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From: Virgil on 22 Oct 2005 16:17 In article <1129996412.551406.155870(a)g49g2000cwa.googlegroups.com>, albstorz(a)gmx.de wrote: > Daryl McCullough wrote: > > > > > No, the number of finite sets of finite naturals is exactly > > the same as the number of finite naturals. 2^N is the number > > of *all* subsets, including infinite and finite subsets. Yes, > > I know you think that N is finite, but you are an idiot. > > I'm really interested to know, what you are think: > what is the result of 2*N. > 2*N = ? > Please. > > regards > AS In Cantor arithmetic (i.e.,Cardinality), 2*Card(N) = Card(N), but N, representing the set of all finite naturals, is not a number at all, any more than {apple, pear, banana} is a number.
From: Virgil on 22 Oct 2005 16:18 In article <1129996854.926208.196560(a)g49g2000cwa.googlegroups.com>, albstorz(a)gmx.de wrote: > David R Tribble wrote: > > Albrecht Storz wrote: > > >> So, if there is an infinite set there is an infinite number. > > > > > > > David R Tribble wrote: > > >> 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)? > > > > > > > Albrecht Storz wrote: > > > 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 > > > > Only for finite sets or numbers. > > > > For infinite sets we must use infinite ordinals and infinite > > cardinals. But these ordinals and cardinals do not equate to any > > naturals because there are no infinite natural numbers. > > Yes. And the cause of this aspect is very good: there are no infinite > sets in the sense of a size. Infinite means just: sizeless big. Modern > math calls this aleph_0. ok. but you can do nothing with this. Else you > will earn contradictions. > > > > > > What is the number for set N = {0,1,2,3,...}? > > > There is no number. > > > If it's w, then how can w be a member of N? > > what is "w"? Nothing more than "unicorn". > > > > If it's w+1, then how can w+1 be a member of N? > > ... > > > > If it's w-1 (whatever that means), then how can w-1 be a member of N? > > Whatever this means? Nothing. > > > Regards > AS That something means nothing to AS does not mean the rest of us are to be bound by his ignorance.
From: Virgil on 22 Oct 2005 16:23 In article <1129997744.485963.27910(a)f14g2000cwb.googlegroups.com>, albstorz(a)gmx.de wrote: > Daryl McCullough wrote: > > albstorz(a)gmx.de says... > > > > >You can biject all constructable subsets of P(N) to N, but not the > > >unconstructable. > > > > By "constructable" do you mean "definable"? Or do you mean "computable"? > > Or do you mean "finite", or what? > > > Let's have a deal: By unconstructable I mean "Unconstructable". May be. Circular definitions are logically null. > > > > > > Let me list the combinations here: > > > > 1. There is no bijection between N and the set of all subsets of N. > > 2. There is no computable bijection between N and > > the set of all computable subsets of N. > > 3. There is no definable bijection between N and the > > set of all definable subsets of N. > > > > Whatever notion of function we want, it is true that there > > is no function mapping N to the set of all functions from > > N into {0,1}. > > No. You are just kidding? Nothing of this is truth. It is truth to those who know how to see truth. Those who choose not to see what is there to be seen are willfully blind. > > > > Oh, it's just mindfucking, nothing more. Show me one computable > infinite subset of N. In totally. Show me a set with exactly a trillion members. In totality! > > Cantor's dream leads to a nondenumerable mass of schwachsinn. Nothing > more. > > Regards > AS To each his own choice of schwachsinn. I prefer Dedekind's and Cantor's to yours.
From: Virgil on 22 Oct 2005 17:04 In article <1130010425.355964.219390(a)z14g2000cwz.googlegroups.com>, albstorz(a)gmx.de wrote: > David Kastrup wrote: > > albstorz(a)gmx.de writes: > > > > > 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. > > > > >> 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. > > > > > > 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. > > > > You are arguing with the guy who thinks that the cardinality of the > > set of even naturals is less than that of the naturals. > > > > The amusing thing is that although Tony is talking rubbish here, > > powersets _do_ have a larger cardinality than the original set, since > > any presumed mapping f:S->P(S) can't cover {x in S| x not in f(x)} > > since that set is different from all f(x) for all x in S. > > This proof needs the expression "x in S| x not in f(x)" has to be > meaningful. But it isn't in all the proofs about infinite sets. For any given function f:S -> P(S), the set {x in S: x not in f(x)} is well-defined, and therefore a legitimate member of P(S). It is equally legitimate to note that claiming that there exists any s in S such that f(s) = {x in S: x not in f(x)} is self-contradictory. > The > antidiagonal is an unicorn. That's all. Anti-diagonals are irrelevant in this situation.
From: David R Tribble on 23 Oct 2005 18:15
Albrecht Storz wrote: >> So, if there is an infinite set there is an infinite number. > David R Tribble wrote: >> 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)? > Albrecht Storz wrote: > Depending on the axiomatic construction and depending on the necessary > of truth (since truth means logic consequence) either there are > infinite natural numbers or there is no infinite set. Well, then by all means, show us the proof for this, because we don't believe it. If you're using any non-standard (non-Peano) axioms, please list those, too. I've got a set S = {0, 2^0, 2^2^0, 2^2^2^0, ...}, which contains all the powers of 2 of the form 2^p, where p=0 or 2^q. 1) If it is not an infinite set, tell me how many members it has. 2) If it is an infinite set, tell me what the smallest (first) infinite number is a member of it. |