An uncountable countable set
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From: Virgil on 5 Jul 2006 14:40 In article <1152100475.267310.35180(a)v61g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > In article <1151865025.167976.75740(a)v61g2000cwv.googlegroups.com> > > mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > > > > you need some limit in order to well-order the set of rationals? > > > > > > > > No, but that is unrelated. Transpositions operate on a set. A > > > > sequence > > > > of transposition hence also operates on a set. There is *no* > > > > definition > > > > how an infinite sequence of transformations operates on a set. > > > > > > These transpositions *are* a set. In my example, they do not destroy > > > the bijective mapping between |N and |Q. > > > > This makes no sense. Yes, there is a set of transpositions, but the > > transposition operate in sequence on an ordered set Q1 (which is a > > well-ordered set to start with). After each finite sequence of > > transpositions there is still a bijective mapping between N and the > > resulting ordered set. You have not proven that that is still the case > > after an infinite sequence of transpositions. Much less that it is > > still the case after an infinite sequence of infinite sequences of > > transpositions. You have first to define what that means. > > All sequences, even all single transpositions which I apply can be > enumerated by natural numbers, just like the lines of Cantor's list. > OK? And as long as I can enumerate, the number reached is finite. OK? But the numbers which are reachable include values larger than any given natural. To assume a limit on what is achievable is counterfactual. > Hence I do not need more than a finite set of transpositions, though > its cardinality cannot be given. You, in fact, need more than any finite number can supply. > This case is in general denoted by > "countably infinite" - but the number oo does *not* appear. Please > consider these facts before you demand something to be proved for "oo". WE do not require that it be proved for "oo", which is not a natural, but that it be proved for "for all n in N". You method produces exceptions to "for all n in N", by implying that there is some remote member of N beyond which it is not necessary to proceed in order to establish "for all n in N".
From: Dave Seaman on 5 Jul 2006 15:08 On Wed, 05 Jul 2006 12:05:01 -0600, Virgil wrote: > In article <e8g7t4$ebo$1(a)mailhub227.itcs.purdue.edu>, > Dave Seaman <dseaman(a)no.such.host> wrote: >> On Tue, 04 Jul 2006 21:31:16 -0600, Virgil wrote: >> >> > Also, in ZF, ZFC and NBG, any countable union of countable sets can be >> > proved countable, so "mueckenh"'s attempts to claim otherwise must fail. >> >> Not in ZF. > Is it the need for C? Yes. > And if any set is known to be countable, doesn't that imply a known, or > at least knowable, bijection from N to that set and thus justify a > bijection from N to any such set of such sets? To say that a set A is countable means that the set of bijections between A and N is nonempty. From a single nonempty set, one can select a member without using AC. But what we have here is a collection A_i (i in N) of countable sets, and for each i we are given that the set of bijections B_i between A_i and N is nonempty. In order to show that U_i A_i is countable, we need to select a particular bijection f_i from each B_i, which is an infinite number of choices. That's where AC comes in. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: Herman Rubin on 5 Jul 2006 16:12 In article <vmhjr2-90C02A.12050005072006(a)news.usenetmonster.com>, Virgil <vmhjr2(a)comcast.net> wrote: >In article <e8g7t4$ebo$1(a)mailhub227.itcs.purdue.edu>, > Dave Seaman <dseaman(a)no.such.host> wrote: >> On Tue, 04 Jul 2006 21:31:16 -0600, Virgil wrote: >> > Also, in ZF, ZFC and NBG, any countable union of countable sets can be >> > proved countable, so "mueckenh"'s attempts to claim otherwise must fail. >> Not in ZF. >Is it the need for C? >And if any set is known to be countable, doesn't that imply a known, or >at least knowable, bijection from N to that set and thus justify a >bijection from N to any such set of such sets? One can obtain a knowable bijection for any given countable set without being able to obtain all at once. We have the classic example of a countable number of pairs of socks; some of choice is needed to prove this is countable. It is consistent that the reals are a countable union of countable sets. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin(a)stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
From: mueckenh on 5 Jul 2006 16:55 Virgil schrieb: > On the other hand, let f map R on P(N) with the only condition on f be > that f map a non-natural to K= {n in N: n !in f(n)}, then f can quite > easily both exist and be a bijection. That is silliest simplicity. Try this: K = {x in R: x notin f(x)},. Regards, WM
From: mueckenh on 5 Jul 2006 17:00
Gc schrieb: > > > Yes it does. For every function N to P(N) there exists K in P(N). > > > > Not for any surjective function. > > Ash. I meant that if there were a surjection then K would have to > exist. If you wan`t to find maximum amount of pairs of some function N > ----> N(P) where there are no K you can just pair each a of N with {a}. All sets of naturals would have to exist. But no set defined as that K can exist. Therefore the set K is not a set of naturals. This would easily be accepted as obvious (though "a bit" counter-intuitive), if it was necessary in order to maintain ZFC. > > You will see it somewhat easier: > > Map R on the set {1, 2, 3, K} with K(f) be { x in R | x is not an > > element of f(x) }. > > What I think is that there has to be a function exactly defined before > K exits and when you have defined that you already have that function, > so K is not in the range of the function. The P(N) is different because > it contains all possible K:s. Insn't that a bit too much naive intuition? Compare | o / \ o o / \ / \ Regards, WM |