Why can no one in sci.math understand my simple point?
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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?
From: Virgil on 23 Jun 2010 16:01
In article <1fe50c9e-496c-401c-aa55-306e93ae8844(a)s9g2000yqd.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 23 Jun., 13:54, Sylvia Else <syl...(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. > > That is a an abbreviation of the construction I proposed. Of course > the "..." stand only for an infinite sequence of well defined digits > at finite places. Then your definition is flawed in one more way, and totally useless. One cannot have an antidiagonal of a list in which the elements are not all sequnces of the same type, and here, LO is not of the same type as A0 or A1 or A2 or any An. > > > You can't > > even form the anti-diagonal of that - what would the first digit of the > > antidiagonal be? > > What would the last digit of a normal Cantor-diagonal? In order to define a "Cantor antidiagonal' for a set of sequences, S, one must first have a bijection f:N -> S. WM has not made clear that he even has such a bijection, much less what it is like. > Why should the > first digit be more important than the last one? Why claim there is a "last one" when infinite sequences of digits don't have any "last ones"? > An infinite sequence > of digits (that is not converging and not defined by a finite formula, > like Cantor's diagonal sequence) is as undefined when the last digit > is missing as it is when the first digit is missing. I think WM's first digit is missing! The whole point of an infinite sequence is that there isn't any such thing as a "last"term. So any demand for one marks the demander as muddled. |