Why can no one in sci.math understand my simple point?
in [Logic]
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From: WM on 22 Jun 2010 07:18 On 22 Jun., 11:49, Sylvia Else <syl...(a)not.here.invalid> wrote: > >> So we have (L0, L1, L2, ...), and corresponding to each Ln we have an > >> antidiagonal An. So we have a sequence (A0, A1, A2, ...). > > >> 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 > > because the antidiagonal of the supposed list would belong to the set > > but not to the list. Therefore it is not countable. > > The antidiagonal of the list of (A0, A1, A2,...) would only belong to > the set if it is also the antidiagnoal of some Ln, which you haven't > proved to be the case. There is a sequence of lists including a sequence of diagonals. Every list Ln includes the diagonals A0 to A(n-1). There is no limit. Not every sequence has a limit, in particular the sequence 1, 2, 3, ... does not have a limit. But as the list is and remains countably infinite, there is no problem if you would like to have a limit. There is none, but if there was one, the list would maintain its same structure. Regards, WM
From: WM on 22 Jun 2010 07:28 On 22 Jun., 11:49, Sylvia Else <syl...(a)not.here.invalid> wrote: > The antidiagonal of the list of (A0, A1, A2,...) would only belong to > the set if it is also the antidiagnoal of some Ln, which you haven't > proved to be the case. Ah, noew I understand. You want to make us believe that there is a limiting list but no limiting antidiagonal. The list containing (L0, A0, A1, A2,...) would only then be a list Ln, if its antidiagonal is the antidiagonal of some Ln. But if (L0, A0, A1, A2,...) is not a list Ln, then something must have been happened in between that was incompatible with the process of my proof. Regards, WM
From: Sylvia Else on 22 Jun 2010 08:34 On 22/06/2010 9:28 PM, WM wrote: > On 22 Jun., 11:49, Sylvia Else<syl...(a)not.here.invalid> wrote: > > >> The antidiagonal of the list of (A0, A1, A2,...) would only belong to >> the set if it is also the antidiagnoal of some Ln, which you haven't >> proved to be the case. > > Ah, noew I understand. You want to make us believe that there is a > limiting list but no limiting antidiagonal. Hadn't even crossed my mind. > > The list containing (L0, A0, A1, A2,...) would only then be a list Ln, > if its antidiagonal is the antidiagonal of some Ln. > > But if (L0, A0, A1, A2,...) is not a list Ln, then something must have > been happened in between that was incompatible with the process of my > proof. > Strictly speaking, by your method of construction, the lists are (L0), (A0, L0), (A1, A0, L0), (... , A2, A1, A0, L0) ... being respectively, L0, L1, L2, L3, ... For (A0, A1, A2, ...) to be a list that should contain its own anti-diagonal, it has to be the same as an Ln, yet I can see no reason to think it would be. The fact that a contradiction would arise seems a powerful indicator that (A0, A1, A2, ...) would not be same as any Ln. Sylvia.
From: Newberry on 22 Jun 2010 11:29 On Jun 21, 9:38 pm, Virgil <Vir...(a)home.esc> wrote: > In article > <2896ff83-7d48-4bcb-80fa-ea38b8e1b...(a)40g2000pry.googlegroups.com>, > > > > > > Newberry <newberr...(a)gmail.com> wrote: > > On Jun 21, 6:11 am, Sylvia Else <syl...(a)not.here.invalid> wrote: > > > On 21/06/2010 1:39 PM, Newberry wrote: > > > > > Not sure why you think you had to tell us how the anti-diagonal is > > > > defined. You claimed you could CONSTRUCT it. Please go ahead and do > > > > so. > > > > I'm sure he will - right after you provide the list of reals. > > > > Sylvia. > > > Dear Sylvia, I did not claim that I could construct a list of reals, > > but Virgil claimed he could construct an anti-diagonal. > > To what list? > > An antidiagonal to a list of decimal representations of reals is simple. > > Ignore any integer digits (to the left of the decimal point) in the > listed numbers and have 0 to the left of the decimal point in the > anti-diagonal. If the nth decimal digit of the nth listed number is 5, > then make the nth decimal digit of the antidiagonal 7, otherwise make it > 3. > > This rule prevents it from being equal to any real in the listing. > > The above is only one of many effective rules for constructing an > antidiagonal different from each listed number. How is this effective if the diagonal has infinite amount of information? > If, as in Cantor's original argument, one has a list of binary > sequences, one takes the nth value of the antidiagonal to be the > opposite value from the nth value of the nth listed sequence.- Hide quoted text - > > - Show quoted text -
From: David R Tribble on 22 Jun 2010 11:49
WM wrote: > A crank suffers from selective perception of reality, doesn't he? Why yes, he does. |