Why can no one in sci.math understand my simple point?
in [Logic]
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From: Virgil on 23 Jun 2010 15:25 In article <f9884d20-096d-4fd6-9f63-eae9eba11bb2(a)w31g2000yqb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> 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)? > 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. Of course no list of reals (or binary sequences) can list all possible nonmembers of that list as there are uncountably many of them. Note that for every permutation of a list one gets a different antidiagonal, and there are uncountably many permutations of an infinite list. >
From: Virgil on 23 Jun 2010 15:38 In article <e98313e8-07a4-4b4f-86a9-cfc392a36616(a)c33g2000yqm.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > 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. Nonsense. Every permutation of a listing creates a list whose antidiagonals are different from those of the original. > Then a list contains also its antidiagonal That is more of WM's nonsense since for every list, its own antidiagonal is constructed so as to be not be a member of THAT list. And if one inserts it into its own list, one then has a different list. So for WM to demand a list containing a nonmember of that list is foolish. , > 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. I can find one quite easily. Arrange the original list together with all the countably many antidiagonals one has generated into a list (which is quite simple) and take the antidiagonal of that list. this last antidiagonal will not be in the original or in any of the derived antidiagnoals. > > 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. WM may find it unlistable, but I do not. Merely prepend that last antidiagonal to the list from which it was generated. > > Think over that only a little bit. It is not difficult to understand. I understand it quite clearly, but WM seems to be having a ton of problems with it.
From: Virgil on 23 Jun 2010 15:42 In article <bde842ea-7152-4476-aa85-3234c67f63a0(a)x21g2000yqa.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > 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? What have "diagonals" to do with anything? > > > > 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. In standard set theories, countable but not listable is impossible. And until WM can give us a complete axiom set (like FOL+ZFC) for his version of set theory, we need not pay any attention to its oddities. If it is listable, however, then it > has no unlisted diagonal. In what set theory? NOt in FOL+ZFC!
From: Virgil on 23 Jun 2010 15:45 In article <87k4ppswxi.fsf(a)dialatheia.truth.invalid>, Aatu Koskensilta <aatu.koskensilta(a)uta.fi> wrote: > WM <mueckenh(a)rz.fh-augsburg.de> writes: > > > Therefore, there is not every subset of an infinite set. > > What on Earth does this mean? WM apparently is not on earth.
From: Virgil on 23 Jun 2010 15:48
In article <88eb0kFrpbU1(a)mid.individual.net>, Sylvia Else <sylvia(a)not.here.invalid> wrote: > 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? One needs a standard list ( or a bijection from N to the set of objects in question) inorder to define an antidiagonal at all. What would the first element of the list itself be? And the second? |