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From: David Kastrup on 19 Oct 2005 08:44 albstorz(a)gmx.de writes: > William Hughes wrote: >> albstorz(a)gmx.de wrote: >> >> <snip> >> >> > >> > If we accept the uncountability as a form of infinity, this leads to >> > the paradoxon that the natural numbers are not countable. >> >> No, the natural numbers are countable precisely because >> they do count themselves. > > This is exact my argument: there are uncountable many natural numbers > (nothing other means infinity many) Whining does not make it so. "countable" has a precise definition in mathematics, and sets that can be placed into bijection with natural numbers are both infinite and countable. > but the natural numbers are shurely countable since they count > themself. You are not able to recognise a paradoxon if you see it. There is no paradoxon. You just have to read the definition of "countable" instead of making up your own and then complaining that it does not correspond to the established one. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: albstorz on 19 Oct 2005 08:45 David Kastrup wrote: > albstorz(a)gmx.de writes: > > > Virgil wrote: > >> In article <1129684276.251366.121150(a)g47g2000cwa.googlegroups.com>, > >> albstorz(a)gmx.de wrote: > >> > >> > David R Tribble wrote: > >> > > >> > > > >> > > Because I can prove it (and it's a very old proof). A powerset of > >> > > a nonempty set contains more elements that the set. Can you prove > >> > > otherwise? > >> > > >> > This argument is stupid. Is there any magic in the powerfunction? > >> > >> "Proofs" are not stupid until they can be refuted. The proof that for an > >> arbitrary set S, Card(S) < Card(P(S)) has not been refuted by anyone. > > > > > > Even if you think that the powersets of finite and infinite sets > > have both a greater cardinality than their starting sets, you would > > not really think it depends on the same cause in both cases. > > > > You must proof it independently for finite and for infinite sets. In > > this sense the argument is stupid. > > Uh, no. The proof depends merely on the fact that some value has to > be either a member of a set, or not. And if some value is in one set, > bur not another, then those two sets are different. > > That's all. Finiteness or infiniteness does not even play into it. > The proof just constructs a set which differs by the membership of at > least one particular value with every target set in the assumedly > complete mapping of set to powerset. > > -- > David Kastrup, Kriemhildstr. 15, 44793 Bochum I don't know what you are talking about. The proof for finite sets needs just a complete induction. This will not hold for infinity I think. Regards AS
From: William Hughes on 19 Oct 2005 09:15 albstorz(a)gmx.de wrote: > William Hughes wrote: > > albstorz(a)gmx.de wrote: > > > > <snip> > > > > > > > > If we accept the uncountability as a form of infinity, this leads to > > > the paradoxon that the natural numbers are not countable. > > > > No, the natural numbers are countable precisely because > > they do count themselves. > > This is exact my argument: there are uncountable many natural numbers > (nothing other means infinity many) This is silly. For a set to be infinite does not mean that it is uncountable (both "infinite" and "countable have precise defintions, and they do not contradict one another). A set is infinite if there is a bijection between the set and a proper subset of itself. A set is countable if there is a bijection between the set and a subset of the natural numbers. Note that using these definitions the set of natural numbers is both infinite and countable. <snip> Try reareading this next bit > > The fact that there is no > > natural number that repsresents this "count" is not a paradox > > because the "count" is defined in terms of bijections. [You > > may not like the use of the terms "count" and "countable" > > because you think they should imply something different. So > > be it. However, you cannot say "you are using a term which > > I think should mean somthing different, so you must mean > > not what you mean but what I mean"] - William Hughes
From: David Kastrup on 19 Oct 2005 09:21 albstorz(a)gmx.de writes: > David Kastrup wrote: >> albstorz(a)gmx.de writes: >> >> > Virgil wrote: >> >> In article <1129684276.251366.121150(a)g47g2000cwa.googlegroups.com>, >> >> albstorz(a)gmx.de wrote: >> >> >> >> > David R Tribble wrote: >> >> > >> >> > > >> >> > > Because I can prove it (and it's a very old proof). A powerset of >> >> > > a nonempty set contains more elements that the set. Can you prove >> >> > > otherwise? >> >> > >> >> > This argument is stupid. Is there any magic in the powerfunction? >> >> >> >> "Proofs" are not stupid until they can be refuted. The proof that for an >> >> arbitrary set S, Card(S) < Card(P(S)) has not been refuted by anyone. >> > >> > >> > Even if you think that the powersets of finite and infinite sets >> > have both a greater cardinality than their starting sets, you would >> > not really think it depends on the same cause in both cases. >> > >> > You must proof it independently for finite and for infinite sets. In >> > this sense the argument is stupid. >> >> Uh, no. The proof depends merely on the fact that some value has to >> be either a member of a set, or not. And if some value is in one set, >> bur not another, then those two sets are different. >> >> That's all. Finiteness or infiniteness does not even play into it. >> The proof just constructs a set which differs by the membership of at >> least one particular value with every target set in the assumedly >> complete mapping of set to powerset. > > I don't know what you are talking about. The proof for finite sets > needs just a complete induction. Nonsense. > This will not hold for infinity I think. If it needed complete induction in any manner, then only countable infinities could be covered. But where do you see complete induction in the following? Given a set A and a presumed complete mapping f(.) from A to P(A), consider the set X={B in A|B not in f(A)}. Now X =/= f(C) for all C in A, since C in A <=> C not in f(A), and so X is not covered by the mapping f(.). This is not complete induction by any means. It is not related to the naturals at all, nor to finiteness or infiniteness. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: stevendaryl3016 on 19 Oct 2005 09:08
albstorz(a)gmx.de says... >I don't know what you are talking about. The proof for finite sets >needs just a complete induction. This will not hold for infinity I >think. There is no induction involved in the finite case, and the infinite case is *exactly* the same proof as the finite case. Here it is once again: Let A be any set whatsoever, finite or infinite, it doesn't matter. Let f be any function from A to P(A). Let w = { x in A | x is not an element of f(x) }. Let x = any set in A. Let u = f(x). We prove that u is not equal to w. By definition of w, we have x in w <-> x is not an element of f(x). So x in w <-> x is not an element of u. That means that there are two cases: Case 1: x in w, and x is not in u. In that case, u cannot equal w. Case 2: x is not in w, and x is in u. In that case, u cannot equal w. So what we have proved is that forall x, w is not equal to f(x). So w is not in the image of f. So f is not a bijection between A and P(A). There's no induction. There's no assumption that A is finite. -- Daryl McCullough Ithaca, NY |