An uncountable countable set
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From: Virgil on 5 Jul 2006 19:47 In article <1152134779.638396.157500(a)b68g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > > Apparently you do not know Cauchy well. According to Cauchy, any number > > that is written decimally has a limit, and that limit is a real number. > > And, no, I have not read much from Cantor, so I have no idea how I should > > interprete your representation here. > > But what you claim above to be Cauchy's is exactly Cantor's (and > before him Meray's) theory of irrational numbers. Not Cauchy's. He is > that guy with the epsilon. Cauchy was there first. > > > > > > The second sequence of transpositions defines the second "infinite > > > > permutation", but that one overlaps the first "infinite permutation". > > > > > > No problem. For a given well-order to start with, every result is > > > determined. > > > > When you use finitely many overlapping sequences of "infinite > > permutations". > > When I stay within the domain of finite numbers. Then "mueckenh" is claiming that permutation groups are commutative, which is easily proven to be false and well known to be false.
From: David Hartley on 5 Jul 2006 20:07 In message <J1yCn4.4qL(a)cwi.nl>, Dik T. Winter <Dik.Winter(a)cwi.nl> writes >In article <1152134107.848676.131670(a)75g2000cwc.googlegroups.com> >mueckenh(a)rz.fh-augsburg.de writes: > > David Hartley schrieb: >... > > > So you acknowledge that your process doesn't give, in the limit, a > > > bijection N -> Q+ ? > > > > Such a bijection does not exist, because there is no smallest positive > > rational. > >There is no need of such for a bijection. Consider the set of positive >numbers P and the set of negative numbers N. Clearly there is a bijection >from P to N: f(x) = -x. But N has no smallest element. A bijection is >not necessarily order preserving. Yes, that was rather sloppy. Throughout the discussion of this process of WM's, is has been implicit that the mappings from N are order-preserving. The essence of his claim is that he has a sequence of w-orderings of the (positive) rationals whose limit is the usual ordering and he assumes that the sequence of associated order-isomorphisms from N must have a limit which is an order-isomorphism from N to the limit-ordering, hence infinity doesn't exist. When he says the bijection doesn't exist, he is presenting the contradiction stage of his argument, not realising that he has never offered a proof that such a bijection does exist. -- David Hartley
From: Herman Jurjus on 6 Jul 2006 05:45 Virgil wrote: > In article <e8g7t4$ebo$1(a)mailhub227.itcs.purdue.edu>, > Dave Seaman <dseaman(a)no.such.host> wrote: > >> On Tue, 04 Jul 2006 21:31:16 -0600, Virgil wrote: >> >>> Also, in ZF, ZFC and NBG, any countable union of countable sets can be >>> proved countable, so "mueckenh"'s attempts to claim otherwise must fail. >> Not in ZF. > > Is it the need for C? Where i come from, Z(F)+countable choice was the standard. So if you used countable choice to prove something, that was accepted as 'not applying AC'. I've always liked this practice, and wondered (still wonder) why the rest of the world doesn't adopt it as well. Countable choice is not implied by ZF, but of course it's much(?) weaker than full AC. -- Cheers, Herman Jurjus
From: mueckenh on 6 Jul 2006 06:10 Virgil schrieb: > > Also, in ZF, ZFC and NBG, any countable union of countable sets can be > proved countable, so "mueckenh"'s attempts to claim otherwise must fail. Feferman and Levy showed that one cannot prove that there is any non-denumerable set of real numbers which can be well ordered. Moreover, they also showed that the statement that the set of all real numbers is the union of a denumerable set of denumerable sets cannot be refuted. Regards, WM
From: mueckenh on 6 Jul 2006 06:27
Dik T. Winter schrieb: > In article <1152099458.014628.279080(a)m79g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > Dik T. Winter schrieb: > > > > > In article <$UHhNrCdjAqEFwc5(a)212648.invalid> David Hartley <me9(a)privacy.net> writes: > > > ... > > > > Indeed, I agree that it is possible to apply a sequence of > > > > transpositions and change a well-ordering of the rationals to the usual > > > > ordering. (I posted my own example last night.) However, I take this to > > > > imply simply that such transformations do not preserve well-ordering, > > > > not that there is a contradiction in standard set theory. > > > > > > Indeed. The simplest example (using indices rather than numbers) is the > > > sequence of transpositions on the natural numbers in there standard order > > > is the sequence > > > for n -> oo (0, 1)(1, 2)(2, 3)(3, 4)...(n, n+1) > > > Applying this will lead (when we properly define what "-> oo" means) to > > > > It means that all natural numbers, which according to set theory do > > exist, are included as indices. Not more and not less. In particular > > there is no number oo. > > Righ, pretty acute. > > > > the ordered set: > > > (1, 2, ..., 0) > > > applying it again will lead to > > > (2, 3, ..., 1, 0) > > > and applying it again and again will lead to > > > (..., 3, 2, 1, 0) > > > So the first sequence will lead to a well-ordered set with a different > > > ordinal number (contradicting what Cantor stated according to the quotes > > > supplied). Applying an infinite number of such sequences will destroy > > > well-orderedness. > > > > It will (and can) not destroy the one-to-one correspondence between q's > > and n's. > > No, of course not. The cardinal number of q will not change. What changes > is ordering of q and also ordinal number and possibly well-ordering. The ordering is omega and remains so, as long as the first index is 1 and there is no limes-ordinalnumber omega between two indices. > > > > Note also that I do not see how to apply transpositions using indices > > > when you start with a set that is not well-ordered. You can only do it > > > with a well-ordered set and they operate only on an initial segment of > > > order type w or smaller. > > > > And that is completely sufficient for the set of all positive > > rationals. > > Under what ordering? The ordering is omega and remains so, as long as the first index is 1 and there is no limes-ordinalnumber omega between two indices. Regards, WM |