ,…,dᵤ/dᵤ} = ⋃(0<d<∞) {0/d!,1/d!,…,d!/d!} =
{n/d ∈ ℚ: 0≤n/d≤1}

{n/d ∈ ℚ: 0≤n/d≤1} is not complete.
Perhaps your line.reals aren't complete,
but you _seem_ to be treating them as complete.
It would be great if you'd say yea or nay.

If you (RF) mean something non.standard here
by 'limit', what is it you mean?

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On 6/14/2024 10:37 PM, Ross Finlayson wrote:

On 06/14/2024 05:31 PM, Jim Burns wrote:

On 6/14/2024 4:12 PM, Ross Finlayson wrote:

So, I'll address further basically that measure is
axiomatized in the standard way and that it arrives
from length assignment of line-reals is the thing,

Are your line.reals Dedekind.complete?

| • If [A,B] is a cut of C, then
| either A has a last element
| or B has a first element.

Are the line.reals in [0,1]
n/d: 0≤n≤d: d → ∞
?

Is  n/d: 0≤n≤d: d → ∞
lim(d → ∞) {0/d,1/d,…,d/d}
?

Standardly,
lim(d → ∞) {0/d,1/d,…,d/d}  =
⋂(0<dᵢ<∞) ⋃(dᵢ<dᵤ<∞) {0/dᵤ,1/dᵤ,…,dᵤ/dᵤ}  =
⋃(0<d<∞) {0/d!,1/d!,…,d!/d!}  =
{n/d ∈ ℚ: 0≤n/d≤1}

{n/d ∈ ℚ: 0≤n/d≤1} is not complete.
Perhaps your line.reals aren't complete,
but you _seem_ to be treating them as complete.
It would be great if you'd say yea or nay.

If you (RF) mean something non.standard here
by 'limit', what is it you mean?

The line-reals are

n/d: 0≤n≤d: d → ∞ ?
lim(d → ∞) {0/d,1/d,…,d/d} ?
⋂(0<dᵢ<∞) ⋃(dᵢ<dᵤ<∞) {0/dᵤ,1/dᵤ,…,dᵤ/dᵤ} ? ⋃(0<d<∞) {0/d!,1/d!,…,d!/d!} ?
{n/d ∈ ℚ: 0≤n/d≤1} ?
None of the above?
All of the above?
The even ones on weekdays and
the odd ones on Sundays?

The line-reals are "iota-complete" or
"Aristotle-complete", vis-a-vis the field-reals,
and "Eudoxus-complete", or "Dedekind-complete",
and the signals-reals, as about
what is in effect the "Fourier-complete" or
"Nyquist-complete".

iota-complete = ?
Aristotle-complete = ?
Eudoxus-complete = ?
Dedekind-complete = ?
Fourier-complete = ?
Nyquist-complete = ?

(Poincare and Dirichlet were
really great both geometers and analysts.)

Three different definitions of completeness,
three different definitions of continuity,
three different definitions of continuous domains,
repleteness, here, is the idea.

Zero definitions here of completeness,
zero definitions here of continuity,
zero definitions here of continuous domains,
repleteness, not defined here, might be the idea.

----

Because iota-values are contiguous, for [0,A]
and (A,1], and [0,A) and [A,1], they're
"complete" both ways, also they're well-ordered.

That's not Dedekind.completeness.

Note that, for each j/k e {n/d ∈ ℚ: 0≤n/d≤1}
splits [0,j/k],(j/k,1] and [0,j/k),[j/k,1] exist.

{n/d ∈ ℚ: 0≤n/d≤1} is not Dedekind.complete.

NOT completeness:
for each point, exists a split the point is between

Completeness:
For each split, exists a point between the split.

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On 6/15/2024 1:31 PM, Ross Finlayson wrote:

On 06/15/2024 09:57 AM, Jim Burns wrote:

On 6/14/2024 10:37 PM, Ross Finlayson wrote:

Because iota-values are contiguous, for [0,A]
and (A,1], and [0,A) and [A,1], they're
"complete" both ways, also they're well-ordered.

That's not Dedekind.completeness.

Note that, for each j/k e {n/d ∈ ℚ: 0≤n/d≤1}
splits [0,j/k],(j/k,1] and [0,j/k),[j/k,1] exist.

{n/d ∈ ℚ: 0≤n/d≤1} is not Dedekind.complete.

NOT completeness:
for each point,  exists a split the point is between

Completeness:
For each split, exists a point between the split.

It reminds me of physics today and
"between relativity and the quantum, where's gravity?"

"Euh..., its mechanism is un-defined."

In mathematics where's continuity?

"Euh...."

NOT completeness:
for each point, exists a split the point is between

Completeness:
For each split, exists a point between the split.

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On 6/15/2024 2:29 PM, Ross Finlayson wrote:

On 06/15/2024 11:12 AM, Ross Finlayson wrote:

On 06/15/2024 10:53 AM, Jim Burns wrote:

NOT completeness:
for each point, exists a split the point is between

Completeness:
For each split, exists a point between the split.

Induction: case 1: not complete, case next: see case 1.

Luckily
we have "multiple modes of inference",
as basically how things connect and flow or
the "-ductive",
the de-ductive, the ab-ductive, the in-ductive,
about helping establishing something better than
a "hypo-crisy", a sort of "juxtapo-crisy".

It is an essential aspect of mathematics that,
when we are discussing a thing or things,
we are not discussing a different thing or things.

That's a principle with wider application, I think,
but mathematics is simply impossible without it.

Consider the set of finite von Neumann ordinals.
None of them are infinite.
Amazing! Fantastic! A tour de force!
Not.

But,
using the knowledge that none of them are infinite,
we explore the infinity of them by means of
our not.first.false telescope.

That is not an exploration of more open theories
about infinite von Neumann ordinals as well,
even though those are theories we could explore.
We _could_ explore them, but we _aren't_
Not right now.


If "we are not discussing a different thing"
is something you decry as hypocrisy,
then you will find plenty of hypocrisy to decry.

We will, despite protestations,
not be discussing a different thing,
until we are discussing that different thing.
In a different discussion.

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On 6/15/2024 2:12 PM, Ross Finlayson wrote:

On 06/15/2024 10:53 AM, Jim Burns wrote:

On 6/15/2024 1:31 PM, Ross Finlayson wrote:

On 06/15/2024 09:57 AM, Jim Burns wrote:

On 6/14/2024 10:37 PM, Ross Finlayson wrote:

Because iota-values are contiguous, for [0,A]
and (A,1], and [0,A) and [A,1], they're
"complete" both ways, also they're well-ordered.

That's not Dedekind.completeness.

Note that, for each j/k e {n/d ∈ ℚ: 0≤n/d≤1}
splits [0,j/k],(j/k,1] and [0,j/k),[j/k,1] exist.

{n/d ∈ ℚ: 0≤n/d≤1} is not Dedekind.complete.

NOT completeness:
for each point,  exists a split the point is between

Completeness:
For each split, exists a point between the split.

It reminds me of physics today and
"between relativity and the quantum, where's gravity?"

"Euh..., its mechanism is un-defined."

In mathematics where's continuity?

"Euh...."

NOT completeness:
for each point,  exists a split the point is between

Completeness:
For each split, exists a point between the split.

Induction:
case 1: not complete, case next: see case 1.

Completeness:
Lines which cross must intersect.
Completeness supplies the intersections.

Cisfinite induction:
⎛ If P(i) changes truth.value from 0 to any k
⎝ then P(i) changes truth.value from some j to j⁺¹
⎛ ∃⟨0,…,k⟩: P(0)∧¬P(k) ⟹
⎝ ∃⟨0,…,j⁺¹⟩: P(j)∧¬P(j⁺¹)

That is more familiar.looking when written as
⎛ P(0) ∧ ∀⟨0,…,j⁺¹⟩: P(j)⇒P(j⁺¹) ⟹
⎝ ∀⟨0,…,k⟩: P(k)

Getting to infinity is not
what cisfinite induction does.
Being true for all finites,
of which there happen to be infinitely.many, is
what cisfinite induction does.

By the way,
Completeness:
Lines which cross must intersect.
Completeness supplies the intersections.

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On 6/15/2024 3:13 PM, Ross Finlayson wrote:

On 06/15/2024 11:49 AM, Jim Burns wrote:

[...]

what's we're talking about is a making a line,
of points, "completion", where here what we
mean exactly in the geometric setting, is
"gaplessness", that "complete" here means
"gapless".

| • If [A,B] is a cut of C, then
| either A has a last element
| or B has a first element.

So, this largely implies that the least
upper bound property is applicable,

Yes.

For a bounded nonempty set S of
any linearly.ordered domain D
[A,B] for
B = {bound of S} and
A = {not.bound of S}
is a cut of D

Either the last element of A
or the first element of B
is the least.upper.bound of S

Thus, from the above, it is implied that,
for any bounded nonempty set,
a least.upper.bound exists.

that it always applies,
that the limit of a convergent series
which exists, is "in", that it is in the set,

The question is:
is "the set" the set you want it to be?

You want "the set" to be the one holding
limit points for each Cauchy sequence.

You describe "the set" as
d/n: 0≤n≤d: d → ∞

I don't think
d/n: 0≤n≤d: d → ∞
is the set you want it to be.

You seem to disagree,
although you seem to avoid saying that outright.

I've told you my reasons. I can repeat them.
What are your reasons?

of the terms which otherwise attain to it,
and which are demonstrated as
readily not so for each finite case,
and, "only in the limit",
thus "only in the infinite limit".

Given then that infinite sequences
converge at all,

Sorry, no.
I can't give you that.
That is what I want you to prove,
that Cauchy sequences
_converge_ in d/n: 0≤n≤d: d → ∞
that Cauchy sequences
_have a limit point_ in d/n: 0≤n≤d: d → ∞

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