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Re: Peer-reviewed arguments against Cantor Diagonalization
Posted:
Oct 28, 2012 11:13 PM
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On Monday, October 29, 2012 2:26:25 AM UTC+2, JRStern wrote:
> On Sun, 28 Oct 2012 15:25:27 -0700 (PDT), Arturo Magidin
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> <magidin@member.ams.org> wrote:
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> >On Saturday, October 27, 2012 4:49:32 PM UTC-5, JRStern wrote:
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> >> Are there any such published?
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> >
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> >You said elsewhere you are interested in "peer-reviewed" criticisms to
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> >Cantor's diagonal argument, but not from the point of view of intuitionism
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> >or some other logical framework, but strictly within the context of ZFC.
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> >
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> >There are no such things, because the argument is a valid argument
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> >within ZF. It is in fact pretty short and clear.
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>
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> Then there can be a book taking on the objections one at a time and
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> knocking them down.
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*** Why do you think there should be such a book? Nobody takes seriously all those clowns with their bogus claims and ""proofs"", which only debunk the idea that all the memebr of the Homo Sapiens specie are intelligent.
As you could easily see if you followed this site, and others, along the years, those cranks are the entertainment jests of the net.
By the way, there are sites in the net disclosing nicks, names and pranks of some of the most well-known and beloved of those cranks. The next one can be lots of fun and perhaps you can recognize several characters:
http://www.crank.net/maths.html *****
>
> >Recall that given a set X, the Axiom of the Power Set states that
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> >there is a set Y such that z in Y if and only if z is a subset of X.
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> >We call this set the "power set of X", an denote it P(X).
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> >
>
> >THEOREM (Cantor) Let X be any set, and let P(X) be the power set of
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> >X. If f:X->P(X) is any function, then f is not onto; that is, there
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> >exists B in P(X) that does not lie in the image of f.
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> >
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> >Proof. Let f:X->P(X) be a function. By the Axiom of Separation,
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> >
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> >B = {x in X | x is not an element of f(X)}
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> >
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> >is a subset of A, hence an element of P(X). We claim that f(y)=/=B for all y\in X.
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> >
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> >Indeed, let y in X. Either f(y)=/=B, or f(y)=B. If f(y)=B,
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> >then y in B -> y in f(y) -> y not in B; since (P->not(P))->not(P) is a tautology,
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> >we conclude that f(y)=/=B. So if y in X, then f(y)=/=B, proving that B is not
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> >in the image of f. QED
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>
>
> ... and not just telling us that there are other theories with the
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> same consequent.
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>
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> J.
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> >
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> >LEMMA 1: There is a bijection g:(0,1)->P(N), where P(N) is the power set of the natural numbers.
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> >
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> >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).
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> >
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> >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.
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> >
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> >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
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> >
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> >
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> >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.
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> >
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> >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.
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> >
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> >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).
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> >
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> >Thus, f is not onto, as claimed. QED
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