On Monday, August 23, 2021 at 10:51:25 PM UTC-7, Jim Burns wrote:
> On 8/23/2021 8:08 AM, WM wrote:
> > Jim Burns schrieb
> > am Montag, 23. August 2021 um 06:55:33 UTC+2:
> >> On 8/22/2021 6:00 PM, WM wrote:
> >>> Jim Burns schrieb
> >>> am Sonntag, 22. August 2021 um 21:45:36 UTC+2:
>
> >>>> *No rational completes any interval*
> >>>
> >>> Then there is no bijection and no cardinality.
> >>> For cardinality completion is required.
> >>>
> >>> There was no objection to a 'potential infinity' in the
> >>> form of an unending process, but an 'actual infinity' in
> >>> the form of a completed infinite set was harder to accept."
> >>> [H.B. Enderton: "Elements of set theory",
> >>> Academic Press, New York (1977) p. 14f]
> >>
> >> "p. 14f" would be the forward. Is that right?
> >
> > Presently I have no acces to this book.
> A preview on Google Books shows the quote starting at the end
> of page 14. It's in "Historical Notes", the last section of
> Chapter 1, "Introduction".
>
> <Enderton, 14-15>
>
> Cantor's work was well received by some of the prominent
> mathematicians of the day, such as Richard Dedekind.
> But his willingness to regard infinite sets as objects
> to be treated in much the same way as finite sets was
> attacked bitterly by others, particularly Kronecker.
> There was no objection to a "potential infinity" in the
> form of an unending process, but an "actual infinity" in
> the form of a completed infinite set was harder to
> accept.
> </Enderton>
>
> Executive summary of "Historical Notes", Ch. 1:
>
> The concept of a set has been used since ancient times.
>
> Georg Cantor was led to the study of set theory by way
> of the study of Fourier series. In 1871 he realized that
> certain operations on sets of real numbers could be
> iterated more than a finite number of times. In 1879
> and subsequent years, he published a series of papers
> setting forth the general concepts of abstract sets
> and "transfinite numbers".
>
> About the turn of the century, attempts were made to
> present the principles of set theory as being
> self-evident truths of deductive thought. But the
> simplicity and directness of Russell's paradox seemed
> to utterly destroy the attempt to base mathematics
> on that sort of set theory.
>
> End summary.
>
> ----
> I was interested in where Enderton _began_ his explanation
> of set theory, so I read his Chapter 1 in Google Books.
>
> This is something I've been picking at for a while now:
> when is it (pedagogically, as material builds on earlier
> material) that posters (such as yourself, WM) with such
> divergent views first begin to diverge? This has given
> me a greater interest in beginnings that I had ages ago
> when I first met these concepts in the classroom.
>
> Enderton's first section is "Baby Set Theory", wherein
> he quickly goes over the concepts "t is a member of A"
> and "t is not a member of A". In my opinion, he should
> have started a little bit earlier, with the concepts
> "t" and "A", that is, with _variables_
>
> With variables, we can partially describe _one of_ the
> steps in an unending process, making claims which are
> true of that step _no matter which_ step is referred
> to. No steps are exempt from that description. But
> no step is last; that's part of its description.
>
> I hear about bitter attacks from Kronecker, and I
> suspect that this is the sort of thing he attacked.
> We, finite beings, claiming to describe _all_ of an
> unending process.
>
> And we do claim that. It doesn't have to take infinitely
> long to say something true about infinitely-many.

Sure, it's usual.

That there are infinitely many either way
as what usually meet in the middle....

That's for example the only possible immediate
rationalization of regular clock arithmetic,
given the only notion that it's a continuous
quantity and continuous quantities are infinitely divisible.

I.e. there's a first and last, first. Then the middle.

Then that results in countable additivity what does
indeed comprise the analytical character of real numbers.

(That the various in the non-measurable and the measurable
in sets, what, are Hardy's points as much as Cauchy the usual,
the sets, because they're defined by their bounds not their neighbors,
that's it's the countably additive measure 1.0 that's topology's
what's relevant to real analysis'.)

Kronecker was a bit more of a sticknoramus and ignored.

Kronecker and Cantor's fights or what, there's Duns Scotus
for that already about 700 years before it. (As an example,
and on the infinitist's side, and, it's since been that way.)

Duns Scotus, who was that guy, Spinoza or the Spinozas,
they have infinity for mathematics pretty well already set out.

McLaurin, is pretty well held-up.

For a lot the linear, and that usual linear algorithms compute,
that's great and fine, but was all invented _after_ plain mental
analysis. And that while efficient and useful and for applications,
ends as the "pure" ends up being more widely applicable.

Anyways I wrote polemics and apologetics in foundations here,
modern ones that serve as reading material.

I, uh, I don't have much else to read....

There isn't much more.

On Monday, August 23, 2021 at 9:33:01 PM UTC+2, WM wrote:
> Greg Cunt schrieb am Montag, 23. August 2021 um 21:06:14 UTC+2:

> > (1, 2, 3, ...) is not a well-ordered set, but a SEQUENCE.
> >
> Everey sequence without repetitions is a well-ordered set.

Nope.

A well-ordered set is an _ordered set_ and hence an ordered pair where the first component is a set and the second component is a well-order defined on that set.

Hint: THIS is an ordered set: (IN, <=).

See: https://www.matheretter.de/wiki/geordnete-mengen

Hint:

> > THE stepwise ANALYSIS is a PROCESS. This does not mean that the set IN or the sequence (1, 2, 3, ...) or the well-orderd set (IN, <=) is a PROCESS too.

> It <bla>

Shut up, you silly crank.

On Monday, August 23, 2021 at 2:36:31 PM UTC-4, WM wrote:
> William schrieb am Montag, 23. August 2021 um 15:11:06 UTC+2:
> > On Monday, August 23, 2021 at 8:02:37 AM UTC-4, WM wrote:
> > > William schrieb am Montag, 23. August 2021 um 05:08:57 UTC+2:
> > > > On Sunday, August 22, 2021 at 6:02:18 PM UTC-4, WM wrote:
> > > > > William schrieb am Sonntag, 22. August 2021 um 21:04:03 UTC+2:
> > > > > > On Sunday, August 22, 2021 at 2:46:49 PM UTC-4, WM wrote:
> > > > > > > > > Here is the proof of incomplete
> > > > > > > > The term "complete" is nonsense. it implies that there are "incomplete" sets,
> > > > > > > No
> > > > > > Yes. All sets contain what they contains. A set can be a proper subset but it cannot be "incomplete". If x is a FISON then x is an element of set set of FISONS.
> > > > > > The set of FISONs is not "incomplete" because the set has no last element.
> > > > > >
> > > > > But a mapping can be incomplete.
> > > > Nope,
> > > It can be incomplete.
> > > >a bijection is a set of ordered pairs.
> > > a complete set.
> > "complete" is a nonsense term. There is no such thing as an "incomplete" set and since a mapping is a set, there is no such thing as an incomplete mapping
> Here is an incomplete mapping: In the interval (1000, 1001] the density of rationals is precisely the same as in the interval (0, 1].

Nope. Let the mapping be f from |N_F to the rationals. Then if r is a rational there exists an element of |N_F, x(r), such that f(x(r))= r. It is nonsense to say the mapping is "incomplete". There is no rational that is not mapped to.

--
William Hughes

On Monday, August 23, 2021 at 12:33:01 PM UTC-7, WM wrote:

> In the interval (1000, 1001] the density of rationals is precisely the same as in the interval (0, 1].

Prove it

On 8/23/2021 2:32 PM, WM wrote:
> Greg Cunt schrieb am Montag, 23. August 2021 um 21:06:14 UTC+2:
>> On Monday, August 23, 2021 at 8:39:52 PM UTC+2, WM wrote:
>>> Greg Cunt schrieb am Montag, 23. August 2021 um 15:40:12 UTC+2:
>>>>
>>>> a set is not a process, you know.
>>>>
>>> The well-ordered set (1, 2, 3, ...)
>> (1, 2, 3, ...) is not a well-ordered set, but a SEQUENCE.
>
> Everey sequence without repetitions is a well-ordered set.

Wrong! you need to get out of the set business.

>>
>>> can be analyzed step by step. [...] This is a process.
>>
>> THE stepwise ANALYSIS is a PROCESS. This does not mean that the set IN or the sequence (1, 2, 3, ...) or the well-orderd set (IN, <=) is a PROCESS too.
>
> It means just this for a non-platonist. But the sequence cannot be analyzed stepby step. That is only possible for the non-dark part.

Do Not Step in the dark thing

>
>>> less than one promille.
>>
>> Are you sure?
>
> Yes, it is an incomplete mapping: In the interval (1000, 1001] the density of rationals is precisely the same as in the interval (0, 1].

It is up to you to mathematically Prove it, but you cant.

(extra credit questions;
take each point on the real line (0,1] add 1000 to it, and you get (1000,1001], is the reverse true?
what about (0,1] to (500,1000] ? take each point on the real line, multiply by 500 and then add 500
what happened to the density of rationals?)

> Alas Cantor enumerates less than one promille. This is an incomplete mapping.

Wrong! prove it is incomplete.

>
> Regards, WM
>
>
>

On 8/23/2021 8:08 AM, WM wrote:
> Jim Burns schrieb
> am Montag, 23. August 2021 um 06:55:33 UTC+2:

>> Should I assume from your use of that quote that
>> you do not accept the assumption that completed
>> infinite sets exist?
>
> No,

I'm more than willing to accept "no" here, but
I expected a different answer, given other things
you have said. I suspect some form of miscommunication
is occurring.

Perhaps my question was mis-read.

Do you accept that a set exists with all k in it such
that steppable 0,...,k exists with adjacent i, j =i+1,
and nothing in it other than k such that steppable
0,...,k with adjacent i, j = i+1
?

That set would contain all the finite natural numbers.
It would not contain some version of what you call
dark numbers. Any "dark number" the set included would
also be a finite natural number.

> Do you understand and accept quantifier logic?
> ~∃n ∈ ℕ: P(n) <==> ∀n ∈ ℕ: ~P(n)
>
> Please apply this to
>
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
>
> In the interval (1000, 1001] the density of rationals
> is precisely the same as in the interval (0, 1].

For each rational in (1000,1001], there is a unique
rational in (0,1].
For each rational in (0,1], there is a unique
rational in (1000,1001].

> Alas Cantor enumerates less than one promille.

For gcd(p,q) = 1, let
index(p/q) = q*q*p*p*/rad(p)

All positive rationals are indexed.

> When going from 1 to 0, then there is no unit fraction
> more later. Either, forced by the linearity of the
> sequence 1, 1/2, 1/3, ..., 0, a last unit fraction
> has been passed. Or there are dark numbers.

When going from 1 to 0, in the negative direction,
there is no unit fraction after 0 (< 0).

If 0 could be stepped to one-by-one along the unit
fractions, then there would be a last unit fraction
before 0 (> 0).

Because no unit fraction 1/k is the unit last fraction
before 0 (> 0), 0 cannot be stepped to one-by-one
along the unit fractions.

Let me emphasize that 1/k such that
steppable 1,...,k with adjacent i j = i+1 exists
is a description true without exception of each
unit fraction.

If we invented more things between 0 and all those
unit fractions they would not be unit fractions,
merely things we inserted.

More to the point I think you're making,
0 cannot be stepped to one-by-one along the unit
fractions _plus whatever you insert_ Call them
dark unit fractions. Why not? But they aren't
unit fractions and they do not change that the
unit fractions cannot be stepped through one-by-one
from first to last.

FredJeffries schrieb am Dienstag, 24. August 2021 um 18:25:07 UTC+2:
> On Monday, August 23, 2021 at 12:33:01 PM UTC-7, WM wrote:
>
> > In the interval (1000, 1001] the density of rationals is precisely the same as in the interval (0, 1].
> Prove it

∀ q ∈ (0, 1] ∃ q+n ∈ (n, n+1] and ∀ q+n ∈ (n, n+1] ∃ q ∈ (0, 1] .

Regards, WM

Gus Gassmann schrieb am Montag, 23. August 2021 um 22:05:06 UTC+2:
> On Monday, 23 August 2021 at 16:21:45 UTC-3, WM wrote:

> > I have taken the limit. There we have the ratio less than 1/1000.
> You took the limit of the ratios of cardinalities, which tells you absolutely *NOTHING* about the problem that you are interested in,

It tells what a mathematician is interested in.

> which is "Did the mapping cover all every positive rational number".

Of course it does disprove this assumption.

> Since it is easy to bound the position at which every positive rational n/m is covered (namely never more than 0.5*(n+m)^2), the answer to that question is an obvious "Yes".

The answer is obviously no. At least half of all indices are issued to the first interval. This is true for every index and for the limit. It proves that more rationals are indexed in the first interval than in any other interval.

> Nothing else matters.

For a matheologians nothimg else matters. For a mathematician my argument is decisive.

> > In the limit the mapping is completed.
> Good. So at last we agree on that. Just to be sure: the mapping is completed after covering every positive rational number, in particular aleph_0 of them in the interval (0,1], aleph_0 in the interval (1, 2], aleph_0 in the interval (1000, 1001], and so on.

Yes, but aleph_0 is a nonsense notion. Neither in the first interval nor in any other interval more than a vanishing subset of rationals is indexed.

Regards, WM

zelos...@gmail.com schrieb am Dienstag, 24. August 2021 um 06:55:34 UTC+2:
> måndag 23 augusti 2021 kl. 14:04:53 UTC+2 skrev WM:

> >The sequence of definable natural numbers works in steps. Every step can be analyzed.
> THere are no steps in it in the way you want there to be.

There are the first thre steps: 1, 2, 3. There are even more. If they cease then it is because the definable numbers cease. Why else should they cease?

> >There are not more FISONs than can be distinguished by FISONs, i.e., there are less than aleph_0 FISONs.
> False, given FISONs are in bijection with N, that means there is aleph_0 FISONs

That means aleoh_0 is a nonsense notion.

Regards, WM

William schrieb am Dienstag, 24. August 2021 um 17:20:26 UTC+2:
> On Monday, August 23, 2021 at 2:36:31 PM UTC-4, WM wrote:

> > Here is an incomplete mapping: In the interval (1000, 1001] the density of rationals is precisely the same as in the interval (0, 1].
> Nope. Let the mapping be f from |N_F to the rationals. Then if r is a rational there exists an element of |N_F, x(r), such that f(x(r))= r. It is nonsense to say the mapping is "incomplete". There is no rational that is not mapped to.

Mathematics tells the contrary. Half of Cantor's indices are issued to the first unit interval. (Compare the so-called first diagonal argument with its zigzag enumeration.) So either there are less rationals in the other intervals or Cantor's enumeration is incomplete. The former assumption is wrong.. In every interval there are precisely same numbers of rationals:

∀ q ∈ (0, 1] ∃ q+n ∈ (n, n+1] und ∀ q+n ∈ (n, n+1] ∃ q ∈ (0, 1].

Hence the latter is true. The mapping is incomplete. There is no *definable* rational that is not mapped to. About dark rationals you cannot know anything concerning the mapping..

Regards, WM

Jim Burns schrieb am Dienstag, 24. August 2021 um 20:29:56 UTC+2:

> All positive rationals are indexed.

How can you utter this silly claim knowing that half of all indexes are issued to the first unit interval? Does it contain half of all rationals? All *definable* rationals are indexed.

Regards, WM

On Tuesday, August 24, 2021 at 10:25:29 PM UTC+2, WM wrote:
> FredJeffries schrieb am Dienstag, 24. August 2021 um 18:25:07 UTC+2:
> > On Monday, August 23, 2021 at 12:33:01 PM UTC-7, WM wrote:
> > >
> > > In the interval (1000, 1001] the density of rationals is precisely the same as in the interval (0, 1].
> >
> > Prove it
> >
> ∀ q ∈ (0, 1] ∃ q+n ∈ (n, n+1] and ∀ q+n ∈ (n, n+1] ∃ q ∈ (0, 1] .

This is not a proof, but a spoof! :-)

On Tuesday, August 24, 2021 at 10:40:27 PM UTC+2, WM wrote:

> aleph_0 is a nonsense notion.

Fascinating, Mückenheim!

On Tuesday, August 24, 2021 at 10:43:27 PM UTC+2, WM wrote:

> There are the first thre steps: 1, 2, 3. There are even more. If they cease then <bla bla>

THE DON'T CEASE, YOU SILLY ASSHOLE!

There is no largest natural number.

On 8/24/2021 3:25 PM, WM wrote:
> FredJeffries schrieb am Dienstag, 24. August 2021 um 18:25:07 UTC+2:
>> On Monday, August 23, 2021 at 12:33:01 PM UTC-7, WM wrote:
>>
>>> In the interval (1000, 1001] the density of rationals is precisely the same as in the interval (0, 1].
>> Prove it
>
> ∀ q ∈ (0, 1] ∃ q+n ∈ (n, n+1] and ∀ q+n ∈ (n, n+1] ∃ q ∈ (0, 1] .
>
> Regards, WM
>

not so;
[∃x∈D,[P(x)⇒[∀y∈D, Q(x, y)]]]∧[∃x∈D, P(x)]∧[∃x∈D,¬P(x)]

On Tuesday, August 24, 2021 at 4:51:46 PM UTC-4, WM wrote:
> William schrieb am Dienstag, 24. August 2021 um 17:20:26 UTC+2:
> > On Monday, August 23, 2021 at 2:36:31 PM UTC-4, WM wrote:
>
> > > Here is an incomplete mapping: In the interval (1000, 1001] the density of rationals is precisely the same as in the interval (0, 1].
> > Nope. Let the mapping be f from |N_F to the rationals. Then if r is a rational there exists an element of |N_F, x(r), such that f(x(r))= r. It is nonsense to say the mapping is "incomplete". There is no rational that is not mapped to.
> Mathematics tells the contrary.

Nope, mathematics says there is no rational that is not mapped to.

>Half...

"Half" has no meaning when applied to infinite cardinalities . The cardinality of a set is not the number of elements in the set. "Number of elements" has no meaning for a set with infinite cardinality like the cardinality of the rationals in an interval, or the cardinality of any Peano set like the set of FISONs.

>There is no *definable* rational that is not mapped to.

Correct there is no rational that you can write down that is not mapped to. Andthere is no rational that you cannot write down that is not mapped to. There are no "dark" numbers of the form x/y, with x an element of |N_F and y a non-zero element of |N_F.

--
William Hughes

On Wednesday, August 25, 2021 at 1:56:46 AM UTC+2, William wrote:

> The cardinality of a set is not the number of elements in the set.

"the cardinality of a set is a measure of the "number of elements" of the set." (Wikipedia)

Go figure... :-P

On 8/24/2021 3:40 PM, WM wrote:
> Gus Gassmann schrieb am Montag, 23. August 2021 um 22:05:06 UTC+2:
>> On Monday, 23 August 2021 at 16:21:45 UTC-3, WM wrote:
>
>>> I have taken the limit. There we have the ratio less than 1/1000.
>> You took the limit of the ratios of cardinalities, which tells you absolutely *NOTHING* about the problem that you are interested in,
>
> It tells what a mathematician is interested in.

646 malformed Ants
647 Master Ants of The Imprecision and Ambiguity Society
648 Math adherent Ants in Sci.math
649 Mathelogician Ants

>
>> which is "Did the mapping cover all every positive rational number".
>
> Of course it does disprove this assumption.

159 Ants that can't do proofs.
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182 Ants with a Deep Misunderstanding of Math

>
>> Since it is easy to bound the position at which every positive rational n/m is covered (namely never more than 0.5*(n+m)^2), the answer to that question is an obvious "Yes".
>
> The answer is obviously no. At least half of all indices are issued to the first interval. This is true for every index and for the limit. It proves that more rationals are indexed in the first interval than in any other interval.

1122 Wrong Ants
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224 Brain-damaged Ants on Math

>
>> Nothing else matters.
>
> For a matheologians nothimg else matters. For a mathematician my argument is decisive.

229 Bullshit Ants that have had Enough
230 Bullshit Ants
231 Bullshit Ants that count rocks but do not number them

>
>>> In the limit the mapping is completed.
>> Good. So at last we agree on that. Just to be sure: the mapping is completed after covering every positive rational number, in particular aleph_0 of them in the interval (0,1], aleph_0 in the interval (1, 2], aleph_0 in the interval (1000, 1001], and so on.
>
> Yes, but aleph_0 is a nonsense notion. Neither in the first interval nor in any other interval more than a vanishing subset of rationals is indexed.

620 Liar Ants
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>
> Regards, WM
>

On 8/24/2021 3:51 PM, WM wrote:
> William schrieb am Dienstag, 24. August 2021 um 17:20:26 UTC+2:
>> On Monday, August 23, 2021 at 2:36:31 PM UTC-4, WM wrote:
>
>>> Here is an incomplete mapping: In the interval (1000, 1001] the density of rationals is precisely the same as in the interval (0, 1].
>> Nope. Let the mapping be f from |N_F to the rationals. Then if r is a rational there exists an element of |N_F, x(r), such that f(x(r))= r. It is nonsense to say the mapping is "incomplete". There is no rational that is not mapped to.
>
> Mathematics tells the contrary. Half of Cantor's indices are issued to the first unit interval. (Compare the so-called first diagonal argument with its zigzag enumeration.) So either there are less rationals in the other intervals or Cantor's enumeration is incomplete. The former assumption is wrong. In every interval there are precisely same numbers of rationals:
>

646 malformed Ants
647 Master Ants of The Imprecision and Ambiguity Society
648 Math adherent Ants in Sci.math
649 Mathelogician Ants

>
> Hence the latter is true. The mapping is incomplete. There is no *definable* rational that is not mapped to. About dark rationals you cannot know anything concerning the mapping..

653 McDuck’s Ants that are QuACkErS
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>
> Regards, WM
>

657 Mind Rotted Ants
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tisdag 24 augusti 2021 kl. 22:43:27 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 24. August 2021 um 06:55:34 UTC+2:
> > måndag 23 augusti 2021 kl. 14:04:53 UTC+2 skrev WM:
>
> > >The sequence of definable natural numbers works in steps. Every step can be analyzed.
> > THere are no steps in it in the way you want there to be.
> There are the first thre steps: 1, 2, 3. There are even more. If they cease then it is because the definable numbers cease. Why else should they cease?
> > >There are not more FISONs than can be distinguished by FISONs, i.e., there are less than aleph_0 FISONs.
> > False, given FISONs are in bijection with N, that means there is aleph_0 FISONs
> That means aleoh_0 is a nonsense notion.
>
> Regards, WM
It doesn't mean that at all. You ASSUME that it shouldn't be so but we prove that it is so. It doesn't fit you so you say "nonsense" then rather than actually construct systems where it isn't so.

tisdag 24 augusti 2021 kl. 22:57:38 UTC+2 skrev WM:
> Jim Burns schrieb am Dienstag, 24. August 2021 um 20:29:56 UTC+2:
>
> > All positive rationals are indexed.
> How can you utter this silly claim knowing that half of all indexes are issued to the first unit interval? Does it contain half of all rationals? All *definable* rationals are indexed.
>
> Regards, WM
ALL rational are indiced. It is trivial to show there is a bijection betwee N and Q+

On 8/24/2021 4:57 PM, WM wrote:
> Jim Burns schrieb
> am Dienstag, 24. August 2021 um 20:29:56 UTC+2:

>> All positive rationals are indexed.
>
> How can you utter this silly claim knowing that
> half of all indexes are issued to the first unit
> interval?

The same way I "know" know that "half" of all finite
indexes are even. Your claim is based on comparing
infinite sets as though they were finite sets.
It is nonsense. *Galileo* pointed out that
it is nonsense. Please keep more up-to-date.

Another claim pretending infinite sets are finite:
Half of all indexes are issued to the first
unit interval.

For gcd(p,q) = 1, *define*
index(p/q) = q*q*p*p/rad(p)

A correct claim:
All positive rationals are finitely indexed.

| For each positive rational,
| there is a pair of non-zero natural numbers p',q'
| for which that rational = p'/q'
| | For each pair of non-zero natural numbers p',q'
| a unique first natural p exists such that,
| for some natural number q, p/q = p'/q'
| | Natural numbers are finite.
| In particular, p and q are finite.
| | q*q*p*p is finite.
| | rad(p) is the radical of p
| https://en.wikipedia.org/wiki/Radical_of_an_integer
| | rad(p) divides p
| p/rad(p) =< p
| q*q*p*p/rad(p) =< q*q*p*p, which is finite.
| q*q*p*p/rad(p) is finite.
| | index(p/q) is finite.

> Does it contain half of all rationals?
> All *definable* rationals are indexed.

| For each positive rational,
| there is a pair of non-zero natural numbers p',q'
| for which that rational = p'/q'

| Natural numbers are finite.

Your argument for the existence of your dark numbers
is that what we mean by rationals, what we mean by
naturals is ambiguous enough that there *might* be
dark numbers.

The problem with this argument is that what we mean
by rationals, what we mean by naturals *is not*
ambiguous enough that there *might* be dark numbers.
*Definable* rationals are all the rationals.
*Definable* naturals are all the naturals.

On Tuesday, August 24, 2021 at 6:11:07 PM UTC-7, Sergio wrote:
> On 8/24/2021 3:40 PM, WM wrote:
> > Gus Gassmann schrieb am Montag, 23. August 2021 um 22:05:06 UTC+2:
> >> On Monday, 23 August 2021 at 16:21:45 UTC-3, WM wrote:
> >
> >>> I have taken the limit. There we have the ratio less than 1/1000.
> >> You took the limit of the ratios of cardinalities, which tells you absolutely *NOTHING* about the problem that you are interested in,
> >
> > It tells what a mathematician is interested in.
> 646 malformed Ants
> 647 Master Ants of The Imprecision and Ambiguity Society
> 648 Math adherent Ants in Sci.math
> 649 Mathelogician Ants
> >
> >> which is "Did the mapping cover all every positive rational number".
> >
> > Of course it does disprove this assumption.
> 159 Ants that can't do proofs.
> 172 Ants that intentionally Stink at Math
> 180 Ants that violate arithmetic
> 181 Ants too stupid and moronic to do mathematics
> 182 Ants with a Deep Misunderstanding of Math
> >
> >> Since it is easy to bound the position at which every positive rational n/m is covered (namely never more than 0.5*(n+m)^2), the answer to that question is an obvious "Yes".
> >
> > The answer is obviously no. At least half of all indices are issued to the first interval. This is true for every index and for the limit. It proves that more rationals are indexed in the first interval than in any other interval.
> 1122 Wrong Ants
> 221 Boo Boo Ants *
> 544 Indexed Ants
> 224 Brain-damaged Ants on Math
> >
> >> Nothing else matters.
> >
> > For a matheologians nothimg else matters. For a mathematician my argument is decisive.
> 229 Bullshit Ants that have had Enough
> 230 Bullshit Ants
> 231 Bullshit Ants that count rocks but do not number them
> >
> >>> In the limit the mapping is completed.
> >> Good. So at last we agree on that. Just to be sure: the mapping is completed after covering every positive rational number, in particular aleph_0 of them in the interval (0,1], aleph_0 in the interval (1, 2], aleph_0 in the interval (1000, 1001], and so on.
> >
> > Yes, but aleph_0 is a nonsense notion. Neither in the first interval nor in any other interval more than a vanishing subset of rationals is indexed.
> 620 Liar Ants
> 240 Card({all ants}) Ants
> 241 cardinality Ants
> 242 Changeling Ants
> 243 Character Flaw Ants
> 244 cheat Ants
> 245 Cheat Ants are back!
> 246 checked Ants that are pissed
> 247 Chicken A seed dropper Ant
> 248 Chuckling Ants
>
> >
> > Regards, WM
> >

Please collect these.

Some have that when there are infinite ants,
each bigger, one's infinite.

It's a turtle all the way down....

It is just like when you apply Russell's argument
to only finite ordinals what according to Russell
contains itself. ZF axiomatizes this away and
non-logically "sets are well-founded and so is omega".

Greg Cunt explained :
> On Wednesday, August 25, 2021 at 1:56:46 AM UTC+2, William wrote:
>
>> The cardinality of a set is not the number of elements in the set.
>
> "the cardinality of a set is a measure of the "number of elements" of the
> set." (Wikipedia)
>
> Go figure... :-P

http://www.cs.bsu.edu/~hfischer/math215/cardinality.pdf

Actually, even the wikipedia article has this.

On Wednesday, August 25, 2021 at 11:49:22 AM UTC+2, FromTheRafters wrote:
> Greg Cunt explained :
> > On Wednesday, August 25, 2021 at 1:56:46 AM UTC+2, William wrote:
> > >
> > > The cardinality of a set is not the number of elements in the set.
> > >
> > "the cardinality of a set is a measure of the "number of elements" of the
> > set." (Wikipedia)
> >
> > Go figure... :-P
> >
> http://www.cs.bsu.edu/~hfischer/math215/cardinality.pdf

Yes, we can! :-)