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Topic: Peer-reviewed arguments against Cantor Diagonalization
Replies: 156   Last Post: Nov 4, 2012 3:01 PM

 Messages: [ Previous | Next ]
 David C. Ullrich Posts: 21,553 Registered: 12/6/04
Re: Peer-reviewed arguments against Cantor Diagonalization
Posted: Oct 29, 2012 1:28 PM

On Sun, 28 Oct 2012 17:27:38 -0700, JRStern <JRStern@foobar.invalid>
wrote:

>On Sun, 28 Oct 2012 15:25:27 -0700 (PDT), Arturo Magidin
><magidin@member.ams.org> wrote:
>

>>On Saturday, October 27, 2012 4:49:32 PM UTC-5, JRStern wrote:
>>> Are there any such published?
>>
>>You said elsewhere you are interested in "peer-reviewed" criticisms to
>>Cantor's diagonal argument, but not from the point of view of intuitionism
>>or some other logical framework, but strictly within the context of ZFC.
>>
>>There are no such things, because the argument is a valid argument
>>within ZF. It is in fact pretty short and clear.

>
>Then there can be a book taking on the objections one at a time and
>knocking them down.

What objections? The fact that you see a lot of confusion here on
sci.math should not lead you to conclude that there are any
actual problems or anything the slightest bit controversial

That book would be like a book knocking down the objections
to the idea that the Earth is round. Except much sillier.

>>Recall that given a set X, the Axiom of the Power Set states that
>>there is a set Y such that z in Y if and only if z is a subset of X.
>>We call this set the "power set of X", an denote it P(X).
>>
>>THEOREM (Cantor) Let X be any set, and let P(X) be the power set of
>>X. If f:X->P(X) is any function, then f is not onto; that is, there
>>exists B in P(X) that does not lie in the image of f.
>>
>>Proof. Let f:X->P(X) be a function. By the Axiom of Separation,
>>
>>B = {x in X | x is not an element of f(X)}
>>
>>is a subset of A, hence an element of P(X). We claim that f(y)=/=B for all y\in X.
>>
>>Indeed, let y in X. Either f(y)=/=B, or f(y)=B. If f(y)=B,
>>then y in B -> y in f(y) -> y not in B; since (P->not(P))->not(P) is a tautology,
>>we conclude that f(y)=/=B. So if y in X, then f(y)=/=B, proving that B is not
>>in the image of f. QED

>
>... and not just telling us that there are other theories with the
>same consequent.
>
>J.
>
>

>>
>>LEMMA 1: There is a bijection g:(0,1)->P(N), where P(N) is the power set of the natural numbers.
>>
>>Proof: There is an injection (0,1) to P(N) as follows: given any number x in (0,1), we can express x in base 2; if there are two expressions for x in base 2, then select the one with finitely many 1s. Map the number x = 0.a_1a_2a_3... to the set S = {n in N | a_n=1}. The map is easily seen to be one-to-one (given that we have specified which expansion to use).
>>
>>And there is an injection from P(N) to (0,1): given a subset X of N, let x be the real number Sum (a_n/10^n), where a_n=5 if n is not in X, and a_n=6 if n is in X. Again, this is easily seen to be an injection.
>>
>>Since we have an injection (0,1)->P(N), and an injection P(N)->(0,1), the Cantor-Bernstein-Schroeder Theorem guarantees (in ZF) the existence of a bijection g:(0,1)->P(N). QED
>>
>>
>>COROLLARY: If f:N->R is any function, where N is the natural numbers and R is the real numbers, then f is not onto.
>>
>>Proof: First, there is a bijection h between R and (-pi/2,pi/2), given by the arctan function; and there is a bijection t between (-pi/2,pi/2) and (0,1), given by t(x) = (x+(pi/2))/pi. And from Lemma 1, we have a bijection g from (0,1) to P(N). Note that the compositum gth:R->P(N) is a bijection.
>>
>>Let f:N->R be any function. Then the compositum gthf is a function N->P(N). By Cantor's Theorem, this function is not onto. In particular, there exists a subset X of P(N) that is not in the image of gthf. Now let r = (gth)^{-1}(X) (the function (gth)^{-1} exists because gth is a bijection); then r is not in the image of f (for if r = f(n), then (gth)^{-1}(X) = f(n), so X = (gth)((gth)^{-1}(X)) = gth(f(n)) = gthf(n), which contradicts the choice of X).
>>
>>Thus, f is not onto, as claimed. QED

Date Subject Author
10/27/12 JRStern
10/27/12 Frederick Williams
10/27/12 JRStern
10/28/12 Tim Little
10/28/12 Peter Webb
10/28/12 JRStern
10/28/12 J. Antonio Perez M.
10/29/12 Michael Stemper
10/29/12 David C. Ullrich
10/29/12 JRStern
10/29/12 JRStern
10/30/12 David C. Ullrich
10/30/12 David C. Ullrich
10/29/12 LudovicoVan
10/27/12 J. Antonio Perez M.
10/28/12 INFINITY POWER
10/28/12 Shmuel (Seymour J.) Metz
10/28/12 Graham Cooper
10/28/12 magidin@math.berkeley.edu
10/28/12 INFINITY POWER
10/28/12 Graham Cooper
10/28/12 Virgil
10/28/12 Graham Cooper
10/28/12 JRStern
10/28/12 magidin@math.berkeley.edu
10/28/12 Graham Cooper
10/28/12 Graham Cooper
10/28/12 magidin@math.berkeley.edu
10/28/12 Graham Cooper
10/28/12 magidin@math.berkeley.edu
10/29/12 Frederick Williams
10/29/12 LudovicoVan
10/29/12 Richard Tobin
10/29/12 LudovicoVan
10/29/12 J. Antonio Perez M.
10/29/12 J. Antonio Perez M.
10/29/12 magidin@math.berkeley.edu
10/29/12 Frederick Williams
10/28/12 Graham Cooper
10/28/12 JRStern
10/28/12 Peter Webb
10/28/12 magidin@math.berkeley.edu
10/28/12 Hercules ofZeus
10/28/12 magidin@math.berkeley.edu
10/28/12 Hercules ofZeus
10/28/12 Arturo Magidin
10/28/12 Hercules ofZeus
10/28/12 Arturo Magidin
10/28/12 Hercules ofZeus
10/29/12 J. Antonio Perez M.
10/29/12 Graham Cooper
10/29/12 JRStern
10/29/12 Frederick Williams
10/29/12 magidin@math.berkeley.edu
10/29/12 Pubkeybreaker
10/29/12 JRStern
10/29/12 magidin@math.berkeley.edu
10/29/12 Graham Cooper
10/29/12 JRStern
10/29/12 magidin@math.berkeley.edu
10/30/12 Graham Cooper
10/30/12 magidin@math.berkeley.edu
10/30/12 Graham Cooper
10/30/12 J. Antonio Perez M.
10/31/12 Peter Webb
10/29/12 Peter Webb
10/29/12 magidin@math.berkeley.edu
10/29/12 Shmuel (Seymour J.) Metz
10/29/12 Virgil
10/29/12 Graham Cooper
10/29/12 Peter Webb
10/29/12 Graham Cooper
10/29/12 Peter Webb
10/29/12 Graham Cooper
10/29/12 Peter Webb
10/29/12 Graham Cooper
10/29/12 Peter Webb
10/30/12 Graham Cooper
10/30/12 Peter Webb
10/30/12 Graham Cooper
10/30/12 Peter Webb
10/30/12 Graham Cooper
10/30/12 MoeBlee
10/30/12 Peter Webb
10/30/12 Graham Cooper
10/31/12 Peter Webb
10/31/12 Graham Cooper
10/31/12 Peter Webb
10/31/12 Graham Cooper
11/2/12 Peter Webb
11/1/12 Peter Webb
11/1/12 Hercules ofZeus
11/2/12 Peter Webb
10/31/12 Peter Webb
10/30/12 Richard Tobin
10/30/12 Graham Cooper
10/30/12 MoeBlee
10/30/12 Graham Cooper
10/30/12 MoeBlee
10/30/12 Graham Cooper
10/31/12 Richard Tobin
10/31/12 Graham Cooper
10/31/12 Graham Cooper
10/31/12 Graham Cooper
10/31/12 Richard Tobin
10/31/12 Graham Cooper
10/31/12 Richard Tobin
10/31/12 Graham Cooper
10/31/12 Graham Cooper
10/29/12 Peter Webb
10/29/12 LudovicoVan
10/29/12 J. Antonio Perez M.
10/29/12 LudovicoVan
10/29/12 Peter Webb
10/29/12 LudovicoVan
10/30/12 Peter Webb
10/30/12 LudovicoVan
10/30/12 Peter Webb
10/30/12 LudovicoVan
10/30/12 Peter Webb
10/30/12 J. Antonio Perez M.
10/30/12 LudovicoVan
10/30/12 Virgil
10/30/12 J. Antonio Perez M.
10/30/12 LudovicoVan
10/31/12 Peter Webb
10/31/12 LudovicoVan
10/31/12 J. Antonio Perez M.
10/31/12 Shmuel (Seymour J.) Metz
11/1/12 LudovicoVan
11/1/12 Jesse F. Hughes
11/2/12 Peter Webb
11/2/12 LudovicoVan
11/2/12 Graham Cooper
11/2/12 Shmuel (Seymour J.) Metz
11/2/12 Virgil
11/2/12 Peter Webb
11/2/12 Peter Webb
11/3/12 J. Antonio Perez M.
11/3/12 Shmuel (Seymour J.) Metz
11/4/12 Peter Webb
11/4/12 Virgil
11/4/12 Shmuel (Seymour J.) Metz
11/1/12 Shmuel (Seymour J.) Metz
10/31/12 Shmuel (Seymour J.) Metz
10/31/12 Shmuel (Seymour J.) Metz
10/29/12 JRStern
10/29/12 Shmuel (Seymour J.) Metz
10/28/12 Graham Cooper
10/28/12 J. Antonio Perez M.
10/29/12 David C. Ullrich
10/29/12 JRStern
10/29/12 LudovicoVan
10/29/12 Fernando Revilla
10/29/12 Shmuel (Seymour J.) Metz