Why can no one in sci.math understand my simple point?
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From: Aatu Koskensilta on 23 Jun 2010 08:57 Tim Little <tim(a)little-possums.net> writes: > In fact it would be a lot easier if instead of reals, we talked about > sets of natural numbers, and instead of computable reals, we talked > about recursively enumerable sets of natural numbers. It would be even more easy if instead of the uncountability of the reals we talked about the standard proof of Cantor's theorem: Let f be a function taking elements of a set A to subsets of A. There is then a subset of A not in the range of f. For consider the set D = {x in A | x not in f(x)}. There is no a in A such that f(a) = D: if there were, we'd have a in f(a) iff a not in f(a), a contradiction. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: WM on 23 Jun 2010 09:10 On 23 Jun., 14:51, Sylvia Else <syl...(a)not.here.invalid> wrote: > Cantor doesn't rely on being able to identify a last digit. He's just > saying that no matter how far down the list you look, you'll find that > the element at that point doesn't match the anti-diagonal. That is wrong. Cantor uses the alleged "fact", that infinity can be completed, i.e., that that the infinite list can be finished. If he would only assume what you say, then the anti-diagonal could remain in the unknown part of the list. You know and appreciate that after *any* line number n there are infinitely many more lines? > But you can't > even begin to formulate his proof if you can't identify the first > element of the list (and hence first digit of the anti-diagonal) either. Isn't it enough, also in my case, to know that every antidiagonal has a first digit? In fact it has. Every antidiagonal is constructed from a list with a first line (that is the previous antidiagonal) and the remaining list. Every line has a finite number n. > > First and last are interchangeable, of course, but with your > construction above, you can't specify either the first or the last. As I told you, my notation is only an abbreviation for the following definition: 1) Take a list L0 of all rational numbers. 2) Construct its antidiagonal A0. 3) Add it at position 0 to get (A0,L0) 4) Construct the antidiagonal A1. 5) and so on. There occurs never a problem, because we know Hilberts hotel, don't we? Regards, WM
From: WM on 23 Jun 2010 09:24 On 23 Jun., 14:57, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > Tim Little <t...(a)little-possums.net> writes: > > In fact it would be a lot easier if instead of reals, we talked about > > sets of natural numbers, and instead of computable reals, we talked > > about recursively enumerable sets of natural numbers. > > It would be even more easy if instead of the uncountability of the reals > we talked about the standard proof of Cantor's theorem: > > Let f be a function taking elements of a set A to subsets of A. There > is then a subset of A not in the range of f. For consider the set > > D = {x in A | x not in f(x)}. > > There is no a in A such that f(a) = D: if there were, we'd have a in > f(a) iff a not in f(a), a contradiction. That could be a good proof, if we knew that all subsets of an infinite set A would exist. But already Fraenkel wrote in the third edition of his famous book (1928, p. 279f) with respect to the axiom of power set, that: der Begriff "Teilmenge eine andere, wesentlich engere Bedeutung hat als in der CANTORschen Mengenlehre. In dieser konnten wir bei der Bildung der Potenzmenge Um eine beliebige Gesamtheit von Elementen aus m zu einer Teilmenge von m zusammenfassen und waren dann sicher, daß diese sich unter den Elementen von Um findet. Jetzt ist uns eine derartige, weitgehende Freiheit gewährende "Bildung einer Teilmenge von m nicht gestattet, also auch ihr Auftreten unter den Elementen von Um keineswegs gesichert. (Contrary to the second edition of 1923, Fraenkel now knew Skolem's proof of the same year and had to explain how it could be circumvented.) Therefore, there is not every subset of an infinite set. Why then should exist the subset of A that contains its pre-image if it does not contain it, and does not contain ist, if it contains it? With no doubt this is a curious result of logic, but it has not the least to do with cardinalities of sets. Regards, WM
From: Aatu Koskensilta on 23 Jun 2010 09:25 WM <mueckenh(a)rz.fh-augsburg.de> writes: > Therefore, there is not every subset of an infinite set. What on Earth does this mean? -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: WM on 23 Jun 2010 09:35
On 23 Jun., 15:25, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > WM <mueck...(a)rz.fh-augsburg.de> writes: > > Therefore, there is not every subset of an infinite set. > > What on Earth does this mean? It is Fraenkels text. I thought that you could understand German? In 1928 Fraenkel had realized Skolems proof (if a first-order-logic theory is consistent, then it has a countable model). In order to circumvent this problem, Fraenkel (and the later interpreters of the axiom of power set) must keep open the possibility that only countably many subsets of omega may exist. Hence, not all subsets exist "automatically". Reagrds, WM |