Examination of Cantor’s proofs for uncountability and axiom for counting infinite sets

[PengKuan Em]

I do a detailed analysis of Cantor’s theory of uncountable sets. The logic of his proofs has some weaknesses. I propose an axiom and a solution to continuum hypothesis.

The main idea is:
Assumption of Cantor’s proofs: All real numbers (set R) are in a list (list L).

This assumption means R=L, considering L as a set. This makes the claim “a real number is created but is not in the list L” wrong. Indeed, if a number is outside L, it is outside R too. So, the statement “the created real number is not in the list L” means it is not in R and is not a real number, which is equivalent to claim that a real number is not real number. This is absurd but Cantor’s diagonal argument and nested intervals proof both claim that a real number is not in the list L and thus, is not a real number, which make them wrong.

On the other hand, Cantor's both proofs search for contradiction. Can “this real number is not a real number” be the contradiction? No. The contradiction of the proofs is in the third step: sout “is not in the list”. By failing to create sout the second step collapses before the third step declares the contradiction.

See the paper
https://www.academia.edu/86410224/Examination_of_Cantors_proofs_for_uncountability_and_axiom_for_counting_infinite_sets

Kuan Peng

[Ross A. Finlayson]

Hi Kuan Peng, have you heard of Ross Finlayson?

[Ben Bacarisse]

PengKuan Em <tita...@gmail.com> writes:

> I do a detailed analysis of Cantor’s theory of uncountable sets. The
> logic of his proofs has some weaknesses.

I think you should revire the proof because it is not at all the proof
you think it is.

> I propose an axiom and a
> solution to continuum hypothesis.
>
> The main idea is:
> Assumption of Cantor’s proofs: All real numbers (set R) are in a list
> (list L).
>
> This assumption means R=L, considering L as a set.

A list of reals is a function (indeed a bijection) between N and R.
It's a set of pairs {(1, r_1), (2, r_2), ...}, so R=L is a bit sloppy.
You mean R=image(L) (sometimes written L(N) since the image is set of
values of L over N).

> This makes the claim “a real number is created but is not in the list
> L” wrong.

How? The proof is that no list (no bijection from N to R) has R as it's
image. From any list, a real can be constructed that is not in the
list.

> Indeed, if a number is outside L, it is outside R too.

Some authors (rather pointlessly) assume that R=image(L) for some L for
the purpose of showing that assumption to be false. But the
construction is always of a real number. Just not one in the list it
was constructed from.

--
Ben.

[PengKuan Em]

I know you here. But, do you have the same idea than me? Please explain.

PK

[PengKuan Em]

Le dimanche 11 septembre 2022 à 22:34:47 UTC+2, Ben Bacarisse a écrit :
> PengKuan Em <tita...@gmail.com> writes:
>
> > I do a detailed analysis of Cantor’s theory of uncountable sets. The
> > logic of his proofs has some weaknesses.
> I think you should revire the proof because it is not at all the proof
> you think it is.

I have explained in the paper where maybe you will find it more correct.

> > I propose an axiom and a
> > solution to continuum hypothesis.
> >
> > The main idea is:
> > Assumption of Cantor’s proofs: All real numbers (set R) are in a list
> > (list L).
> >
> > This assumption means R=L, considering L as a set.
> A list of reals is a function (indeed a bijection) between N and R.
> It's a set of pairs {(1, r_1), (2, r_2), ...}, so R=L is a bit sloppy.
> You mean R=image(L) (sometimes written L(N) since the image is set of
> values of L over N).

I mean that when seeing L as only the set of it members, L have the same members than R.

> > This makes the claim “a real number is created but is not in the list
> > L” wrong.
> How? The proof is that no list (no bijection from N to R) has R as it's
> image. From any list, a real can be constructed that is not in the
> list.

This is what Cantor said. But if L=R, then if r (real number) is not in L it is not in R. I have constructed a list L in my paper

> > Indeed, if a number is outside L, it is outside R too.
> Some authors (rather pointlessly) assume that R=image(L) for some L for
> the purpose of showing that assumption to be false. But the
> construction is always of a real number. Just not one in the list it
> was constructed from.
>
> --
> Ben.

I have cited this example in my paper:
Let us explain this logical hole with a simple example. Let N3 = {2, 3, 1} be the set of all integers smaller than 4. Let L3 = {1, 2, 3} be a list of all the members of N3. We propose to create an integer in N3 which is not in L3. Suppose that we have created 5, but 5 is not in N3. The hole in this reasoning is that the created integer can only be 1 or 2 or 3 and then, it must be in L3. In conclusion, a member of N3 cannot be out of the list L3.

Also, a list of all the n-bits contains 2^n members, but they have only n bit. The constructed number is created with n numbers only, because of the diagonal. So, the constructed number is not in the first n number, but is in the remaining 2^n-n members of the list. when n goes to infinity, this is always true.

I have explained this in detail in the paper.

PK

[Ben Bacarisse]

PengKuan Em <tita...@gmail.com> writes:

> Le dimanche 11 septembre 2022 à 22:34:47 UTC+2, Ben Bacarisse a écrit :
>> PengKuan Em <tita...@gmail.com> writes:
>>
>> > I do a detailed analysis of Cantor’s theory of uncountable sets. The
>> > logic of his proofs has some weaknesses.
>> I think you should revire the proof because it is not at all the proof
>> you think it is.
>
> I have explained in the paper where maybe you will find it more
> correct.

I read sci.logic. If you can't explain here, that's fine. I wish you
luck and I'll read you published paper (you know, properly published)
when it comes out.

>> > I propose an axiom and a
>> > solution to continuum hypothesis.
>> >
>> > The main idea is:
>> > Assumption of Cantor’s proofs: All real numbers (set R) are in a list
>> > (list L).
>> >
>> > This assumption means R=L, considering L as a set.
>> A list of reals is a function (indeed a bijection) between N and R.
>> It's a set of pairs {(1, r_1), (2, r_2), ...}, so R=L is a bit sloppy.
>> You mean R=image(L) (sometimes written L(N) since the image is set of
>> values of L over N).
>
> I mean that when seeing L as only the set of it members, L have the
> same members than R.

I knew what you meant. I was hoping you'd start to used terms
correctly. But no L can't "have the same members" as R. If anyone
assumes it does, it is simply to show that assumption leads to a
contradiction.

>> > This makes the claim “a real number is created but is not in the list
>> > L” wrong.
>> How? The proof is that no list (no bijection from N to R) has R as it's
>> image. From any list, a real can be constructed that is not in the
>> list.
>
> This is what Cantor said. But if L=R, then if r (real number) is not
> in L it is not in R.

Yes, but L =/= R for all L. Assuming something does not make it so.

> I have constructed a list L in my paper

This is called burying the lede! You should start with this and explain
how you think it is not possible to construct, from your L, an r in R
such that r not in image(L). (But from what you say below, you don't
know enough about R to understand the proof you claim to be
undermining.)

> I have cited this example in my paper:
> Let us explain this logical hole with a simple example. Let N3 = {2,
> 3, 1} be the set of all integers smaller than 4. Let L3 = {1, 2, 3} be
> a list of all the members of N3. We propose to create an integer in N3
> which is not in L3. Suppose that we have created 5, but 5 is not in
> N3. The hole in this reasoning is that the created integer can only be
> 1 or 2 or 3 and then, it must be in L3. In conclusion, a member of N3
> cannot be out of the list L3.

This is a pointless analogy because N (and specifically the subset of N
less that 4) does not have the key property. R is closed under taking
the least upper bound, that means that any countable list of reals can
be used to construct a real not in the list.

I find it extraordinary that people who don't know much about a subject
feel empowered to claim that others have got it all wrong. You don't
know the key property that distinguishes R from, say, Q but you don't
let that hold you back. And I am 100% sure that someone telling you
that you are missing a key fact won't make you go and read about how R
is defined. You'll just keep posting the same stuff.

> Also, a list of all the n-bits contains 2^n members, but they have
> only n bit. The constructed number is created with n numbers only,
> because of the diagonal. So, the constructed number is not in the
> first n number, but is in the remaining 2^n-n members of the
> list. when n goes to infinity, this is always true.

Same problem.

> I have explained this in detail in the paper.

More detail about a fundamental understanding is not going to help.

--
Ben.

[Ross A. Finlayson]

Can you provide two counterexamples?

Define a function with domain natural integers and range [0,1],
i.e. that is 1-1, and onto, from N to R[0,1].

I describe a reductio where an example is defined,
then what are its mathematical properties.



Then also there's apologetics to explain why Cantor's otherwise
theorem(s) don't apply, i.e. why it's a "counterexample",
that otherwise Cantor provides any number of "examples",
of an element not on his list.

Here my usual examples are f(n) = n/d, d->oo, n->d,
then a in the space of the signal domain, point out
what is the usual abstract source of reconstruction,
with a doubling space and an infinite domain.

Simply though it's particularly that one counterexample
there, or, "sweep", these days about writing same in
generalized functions and modeling functions as
limts of standard functions. (Much simpler geometrically.)

This is the most sort of glaring example, but it's apologetics
that all functions in Cantor's theory are Cartesian, this is not.

So, Ross Finlayson wrote it.

[Jeff Barnett]

Should that be, instead "More detail about a fundamental
MISunderstanding is not going to help."?
--
Jeff Barnett

[Ben Bacarisse]

Yup. Thanks for spotting that. I make lots of typos, but the ones that
reverse the meaning are bad!

--
Ben.

[Fritz Feldhase]

On Sunday, September 11, 2022 at 8:48:42 PM UTC+2, tita...@gmail.com wrote:

> The main idea is:
> Assumption of Cantor’s proofs: All real numbers (set R) are in a list (list L).

Actually, there is no need to assume that. Moreover Cantor never proved anything in this way.

It suffices to consider an arbitrary subset L c IR. If we assume that L is countable, we can prove that there is a number r e IR which is not in L. Case closed.

It seems to me that you do not understand indirect proofs (proofs by contradiction). Hence the approach mentioned above should be less problematic for you.

Note the differences marked with "[...]":

> This assumption means R = L,

We don't start with that assumption. (See comment above.) Hence our approach

> [doesn't make] the claim “a real number is [defined which] is not in the [set] L” wrong.

> if a [real] number is outside [of] L, it is [just] outside [of L].

> So, the statement “the [defined] real number is not in the [set] L” [just] means it is not in [the set L]

> [Now the] claim that a [real] number [that's not in L] is not [in L] is [not] absurd [but a tautology].

> Cantor’s diagonal argument [...] [shows/proves] that [there is] a real number [which] is not in [the set] L

Right.

Actually, the argument is quite simple.

[Mike Terry]

On 11/09/2022 19:48, PengKuan Em wrote:
> I do a detailed analysis of Cantor’s theory of uncountable sets. The logic of his proofs has some weaknesses. I propose an axiom and a solution to continuum hypothesis.
>
> The main idea is:
> Assumption of Cantor’s proofs: All real numbers (set R) are in a list (list L).
>
> This assumption means R=L, considering L as a set.

I think you're not really familiar with everyday set theory, where "lists", "functions",
"countability" etc. are formally defined.

A better (for mathematicians) way of saying what you try to say in your PDF is:

Set R is the set of all real numbers.
List L is [by definition of "list"] a function N: --> R, and its range is (precisely) the whole of
R. In this post I use N for set of natural numbers {1,2,3,...}

It's also ok to summarise L as something like "assume L is an enumeration of R" or "L is a list
containing all members of R", but you have to understand that they are all saying the same thing - L
is actually a function mapping N to R! (And hence L =/= R )

Sets are not a priori ordered, and L is not simply the same set as R - it has a "function"
structure. Mostly mathematicians define functions (using set theory) as sets of ordered pairs, so
function f is the set { (x, f(x)) : x in domain of f}

I understand what you mean, however. I would say R = Range(L). This is not the biggest problem
with your paper, but your paper very early on (before L is introduced) says:

R = {r_1, r_2, r_3, ...}, i in N, r_i in R, [1]

so you are already pre-assuming that R is countable! R cannot be written in this form, as shown by
Cantor's argument. (If nothing else, this alerts readers to the lack of maths background of the
author, and primes them to expect a flurry of similar basic level mistakes... i.e. although they
don't say it out loud, they're already thinking "duffer"? You only get 3 strikes and you're out! :) )

Your paper should just say R = set of real numbers, define L by one of the wordings I used, and
conclude R = Range(L).

> This makes the claim “a real number is created but is not in the list L” wrong. Indeed, if a number is outside L, it is outside R too. So, the statement “the created real number is not in the list L” means it is not in R and is not a real number, which is equivalent to claim that a real number is not real number. This is absurd but Cantor’s diagonal argument and nested intervals proof both claim that a real number is not in the list L and thus, is not a real number, which make them wrong.

This is the key problem with your paper. I think the cause is that you don't understand proof by
contradiction, but maybe it's more subtle.

Quote from paper:
1. Assumption A: All real numbers are in a list.
2. sflip is created with the diagonal construction and is a real number.
3. sflip is not in the list, contradicting thus the Assumption A.
4. Conclusion: the set of all real numbers cannot be put into a list.

So Assumption A is where we ASSUME the existence of the list L above. L is not "constructed", just
assumed to exist, i.e. we ASSUME there is a function L: N --> R with Range(L) = R.

The point is that HAVING ASSUMED assumption A, IT FOLLOWS by solid logic, that what you call sflip
IS NOT in Range(L), AND ALSO that s_flip IS in R.

Above (and in your paper) you're saying the claim is "wrong", and "absurd" - but the claim is
logically valid GIVEN THE ASSUMPTION A. There is nothing "wrong" with the /argument/ in the proof!
It has just derived a contradiction from assumption A. Therefore we conclude assumption A is false.

That is the pattern for all proof by contradiction, and it's clear in your paper you are horribly
confused by this - but it's one of the most common proof patterns. Nobody claims that it is TRUE
that sflip is both in R and not in R. The only claim is that IF ASSUMPTION A WERE TRUE, then it
would follow that sflip is both in R and not in R. But assumption A is in fact not true, so no
problem. (Unless you just don't accept proof by contradiction...)

Hmm, also you use the word "created" for sflip. sflip is /constructed/ from L, and was always in
the set R (which contains ALL real numbers), so "created" is a bad choice of words - hopefully
that's just a language issue, not a real misunderstanding.

> On the other hand, Cantor's both proofs search for contradiction. Can “this real number is not a real number” be the contradiction? No.

Yes, that's one way of realising a concrete contradiction. When we have made an inconsistent
assumption, there will be MANY possible contradiction statements that we could derive - deriving any
one of them is sufficient to reject the original assumption.

> The contradiction of the proofs is in the third step: sout “is not in the list”.

Well, it comes out the same. "sflip is not in the list" means sflip is not in Range(L) = R, i.e.
it's not a real number. And yet step 2 ensures it is in R: contradiction! Any contradiction we
can deduce from the starting assumption suffices equally well.

> By failing to create sout the second step collapses before the third step declares the contradiction.

But the second step DOESN'T FAIL. It validly CONSTRUCTS sflip from the assumed list L, and the
construction by its nature guarantees sflip is a real number. [...under the assumption A, of course...]

You just don't understand how proof by contradiction works.


Mike.

[Ross A. Finlayson]

-0 = 0

[Ross A. Finlayson]

In Cartesian functions there are everywhere-non-diagonals,
here there is only antidiagonal then that the function is only
defined while the antidiagonal is at the end, where it goes.

It's always so contrived that, ....

Assume you have an antidiagonal algorithm,
but the equivalency function already has exactly one,
an antidiagonal algorithm, and its end isn't missing.

Apply to nested intervals, no worries there,
it's a counterexample to the existence.

All, proofs by contradiction, where "the diagonal argument",
is, ..., not constructive in the manner of not being the usual
deliberative constructible, the proof by contradiction.

Funny that in a formalist's world such an intuitionist notion
is what for formalists should be all "profound".

"Hodges' hopeless" is the usual collection of unsound arguments
against uncountability, or rather for the countable.

[Mike Terry]

In the paper, there is an attempt at list construction, but it totally fails in a common mistake.

Basically, PKE constructs finite length lists of all n bit binary numbers. PKE correctly observes
that the sflip construction (on just the first n entries) constructs a number beyond the first n
entries. So far so good!

Then the "magic": [note: m=2^n, where n is the number of bits]

----- quote:
When m increases indefinitely to reach infinity, we write m=∞ and the number of bits equals
log_2(∞), see (9). In this case, the numbers f_i in Table 3 will have infinitely many bits because
log_1(∞) is the numbers of bits. Then f_i become the real numbers ri, which equal 0.b1b2b3 …, see
(10). Equation (10) is in fact (6) with n=∞.
----- endquote

so, no list is actually constructed! It's just a series of examples of finite lists - and even
these are presented in what looks like random order - followed by a "now let n=∞" suggestion.
The conclusion is also just trying to state the conclusion above for n (correct for finite n)
changing n or m to ∞ :

----- quote:
The flipped number for ri is named rflip and equals fflip with m=∞. rflip is a real number but does
not equal any number of the first list Llog2(∞), which respects Cantor’s diagonal argument. But the
list L∞ being longer than Llog2(∞), we would find rflip at the position log2(∞)+p, rflip =
rlog2(∞)+p. So, rflip belongs to the list L∞, see Table 4.
----- endquote

sort of like saying sflip IS a real number, and its not entry 1 in the list, nor entry 2,3,4,5,6,...
but it IS still there in the list at some entry beyond all the entries indexed by all those finite
numbers in N. (IIUC)

Umm, obviously there are no entries in the list beyond those indexed by one of 1,2,3,4,5.... I
don't think PKE is going to be convinced by anything anyone says here.


Mike.

[Ben Bacarisse]

Mike Terry <news.dead.p...@darjeeling.plus.com> writes:

> Above (and in your paper) you're saying the claim is "wrong", and
> "absurd" - but the claim is logically valid GIVEN THE ASSUMPTION A.
> There is nothing "wrong" with the /argument/ in the proof! It has
> just derived a contradiction from assumption A. Therefore we conclude
> assumption A is false.

Proof by contradiction seems to trip lots of people up, and it's
annoying that it's so often used for this theorem because it's absolutely
not needed. It is simple to prove that for any L: N -> R, there is an r
in R not in range(L).

Proof by contradiction goes

Exists X such that X is what we want
-> bad things happen -> ~Exists X such that X is what we want

where the direct version is just a proof that

For all X, X is not what we want

(from which "~Exists X such that X is what want" follows, but it often
unsaid.)

--
Ben.

[Mike Terry]

On 12/09/2022 14:44, Ben Bacarisse wrote:
> Mike Terry <news.dead.p...@darjeeling.plus.com> writes:
>
>> Above (and in your paper) you're saying the claim is "wrong", and
>> "absurd" - but the claim is logically valid GIVEN THE ASSUMPTION A.
>> There is nothing "wrong" with the /argument/ in the proof! It has
>> just derived a contradiction from assumption A. Therefore we conclude
>> assumption A is false.
>
> Proof by contradiction seems to trip lots of people up, and it's
> annoying that it's so often used for this theorem because it's absolutely
> not needed. It is simple to prove that for any L: N -> R, there is an r
> in R not in range(L).
>
> Proof by contradiction goes
>
> Exists X such that X is what we want
> -> bad things happen -> ~Exists X such that X is what we want
>

Just to be devil's advocate, I like this, because what we actually want to prove is

Not (exists X such that...)

and to prove a "not anything..." it is natural to assume the negative and reach a contradiction.

I know from many of your posts that you (mildly?) dislike this for some reason, or at least believe
your alternative is better, but I don't /really/ get it. (I get that you can do argue a different
way, but not why so many seem to consider that to be preferable, as though there's an unspoken
/flaw/ in proof by contradiction. TBC I've no objection to the other way of arguing.)

> where the direct version is just a proof that
>
> For all X, X is not what we want
>
> (from which "~Exists X such that X is what want" follows, but it often
> unsaid.)
>

Sure, no problem. How might we justify the unspoken final step "~Exists X such that X is what want"
from "For all X, X is not what we want"? Probably something like:

Assume Exists x such that x is what we want.
Then some individual m has the properties we want. [existential um whatnot]
Then from the hypothesis, m does not have the properties we want. [universal instantiation]
Contradiction, hence not (Exists X which is what we want)

Of course we can just say "it's a basic law of logic", but that's just a way of side-stepping the
natural semantic justification which is a proof by contradiction. Or so it seems to me (but what do
I know!?). Bottom line for me I suppose is that I consider it a total non-issue. :) [Same with HP
proof starting by assuming there exists a halt decider...] And surely proof by contradiction is
something every student HAS to get to grips with at some point? So if there's a problem, best to
confront it head on and sort it.


Mike.

[Fritz Feldhase]

On Monday, September 12, 2022 at 6:40:59 PM UTC+2, Mike Terry wrote:

> I know from many of your posts that you (mildly?) dislike this for some reason, or at least believe
> your alternative is better, but I don't /really/ get it. (I get that you can do argue a different
> way, but not why so many seem to consider that to be preferable, as though there's an unspoken
> /flaw/ in proof by contradiction.

I'd say the idea is that the direct proof is "constructive". Given any countable subset S of IR we can "construct" a real number d which is not in S. Concerning your comment above, especially see (*) below.

"A proof is nonconstructive if it asserts the existence of some object without actually constructing or finding that object. Such proofs are used freely in mainstream ("classical") mathematics. Constructivism is the practice of avoiding such proofs or at least pointing them out explicitly. Different philosophical views lead to different kinds of constructivism:

- A mathematician who accepts nonconstructive proofs may nevertheless prefer constructive ones when available, because they are more informative. Indeed, most mathematicians are part-time constructivists, at least in this respect: When we are teaching, we try to illustrate abstract concepts with concrete examples. (*)

- On the other hand, some mathematicians consider all nonconstructive proofs unacceptable, for philosophical reasons. At issue are the meaning of "existence" of an object and the meaning of "proof" of a statement."

Source: https://math.vanderbilt.edu/schectex/papers/difficult.pdf

[PengKuan Em]

Le lundi 12 septembre 2022 à 01:16:43 UTC+2, Ben Bacarisse a écrit :
> PengKuan Em <tita...@gmail.com> writes:
> > I mean that when seeing L as only the set of it members, L have the
> > same members than R.

> But no L can't "have the same members" as R. If anyone
> assumes it does, it is simply to show that assumption leads to a
> contradiction.

You think that you have the right to say “But no L can't have the same members as R”. However, you don’t because “L can't have the same members as R” is the result of Cantor's nested intervals proof and diagonal argument. In our case, we are discussing whether Cantor is right or wrong. So, we have to discuss on the ground where Cantor’s nested intervals proof and diagonal argument do not exist.

If you use “L can't have the same members as R”, then you are saying:
Because “Cantor is right”, then “L can't have the same members as R”, so “Cantor is right”. This is cycling argument which is a logical fallacy.

So, starting from the ground where Cantor does not exist, I can assume “L has the same members as R”, then derive what its consequence is. So, the derivation below is correct in logic :
Assumption: “L has the same members as R”.
r being not in L, then r is not in R because of the assumption.

> > This is what Cantor said. But if L=R, then if r (real number) is not
> > in L it is not in R.
> Yes, but L =/= R for all L. Assuming something does not make it so.

Again, “L =/= R for all L” cannot be used on the ground where we discuss.

“Assuming something does not make it so”. Yes, Assumption is only the first step of the proof by contradiction.

> > I have constructed a list L in my paper
> This is called burying the lede! You should start with this and explain
> how you think it is not possible to construct, from your L, an r in R
> such that r not in image(L). (But from what you say below, you don't
> know enough about R to understand the proof you claim to be
> undermining.)

On the ground where Cantor does not exist, on have the right to construct the list L because the set R is not proven uncountable.

> > I have cited this example in my paper:

> This is a pointless analogy because N (and specifically the subset of N
> less that 4) does not have the key property. R is closed under taking
> the least upper bound, that means that any countable list of reals can
> be used to construct a real not in the list.

Why does “R is closed under taking the least upper bound” implies “any countable list of reals can be used to construct a real not in the list”?

I put the link to my paper just in case where someone need it.
https://www.academia.edu/86410224/Examination_of_Cantors_proofs_for_uncountability_and_axiom_for_counting_infinite_sets

[PengKuan Em]

Le lundi 12 septembre 2022 à 01:16:43 UTC+2, Ben Bacarisse a écrit :
> This is a pointless analogy because N (and specifically the subset of N
> less that 4) does not have the key property. R is closed under taking
> the least upper bound, that means that any countable list of reals can
> be used to construct a real not in the list.

> --
> Ben.

Cantor’s diagonal argument is a proof by contradiction which we summarize as the 4 steps deduction below:

1. Assumption A: All real numbers are in a list.
2. sflip is created with the diagonal construction and is a real number.
3. sflip is not in the list, contradicting thus the Assumption A.
4. Conclusion: the set of all real numbers cannot be put into a list.

Under the assumption “all real numbers are in L”, L includes R
Every real number is necessarily member of L.
So, real number that is out of the list L does not exist under the assumption.
Under the assumption, real number that is out of the list L cannot be constructed.

Cantor constructs 0.(flipped bits) from a list Ld
0.(flipped bits) is a real number.
Because the bits are flipped, 0.(flipped bits) is not in the list from which 0.(flipped bits) is constructed, Ld
Is Ld the list L mentioned by the Assumption?

0.(flipped bits) is not in Ld. On the other hand, 0.(flipped bits) is a real number and thus, under the assumption, is in the list L.
Because 0.(flipped bits) is in L, 0.(flipped bits) is not in Ld:
Ld is not L.

The assumption about L is not contradicted by the result about Ld. That is, “0.(flipped bits) is not in Ld” is not against “0.(flipped bits) is in L”. As 0.(flipped bits) can be in L, Cantor’s proofs fail.

I put the link to my paper just in case where someone want to read it.
https://www.academia.edu/86410224/Examination_of_Cantors_proofs_for_uncountability_and_axiom_for_counting_infinite_sets

[PengKuan Em]

Le lundi 12 septembre 2022 à 01:45:04 UTC+2, Ross A. Finlayson a écrit :
> Can you provide two counterexamples?
>
> Define a function with domain natural integers and range [0,1],
> i.e. that is 1-1, and onto, from N to R[0,1].
>

I gave this example:
(all natural numbers with n bits from 0 up to 2^n,) /2^n, n>oo.

Which is a function N to R[0,1] , in my paper section 4.b
https://www.academia.edu/86410224/Examination_of_Cantors_proofs_for_uncountability_and_axiom_for_counting_infinite_sets

I have also given (all n-digits natural numbers) /10^n, n>oo in
https://www.academia.edu/23155464/Cardinality_of_the_set_of_decimal_numbers

>
> Then also there's apologetics to explain why Cantor's otherwise
> theorem(s) don't apply, i.e. why it's a "counterexample",
> that otherwise Cantor provides any number of "examples",
> of an element not on his list.
>

> So, Ross Finlayson wrote it.

I have just given this in this thread.

Cantor’s diagonal argument is a proof by contradiction which we summarize as the 4 steps deduction below:
1. Assumption A: All real numbers are in a list.
2. sflip is created with the diagonal construction and is a real number.
3. sflip is not in the list, contradicting thus the Assumption A.
4. Conclusion: the set of all real numbers cannot be put into a list.

Under the assumption “all real numbers are in L”, L includes R
Every real number is necessarily member of L.
So, real number that is out of the list L does not exist under the assumption.
Under the assumption, real number that is out of the list L cannot be constructed.

Cantor constructs 0.(flipped bits) from a list Ld
0.(flipped bits) is a real number.
Because the bits are flipped, 0.(flipped bits) is not in the list from which 0.(flipped bits) is constructed, Ld
Is Ld the list L mentioned by the Assumption?

0.(flipped bits) is not in Ld. On the other hand, 0.(flipped bits) is a real number and thus, under the assumption, is in the list L.
Because 0.(flipped bits) is in L, 0.(flipped bits) is not in Ld:
Ld is not L.

The assumption about L is not contradicted by the result about Ld. That is, “0.(flipped bits) is not in Ld” is not against the assumption “0.(flipped bits) is in L”. As 0.(flipped bits) can be in L, Cantor’s proofs fail.

PK

[PengKuan Em]

Le lundi 12 septembre 2022 à 02:20:31 UTC+2, Fritz Feldhase a écrit :

> On Sunday, September 11, 2022 at 8:48:42 PM UTC+2, wrote:
>
> > The main idea is:
> > Assumption of Cantor’s proofs: All real numbers (set R) are in a list (list L).
> Actually, there is no need to assume that. Moreover Cantor never proved anything in this way.
>
> It suffices to consider an arbitrary subset L c IR. If we assume that L is countable, we can prove that there is a number r e IR which is not in L. Case closed.

I have just given this in this thread

[Fritz Feldhase]

On Monday, September 12, 2022 at 7:04:28 PM UTC+2, tita...@gmail.com wrote:

> So, starting from the ground [...] I can assume “L has the same members as R”, then derive [a contradiction].

Right - almost. If L is a "list" (in mathematical terms a /sequence/) then the assumption would be that there exists a sequence (L_n)_(n e IN) of real numbers such that for all r e IR there is (exactly) an n e IN such that r = L_n.

> So, the derivation below is correct in logic :
> Assumption: “L has the same members as R”.

If you can find (construct/define) an r0 e IR such that there is no n e IN with r0 = L_n then this implies a contradiction, since from our ASSUMPTION we get that for all r e IR there is (exactly) an n e IN such that r = L_n.

> Yes, [the] assumption is only the first step of the proof by contradiction.

Indeed! Now we can derive a contradiction! See comment above.

Go figure!

[Mike Terry]

On 12/09/2022 17:52, Fritz Feldhase wrote:
> On Monday, September 12, 2022 at 6:40:59 PM UTC+2, Mike Terry wrote:
>
>> I know from many of your posts that you (mildly?) dislike this for some reason, or at least believe
>> your alternative is better, but I don't /really/ get it. (I get that you can do argue a different
>> way, but not why so many seem to consider that to be preferable, as though there's an unspoken
>> /flaw/ in proof by contradiction.
>
> I'd say the idea is that the direct proof is "constructive". Given any countable subset S of IR we can "construct" a real number d which is not in S. Concerning your comment above, especially see (*) below.
>
> "A proof is nonconstructive if it asserts the existence of some object without actually constructing or finding that object. Such proofs are used freely in mainstream ("classical") mathematics. Constructivism is the practice of avoiding such proofs or at least pointing them out explicitly. Different philosophical views lead to different kinds of constructivism:
>

So... in our case we're looking to prove that there is no X meeting our requirements; a statement of
the form ¬∃x P(x). So how would a constructive proof of this go?

I'll suggest it would be along the lines of demonstrating that a proof of P(a) leads to a
contradiction. I.e. that ASSUMING P(a) [..has been proved..] we can make further [constructively
acceptable] proof steps to reach a contradiction.

That's exactly what happens when we assume L is a list covering all of R, then use it to CONSTRUCT
the sflip number which is proved missing from L. Of course, we haven't "constructed" L which is
effectively given, but we can argue constructively from its existence to reach the required
contradiction.

Well, I may have got all the above wrong as I'm not a constructivist. I think it would be the
position taken by an intuitionist at least, but I may well have got wrong too, as it's just the
result of background reading from many years ago. [I would genuinely be interested in how such
proofs are handled by constructivists, from some reasonably definitive source rather than people
just saying what they imagine.]

Bottom line is I'm not a constructivist, and I don't think the people frowning at "proof by
contradiction" proofs, instead prefering e.g. "given any list of elements of R we can find an
element not in the list" are constructivists! So I'm doubting whether that's really what's going on
here...

> - A mathematician who accepts nonconstructive proofs may nevertheless prefer constructive ones when available, because they are more informative. Indeed, most mathematicians are part-time constructivists, at least in this respect: When we are teaching, we try to illustrate abstract concepts with concrete examples. (*)

Yes, I 100% agree with at least the sentiment behind that. Proofs can sometimes be quick, but
leaving you feeling uninformed, where a longer constructive (or more "direct") proof seems to give a
better understanding.

An example would be where we come to proofs of Halting Problem (no TM decides Halting). Books like
Linz often give the "usual" proof where H is assumed to be a halt decider, and the proof constructs
from H a new TM H_Hat and proves that H incorrectly decides the input (<H_Hat>, <H_Hat>). Of
course, we have the same issue here as to whether this kind of proof is "constructive" given that
existence of H is merely assumed within the proof as a working hypothesis, but the construction of
H_Hat from H is a concrete set of steps that can be applied to any decider H, and the proof is
constructive in that sense. So for students I'd say the proof is informative, and "convinving" in
the sense you were referring to.

In contrast, Linz also gives a quicker proof, based on a proof that there exists a recursively
enumerable language which is not decideable (its complement is not recursively enumerable). Well
this is fine, I agree, and having done the pre-req work it is shorter. But I imagine to many it
seems to lack the convincing "constructive" demonstration provided by the "usual" H/H_Hat proof. I
actually don't know whether this second proof is constructive!! (I suspect not, but if it is, there
would need to be a fair bit of careful unwinding of proofs and definitions to properly convince me.
And as I'm not a constructivist there's not much motivation for me to investigate this...)

So which HP proof is "better"? I accept both and can see different merits in each. I would expect
students get more from the "constructive" style H/H_Hat proof at least on first meeting them, but
I'm really just guessing there.

>
> - On the other hand, some mathematicians consider all nonconstructive proofs unacceptable, for philosophical reasons. At issue are the meaning of "existence" of an object and the meaning of "proof" of a statement."
>
> Source: https://math.vanderbilt.edu/schectex/papers/difficult.pdf
>

Yes, all these concerns can be quite tricky. Another one I struggle with is whether proofs are
considered "finitary" and exactly why that is. All seems slightly wooly to me! :) Perhaps it's
better to just get on with doing maths rather than talking about doing it, hehe.

Mike.

[Fritz Feldhase]

On Monday, September 12, 2022 at 8:54:20 PM UTC+2, tita...@gmail.com wrote:
> Le lundi 12 septembre 2022 à 02:20:31 UTC+2, Fritz Feldhase a écrit :
> > On Sunday, September 11, 2022 at 8:48:42 PM UTC+2, wrote:
> >
> > > The main idea is:
> > > Assumption of Cantor’s proofs: All real numbers (set R) are in a list (list L).

I SAID:

> > Actually, there is no need to assume that. Moreover Cantor never proved anything in this way.
> >
> > It suffices to consider an arbitrary subset L c IR. If we assume that L is countable, we can prove that there is a number r e IR which is not in L. Case closed.

NO PROOF BY CONTRADICTION NEEDED!

> Cantor’s diagonal argument is a proof by contradiction which we summarize as the 4 steps deduction below:

1. Assumption A: All real numbers are in a list. <bla bla bla>"

CAN'T YOU READ?!

I SAID:

(a) Actually, there is no need to assume that.

(b) Moreover Cantor never proved anything in this way

...especially NOT the uncountability of "the real numbers".

[Ben Bacarisse]

Mike Terry <news.dead.p...@darjeeling.plus.com> writes:

> On 12/09/2022 14:44, Ben Bacarisse wrote:
>> Mike Terry <news.dead.p...@darjeeling.plus.com> writes:
>>
>>> Above (and in your paper) you're saying the claim is "wrong", and
>>> "absurd" - but the claim is logically valid GIVEN THE ASSUMPTION A.
>>> There is nothing "wrong" with the /argument/ in the proof! It has
>>> just derived a contradiction from assumption A. Therefore we conclude
>>> assumption A is false.
>>
>> Proof by contradiction seems to trip lots of people up, and it's
>> annoying that it's so often used for this theorem because it's absolutely
>> not needed. It is simple to prove that for any L: N -> R, there is an r
>> in R not in range(L).
>>
>> Proof by contradiction goes
>>
>> Exists X such that X is what we want
>> -> bad things happen -> ~Exists X such that X is what we want
>
> Just to be devil's advocate, I like this, because what we actually
> want to prove is
>
> Not (exists X such that...)
>
> and to prove a "not anything..." it is natural to assume the negative
> and reach a contradiction.
>
> I know from many of your posts that you (mildly?) dislike this for
> some reason, or at least believe your alternative is better, but I
> don't /really/ get it.

It's absolutely fine for mathematicians. But a few non-mathematicians
can get into a state about it, and I used to teach non-mathematicians.

I think the notion of /assuming/ as opposed to /asserting/ might be part
of what trips some students up. Another is that the assumption itself
can play into a deep unspoken belief. If you come to CS as a
programmer, the idea of a program that can't exist is very odd. Surely,
the student imagines, every reasonable sounding specification can be
met?

Specification: determine if a context free grammar (give in, say, BNF)
is ambiguous or not. The result is 100% pinned down. No ambiguity.
How could there /not/ be an algorithm to do this?

I'm not saying it's a major problem, but when you spend a few years
re-writing a lecture course, you end up making changes based on all sort
of seemingly minor hiccups from the past.

--
Ben.

[Ben Bacarisse]

Indeed. Both of the famous proofs are of the form "for any list L, L
fails because...".

--
Ben.

[Ben Bacarisse]

PengKuan Em <tita...@gmail.com> writes:

> Le lundi 12 septembre 2022 à 01:16:43 UTC+2, Ben Bacarisse a écrit :
>> PengKuan Em <tita...@gmail.com> writes:
>> > I mean that when seeing L as only the set of it members, L have the
>> > same members than R.
>> But no L can't "have the same members" as R. If anyone
>> assumes it does, it is simply to show that assumption leads to a
>> contradiction.
>
> You think that you have the right to say “But no L can't have the same
> members as R”. However, you don’t because “L can't have the same
> members as R” is the result of Cantor's nested intervals proof and
> diagonal argument. In our case, we are discussing whether Cantor is
> right or wrong. So, we have to discuss on the ground where Cantor’s
> nested intervals proof and diagonal argument do not exist.

No we don't. You'd like to, but I don't have to. You have shown you
know very little about the real numbers and what Cantors proofs actually
are (neither uses proof by contradiction, at least no explicitly).
There is very little reason for me to do more than read the outline you
gave, note that you don't know what you are talking about, and post a
correction.

In fact I went further, but that may yet prove to be a mistake.

> If you use “L can't have the same members as R”, then you are saying:
> Because “Cantor is right”, then “L can't have the same members as R”,
> so “Cantor is right”. This is cycling argument which is a logical
> fallacy.

No it's not. If the proof is sound, I may assert it. To be explicit: I
am /not/ saying "let's assume Cantor is right, therefore you are wrong"
-- that would be circular. I am saying "Cantor is right. I and 7,815
others have checked the proof and it is sound. You are wrong". Note
that there is no "therefore". I am not making /any/ kind of logical
argument as to why you are wrong.

> So, starting from the ground where Cantor does not exist, I can assume
> “L has the same members as R”, then derive what its consequence is.

Indeed you can (barring the "detail" that L and R don't even have the
same sort if members).

> So, the derivation below is correct in logic :
> Assumption: “L has the same members as R”.
> r being not in L, then r is not in R because of the assumption.

Yes. But if, (as is the case), r is also provably in R (by the
definition or R and the construction of r), we have a contradiction.
Something has to go: r not in R by the assumption; r in R by the
construction. The only candidate is the assumption.

>> > I have cited this example in my paper:
>>
>> This is a pointless analogy because N (and specifically the subset of N
>> less that 4) does not have the key property. R is closed under taking
>> the least upper bound, that means that any countable list of reals can
>> be used to construct a real not in the list.
>
> Why does “R is closed under taking the least upper bound” implies “any
> countable list of reals can be used to construct a real not in the
> list”?

I'm sorry but this is the sort of thing you should know before wading
into a topic like this. You should be able to explain exactly how the
proof /claims/ to work, even if you think it has a flaw.

This property of R (which is equivalent to the definition using Dedekind
cuts and the definition using Cauchy sequences) is available all over
the web and in every textbook on the subject. I don't want to write ip
up for you here. But if you want to one-minute summary I'd give to a
student who stops me in the corridor it would be:

In the proof using decimal digits (one that Cantor never gave) the value
of the anti-diagonal is given by a Cauchy sequence -- a bounded
monotonic sequence of rationals. The limit so denoted is therefore a
real number, but it is also, by construction, not in range(L).

The other proofs all use a version of this property, but the details
vary.

--
Ben.

[Ben Bacarisse]

PengKuan Em <tita...@gmail.com> writes:

> Le lundi 12 septembre 2022 à 01:16:43 UTC+2, Ben Bacarisse a écrit :
>> This is a pointless analogy because N (and specifically the subset of N
>> less that 4) does not have the key property. R is closed under taking
>> the least upper bound, that means that any countable list of reals can
>> be used to construct a real not in the list.
>

> Cantor’s diagonal argument is a proof by contradiction which we
> summarize as the 4 steps deduction below:
> 1. Assumption A: All real numbers are in a list.
> 2. sflip is created with the diagonal construction and is a real number.
> 3. sflip is not in the list, contradicting thus the Assumption A.
> 4. Conclusion: the set of all real numbers cannot be put into a list.

Well, it's not Cantor's proof (either of them) but it is how the theorem
is often presented these days. Cantor's proof are not by contradiction.
(This is an over-simplification. Detail available in anyone cares.)

> Under the assumption “all real numbers are in L”, L includes R

Yes. (I hate the imprecision, but I don't want to come over as fussing
over "details" like L and R and not the same kind of thing at all.)

> Every real number is necessarily member of L.

Since that was assumed, yes.

> So, real number that is out of the list L does not exist under the
> assumption.

Yes.

> Under the assumption, real number that is out of the list L cannot be
> constructed.

Yes.

> Cantor constructs 0.(flipped bits) from a list Ld
> 0.(flipped bits) is a real number. Because the bits are flipped,
> 0.(flipped bits) is not in the list from which 0.(flipped bits) is
> constructed, Ld Is Ld the list L mentioned by the Assumption?
>
> 0.(flipped bits) is not in Ld. On the other hand, 0.(flipped bits) is
> a real number and thus, under the assumption, is in the list L.
> Because 0.(flipped bits) is in L, 0.(flipped bits) is not in Ld: Ld is
> not L.
>
> The assumption about L is not contradicted by the result about
> Ld. That is, “0.(flipped bits) is not in Ld” is not against
> “0.(flipped bits) is in L”. As 0.(flipped bits) can be in L, Cantor’s
> proofs fail.

None of this is in Cantor[1]. But other proofs do use bits, often
incorrectly. You can't just flip bits because many reals in [0,1] have
two binary representations. If you see a proof flipping bits, it's
probably wrong. There are lots of fixes, so let's not get hung up more
of those details mathematicians are so obsessed by.

Unfortunately I can't make out what your construction is. Suffice it to
say that the usual construction makes are real, r, not in range(L) for
any function L: N -> R.

By assumption, since r is a real, r is in range(L) for the assumed L,
and by construction r is /not/ in range(L). We have to ditch the
assumption.

[1] I think some people think Cantor's m/w proof is about bits, but it's
not.

--
Ben.

[Fritz Feldhase]

On Monday, September 12, 2022 at 11:26:23 PM UTC+2, Ben Bacarisse wrote:

> [1] I think some people think Cantor's m/w proof is about bits, but it's not.

Yeah.

But since you mentioned his m/w proof: Isn't his proof a real beauty? I'd tend to say _yes_.

Funny thing, that *Cantor* never proved the uncountability of the reals "this way" (i. e. by that diagonal argument). ("Funny" because the proof is "practically" credited to him.)

(I'm quite sure you know that there are some subtle points when using the diagonal argument for proving the uncountability of the reals and that these points do not arise when just considering w/m sequences.)

[Ben Bacarisse]

Fritz Feldhase <franz.fri...@gmail.com> writes:

> On Monday, September 12, 2022 at 11:26:23 PM UTC+2, Ben Bacarisse wrote:
>
>> [1] I think some people think Cantor's m/w proof is about bits, but it's not.
>
> Yeah.
>
> But since you mentioned his m/w proof: Isn't his proof a real beauty?
> I'd tend to say _yes_.

Oh, yes. It's extraordinary. Truly simple and elegant.

> Funny thing, that *Cantor* never proved the uncountability of the
> reals "this way" (i. e. by that diagonal argument). ("Funny" because
> the proof is "practically" credited to him.)

Yes, just as Turing never proved the halting theorem! (He proved
something similar that fitted his purpose -- addressing the
Entscheidungsproblem.)

--
Ben.

[Khong Dong]

On Monday, 12 September 2022 at 17:16:35 UTC-6, Ben Bacarisse wrote:
> Fritz Feldhase <franz.fri...@gmail.com> writes:
>
> > On Monday, September 12, 2022 at 11:26:23 PM UTC+2, Ben Bacarisse wrote:
> >
> >> [1] I think some people think Cantor's m/w proof is about bits, but it's not.

> > Yeah.
> >
> > But since you mentioned his m/w proof: Isn't his proof a real beauty?
> > I'd tend to say _yes_.

> Oh, yes. It's extraordinary. Truly simple and elegant.

Except it's logically _invalid_ -- just as Euclid's "proof" of the infinitude of primes!

> > Funny thing,

> Yes,

That's why there's this caveat from a SME mathematician that there is a:

"need for researchers to deactivate the thought patterns that they have installed in their brains and taken for granted for so many years".

(https://www.scientificamerican.com/article/math-mystery-shinichi-mochizuki-and-the-impenetrable-proof).

[Khong Dong]

On Monday, 12 September 2022 at 20:22:43 UTC-6, Khong Dong wrote:
> On Monday, 12 September 2022 at 17:16:35 UTC-6, Ben Bacarisse wrote:
> > Fritz Feldhase <franz.fri...@gmail.com> writes:
> >
> > > On Monday, September 12, 2022 at 11:26:23 PM UTC+2, Ben Bacarisse wrote:
> > >
> > >> [1] I think some people think Cantor's m/w proof is about bits, but it's not.
>
> > > Yeah.
> > >
> > > But since you mentioned his m/w proof: Isn't his proof a real beauty?
> > > I'd tend to say _yes_.
>
> > Oh, yes. It's extraordinary. Truly simple and elegant.

> Except it's logically _invalid_ -- just as Euclid's "proof" of the infinitude of primes!

In Euclid's case, he assumed -- one way or the other -- there are infinitely many primes to prove there can't be just a finitude of.

In Cantor's "proof", he (and we) assumed -- one way or the other [via the definition of "countable" infinity] -- that there can't be an 1-1 mapping from R to N.

Did Cantor know anything about _UPPER_ Löwenheim–Skolem theorem (https://en.wikipedia.org/wiki/L%C3%B6wenheim%E2%80%93Skolem_theorem)?

What did Cantor know about the _transcendency_ of the well-ordering of prime-numerals? About the _invalidity_ of Henkin's theory-*assumption*?

[Ben Bacarisse]

Khong Dong <khongdo...@gmail.com> writes:

> On Monday, 12 September 2022 at 20:22:43 UTC-6, Khong Dong wrote:
>> On Monday, 12 September 2022 at 17:16:35 UTC-6, Ben Bacarisse wrote:
>> > Fritz Feldhase <franz.fri...@gmail.com> writes:
>> >
>> > > On Monday, September 12, 2022 at 11:26:23 PM UTC+2, Ben Bacarisse wrote:
>> > >
>> > >> [1] I think some people think Cantor's m/w proof is about bits, but it's not.
>>
>> > > Yeah.
>> > >
>> > > But since you mentioned his m/w proof: Isn't his proof a real beauty?
>> > > I'd tend to say _yes_.
>>
>> > Oh, yes. It's extraordinary. Truly simple and elegant.
>
>> Except it's logically _invalid_ -- just as Euclid's "proof" of the
>> infinitude of primes!
>
> In Euclid's case, he assumed -- one way or the other -- there are
> infinitely many primes to prove there can't be just a finitude of.

No, he did not.

> In Cantor's "proof", he (and we) assumed -- one way or the other [via
> the definition of "countable" infinity] -- that there can't be an 1-1
> mapping from R to N.

No, he did not.

--
Ben.

[PengKuan Em]

Le lundi 12 septembre 2022 à 04:20:08 UTC+2, Mike Terry a écrit :
> On 11/09/2022 19:48, PengKuan Em wrote:
> I think you're not really familiar with everyday set theory, where "lists", "functions",
> "countability" etc. are formally defined.
>
> A better (for mathematicians) way of saying what you try to say in your PDF is:
>
> Set R is the set of all real numbers.
> List L is [by definition of "list"] a function N: --> R, and its range is (precisely) the whole of
> R. In this post I use N for set of natural numbers {1,2,3,...}

Thank you. Your reply is kind and informative. I have corrected the paper with some of your suggestions.

> Yes, that's one way of realising a concrete contradiction. When we have made an inconsistent
> assumption, there will be MANY possible contradiction statements that we could derive - deriving any
> one of them is sufficient to reject the original assumption.
> > The contradiction of the proofs is in the third step: sout “is not in the list”.
> Well, it comes out the same. "sflip is not in the list" means sflip is not in Range(L) = R, i.e.
> it's not a real number. And yet step 2 ensures it is in R: contradiction! Any contradiction we
> can deduce from the starting assumption suffices equally well.
> > By failing to create sout the second step collapses before the third step declares the contradiction.
> But the second step DOESN'T FAIL. It validly CONSTRUCTS sflip from the assumed list L, and the
> construction by its nature guarantees sflip is a real number. [...under the assumption A, of course...]
>
> You just don't understand how proof by contradiction works.
>
>
> Mike.

Let me explain my corrected reasoning.

Cantor’s Assumption A: All real numbers are in a list.
Then, R belongs to L
sout is constructed with diagonal, nested intervals or whatever others method
sout is a real number.

The Assumption A is “All real numbers are in a list” where “All real numbers” means the set R and the “list” is an infinite list of all real numbers which we name L. From the Assumption A we deduce that if r is a real number, r belongs to the list L.

Cantor claims sout is a real number. We derive from Assumption A that sout is a real number that is, sout belongs to R. Then, sout belongs to L.

In the opposite way, we derive from Assumption A that if r is not in L, r is not in R and is not a real number.
Cantor's both proofs claim that sout is not in the list L and thus, is not a real number.
If the number he constructs is not a real number, his proofs do not work within R.

Either sout is a real number and belongs to L, then there is no contradiction
Or the number he constructs is not a real number and is outside R, then no contradiction

In fact, Cantor constructs sout from a list which he did not prove to be L. Let it be Ld.
Cantor’s claim “sout is not in the list” is in reality “sout is not in Ld”
Ld is different from L the list mentioned by the Assumption A
The statement “sout is not in Ld” does not contradict that “sout is in L”

So, “sout is not in Ld” does not contradict the Assumption A and Cantor fails to prove R is uncountable.

PK

I put the link to my paper just in case where someone wants to read it.
https://www.academia.edu/86410224/Examination_of_Cantors_proofs_for_uncountability_and_axiom_for_counting_infinite_sets

[PengKuan Em]

Let me explain my corrected reasoning.

Cantor’s Assumption A: All real numbers are in a list.
Then, R belongs to L
sout is constructed with diagonal, nested intervals or whatever others method
sout is a real number.

The Assumption A is “All real numbers are in a list” where “All real numbers” means the set R and the “list” is an infinite list of all real numbers which we name L. From the Assumption A we deduce that if r is a real number, r belongs to the list L.

Cantor claims sout is a real number. We derive from Assumption A that sout is a real number that is, sout belongs to R. Then, sout belongs to L.

In the opposite way, we derive from Assumption A that if r is not in L, r is not in R and is not a real number.
Cantor's both proofs claim that sout is not in the list L and thus, is not a real number.
If the number he constructs is not a real number, his proofs do not work within R.

Either sout is a real number and belongs to L, then there is no contradiction
Or the number he constructs is not a real number and is outside R, then no contradiction

In fact, Cantor constructs sout from a list which he did not prove to be L. Let it be Ld.
Cantor’s claim “sout is not in the list” is in reality “sout is not in Ld”
Ld is different from L the list mentioned by the Assumption A
The statement “sout is not in Ld” does not contradict that “sout is in L”

So, “sout is not in Ld” does not contradict the Assumption A and Cantor fails to prove R is uncountable.

I put the link to my paper just in case where someone wants to read it.
https://www.academia.edu/86410224/Examination_of_Cantors_proofs_for_uncountability_and_axiom_for_counting_infinite_sets

PK

[Mike Terry]

On 13/09/2022 13:40, PengKuan Em wrote:
> Le lundi 12 septembre 2022 à 04:20:08 UTC+2, Mike Terry a écrit :
>> On 11/09/2022 19:48, PengKuan Em wrote:
>> I think you're not really familiar with everyday set theory, where "lists", "functions",
>> "countability" etc. are formally defined.
>>
>> A better (for mathematicians) way of saying what you try to say in your PDF is:
>>
>> Set R is the set of all real numbers.
>> List L is [by definition of "list"] a function N: --> R, and its range is (precisely) the whole of
>> R. In this post I use N for set of natural numbers {1,2,3,...}
>>

>> It's also ok to summarise L as something like "assume L is an enumeration of R" or "L is a list
>> containing all members of R", but you have to understand that they are all saying the same thing - L
>> is actually a function mapping N to R! (And hence L =/= R )
>>
>> Sets are not a priori ordered, and L is not simply the same set as R - it has a "function"
>> structure. Mostly mathematicians define functions (using set theory) as sets of ordered pairs, so
>> function f is the set { (x, f(x)) : x in domain of f}
>>
>> I understand what you mean, however. I would say R = Range(L). This is not the biggest problem
>> with your paper, but your paper very early on (before L is introduced) says:
>>
>> R = {r_1, r_2, r_3, ...}, i in N, r_i in R, [1]
>>
>> so you are already pre-assuming that R is countable! R cannot be written in this form, as shown by
>> Cantor's argument. (If nothing else, this alerts readers to the lack of maths background of the
>> author, and primes them to expect a flurry of similar basic level mistakes... i.e. although they
>> don't say it out loud, they're already thinking "duffer"? You only get 3 strikes and you're out! :) )
>>
>> Your paper should just say R = set of real numbers, define L by one of the wordings I used, and
>> conclude R = Range(L).

> Thank you. Your reply is kind and informative. I have corrected the paper with some of your suggestions.
>
>> Yes, that's one way of realising a concrete contradiction. When we have made an inconsistent
>> assumption, there will be MANY possible contradiction statements that we could derive - deriving any
>> one of them is sufficient to reject the original assumption.
>>> The contradiction of the proofs is in the third step: sout “is not in the list”.
>> Well, it comes out the same. "sflip is not in the list" means sflip is not in Range(L) = R, i.e.
>> it's not a real number. And yet step 2 ensures it is in R: contradiction! Any contradiction we
>> can deduce from the starting assumption suffices equally well.
>>> By failing to create sout the second step collapses before the third step declares the contradiction.
>> But the second step DOESN'T FAIL. It validly CONSTRUCTS sflip from the assumed list L, and the
>> construction by its nature guarantees sflip is a real number. [...under the assumption A, of course...]
>>
>> You just don't understand how proof by contradiction works.
>>
>>
>> Mike.
>

> Let me explain my corrected reasoning.
>
> Cantor’s Assumption A: All real numbers are in a list.
> Then, R belongs to L

> The number sout is constructed with diagonal, nested intervals or whatever others method

> sout is a real number.

Yes.

>
> The Assumption A is “All real numbers are in a list” where “All real numbers” means the set R and the “list” is an infinite list of all real numbers which we name L. From the Assumption A we deduce that if r is a real number, r belongs to the list L.

Yes.

>
> Cantor claims sout is a real number. We derive from Assumption A that sout is a real number that is, sout belongs to R. Then, sout belongs to L.

Yes.

>
> In the opposite way, we derive from Assumption A that if r is not in L, r is not in R and is not a real number.

Yes.

> Cantor's both proofs claim that sout is not in the list L and thus, is not a real number.

Yes. This is a contradiction, from which we conclude assumption A is false.

> If the number he constructs is not a real number, his proofs do not work within R.

Eh? Statements like "sout is a real number" are claims that WOULD hold, IF Assumption A were true.
sout is shown to be a real number, and so is within R so the proof works. (It is also shown to not
be in R - a contradiction. THIS IS HOW PROOF BY CONTRADICTION WORKS.)

>
> Either sout is a real number and belongs to L, then there is no contradiction

> Or the number he constructs is not a real number and is outside R and then, no contradiction

No. sout (IF IT EXISTED) would be both inside R and not inside R. A contradiction.

>
> In fact, Cantor constructs sout from a list which he did not prove to be L. Let it be Ld.

The list is L. That is how the proof works - it assumes the existence of L, and using that L it
constructs sout. There is no second list anywhere - you're just confused here.

Your problem is not understanding "proof by contradiction" proofs - in particular, you seem to
misunderstand the nature of statements introduced by wordings like "SUPPOSE L is a list containing
all real numbers." It turns out that in fact there is no such list, but that's ok - the proof is
inviting the reader to CONSIDER THE CONSEQUENCES that would follow IF such a list L did exist,
without claiming that L does in fact exist. (That's what the "suppose" word means. Same for other
wordings that just ask the reader to CONSIDER consequences of some statement being true.)

Just about every single maths proof in existence will use this pattern of logic at some point. So
misunderstanding it will mean even the simplest proofs are not understood.

Regards,
Mike.

[Ross A. Finlayson]

Oh, that's about what's "constructible", what's constructible is countable.

Also called "define-able", "discern-ible", running out all the finite words,
the usual problem is "not including any infinite words", or, "not including
all these infinite words".

Each infinite sequence is, ..., "define-able", by the value of the elements of
the sequence, though then in real numbers that gets to dual representation.

Here it's instead that two copies of the integers essentially make a square,
then that there's only one antidiagonal, and no other everywhere-non-diagonal,
and it so happens that the one antidiagonal is always at the end of the list.

This is due the properties of the function and linearity and so on, as a limit of
functions of integers and it's n/d for, numerator and denominator, where the
denominator goes to infinity.

Can you say exactly where in your list the antidiagonal or some everywhere-non-diagonal
is? In a thory where all functions are Cartesian, Virgil will show exists his algorithm,
that as well-defined as the sequences are, is the antidiagonal.

As far as these being "real numbers R[0,1]" it results that it's a different model
(in model theory for set theory) than the usual model "R the complete ordered
field, and a proper subset R[0,1], bounded by 0 and 1", instead it's "R these iota-values
according to function theory that fills [0,1[, has a total natural well orderingsame as
the integers, and is a continuous domain, is countable, and is unique among functions
that are 1-1 and onto a continuous domain". I.e. a different model of a continuous
domain for set theory, function theory, and what results all else the rest of mathematics,
has all the properties of a continuous domain its elements, then with regards to a
constructivist's "rather restricted transfer principle" or "Schmieden and Laugwitz,
who are constructivists and like whole things countable".

Then these days "metrizing ultrafilter", "Schwarz function support", complementary topoi,
about Vitali and Hausdorff geometers and Banach and Tarski algebraists, it just results
that after apologetics and definitions in function theory and topology, that it's very
simple again that "R[0,1] is as much a whole set clock arithmetic according to any
granularity of for example time", while, "R[0,1] is only a subset of the complete ordered
field reduced to the unit interval".

That otherwise all those things have pretty directly ways to apply arguments, ...,
"either well-defined and founded by multiple models of continuous domains,
or, inconsistent with fundamental ordinary relations, set theory".


Then, after scale in b^p and numerical precision in the algebraic according to
arithmetic coding, all words, each define-able and construct-ible, it's very simple
again and much better for mathematics in terms of that relevant foundations for
all sorts analysis have much more brief and closed derivations, that establish most
all usual definition.


You might axiomatize "there is a big infinity and confoundingly it exhausts
these regions in sequences as well as a sequence little infinity exhausts the
sequence", you might axiomatize that: but then it just results that you have
your own theory, and to say anything at all in terms of theories that are otherwise
totally blind to each other, in model theory it's in terms of transfer principle,
transfer of valid inference about properties, that then going about building
the theories soundly together, is what's called foundations of mathematics,
that alone are instead planks or platforms, of mathematics: here though
that involves solving that the other theory has the opposite conclusion as
a more-than-less direct, inference.

To solve the theories together I arrived at there are two models of continuous
domains at least, then it involves all technical philosophy if though to reduce
to extra-ordinary set theory: that results for that besides the (four or five, ...)
proofs of "uncountability of the reals, ...", is about the one proof "uncountability
of the powerset of integers", that to solve the paradoxes, is also an exercise
in showing that ordinals the objects fulfill showing the exponentiation as increment.

So, there are four or five "Cantor's proofs of uncountability of reals" like antidiagonal
argument and nested intervals, then variously "Cantor's proof of uncountability of
powerset", in "a set theory where all models of functions are Cartesian and not all
models of ordinals are compact", in larger theory there's a less-than-Cartesian function
in terms of its space or the support, and ordinals are ubiquitous and make order theory,
first.

Then, that it works out, "it's the same unique counterexample for all those, this
natural/unit equivalency function or after the slate, and for powerset, the modular
and clock", this then I called "sweep" so the function and principle about make it so
that both it's simple and it's foundations and it's all modern foundations.


So, ..., I provide _one_ example.

[Khong Dong]

On Tuesday, 13 September 2022 at 04:09:00 UTC-6, Ben Bacarisse wrote:
> Khong Dong <khongdo...@gmail.com> writes:
>
> > On Monday, 12 September 2022 at 20:22:43 UTC-6, Khong Dong wrote:
> >> On Monday, 12 September 2022 at 17:16:35 UTC-6, Ben Bacarisse wrote:
> >> > Fritz Feldhase <franz.fri...@gmail.com> writes:
> >> >
> >> > > On Monday, September 12, 2022 at 11:26:23 PM UTC+2, Ben Bacarisse wrote:
> >> > >
> >> > >> [1] I think some people think Cantor's m/w proof is about bits, but it's not.
> >>
> >> > > Yeah.
> >> > >
> >> > > But since you mentioned his m/w proof: Isn't his proof a real beauty?
> >> > > I'd tend to say _yes_.
> >>
> >> > Oh, yes. It's extraordinary. Truly simple and elegant.
> >
> >> Except it's logically _invalid_ -- just as Euclid's "proof" of the
> >> infinitude of primes!
> >
> > In Euclid's case, he assumed -- one way or the other -- there are
> > infinitely many primes to prove there can't be just a finitude of.

> No, he did not.

Wrong. Do you know what the _correct_ ( _FTA compliant_ ) definition of the unary-predicate "prime" is?

[PengKuan Em]

Let me explain without bits and flipping bits.

Cantor’s Assumption A: All real numbers are in a list.
Then, R belongs to L

sout is constructed with diagonal, nested intervals or whatever others method
sout is a real number.

The Assumption A is “All real numbers are in a list” where “All real numbers” means the set R and the “list” is an infinite list of all real numbers which we name L. From the Assumption A we deduce that if r is a real number, r belongs to the list L. As sout is a real number sout belongs to L.

Cantor claims sout is a real number. Then, Cantor claims that sout belongs to L.

In the opposite way, we derive from the Assumption A that if r is not in L, r is not in R and is not a real number.
Cantor's both proofs claim that sout is not in the list and thus, is not a real number. Does the number he constructs is not a real number? Do his proofs go out of R?

In fact, Cantor constructs sout from a list which he has not proved to be L. Let it be Ld. Since sout is not in the list Ld, but is in L, Ld and L do not have the same members, which shows that they are different. So, Cantor’s claim “sout is not in the list” is in reality “sout is not in Ld”, but not “sout is not in L”. In consequence, even sout is not in Ld, sout is still in L which is the Assumption A. So, the Assumption A is not contradicted by “sout is not in Ld”.

On the other hands, when sout is a real number, it belongs to L and there is no contradiction either. In both cases, sout does not contradict the Assumption A and Cantor fails to prove R is uncountable.

I put the link to my paper just in case where someone wants to read.
https://www.academia.edu/86410224/Examination_of_Cantors_proofs_for_uncountability_and_axiom_for_counting_infinite_sets

PK

[PengKuan Em]

Le mardi 13 septembre 2022 à 19:05:46 UTC+2, Ross A. Finlayson a écrit :


> Then, that it works out, "it's the same unique counterexample for all those, this
> natural/unit equivalency function or after the slate, and for powerset, the modular
> and clock", this then I called "sweep" so the function and principle about make it so
> that both it's simple and it's foundations and it's all modern foundations.
>
>
> So, ..., I provide _one_ example.

I just explain without bits and flipping bits, no binary or digital numbers that go forever. Each real number is a point on the number line. Let's go.

[Fritz Feldhase]

On Tuesday, September 13, 2022 at 10:49:51 PM UTC+2, tita...@gmail.com wrote:

> Cantor <bla bla bla>

Actually, the argument is extremely simple, elegant and easy (almost trivial).

We show: For any list (infinite sequence) of real numbers there is a real number which is not in the list.
This implies: There is no "complete" list of real nmbers (i. e. a list which contains _all_ real numbers).

No need to assume counterfactually "All real numbers are in a list."

[Ben Bacarisse]

PengKuan Em <tita...@gmail.com> writes:

> Let me explain without bits and flipping bits.

What you should be doing is asking intelligent questions that might help
you understand what other people are saying. The basic problem is that
you don't understand the proof method so you think it's wrong.

> Cantor’s Assumption A: All real numbers are in a list.

It's not Cantor's, it's yours. There's nothing wrong with a proof by
contradiction, but Cantor did not publish one that goes like this.

> Then, R belongs to L

Yes, except for the details that L and R are sets of different kinds of
objects.

> sout is constructed with diagonal, nested intervals or whatever others method
> sout is a real number.

Yes.

> The Assumption A is “All real numbers are in a list” where “All real
> numbers” means the set R and the “list” is an infinite list of all
> real numbers which we name L. From the Assumption A we deduce that if
> r is a real number, r belongs to the list L. As sout is a real number
> sout belongs to L.

Yes.

> Cantor claims sout is a real number. Then, Cantor claims that sout
> belongs to L.

Yes.

> In the opposite way, we derive from the Assumption A that if r is not
> in L, r is not in R and is not a real number.

You can do this, but there is a simpler way to get the contradiction.

> Cantor's both proofs claim that sout is not in the list and thus, is
> not a real number. Does the number he constructs is not a real number?
> Do his proofs go out of R?

Your proof, remember, not Cantor's. Cantor's proof is not by
contradiction. His argument is much simpler: from any list of reals
(not necessarily a full list -- any list), a real can be constructed not
in that list.

But back to you version of the proof... What do you do now that you can
show these things from the assumption that L includes all of R:

(a) r is in R
(b) r is in L
(c) r is not in L (by construction)
(d) r is not in R

The last step us up to you... Maybe it's time to conclude that the
assumption is untenable?

--
Ben.

[Ben Bacarisse]

Euclid did not assume what you said he assumed. If you were not a
crank, I'd ask for evidence, but you just want to chat so this reply
with do just as well as any other, yes?

--
Ben.

[Fritz Feldhase]

On Wednesday, September 14, 2022 at 3:04:35 AM UTC+2, Ben Bacarisse wrote:

> Euclid did not assume what you said he assumed.

The following might be a good starting point for people willing to learn:

"The proof above is actually quite a bit different from what Euclid wrote. [...] Below is a proof closer to that which Euclid wrote, but still using our modern concepts of numbers and proof.

Theorem. There are more primes than found in any finite list of primes.

Proof. Call the primes in our finite list p1, p2, ..., pr. Let P be any common multiple of these primes plus one (for example, P = p1p2...pr + 1). Now P is either prime or it is not. If it is prime, then P is a prime that was not in our list. If P is not prime, then it is divisible by some prime, call it p. Notice p can not be any of p1, p2, ..., pr, otherwise p would divide 1, which is impossible. So this prime p is some prime that was not in our original list. Either way, the original list was incomplete.

See David Joyce's pages for an English translation of Euclid's actual proof."

Source: https://primes.utm.edu/notes/proofs/infinite/euclids.html

[PengKuan Em]

Le mardi 13 septembre 2022 à 16:56:06 UTC+2, Mike Terry a écrit :
> On 13/09/2022 13:40, PengKuan Em wrote:
> >
> > In fact, Cantor constructs sout from a list which he did not prove to be L. Let it be Ld.
> The list is L. That is how the proof works - it assumes the existence of L, and using that L it
> constructs sout. There is no second list anywhere - you're just confused here.
>

> Mike.

In Cantor’s proofs there are two lists.
Cantor’s Assumption A: All real numbers are in a list. This is list L.
sout is constructed from a list, let it be Ld. sout does not belongs to Ld.
Is Ld the same as L?
R belongs to L, so sout belongs to L. But sout does not belongs to Ld. So, Ld and L do not have the same members. Ld and L are different lists.
We know that sout is out of Ld by construction.
But is sout out of L? No, because R belongs to L and sout is a real number.

[Mike Terry]

On 14/09/2022 20:09, PengKuan Em wrote:
> Le mardi 13 septembre 2022 à 16:56:06 UTC+2, Mike Terry a écrit :
>> On 13/09/2022 13:40, PengKuan Em wrote:
>>>
>>> In fact, Cantor constructs sout from a list which he did not prove to be L. Let it be Ld.
>> The list is L. That is how the proof works - it assumes the existence of L, and using that L it
>> constructs sout. There is no second list anywhere - you're just confused here.
>>
>> Mike.
>
> In Cantor’s proofs there are two lists.
> Cantor’s Assumption A: All real numbers are in a list. This is list L.

Yes.

> sout is constructed from a list, let it be Ld. sout does not belongs to Ld.
> Is Ld the same as L?

Yes.

Dude, there's no mention anywhere of a second list of real numbers - you're just imagining it somehow!

To construct sout, we utilise:
digit 1 of entry 1 OF list *L* [not entry 1 of list Ld!]
digit 2 of entry 2 OF list *L* [not entry 2 of list Ld!]
digit 3 of entry 3 OF list *L* [not entry 3 of list Ld!]
...
The digits are flipped and used to create the new number sout.

IE we have constructed sout from the list *L*.

> R belongs to L, so sout belongs to L. But sout does not belongs to Ld.

There is no Ld. sout belongs to L, because sout is real, and all real numbers are in L. Also sout
does not belong to L, by construction.

THIS IS THE CONTRADICTION which leads to the conclusion that assumption A is incorrect. Proof by
contradiction.

> So, Ld and L do not have the same members. Ld and L are different lists.
> We know that sout is out of Ld by construction.
> But is sout out of L? No, because R belongs to L and sout is a real number.

Um, YES, because by construction sout is not L(1), since its first digit differs from L(1). Also
sout is not L(2), since its seconde digit differs. And so on: sout is not equal to L(n) for ANY n.

In other words we have proved sout is NOT IN L.

You seem to be stuck on the fact that we can also prove that sout IS in L. We started the proof
with "SUPPOSE there exists a list L containing every real number...". Note this is not saying such
a list L actually exists - at this point in the proof, maybe it does or maybe it does not...

*IF* such an L really does exist, there would be two unavoidable consequences:
- there would exist a real number (sout) which is in L
- also that same number (sout) is NOT in L.

Yes, you can say that's "absurd" or whatever - maths people calmly call it a contradiction. sout
cannot be both IN L and NOT IN L. The assumption that led inevitably to this contradiction was that
such an L exists. Therefore, that assumption is shown to be false - there is no such L. Simples!

(Simple for most people, but there's no point in my explaining this over and over, so it will be the
last time I explain it.)


Regards,
Mike.

[PengKuan Em]

Le mercredi 14 septembre 2022 à 21:56:48 UTC+2, Mike Terry a écrit :
> (Simple for most people, but there's no point in my explaining this over and over, so it will be the
> last time I explain it.)
>
> Regards,
> Mike.

Thank you for your kindness to explain so much.

> Yes, you can say that's "absurd" or whatever - maths people calmly call it a contradiction.

Please do not feel offended by the word “absurd”. I’m French and have learned to use this word to refer to contradiction in maths.

> To construct sout, we utilise:
> digit 1 of entry 1 OF list *L* [not entry 1 of list Ld!]
> digit 2 of entry 2 OF list *L* [not entry 2 of list Ld!]
> digit 3 of entry 3 OF list *L* [not entry 3 of list Ld!]
> ...
> The digits are flipped and used to create the new number sout.
>
> IE we have constructed sout from the list *L*.

Yes. The entries 1, 2, 3 … are from the list *L*.
By doing the construction from 1 to n, you construct sout(n) from the entries 1st to nth.

> The digits are flipped and used to create the new number sout.
>
> IE we have constructed sout from the list *L*.

Are you sure to have constructed sout from the whole list *L*?

> because by construction sout is not L(1), since its first digit differs from L(1). Also
> sout is not L(2), since its seconde digit differs. And so on: sout is not equal to L(n) for ANY n.

So, for you “sout is not equal to L(n) for ANY n” proves that sout is not in the whole list *L*.
Maybe you will say, when n=infinity, sout is out of the whole list L.
But I have the feel that maths people do not like n=infinity.

Let us do the construction with n first entries from L and we consider the entries n+1st to the 2*nth.
The sout(n) is constructed from the n first entries from L, but there are n entries that are not use for the construction: the entries n+1st to the 2*nth.
Then n increases. When n=infinity, “sout is not equal to L(n) for ANY n”, but sout(n) can be in the part of entries of L from infinity to 2*infinity. You can never reach the end of L.

May be you will think “end of L”is a stupid idea because L does not have end.
Correct. L does not have end. In this case, the n first entries from L will be the list Ld(n). Even n=infinity, the list Ld(infinity) does not equal L because L does not have end.
We have well 2 lists: Ld(infinity), used to construct sout; L that includes the whole R.

> In other words we have proved sout is NOT IN L.

You have constructed sout(n) that does not belong to Ld(n), even for n=infinity. But you have not “proved sout is NOT IN L” because L does not have end.

> Dude, there's no mention anywhere of a second list of real numbers - you're just imagining it somehow!

Is the list Ld(infinity) a second list in the real numbers? Does Ld(infinity)=L?

> There is no Ld. sout belongs to L, because sout is real, and all real numbers are in L. Also sout
> does not belong to L, by construction.

“Also sout does not belong to L, by construction.” Are you sure?
Does the construction from 1st to infinity th entries used the whole R?

[Ross A. Finlayson]

I agree you make all the finite initial segments, but, these are all infinite sequences.

So, besides a) showing the existence and value of 1-1 and onto, then also b) applying
Cantor's theorems to your example and showing it fall out or through, that's what you
need, according to the burden of proof.

I had a pretty good mathematics education then when I started researching foundations
for continuity, I only learned set theory five years already after "Calc III", say.


Then I studied it and got a mathematics degree, but really since I started studying
foundations of continuity, the first example (or, counterexample, ...), of a function
in integers of a continuous domain: is this sweep function.

Then I had to learn all the set theory and model theory and geometry and part
theory and modern set theory and ZFC, all about Dedekind after Cauchy/Weierstrass,
fundamental theorem of calculus its derivation, to explain why it's so.

Reading Boyce or Motz and Weaver, also helps learning all mathematics.

Then, it helps that it's also fundamental for continuum analysis, meaning
also all about geometry and Vitali and Hausdorff, and topology, all the way
up to metrization after measure, signal analysis and reconstruction in frequency
analysis often (waves, signals), helps to make a fundamental geometry,
matches all the great tenets and goals of foundation, that you know.

So, I also agree that for any finite sequences, there are b^p of them,
for base b and precision p, and that sometimes b^2 many are enough,
about the quadratic and exponential, besides that in the infinite case,
there are two forms of the exponentiation arithmetic in the space of values,
that are trichotomous, the cardinal and the ordinal arithmetic, that most
people's usual estimate of the arithmetic in the unbounded is the ordinal,
but according to theory that's yet small to the cardinal, how many of things
there are in terms of real numbers or words in a formal language.

[Ross A. Finlayson]

Euclid was more than one guy, like the Bourbaki of its day.

[Ross A. Finlayson]

(Excuse me I meant Carl Boyer's history of the calculus.)

[Mike Terry]

On 15/09/2022 15:22, PengKuan Em wrote:
> Le mercredi 14 septembre 2022 à 21:56:48 UTC+2, Mike Terry a écrit :
>> (Simple for most people, but there's no point in my explaining this over and over, so it will be the
>> last time I explain it.)
>>
>> Regards,
>> Mike.
>
> Thank you for your kindness to explain so much.
>
>> Yes, you can say that's "absurd" or whatever - maths people calmly call it a contradiction.
>
> Please do not feel offended by the word “absurd”. I’m French and have learned to use this word to refer to contradiction in maths.
>
>> To construct sout, we utilise:
>> digit 1 of entry 1 OF list *L* [not entry 1 of list Ld!]
>> digit 2 of entry 2 OF list *L* [not entry 2 of list Ld!]
>> digit 3 of entry 3 OF list *L* [not entry 3 of list Ld!]
>> ...
>> The digits are flipped and used to create the new number sout.
>>
>> IE we have constructed sout from the list *L*.
>
> Yes. The entries 1, 2, 3 … are from the list *L*.
> By doing the construction from 1 to n, you construct sout(n) from the entries 1st to nth.
>
>> The digits are flipped and used to create the new number sout.
>>
>> IE we have constructed sout from the list *L*.
>
> Are you sure to have constructed sout from the whole list *L*?

Yes.

The entries of L are indexed by the natural numbers 1, 2, 3, .... Every entry in L has been used as
input to the selection of digits for sout.

For example, entry 7843 in L has been used in the selection of sout's 7843rd digit.

There are no entries in L other than those indexed by natural numbers 1,2,3,...,7843,...

>
>> because by construction sout is not L(1), since its first digit differs from L(1). Also
>> sout is not L(2), since its seconde digit differs. And so on: sout is not equal to L(n) for ANY n.
>
> So, for you “sout is not equal to L(n) for ANY n” proves that sout is not in the whole list *L*.

Of course. That's what "list" means, mathematically, and specifically for the purposes of the proof.

The list has entries indexed 1,2,3,... and so any number x which is in the list has one of those
numbers for its index in the list. If none of those numbers indexes x, then x is not in the list.

> Maybe you will say, when n=infinity, sout is out of the whole list L.

No. n is a natural number which never equals infinity.

> But I have the feel that maths people do not like n=infinity.

That's true - because n only takes (finite) natural number values 1,2,3,... and infinity is not a
natural number. Maths people do not like mistakes.

>
> Let us do the construction with n first entries from L and we consider the entries n+1st to the 2*nth.
> The sout(n) is constructed from the n first entries from L, but there are n entries that are not use for the construction: the entries n+1st to the 2*nth.
> Then n increases. When n=infinity, “sout is not equal to L(n) for ANY n”, but sout(n) can be in the part of entries of L from infinity to 2*infinity. You can never reach the end of L.

There are no entries in L beyond those indexed 1,2,3,...
There is no entry indexed by infinity, or infinity+7, or 2*infinity etc.

THAT'S JUST THE DEFINITION OF "LIST" USED IN THE PROOF.

You can invent your own definition for "list", but to avoid confusion you should choose a new name
for it, such as "PK-List". That's fine, and maybe with your definition it will be possible for a
PK-List to contain every real. There are certainly mathematical objects which extend the definition
of list in a reasonable way, and are able to include every real number!

But "Cantor's" proof which we're discussing uses the definition of list which means every entry is
indexed by a FINITE NATURAL NUMBER. With THAT definition of list, the proof is sound.

It's true that you can never reach the end of L. The sequence 1,2,3,... has no last entry, but all
entries are finite.

>
> May be you will think “end of L”is a stupid idea because L does not have end.
> Correct. L does not have end. In this case, the n first entries from L will be the list Ld(n). Even n=infinity, the list Ld(infinity) does not equal L because L does not have end.

This begs the question of what you actually mean by Ld(infinity). You have made clear what you mean
by L(1), L(2), L(3), ... : they are all the finite "prefix" sub-lists of L. For your Ld(infinity)
all that is left for it to mean is the whole list L. Or do you mean something else?

> We have well 2 lists: Ld(infinity), used to construct sout; L that includes the whole R.

The proof does not make any mention of Ld(infinity). There is just the one list L, which was used
to construct sout.

>
>> In other words we have proved sout is NOT IN L.
>
> You have constructed sout(n) that does not belong to Ld(n), even for n=infinity. But you have not “proved sout is NOT IN L” because L does not have end.

No, sout is not in L, because none of the entries of L is equal to sout.

>
>> Dude, there's no mention anywhere of a second list of real numbers - you're just imagining it somehow!
>
> Is the list Ld(infinity) a second list in the real numbers? Does Ld(infinity)=L?

You are the person who introduced Ld(infinity) so you should decide what it means! Since you have
included a word "infinity" in the term, you might also want to explain what your "infinity" means,
and what the notation "L(x)" means generally for a given x! You have explained:

Ld(1) means "L truncated after entry 1"
Ld(2) means "L truncated after entry 2"
Ld(3) means "L truncated after entry 3"
...

All this suggests L(x) requires x to be a natural number, but then you say Ld(infinity)! Presumably
that /should/ mean:
Ld(infinity) means "L truncated after entry infinity"

but that is nonsense - there is no "entry infinity". All that I can see that it could mean is
Ld(infinity) means "L not truncated"

...which of course is just L.

By all means explain what you actually mean if you mean something else!

>
>> There is no Ld. sout belongs to L, because sout is real, and all real numbers are in L. Also sout
>> does not belong to L, by construction.
>
> “Also sout does not belong to L, by construction.” Are you sure?

Yes.

> Does the construction from 1st to infinity th entries used the whole R?

There is no infinity th entry in L, just entries indexed by the finite numbers 1,2,3...

They are assumed to contain the whole of R as the first step in the proof.

Mike.

[PengKuan Em]

Le jeudi 15 septembre 2022 à 17:40:09 UTC+2, Ross A. Finlayson a écrit :
> I agree you make all the finite initial segments, but, these are all infinite sequences.

Do you mean that there are infinite sequences/numbers in one segment? This is interesting.
Take a segment of number line. Say [0,1]. Before we use it, is it filled with all the real numbers? Or is it void?
When we use a function, like b^p, we are in fact filling it with our numbers. Between 2 numbers, is the segment void? If yes, then is the segment [0,1] continuous?
Will it be continuous if we fill it forever?
Since a number on the line has 0 length, there will never be 2 numbers that touch each other and we cannot fill the segment completely. So, there will be not continuity and the continuum does not exist.
What is continuum on a number line if there will never be 2 numbers that touch each other?

>
> I had a pretty good mathematics education then when I started researching foundations
> for continuity, I only learned set theory five years already after "Calc III", say.

I see that your mathematics education is really strong.
Do you think Cantor’s proofs about uncountability correct? Precise enough?

PK

[Ross A. Finlayson]

I tihnk we should start with Zeno before getting into Cantor.

[Ross A. Finlayson]

For Zeno are named these "paradoxes", of relative motion, distance, and time.

Cantor's paradox is "set of all sets would be its own powerset, counterexample".
This is where a universe has to exist so it's also the "set of all sets", or what Cantor
calls "domain principle".


Before trans-finite set theory there was Gottlob Frege, he about had a system of
rules of relation, which is all set theory is with one relation, but then Russell stuck
him with Russell's paradox, that though it seems hypocritical not to apply it to
other what would be trans-finite-yet-axiomatically-regular, there's, like "trans-finite
ordinary extensions of Frege" that are, ..., exactly same as ZF set theory, or,
"equi-interpretable". I.e., the entire "finite and unbounded" theory exists in
something like Frege, before adding an axiom that there's an infinite set.

Then, making the infinite set axiomatized as ordinary and regular one way,
there's though that from the universe or space, instead it's ordinary or
regular in partitions, moduli, as much as "ZF's inductive set" is regular in
increment: in fact that it's required to attach this fact to integers that otherwise
they are only ordered, not regular in increment, or uniform.

(Also "ZF's finite fragment would suffer any paradox as would finite Frege's".
This can be read as "ZF's infinity must be actually infinite and not just an inductive
set or abitrarily large size", or it's just this plain take on an inequality in words in
arithmetization.)

So, "the strength of ZF(C)", the theory, with all the models for abstract and symbolic
logic or the practice of mathematics written in it, it's as strong as Frege's and the
classical, then that its extension about the infinite, or the axiom of infinity,
could be seen as one of two things. One idea is that "the axiom of infinity
adds the infinite set to the theory", some, "expansion of comprehension",
according to otherwise the other already existing axioms in the theory. But,
when according to the existing theorems, already both ordinary and extra-ordinary
infinite sets exist and include being indiscernibiles, it works out that ZF's axiom
is really a "restriction of comprehension", that really is only reading off the "axiom
of foundation" rather "axiom of well-foundedness", "axiom of regularity",
restricting the comprehension of what would be infinite collections to ordinary
infinite collections.

It involves then geometry, numbers, real numbers, continuity, and so on, if it's
to be a "foundation", where a main goal of set theory is that it's ubiquitous and
simple and clear and people automatically attribute hte symbolic properties of
mathematical objects to labels of them as by their names as sets. Of course this
involves an "arithmetization" of all the structure mathematically, or the
"always granular, but ever fine".

Then, geometrically, it works out that the space of words is digital while the space
of geometry is analog. Then, those go together into one theory that results that
all the useful properties of mathematics get along, because "paradox", ..., only
reveals a mathematical structure where there is no paradox, combined with limited
means.

[Ross A. Finlayson]

On Thursday, September 15, 2022 at 1:36:52 PM UTC-7, tita...@gmail.com wrote:

Basically the idea here is that there's only one function from
integrs, and rather: two "copies in numerical resources",
the integers, to a unit interval of reals.

This is where mathematics has the symbolic definition that
the symbol always is immediately identical to other utterances
of the same symbol. But, actually implementing algorithms in
mathematics in terms of a space of structures that support
perspective, all this various bookkeeping is involved where much
of mathematics is about trying to find elementary and special
formulaic relations and identities, which result then addition
formula, which result a regular scale of computation.

So, this idea of "two copies of the integers" is about taking the limit,

f(n) = n/d, d->oo, n->s

You can read that's a very simple definition of a function,
that with n and d as integers, then there's a much less simple
"goes", the arrow, that "n _goes_ to d", and "d _goes_ to infinity".
The goes is the much less simple part, then here to be utterly
atomized by d the denominator, is that it's infinite or: "in the world
of all the other variables, there is only one that reach infinite values",
or "eppi si mueve" (eppur si muove).

This way then there is a careful application of the formula as holding,
in terms of what are the elements of the granular and fine, that
establish together what is smooth.

Of course, it's necessary to show two things: 1) that the function is
1-1 and onto, and 2) that Cantor's counterexamples don't apply.


Then, in, "set theory", these "line reals" and the "utterly standard
modern mathematics' complete ordered field with uncountable
irrationals and field reals", are two different sets. Then, that they
share values in the integer lattice, is where that yes between zero
and one, for that there is one and then all the rest of the numbers,
also between zero and one is a microcosm of same.

Then, that the "line reals" are "continuous, ..., and contiguous" while
the "field reals" are "continuous, ..., not contiguous", result two different
models of those structures of continuous domains. They're entirely
different "sets". The entire "world of sets" with "not just some ordinals
a model of integers but all the models of structures", they're distinct
and not interchangeable, yet, contain values in the integers that they
share in the integers, and their integer part and non-integer part.

Anyways then there are simply "restrictions of application" of the properties
of the one model of a continuous domain and the properties of another
model of a continuous domain, "rather restricted transfer principle",
but that's just a matter of care, and it also works out that in the larger
theory, both "line reals" and "field reals" work out to: "exist".


Here then it further goes on and besides just having more and better
foundations of real analysis, gets into all theory and all the requirements
of any foundations: natural axiomless deduction perhaps surprisingly
arrives via "truth is regular" that "geometry is motion".

[PengKuan Em]

Le jeudi 15 septembre 2022 à 18:36:14 UTC+2, Mike Terry a écrit :
> You are the person who introduced Ld(infinity) so you should decide what it means! Since you have
> included a word "infinity" in the term, you might also want to explain what your "infinity" means,

> Maths people do not like mistakes.

Let’s discard the list Ld and infinity to avoid confusion. Instead, I will use L(n).

> The entries of L are indexed by the natural numbers 1, 2, 3, .... Every entry in L has been used as
> input to the selection of digits for sout.

I agree that we construct sout as you said below

>> To construct sout, we utilise:
>> digit 1 of entry 1 OF list *L*

>> digit 2 of entry 2 OF list *L*

>> digit 3 of entry 3 OF list *L*

>> ...
>> The digits are flipped and used to create the new number sout.

Let L(n) be the elements 1 to n of the list L and sout(n) be the diagonally constructed real number from L(n).
sout(n) is not in L(n).

> But "Cantor's" proof which we're discussing uses the definition of list which means every entry is
> indexed by a FINITE NATURAL NUMBER. With THAT definition of list, the proof is sound.

Let’s define “Statement”.
Statement 1 : sout(n) is not in L(n).

> It's true that you can never reach the end of L. The sequence 1,2,3,... has no last entry, but all
> entries are finite.

Statement 2 :“all entries are finite” means all used n have finite value.
Statement 3 : Statement 1 is true for any finite n

> No, sout is not in L, because none of the entries of L is equal to sout.

Deduction
Condition: The sequence 1,2,3,... n, … has no last entry
Because “sout(n) is not in L(n)” is true for any finite n.
Conclusion: none of the entries of L is equal to sout. sout is not in L.

>
> Mike.

Can we abstract the deduction from Statement 1 to Conclusion into a general pattern?
1. A Statement is true for any finite n, n=1,2,3,... with no end
2. Then Statement is true for the whole set N at once

Does this method of deduction have an existing name? Let’s call it method A-Statement

What can we obtain if we apply the method A–Statement to other operations? For example:

Let s1=n+1
Let the Statement be: s1>n
Because “s1>n” is true for any finite n.
Conclusion: none natural number is bigger than s1
Then, s1 is bigger than all members of N. s1 is not in N.

I think that the weakness of the method A-Statement is that when you say “because none of the entries of L is equal to sout”, the valid statement you have is in fact “none of the entries of L(n) is equal to sout for any finite n”.

In fact, for any n, the entries from n+1 to 2*n are not used to construct sout and sout can well be one of them. Because n has no end, the entries from n+1 to 2*n exist for any n, which does not allow to generalize “none of the entries of L(n) is equal to sout” into “none of the entries of L is equal to sout”.


The link to my paper just in case where someone want to read it.
https://www.academia.edu/86410224/Examination_of_Cantors_proofs_for_uncountability_and_axiom_for_counting_infinite_sets

PK

[Mike Terry]

On 16/09/2022 23:09, PengKuan Em wrote:
> Le jeudi 15 septembre 2022 à 18:36:14 UTC+2, Mike Terry a écrit :
>> You are the person who introduced Ld(infinity) so you should decide what it means! Since you have
>> included a word "infinity" in the term, you might also want to explain what your "infinity" means,
>> Maths people do not like mistakes.
>
> Let’s discard the list Ld and infinity to avoid confusion. Instead, I will use L(n).
>
>> The entries of L are indexed by the natural numbers 1, 2, 3, .... Every entry in L has been used as
>> input to the selection of digits for sout.
>
> I agree that we construct sout as you said below
>>> To construct sout, we utilise:
>>> digit 1 of entry 1 OF list *L*
>>> digit 2 of entry 2 OF list *L*
>>> digit 3 of entry 3 OF list *L*
>>> ...
>>> The digits are flipped and used to create the new number sout.
>
> Let L(n) be the elements 1 to n of the list L and sout(n) be the diagonally constructed real number from L(n).
> sout(n) is not in L(n).

Yes, sout(n) is not in L(n), although of course it could be in L further down the list.

>
>> But "Cantor's" proof which we're discussing uses the definition of list which means every entry is
>> indexed by a FINITE NATURAL NUMBER. With THAT definition of list, the proof is sound.
>
> Let’s define “Statement”.
> Statement 1 : sout(n) is not in L(n).
>
>> It's true that you can never reach the end of L. The sequence 1,2,3,... has no last entry, but all
>> entries are finite.
>
> Statement 2 :“all entries are finite” means all used n have finite value.
> Statement 3 : Statement 1 is true for any finite n

Yes. [Since there is no question of n not being finite, normally we'd just say Statement 1 is true
for all n. Infinite n don't make sense in the statement.]

>
>> No, sout is not in L, because none of the entries of L is equal to sout.
>
> Deduction
> Condition: The sequence 1,2,3,... n, … has no last entry
> Because “sout(n) is not in L(n)” is true for any finite n.
> Conclusion: none of the entries of L is equal to sout. sout is not in L.

No, you've suddenly switched from talking about sout(n) to sout, and L(n) to L, with no connecting
justification.

It is true that none of the entries of L is equal to sout.
Conclusion: sout is not in L.

There is no need to introduce the sub-lists - that just complicates things! And if you want to
argue from statements about the sout(n), L(n), there would need to be some argument linking
properties of sout and L to sout(n) and L(n). (Maybe some kind of limit argument? I don't know,
but anyway it's your idea not mine.)

>
>>
>> Mike.
>
> Can we abstract the deduction from Statement 1 to Conclusion into a general pattern?
> 1. A Statement is true for any finite n, n=1,2,3,... with no end
> 2. Then Statement is true for the whole set N at once

No, that doesn't make sense - Statement is a property of individual natural numbers. N is the SET
of ALL natural numbers, and is out of place in Statement.

What we /can/ say is "Then Statement is true for every element of N". Well, that's not much help
because it's obviously just 1. reworded.

>
> Does this method of deduction have an existing name? Let’s call it method A-Statement

As explained above, method A-statement makes no sense.

>
> What can we obtain if we apply the method A–Statement to other operations? For example:
>
> Let s1=n+1
> Let the Statement be: s1>n
> Because “s1>n” is true for any finite n.

s1 depends in n. You seem to be going to some effort to hide this.
Anyway, the corrected (by me, above) method A-Statement now yields:

"Then n+1 > n for every element of N"

which is correct.

> Conclusion: none natural number is bigger than s1
> Then, s1 is bigger than all members of N. s1 is not in N.

I'm sure you know full well that s1 isn't a fixed number. And this has nothing to do with the
diagonal proof.

>
> I think that the weakness of the method A-Statement is that when you say “because none of the entries of L is equal to sout”, the valid statement you have is in fact “none of the entries of L(n) is equal to sout for any finite n”.

No, the valid statement is "because none of the entries of L is equal to sout". Nobody but you has
ever talked about L(n) and sout(n).

Since none of the entries of L is equal to sout, that means sout isn't in the list. But hey, the
list covers all of R : so we have a contradiction. That means the starting assumption was false,
and there is no list L containing all of R.

Or perhaps you believe that one of the entries in L CAN be equal to sout? If you do, you need to
explain what the index of that entry might be! Could it be entry 230, perhaps? What about 10^666 +
83 ? What entry, exactly?

>
> In fact, for any n, the entries from n+1 to 2*n are not used to construct sout and sout can well be one of them.

Nonsense, I explained in a previous post that ALL entries in L contribute to the construction of
sout, and sout cannot match any entry in L. [including those from n+1 to 1*n]

> Because n has no end, the entries from n+1 to 2*n exist for any n, which does not allow to generalize “none of the entries of L(n) is equal to sout” into “none of the entries of L is equal to sout”.

There is no need to talk about L(n), so this is all irrelevant.

The key point which you don't get is that sout is not equal to ANY of the entries in L. The proof
of this does NOT involve L(n) or sout(n), but is just a direct consequence of the construction of
sout from L. It is not proved by "generalising" statements about L(n) to L, regardless of whether
or not that would be justifiable!

It seems you have a problem generally with what an infinite list is, but I don't know how to explain
that in a way that you would understand...


Regards,
Mike.

[Julio Di Egidio]

On Saturday, 17 September 2022 at 02:57:10 UTC+2, Mike Terry wrote:
> On 16/09/2022 23:09, PengKuan Em wrote:

> I'm sure you know full well that s1 isn't a fixed number.

While I am sure you are too much of a retarded asshole
not to keep feeding yet another patent troll. Go figure.

Get extinguished already---

*Plonk*

Julio

[PengKuan Em]

Le samedi 17 septembre 2022 à 02:57:10 UTC+2, Mike Terry a écrit :
> No, you've suddenly switched from talking about sout(n) to sout, and L(n) to L, with no connecting
> justification.

In one of your previous posts, you said:
> To construct sout, we utilise:

> digit 1 of entry 1 OF list *L* [not entry 1 of list Ld!]
> digit 2 of entry 2 OF list *L* [not entry 2 of list Ld!]
> digit 3 of entry 3 OF list *L* [not entry 3 of list Ld!]

> ...
> The digits are flipped and used to create the new number sout.

As you write

> digit 1 of entry 1 OF list *L* [not entry 1 of list Ld!]
> digit 2 of entry 2 OF list *L* [not entry 2 of list Ld!]
> digit 3 of entry 3 OF list *L* [not entry 3 of list Ld!]
> ...

you are taking the entries one by one, flipping the bits one by one. This one by one procedure is shown by your argument later in the same post :

> because by construction sout is not L(1), since its first digit differs from L(1). Also

> sout is not L(2), since its seconde digit differs. And so on: sout is not equal to L(n) for ANY n.

The address of the post is :
https://groups.google.com/g/sci.logic/c/ztJftzO9qaU/m/haF2HPykAQAJ

So, you have to chose/flip/compare the first bit with your L(1), the second with your L(2), …, then, you arrive to your L(n). You construct sout bit by bit with one entry of L by one entry of L, as everyone does.

So, you do it one by one but not by magic with only one action for all the entries of L without passing by 1, 2, … , n, …

> sout is not L(2), since its seconde digit differs. And so on: sout is not equal to L(n) for ANY n.

Here you generalize to “ANY n”. The “And so on” is the act of generalization.

> No, you've suddenly switched from talking about sout(n) to sout, and L(n) to L, with no connecting
> justification.

So do you, as you argue with “L(1), L(2), L(n), for ANY n” in the sentences below:

> because by construction sout is not L(1), since its first digit differs from L(1). Also

> sout is not L(2), since its seconde digit differs. And so on: sout is not equal to L(n) for ANY n.

“And so on” is a switch from finite number 1, 2, … n to any n which is done “with no connecting justification.”

Then you said:
> IE we have constructed sout from the list *L*.

You have constructed sout from the list *L* not by magic at once, but one element by one element. This is where you need n, sout(n), L(n). And as I have shown above, you have used them as I did.

Sorry if my method of arguing offended you. But what I did was to show the logic of the construction of sout and the comparison of the bits of sout with those of the entries of L. I use your language because this may make you more receptive and then see the logic of one by one procedure.

The logic of one by one procedure is in the heart of one-to-one correspondence because one-to-one correspondence would be proven only after all the members are checked one by one. The check cannot be done with only one action.

> What we /can/ say is "Then Statement is true for every element of N". Well, that's not much help
> because it's obviously just 1. reworded.

“true for every element of N” is a check of each element one by one.

> > Let s1=n+1
> > Let the Statement be: s1>n
> > Because “s1>n” is true for any finite n.
> s1 depends in n. You seem to be going to some effort to hide this.

Yes, s1 depends in n. But I do not hide this. In the contrary, I want to show it clearly by using sout(n), L(n). sout(n) is an entity that depends in n. sout(n) is created at n th element, s1 the is created at n th element too. Both vary with n. This is why the logic of s1=n+1 matches that of sout(n).

> I'm sure you know full well that s1 isn't a fixed number. And this has nothing to do with the
> diagonal proof.

s1 isn't a fixed number, sout(n) isn't a fixed number either. One by one construction is the relation between s1 and the diagonal proof.

> No, the valid statement is "because none of the entries of L is equal to sout".

"none of the entries of L is equal to sout", this can be asserted only after the whole L has been checked. But this check cannot be done because L has no end and such sout cannot be constructed in one by one procedure.

> There is no need to introduce the sub-lists - that just complicates things!

You did mention sub-lists, see below.

> Nobody but you has ever talked about L(n) and sout(n).

You did talk about L(n) and sout(n) and have used them in your way to prove "none of the entries of L is equal to sout" because you said

> because by construction sout is not L(1), since its first digit differs from L(1). Also

> sout is not L(2), since its seconde digit differs. And so on: sout is not equal to L(n) for ANY n.

However, as one cannot reach the end of L, the proof of the assertion "none of the entries of L is equal to sout" cannot be done in one by one procedure.

Again, trying not to offend you, I’m not saying that you are wrong, but the logic that leads to the conclusion “none of the entries of L is equal to sout” is false, because the task “after the whole L be checked” cannot be achieved in one by one procedure.

> Or perhaps you believe that one of the entries in L CAN be equal to sout?

Yes. Because under the assumption of Cantor sout is a real number, sout is in the list L. Its index depends in the manner L is constructed. At n th step, sout(n) has an index, at n+p th step, sout(n+p) has another index.

As the construction of sout(n) is done one by one and never ends, sout(n) has an index forever but with different value.

> Nonsense, I explained in a previous post that ALL entries in L contribute to the construction of
> sout, and sout cannot match any entry in L. [including those from n+1 to 1*n]

“ALL entries in L contribute to the construction of sout”. This cannot be true because the construction cannot be completed in one by one procedure.

> The key point which you don't get is that sout is not equal to ANY of the entries in L.

“sout is not equal to ANY of the entries in L” except that the construction of sout cannot be finished in one by one procedure.

[Ross A. Finlayson]

This way for example "theories of omnipotence"
and what results "theorems of G-d" are part of
the theory, what natural symbolic objects are those things.

While it's all a science itself, ....

This is just a science - it's not really that complicated
without bringing equivocation into it.

Here the main point of platonism is "if Platonism is
real then also omnifically, omnific platonism is real".

That's always so in any simple accordance of theory
and "being wrong", for example "judged by G-d".

.So, this sort of approach to all mathematics can really
help understand the higher level apparatus, espcially
when in terms of its the lower level apparatus, what
results "foundations: lowest".

Of course foundations are still "replete", just as there
is dividing one into halves, there is adding one to one.
Here "replete" means in different elementary theories,
then also for continuity "replete" means for "complete",
that eventually foundations is both singular and all
(the theories).

Theories can be internally consistent and externally
for example "either right, or, wrong". That's the external
theory of a science of theories, for example in a theory
of stimulus or unconscious, or, as much as theory fits
into thinking, that it's usually right before and after.

Or, ..., was wrong.


Here "completeness" is much the criterion for
continuity, ..., for topology it's gaplessness,
set theory has least upper bound, this [0,1]
unit real character completeness, is what
results a length assignment "length 1.0",
just to make sure that according to the sigma algebras
and "measure theory", "measure 1.0" is most the
basis and of measure theory, the "real character"
is the "real analytical character".



So, my counterexample here is part of the "foundations".
But, by itself it's also a strong platform.


This is what I've found myself on my own direction for myself,
true foundations, foundations of truth, ..., I got into it though
defending line-drawing after trans-finite set theory. Line-drawing
itself is of course about simple, while usual calculus in differentiable
functions with "Int dx = x+c", has that the line integral
of length 1.0, as a point-set is [0,1].

So, it's a pretty good theory.

[Mike Terry]

On 17/09/2022 21:12, PengKuan Em wrote:
> Le samedi 17 septembre 2022 à 02:57:10 UTC+2, Mike Terry a écrit :
>> No, you've suddenly switched from talking about sout(n) to sout, and L(n) to L, with no connecting
>> justification.
>
> In one of your previous posts, you said:
>> To construct sout, we utilise:
>> digit 1 of entry 1 OF list *L* [not entry 1 of list Ld!]
>> digit 2 of entry 2 OF list *L* [not entry 2 of list Ld!]
>> digit 3 of entry 3 OF list *L* [not entry 3 of list Ld!]
>> ...
>> The digits are flipped and used to create the new number sout.
>
> As you write
>> digit 1 of entry 1 OF list *L* [not entry 1 of list Ld!]
>> digit 2 of entry 2 OF list *L* [not entry 2 of list Ld!]
>> digit 3 of entry 3 OF list *L* [not entry 3 of list Ld!]
>> ...
> you are taking the entries one by one, flipping the bits one by one. This one by one procedure is shown by your argument later in the same post :

Nothing I said above implies any sort of "procedure" with discrete steps that produce intermediate
results. This is not a computer program, its a mathematical proof. The definition of sout is
produced from the whole list L, without any intermediary calculation results!

What I said above is just multiple true statements. I.e. it's true that entry 1 OF list L is is
used in the construction of sout. Also it's true for entry 2, and for entry 3. None of that
implies a sequence of calculations where first one is used producing result 1, then 2 is used
producing result 2, etc..

Most of the rest of your post just continues your mistake, and I see no point responding to it.

There was a question I asked, that you only partially answered, so I'll respond to that bit...

<..snip..>

>> Or perhaps you believe that one of the entries in L CAN be equal to sout?

Ah, you've somehow lost the rest of the paragraph - here it is again:

Or perhaps you believe that one of the entries in L CAN be equal to sout? If you do, you need to
explain what the index of that entry might be! Could it be entry 230, perhaps? What about 10^666 +
83 ? What entry, exactly?


>

> Yes. Because under the assumption of Cantor sout is a real number, sout is in the list L. Its index depends in the manner L is constructed. At n th step, sout(n) has an index, at n+p th step, sout(n+p) has another index.

But we're talking about sout. YOU JUST SAID *SOUT* HAS AN INDEX IN L. (I have previously agreed
that sout(n) can appear later in the list.)

So you need to say what index in L you think SOUT might have (not sout(n)). Is it 230? Perhaps
it's 5923? I don't expect you to give a fixed numerical answer, just to say what is the nature of
this index you say exists. E.g. IS THE INDEX A FINITE NATURAL NUMBER?

Mike.

[Ben Bacarisse]

The proof does not involve any one by one procedure. Nor (as originally
presented) does it involve a contradiction. Written in a more modern
style the core of it goes like this:

For any s : N -> N -> {0, 1}, the function s'(n) = 1 - s(n)(n) is not in
the image of s, i.e. there is no k in N such that s(k) = s'.

Short and sweet. No assumption. No contradiction. No process.

Notes: A sequence is just a function of N so a sequence of bits is a
function from N to {0, 1} and a sequence of bit sequences is therefore a
function s : N -> N -> {0, 1} -- a function-valued function.

And of course we must prove that for all k in N s(k) =/= s', but that
follows trivially from the fact that s(k)(k) =/= s'(k).

--
Ben.

[Khong Dong]

On Tuesday, 13 September 2022 at 20:30:38 UTC-6, Fritz Feldhase wrote:
> On Wednesday, September 14, 2022 at 3:04:35 AM UTC+2, Ben Bacarisse wrote:
>
> > Euclid did not assume what you said he assumed.
> The following might be a good starting point for people willing to learn:
>
> "The proof above is actually quite a bit different from what Euclid wrote. [...] Below is a proof closer to that which Euclid wrote, but still using our modern concepts of numbers and proof.

You meant "using our modern concepts of [...] proof" -- like Shoenfield's concepts of proof and theorems (pg. 3-4 https://www2.karlin.mff.cuni.cz/~krajicek/shoenfield.pdf)?

> Theorem. There are more primes than found in any finite list of primes.

Theorem. There are infinitely primes.

Why don't you (or Ben) prove this theorem in Shoenfield's PA (pg. 22, 204). Can you?

(Hint: Graduate level Shoenfield's PA can't even prove the notion that 2 is a prime -- since it lacks an axiom for primes!)

> Proof. Call the primes in our finite list p1, p2, ..., pr. Let P be any common multiple of these primes plus one (for example, P = p1p2...pr + 1). Now P is either prime or it is not. If it is prime, then P is a prime that was not in our list. If P is not prime, then it is divisible by some prime, call it p. Notice p can not be any of p1, p2, ..., pr, otherwise p would divide 1, which is impossible. So this prime p is some prime that was not in our original list. Either way, the original list was incomplete.
>
> See David Joyce's pages for an English translation of Euclid's actual proof."
>
> Source: https://primes.utm.edu/notes/proofs/infinite/euclids.html

If you and Ben can't prove it in the contemporary Shoenfield's PA, forget about Euclid's "proof".

[Jeff Barnett]

On 9/18/2022 9:10 PM, Khong Dong wrote:
> On Tuesday, 13 September 2022 at 20:30:38 UTC-6, Fritz Feldhase wrote:
>> On Wednesday, September 14, 2022 at 3:04:35 AM UTC+2, Ben Bacarisse wrote:
>>
>>> Euclid did not assume what you said he assumed.
>> The following might be a good starting point for people willing to learn:
>>
>> "The proof above is actually quite a bit different from what Euclid wrote. [...] Below is a proof closer to that which Euclid wrote, but still using our modern concepts of numbers and proof.
>
> You meant "using our modern concepts of [...] proof" -- like Shoenfield's concepts of proof and theorems (pg. 3-4 https://www2.karlin.mff.cuni.cz/~krajicek/shoenfield.pdf)?
>
>
>> Theorem. There are more primes than found in any finite list of primes.
>
> Theorem. There are infinitely primes.
>
> Why don't you (or Ben) prove this theorem in Shoenfield's PA (pg. 22, 204). Can you?
>
> (Hint: Graduate level Shoenfield's PA can't even prove the notion that 2 is a prime -- since it lacks an axiom for primes!)

That's like saying set theory can not be used to develop the theories of
the natural or real numbers since it doesn't contain the definitions and
axioms that define them. That's utter nonsense.

Hint: A base theory is developed in such a way that it can be enhanced
in various ways to define and develop other theories. If one is going to
take your statement seriously (rather than just having a good chuckle),
we could say that:

PA can't even prove the existence of 2 since it doesn't contain that
definitions and axioms that define 2!

PA can't even prove the existence of 3 since it doesn't contain that
definitions and axioms that define 3!

..

So now we have a countable set of can't do properties of PA!

Is this valid reasoning? The curious want to know.

>> Proof. Call the primes in our finite list p1, p2, ..., pr. Let P be any common multiple of these primes plus one (for example, P = p1p2...pr + 1). Now P is either prime or it is not. If it is prime, then P is a prime that was not in our list. If P is not prime, then it is divisible by some prime, call it p. Notice p can not be any of p1, p2, ..., pr, otherwise p would divide 1, which is impossible. So this prime p is some prime that was not in our original list. Either way, the original list was incomplete.
>>
>> See David Joyce's pages for an English translation of Euclid's actual proof."
>>
>> Source: https://primes.utm.edu/notes/proofs/infinite/euclids.html
>
> If you and Ben can't prove it in the contemporary Shoenfield's PA, forget about Euclid's "proof".

--
Jeff Barnett

[Khong Dong]

Sorry but I didn't write the graduate level "Mathematical Logic" (https://www2.karlin.mff.cuni.cz/~krajicek/shoenfield.pdf). Shoenfield did!

[Ross A. Finlayson]

Nope, you just are missing the profound.

Axiomless natural deduction for the omnific, is that a real theory
like platonism is the same as the omnific platonism, then that
it results that just to get some counting integers, is finger-counting,
while, the theory of all of them has all the traditional apparatus
like the foundational and anti-foundational, it results that axiomatics
is only a fragment of this bigger, better theory.

(With less lies.)


Having that the integers result a lattice, for example, i.e.
not just some counting but a vector space in R, then R^2, ...,
there's no putting it back.


Anyway's you two's burbling on forms of proof including
"can't be right"and "can't be wrong", is not much relevant
to here these types issues in foundations, what thinking people
have that provide the frameworks for theories and sound theory.

And an _objective_ theory.

I never rely on "Nam" except to do Schoenfield wrong,
and pretty much "Barnett"'s only good for nothing.


You stubs down on Peano arithmetic,
there's infinite numbers here.

My favorite of Peano's is his infinitesimals,
it's about something you'd add to his integers,
to have a common distance betwen his integers
and that they're (in a ) vector space.

Where "Shoenfield is just model theory, stupid".

[PengKuan Em]

Le dimanche 18 septembre 2022 à 00:33:15 UTC+2, Mike Terry a écrit :
> So you need to say what index in L you think SOUT might have (not sout(n)). Is it 230? Perhaps
> it's 5923? I don't expect you to give a fixed numerical answer, just to say what is the nature of
> this index you say exists. E.g. IS THE INDEX A FINITE NATURAL NUMBER?
>
> Mike.

2. Construction of the list L
We write the natural numbers 1,2,3, … in column 1, see Table 1,write them in binary form in column 2, invert the bits of all the numbers of column 2, the leftmost bit becomes the rightmost bit etc, then add 0. on the left of each inverted number in the column 3 to make them smaller than 1. The column 3 is the list L and does not end. As the bits of all these numbers do not ends, each number in the list L is a real number. So, L contains all the real numbers in the interval [0, 1].

Column 1 Column 2 Column 3
0 0 0.00000000000...
1 1 0.10000000000...
2 10 0.01000000000...
3 11 0.11000000000...
4 100 0.00100000000...
5 101 0.10100000000...
6 110 0.01100000000...
7 111 0.11100000000...
⋮ ⋮ ⋮
999999999... 1111111111... 0.11111111111...
Table 1


I have put the table here for easier reading.
https://www.academia.edu/86917528/Construction_of_the_diagonal_flipped_number

3. Construction of the diagonal flipped number
We notice that the bits of the diagonal of the column 3 are all zero. Indeed, the number of the cell column 2 line n is n-1, the number of bits of this number is log2(n). As log2(n)<n, the nth bit of the nth line column 3 is always zero. Flipped 0 is 1, then the diagonal flipped number is 0.11111… which is put in the last line of the Column 3.

0.11111… does not equal any entry of the column 3 except itself. The column 3 is a list of smaller than 1 real numbers which we name L. The list in the column 3 is then in one-to-one correspondence with the set of natural numbers ℕ which is in column 1.

[PengKuan Em]

Le vendredi 16 septembre 2022 à 15:42:20 UTC+2, Ross A. Finlayson a écrit :
> I tihnk we should start with Zeno before getting into Cantor.

Ross A. Finlayson,

You have put the finger on it. Zeno and Cantor are on the same subject but 2500 years apart.

But Can you write about a point with less complex language? I feel sorry for you to see that you have written so much with so much effort, while I cannot understand what you are saying. I think I must not be the only one to think this I'm sure.

PK

[Ben Bacarisse]

PengKuan Em <tita...@gmail.com> writes:

> 2. Construction of the list L
> We write the natural numbers 1,2,3, … in column 1, see Table 1,write
> them in binary form in column 2, invert the bits of all the numbers of
> column 2, the leftmost bit becomes the rightmost bit etc, then add
> 0. on the left of each inverted number in the column 3 to make them
> smaller than 1. The column 3 is the list L and does not end. As the
> bits of all these numbers do not ends, each number in the list L is a
> real number. So, L contains all the real numbers in the interval [0,
> 1].

No. Every fraction has a finite number non-zero bits so L does not even
contain 1/3 let along sqrt(2) or pi. L contains only what are called
the diadic rationals.

> Column 1 Column 2 Column 3
> 0 0 0.00000000000...
> 1 1 0.10000000000...
> 2 10 0.01000000000...
> 3 11 0.11000000000...
> 4 100 0.00100000000...
> 5 101 0.10100000000...
> 6 110 0.01100000000...
> 7 111 0.11100000000...
> ⋮ ⋮ ⋮
> 999999999... 1111111111... 0.11111111111...

This entry does not exist as 999999999... is not in N, but that's a
detail.

--
Ben.

[Jeff Barnett]

Since you "quoted" it, I thought that you might understand it so would
be able to address my probe and elaborate. If you did not understand it,
what was your purpose in introducing it in this thread? Once again, the
curious want to know.

[Mike Terry]

On 19/09/2022 17:33, PengKuan Em wrote:
> Le dimanche 18 septembre 2022 à 00:33:15 UTC+2, Mike Terry a écrit :
>> So you need to say what index in L you think SOUT might have (not sout(n)). Is it 230? Perhaps
>> it's 5923? I don't expect you to give a fixed numerical answer, just to say what is the nature of
>> this index you say exists. E.g. IS THE INDEX A FINITE NATURAL NUMBER?
>>
>> Mike.
>
>
> 2. Construction of the list L
> We write the natural numbers 1,2,3, … in column 1, see Table 1,write them in binary form in column 2, invert the bits of all the numbers of column 2, the leftmost bit becomes the rightmost bit etc, then add 0. on the left of each inverted number in the column 3 to make them smaller than 1. The column 3 is the list L and does not end. As the bits of all these numbers do not ends, each number in the list L is a real number.

OK so far.

> So, L contains all the real numbers in the interval [0, 1].

No it doesn't. All the entries in your list L contain only a finite number of 1s. Many real
numbers have a binary representation with infinitely many 1s. E.g. 1/3 = 0.01010101010101....

>
> Column 1 Column 2 Column 3
> 0 0 0.00000000000...
> 1 1 0.10000000000...
> 2 10 0.01000000000...
> 3 11 0.11000000000...
> 4 100 0.00100000000...
> 5 101 0.10100000000...
> 6 110 0.01100000000...
> 7 111 0.11100000000...
> ⋮ ⋮ ⋮
> 999999999... 1111111111... 0.11111111111...
> Table 1
>
>
> I have put the table here for easier reading.
> https://www.academia.edu/86917528/Construction_of_the_diagonal_flipped_number
>
> 3. Construction of the diagonal flipped number
> We notice that the bits of the diagonal of the column 3 are all zero. Indeed, the number of the cell column 2 line n is n-1, the number of bits of this number is log2(n). As log2(n)<n, the nth bit of the nth line column 3 is always zero. Flipped 0 is 1, then the diagonal flipped number is 0.11111… which is put in the last line of the Column 3.
>
> 0.11111… does not equal any entry of the column 3 except itself.

Right - it's not in L, just like the proof says! L does not contain all real numbers in [0,1].

> The column 3 is a list of smaller than 1 real numbers which we name L.

First, L is already defined above, so lets call this new creation L2.

I assume you want L2 to be a list matching L, but then having one more line (starting 999999999...)
appended? This L2 IS NOT A LIST!

Remember, lists have entries, and each entry is at a FINITE index into the list. Lists do not have
entries with infinite indexes.

You could put the 999999999... line as the first line in your L2, and then have all the L lines
after it (i.e. moved down one entry). But then the diagonal is different, so the sout number is
different, and will still be missing from L2, so this approach goes nowhere.

> The list in the column 3 is then in one-to-one correspondence with the set of natural numbers ℕ which is in column 1.

Eh? If you ignore the line starting 9999999... then you are right. You have created a list L which
is missing loads of real numbers, namely all real numbers requiring infinitely many 1s in their
binary representations, including sout. No prize awarded for this :)

If you include the line starting 9999999... then you are wrong - the last line has no natural number
association. (I hope you realise 9999999... is NOT a natural number?)

Regards,
Mike.

[Ross A. Finlayson]

Sure, I have a geometry after an axiomless natural deduction that is
a geometry of points and spaces, that results that instead of axioms
for lines, ..., then about inferences of point-sets in definitions of lines,
that there are points for geometry (like Hardy's real numbers) that come
from a natural geometry that points form a space-filling curve, that defines
the entire space, that for example is a line segment, vector basis of a space R,
real numbers.



Then, there are points: "in" a line, that they are contiguous points, or,
there are points: "on" a line, that according to the field what defines
the point falls in the line where the values of formulas that in sequences
define what value exists of the convergent.

The points "in" the line are very automatically convergent and so on:
they have properties all defined by this remedial line-drawing as
a plain object of the theory.


The, "points" then in their spaces as dimensioned, dimensions, these
are the "dimensional" points, then about that according to the subspaces
and projection, it is simply the "pan-" and "poly-" dimensional, points,
about that there are points 1-sided in a line and 2-sided on a line.


Then, for some very fundamental derivations of measure and Vitali,
and analytical character of course all defined by the sides of these points
or what is aside them in spaces, I define measure theory right on top of
that, so making sure that points are geometry's, and also real numbers
for Hardy, have they're also geometry's.

[PengKuan Em]

Hi Mike,

Here is "Real numbers and points on the number line with regard to Cantor’s diagonal argument" at the thread
https://groups.google.com/g/sci.logic/c/JC3Jt7njakU/m/DBVwdEfiAwAJ

Where I explain that when we see real numbers as points on the number line, we can put a name on each point and put the names into a list without contravening Cantor’s diagonal argument because we cannot create a diagonal from a list of names.

However, we do not need such a impossible list, but just to split R into two parts, S2 and S10. the members of S2 are real numbers expressed in binary, those of S10 in decimal. We create a list of real numbers by picking one member from S2 and one member from S10 alternately and forever. This list is a composite list whose members are in binary and decimal alternately. The diagonal of this list is a sequence of binary and decimal digits alternately and out-of-the-list-number cannot be constructed from it.

In fact, composite list can be created in splitting R into many subsets in numeral systems of different bases from which no out-of-the-list-number can be created and there is no real number excluded from the composite list. Because the composite list is constructed from the whole R and no real number is found outside, the composite list contains R.

If there is one list that contains R we can already conclude that R is countable. But the permutation of the subsets of R can create a huge number of different composite lists which all contain R. So, we conclude with confidence that R is countable. Then Cantor’s diagonal argument fails.

Cantor’s diagonal argument expresses real numbers only in one numeral system, which restricts the used list. If a binary list is shown not to contain R, this can be caused either by “list” or by “binary”. Because Cantor has focused only on “list” overlooking “binary”, this is the flaw that breaks Cantor’s diagonal argument which then does not prove ℝ uncountable.

For more detail of this study please read the complete paper here:
«Real numbers and points on the number line with regard to Cantor’s diagonal argument»
https://www.academia.edu/88279926/Real_numbers_and_points_on_the_number_line_with_regard_to_Cantors_diagonal_argument

Kuan Peng

[PengKuan Em]

Hi Ross A. Finlayson,

[Ross A. Finlayson]

There's only one counterexample, it's special, then, there are
the dense rationals and signal continuity, what before and after
"field continuity", define "line continuity" and "signal continuity".