Why can no one in sci.math understand my simple point?
in [Logic]
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From: WM on 23 Jun 2010 06:34 On 23 Jun., 04:18, Sylvia Else <syl...(a)not.here.invalid> wrote: > On 23/06/2010 11:33 AM, Sylvia Else wrote: > > > > > > > On 23/06/2010 11:03 AM, Virgil wrote: > >> In article > >> <2000e81b-7c5a-41be-b6af-98f96f2fb...(a)w31g2000yqb.googlegroups.com>, > >> WM<mueck...(a)rz.fh-augsburg.de> wrote: > > >>> On 22 Jun., 21:34, Virgil<Vir...(a)home.esc> wrote: > > >>>>>> But (A0, A1, A2, ...) is obviously countable. Above you say it's > >>>>>> "certainly > >>>>>> not countable", but it is. > > >>>>> The set is certainly countable. But it cannot be written as a list > > >>>> But it HAS been written as a list (A0, A1, A2, ...), > > >>> Does this list contain the anti-diagonal of (..., An, ... A2, A1, A0, > >>> L0)? > > >> Since (..., An, ... A2, A1, A0, L0) does not appear to be a list, why > >> should there be any antidiagonal for it? > > > Ach! Let's scrap A0 - it's confusing. > > > If we let L_n be the nth element in the list L0, and An the > > anti-diagonal of the An-1, An-2,...., A1, L_1, L_2, L_3,... > > > then > > > L_1 > > A1 > > L_2 > > A2 > > L_3 > > A3 > > L_4 > > ... > > > is a list. I'm still thinking about that. > > > Sylvia. > > Hmm... > > A1 is the antidiagonal of L1 L2 L3... > > A2 is the antidiagonal of L1 A1 L2 L3... > > A3 is the antidiagonal of L1 A1 L2 A2 L3 L4... > > Each An is thus constructed from a list that is different from the list > into which it is inserted. So the construction does not lead to a list > that should contain its own anti-diagonal, and it doesn't. Ln = An .... A2 A1 A0 L0 Does your bijection contain the anti-diagonal of (..., An, ... A2, A1, A0, L0)? If yes, then Cantor's argument fails as a list contains its antidiagonal. Reason: you list in your listing above only the lines of lists. That is so because the antidiagonal of every list Ln belongs to another list L(n+1)). b) Does it not? Then a list cannot list all anti-diagonals that belong to the countabe set constructed in my argument. Regards, WM
From: WM on 23 Jun 2010 06:44 On 23 Jun., 06:47, Sylvia Else <syl...(a)not.here.invalid> wrote: > > Rather than argue that VMs proposition fails on that point, I wanted to > address the flaw, in order to find a more substantial objection. My initials ar WM. There is no flaw in the argument: Your bijection either contains all constructed antidiagonals. Then a list contains also its antidiagonal, because every list has an antidiagonal and your bijection (that is only a permutation of my list) contains all lines and antidiagonals. I.e., there is no missing antidiagonal of a "limit" list outside of the bijection. Or there is a last diagonal of the limit list that does not belong to your bijection (and to my list). Then there is a countable set (the set that includes this last diagonal) that is not listable. Think over that only a little bit. It is not difficult to understand. Regrads, WM
From: WM on 23 Jun 2010 06:48 On 23 Jun., 08:39, Sylvia Else <syl...(a)not.here.invalid> wrote: > > Given that "ultimate list" {...a2,a1,a0,q0,q1,q2,...} merely rearrange > > it to {q0, a0, q1, a1, q2, a2, ...} and you have comprehensive list of > > everything with a findable antidiagonal which is not listed. With or without all diagonals? > > I thought that was what I did. It would not help, because then the set of all lines (of my construction) including the antidiagonal of this construction is a countable set but not listable. If it is listable, however, then it has no unlisted diagonal. Regards, WM
From: Sylvia Else on 23 Jun 2010 07:54 On 23/06/2010 8:34 PM, WM wrote: > On 23 Jun., 04:18, Sylvia Else<syl...(a)not.here.invalid> wrote: >> On 23/06/2010 11:33 AM, Sylvia Else wrote: >> >> >> >> >> >>> On 23/06/2010 11:03 AM, Virgil wrote: >>>> In article >>>> <2000e81b-7c5a-41be-b6af-98f96f2fb...(a)w31g2000yqb.googlegroups.com>, >>>> WM<mueck...(a)rz.fh-augsburg.de> wrote: >> >>>>> On 22 Jun., 21:34, Virgil<Vir...(a)home.esc> wrote: >> >>>>>>>> But (A0, A1, A2, ...) is obviously countable. Above you say it's >>>>>>>> "certainly >>>>>>>> not countable", but it is. >> >>>>>>> The set is certainly countable. But it cannot be written as a list >> >>>>>> But it HAS been written as a list (A0, A1, A2, ...), >> >>>>> Does this list contain the anti-diagonal of (..., An, ... A2, A1, A0, >>>>> L0)? >> >>>> Since (..., An, ... A2, A1, A0, L0) does not appear to be a list, why >>>> should there be any antidiagonal for it? >> >>> Ach! Let's scrap A0 - it's confusing. >> >>> If we let L_n be the nth element in the list L0, and An the >>> anti-diagonal of the An-1, An-2,...., A1, L_1, L_2, L_3,... >> >>> then >> >>> L_1 >>> A1 >>> L_2 >>> A2 >>> L_3 >>> A3 >>> L_4 >>> ... >> >>> is a list. I'm still thinking about that. >> >>> Sylvia. >> >> Hmm... >> >> A1 is the antidiagonal of L1 L2 L3... >> >> A2 is the antidiagonal of L1 A1 L2 L3... >> >> A3 is the antidiagonal of L1 A1 L2 A2 L3 L4... >> >> Each An is thus constructed from a list that is different from the list >> into which it is inserted. So the construction does not lead to a list >> that should contain its own anti-diagonal, and it doesn't. > > Ln = > > An > ... > A2 > A1 > A0 > L0 > > Does your bijection contain the anti-diagonal of > (..., An, ... A2, A1, A0, L0)? I don't understand why you've recast it back to that form. You can't even form the anti-diagonal of that - what would the first digit of the antidiagonal be? Sylvia.
From: Aatu Koskensilta on 23 Jun 2010 08:13
Transfer Principle <lwalke3(a)lausd.net> writes: > But this proof that there is no list of all real numbers must use > the axioms of some _theory_ (such as ZFC) -- a theory which > Newberry isn't required to accept just because Hughes says so. Cantor's theorem uses essentially nothing but predicative comprehension. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus |