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From: Virgil on 19 Oct 2005 15:36 In article <MPG.1dc0400524b395cd98a4e3(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> 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? The "magic" is in the proof that from any set to its power set no surjection ( and therefore no bijection ) is possible. > > > Why should all operations with transfinite numbers lead to results > > with the same "level" of infinity, but only powerfunction beams up > > to the next level? Creation of powersets is not a standard arithmetical operation, so need not operate that way. > As far as bijections go, none. Bijections alone are insufficient with > infinite sets. Certainly, you don't disagree that any set's power set > is larger than itself? Or, perhaps you do. "Size" is not the issue with sets and their power sets. The issue is existence of a surjection from a set to its power set. And the proven conclusion is that there never is one.
From: Virgil on 19 Oct 2005 15:37 In article <MPG.1dc040cc19fad5ff98a4e4(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Did you have an objection to the bijection between *N and P(*N)? What > did I do wrong there, bijection-wise? You assume your result in order to prove it.
From: Virgil on 19 Oct 2005 15:40 In article <MPG.1dc0411e718b821798a4e5(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Virgil said: > > In article <MPG.1dbf2de38cf6ec898a4da(a)newsstand.cit.cornell.edu>, > > Tony Orlow <aeo6(a)cornell.edu> wrote: > > > > > David R Tribble said: > > > > But that's an incomplete mapping, because there are not enough infinite > > > > binary strings in *N to enumerate all of the subsets of *N. Try it, > > > > if you don't believe me. > > > > > > > > > > > Not enough infinite binary strings? > > > > Precisely. > > > So, you need more than an infinite amount? You need "more" then you have, in the same sense that there are "more" reals than rationals even though there is an endless supply of both.
From: Virgil on 19 Oct 2005 15:49 In article <MPG.1dc042163fd02d9198a4e6(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > Virgil said: > > 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. > > > Except for the obvious bijection between *N and P(*N). But, hey, what's one > measly counterexample? In this case it is quite "measly", since it specifically requires that one assume most of the result in order to "prove" the result. If *N bijects to the set of all endless sequences of binary digits, what is the mechanism by which TO maps these to the set of finite naturals so that he can then represent sets of finite naturals by endles sequences of binary digits. Without some such countable well ordering of the set of endless binary sequences, no such representation of subsets is possible. But any such countable well-ordering would prove that the set of endless sequences of binary digits os TO-finite.
From: Virgil on 19 Oct 2005 15:52
In article <MPG.1dc044c61d46b15798a4e8(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > albstorz(a)gmx.de said: > > > > 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. That is only true if 'size' means cardinality, since it is based entirely on injection/surjection/bijection arguments. And TO correcting albstorz is a perfect illustration of the blind leading the blind. |