An uncountable countable set
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From: Virgil on 3 Jul 2006 21:44 In article <1151962360.076787.291910(a)j8g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Try the same with Canor's diagonal. Who is this Canor person?
From: Virgil on 3 Jul 2006 21:54 In article <1151962633.174580.226060(a)b68g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > The given well-order to start with is also known, and every result is > determined. If, as "mueckenh" claims, one starts with a well ordered set set order isomorphic to the naturals, and applies a well ordered set of transpositions to it to produce a non-well ordered set, there must be (by the well-orderedness of everything involved) a FIRST transposition whose effect is to change from a well-ordered set to a non-well ordered set. But as "mueckenh" cannot provide us with such a first, we may properly assume it does not exist.
From: David Hartley on 4 Jul 2006 03:53 In message <vmhjr2-17B81C.19542203072006(a)news.usenetmonster.com>, Virgil <vmhjr2(a)comcast.net> writes >In article <1151962633.174580.226060(a)b68g2000cwa.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > >> The given well-order to start with is also known, and every result is >> determined. > >If, as "mueckenh" claims, one starts with a well ordered set set order >isomorphic to the naturals, and applies a well ordered set of >transpositions to it to produce a non-well ordered set, there must be >(by the well-orderedness of everything involved) a FIRST transposition >whose effect is to change from a well-ordered set to a non-well ordered >set. > >But as "mueckenh" cannot provide us with such a first, we may properly >assume it does not exist. You are attacking the wrong part of WM's "argument" At least three people have posted examples of sequences that change a well-ordering in this way. The simplest is to apply, to N with the usual ordering, the transpositions (n 1), (n 2), .., (n n-1) for each n from 2 up. After the nth stage, the ordering is n n-1 n-2 .... 2 1 n+1 n+2 ... and the limit ordering is the reversal of the usual one. WM's fallacy is not that his sequence (of sequences) of transpositions alters the type of ordering, but his claim, that you are endorsing, that it cannot do so. -- David Hartley
From: Virgil on 4 Jul 2006 23:31 In article <1151505625.358363.7610(a)m73g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Daryl McCullough schrieb: > > > mueckenh(a)rz.fh-augsburg.de says... > > > > >Virgil schrieb: > > > > > > > > >> It is not "Cantor's list" but ANY list. > > >> You are challenged to find a list which lists all reals, then Cantor > > >> presented a rule for showing any list you provide is incomplete. > > > > > >I have already shown that any list of all or even some rationals and > > >its diagonal number is uncountable. > > > > No, you didn't, since it is false. > > > > 1. Start with a function f that is a surjection from > > N to Q (Q is the rationals). > > 2. Compute the diagonal d(f). This is a real number that is > > guaranteed to not be in the image f. > > 3. Form a new set M(f) = Q union { d(f) }. > > 4. This new set is *countable* > > like any subset of P(N), unless you require the *definition* of K being > included Since any set which is countable allows uncountably many bijections from N onto it, here is no sort of K which can simultaneously prohibit all of them. 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.
From: Dave Seaman on 5 Jul 2006 07:30
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. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/> |