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From: Tony Orlow on 19 Oct 2005 12:56 albstorz(a)gmx.de said: > > 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. > > > Regards > AS > > Albrecht, do you accept the axiom of induction? If so, it is easily provable inductively that the power set of a set of size n has size 2^n, and since this is an equality property, it holds for the infinite case. The power set of an infinite set is infinite, but a larger infinity than the set. -- Smiles, Tony
From: Tony Orlow on 19 Oct 2005 12:58 David Kastrup said: > 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. Oh! So, N is a different size from N/{1}? Finally we reach agreement. Welcome to reality, David. > > That's all. Finiteness or infiniteness does not even play into it. That's right. Proper superset relation trumps any special rules for the finites or infinite. Right David? > 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. And we all know, that one element makes a world of difference! Good job, David! :D > > -- Smiles, Tony
From: David R Tribble on 19 Oct 2005 12:59 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? > Albrecht Storz wrote: > This argument is stupid. Is there any magic in the powerfunction? A > hidden megabooster for transcendental overflow? What is the very > special aspect of the powerfunction to be so magic? [...] > > I'm very sensible about this because this argument is found in very > much books although it's total meaningless. > > (Weak minds might be impressed by the big numbers which are easily > produced by powerfunction.) > > What in finity holds may not (or do not) hold in infinity. So what is your proof?
From: imaginatorium on 19 Oct 2005 13:08 Tony Orlow wrote: > albstorz(a)gmx.de said: > > 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? A > > hidden megabooster for transcendental overflow? What is the very > > special aspect of the powerfunction to be so magic? > Why do you think there is magic in the notion that the set of all possible > combinations of elements is larger than the set of single elements, especially > when that power set essentially includes the set, in the form of the singleton > subsets? Because there are all sorts of ways one might define "larger" when talking about comparing (infinite) sets. One very obvious one is to say that if A is a proper subset of B, then B is larger than A. A totally well-defined partial ordering on all sets. (Can I use 'CC' for "is a proper subset of"...) For infinite sets, even when A CC B, there may well still be a bijection between the two sets - as in the example of the even naturals E which can be bijected to the naturals N (by the function f: x -> 2x). Incidentally, I'm talking about normal sets of all of the naturals, not just some all of them, but actually all of all of them. The union of all the sets you could ever think of that you called "all". The naive guess might be that taking a bijection as a measuring stick - which it is, though contrary to what sometimes gets said here, it is not "the unique" measuring stick - anyway to continue the sentence I started, one would guess that finite sets can only be bijected if when counted the ditty stops at the same number-name, whereas all unending sets can be bijected, since the ditty just goes on and never stops. This is not true, since as demonstrated in a short proof you haven't managed to grok yet, no set can be bijected to its power set. Incidentally, you referred to this proof as a "largest-finite type argument" (or similar words). There really is no similarity at all, but it occurs to me it's quite possible you see it as similar, because this is the first time you've actually read a mathematical argument. Anyway, keep working on it. Brian Chandler http://imaginatorium.org
From: David R Tribble on 19 Oct 2005 13:11
Albrecht Storz wrote: > So, if there is an infinite set there is an infinite number. Do you mean that an infinite set (or natural numbers) must contain an infinite number as a member (which is false)? Or do you mean that the size of an infinite set is represented by an infinite number (which is partially true)? |