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From: albstorz on 21 Oct 2005 05:08 Virgil wrote: > In article <1129721500.027651.21470(a)f14g2000cwb.googlegroups.com>, > albstorz(a)gmx.de wrote: > > > Virgil wrote: > > > In article <1129684564.158859.119300(a)o13g2000cwo.googlegroups.com>, > > > albstorz(a)gmx.de wrote: > > > > > So biject two non-well-orderable sets. > > > > > > > > > With or without an axiom of choice? > > > > > > I think it is more interesting for the most people without AC, since AC > > is not widely loved by the mathematics. > > > > But you may show both if you want. > > > > Regards > > AS > > That shows how little AS knows about things, since given the axiom of > choice, there is no such thing as a non-well-orderable set (though > admittedly there are some sets known that have yet to be well-ordered). Würgül
From: albstorz on 21 Oct 2005 05:16 David R Tribble wrote: > 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)? Not only partially. If there is no infinite number there is no infinite set. If sets consist of discret, distinguishable, individual elements, and sets are definde like this, the natural numbers are just representative for the elements and also for the sets. {1,2,3} means a set with element #1, element #2, element #3, this is the ordinal aspect of numbers. The set with cardinality 3 is just this set {1,2,3}, and at the same time it represents all sets with 3 elements. That's the open secret of the numbers. ordinal = cardinal = natural Regards AS
From: albstorz on 21 Oct 2005 05:22 David R Tribble 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? > > > > 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? The Gedanken in the context is, if finite powersets are always greater than there sarting sets, is there any connectivity to the idea that infinite powersets are greater than there starting sets. There is no connectivity. Regards AS
From: David Kastrup on 21 Oct 2005 05:26 albstorz(a)gmx.de writes: > David R Tribble 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? >> > >> >> 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? > > The Gedanken in the context is, if finite powersets are always > greater than there sarting sets, is there any connectivity to the > idea that infinite powersets are greater than there starting sets. > There is no connectivity. You are babbling. The standard proof about the larger cardinality of powersets makes no use of finiteness, infiniteness or, in fact, the power function at all. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: albstorz on 21 Oct 2005 05:34
Tony Orlow wrote: > 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 No Tony. If you think so, you should also argue, that since the cartesian product of two sets like {1,2}x{3,4}={{1,3},{1,4},{2,3},{2,4}} in the finite case is always of graeter cardinality than the starting sets, it should be also in the infinite case. But it isn't. Regards AS |