Why can no one in sci.math understand my simple point?
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From: Virgil on 22 Jun 2010 16:29 In article <994c8353-9f4b-4697-bf57-b39fbf33506c(a)y11g2000yqm.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 22 Jun., 14:34, Sylvia Else <syl...(a)not.here.invalid> wrote: > > On 22/06/2010 9:28 PM, WM wrote: > > > > > On 22 Jun., 11:49, Sylvia Else<syl...(a)not.here.invalid> wrote: > > > > > 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, ... > > Correct. > > > > For (A0, A1, A2, ...) to be a list that should contain its own > > anti-diagonal, > > I do not argue about (A0, A1, A2, ...). The list that I consider has > the form > > Ln = > > An > ... > A2 > A1 > A0 > L0 > > where L0 is an infinite list of all rational numbers. > This list Ln does not contain its own antidiagonal. Lets see if I have this correctly, with only a slight change in notation. L_0 is a listing of the rationals. L_(n+1) is a listing in which A_N, an antidiagonal of L_N, is prepended to list L_n to form a new list. > > > it has to be the same as an Ln, yet I can see no reason > > to think it would be. > > It is not, need not cannot and should not. If I have it right above, there there is clearly a listing of the A_n. > > > The fact that a contradiction would arise seems a > > powerful indicator that (A0, A1, A2, ...) would not be same as any Ln. > > A contradiction arises for every Cantor list. Cantor thought that he > had proved the uncountability of all definable binary sequences > (because he used only definable binary sequences in his list and > obtained a definable binary sequence as its antidiagonal). Actually, nothing in Cantor's argument requires any of the binary sequences in his argument to be "definable". What Cantor says is "for ANY list of binary sequences". The > contradiction arose when the definable sequences were recognized to be > countable. Except that Cantor made no such restriction to definable sequences as WM does, so it is only WM's statements, not Cantors, which don't work as Cantor said they did.
From: WM on 22 Jun 2010 16:38 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)? Then Cantor's argument fails. Does it not? Then a list can not list all anti-diagonals that belong to the countabe set constructed in my argument. Regards, WM
From: Virgil on 22 Jun 2010 16:44 In article <2af854b2-1a9a-423f-8f6a-f831e78f584f(a)t10g2000yqg.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 22 Jun., 18:27, "Mike Terry" > <news.dead.person.sto...(a)darjeeling.plus.com> wrote: > > "WM" <mueck...(a)rz.fh-augsburg.de> wrote in message > > > > news:b71db24f-d637-4afd-a717-d2b5055f4fbf(a)a30g2000yqn.googlegroups.com... > > > > > 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. > > > > L0 is not a number. It's a list, and so is not eligible for belonging to a > > list of numbers. (I.e. this response makes no sense.) > > > > ----------------- > > > > So this is where I believe this sub-thread has got to as of Tue 16:00 UTC: > > (to stop us going round and around) > > > > WM has defined a sequence of lists (L0, L1, ...) with corresponding > > antidiags (A0, A1, ...). There is a claim that "the set of antidiagonals" > > appears to be uncountable, but it was not clear which set this was. > > > > WM seemed to suggest he meant the set (A0, A1, ...). > > No, I added the A's to a list. > > > > I pointed out this was obviously countable. > > > > WM agreed but said it can't be written as a list, since it's antidiag is in > > the list. > > > > Sylvia pointed out why this is obviously wrong. > > No, she misunderstood. > > > > WM has now suggested an alternative list (L0, A0, A1,...) but that is not a > > valid list of numbers. > > I have, from the beginning, used the following list > > Ln = > > An > ... > A2 > A1 > A0 > L0 > > where L0 is thelist of all rationals. > > This list Ln contains a countable set of numbers but the set of its > diagonals is not listable, because An is not in the list. > > > > It is true none the less that the image of L0 is countable, and if we append > > all the A0, A1,... it is still countable, so the combined set IS listable. > > Suppose LW lists this new set. Of course LW has a NEW antidiagonal which is > > not in the list LW, so this isn't going anywhere. > > All possiblke diagonals of this set of lists Ln belong to a coutable > set, but there is no list of all of them. Sure there is. Let L_0 = {q_0, q_1, q_2 ...} be a listing of the rationals. Let a_0 be an antidiagonal of L_0. Let L1 = { a_0, q_0, q_1, q_2 ...} be a new list. Extend this recursively with a_n being an antidiagnal to L_n, and L_(n+1) = L_n with a_n prepended to it = { a_n, a_(n-1), ...a_0, q_0, q_1, q_2, ...} If this is alt all correct then WM is totally incorrect. The set {a_n : n in N} is already a list as it is indexed by N. And it is still the case in standard mathematics that listability is equivalent to countability. > > It is the same with all Cantor-list. All diagonals of all Cantor-lists > belong to a countable set. Actually not. To start with, there exist uncountably many "Cantor-lists". And WM's definition of countability is fatally flawed. > > Regards, WM
From: Mike Terry on 22 Jun 2010 17:21 "WM" <mueckenh(a)rz.fh-augsburg.de> wrote in message news:103cc56f-746e-482f-9315-0c9e8eaf45e1(a)c10g2000yqi.googlegroups.com... > On 22 Jun., 21:05, "Mike Terry" > <news.dead.person.sto...(a)darjeeling.plus.com> wrote: > > > > > Ln = > > > > > An > > > ... > > > A2 > > > A1 > > > A0 > > > L0 > > > > > where L0 is thelist of all rationals. > > > > > This list Ln contains a countable set of numbers > > > > ..correct, {An, ...A0, L0(0), L0(1), ...L0(n),...} > > is obviously countable. [L0(k) is the k'th element in list L0] > > > > > but the set of its > > > diagonals is not listable, because An is not in the list. > > > > The "set of its diagonals" = {An}. A list has just one diagonal. Every set > > of one element is listable. Like Sylvia I must be misunderstanding what you > > mean. (But I'm not misunderstanding what you actually say. :-) > > To spell it out clearly: The set of all diagonals (including or > exluding all rationals - that does not matter) > {..., An, ...A0, L0(0), L0(1), ...L0(n),...} > that are constrcuted according to my prescription cannot be listed > although it is countable. Yes, that's clear, thanks! But of course the set can be listed: (A0, L0(0), A1, L0(1), A2, L0(2), A3, L0(3), ...) and the set is countable. > > If we use Cantor's definiton of "countable", then the set > {..., An, ...A0, L0(0), L0(1), ...L0(n),...} > is uncountable. Cantor's definition of countable is that there is an injection from the set to N (set of natural numbers). E.g.: A0 ---> 1 L0(0) ---> 2 A1 ---> 3 L0(1) ---> 4 A2 ---> 5 L0(2) ---> 6 ... Right? (I'm thinking you must have a different idea of Cantor's definition of countability - if so it would be useful if you could provide a link to your source, as the definition above was from Wikipedia which says the definition is due to Cantor!) (But of course Wikipedia isn't the ultimate reference...) > If we use the definition that a subset of a countable set is > countable, then the set > {..., An, ...A0, L0(0), L0(1), ...L0(n),...} > is countable. That's not a definition, it's a theorem. > > > > > > > > > This is wrong. An obvious listing is (A0, A1, ...) > > The set > {..., An, ...A0, L0(0), L0(1), ...L0(n),...} > cannot be listed. Obviously it can. (I'm completely missing why you could possibly be thinking it couldn't) > > > > > there exists a countable set M, such that > > If L is a Cantor-list, then > > (anti-?)diagonal of L belongs to M. > > > > That is so obviously false that its banal. > > No it is not. If there exists a Cantor-list, i.e., that what Cantor > really understood by the term list, then it is a list of *defined* > reals. And then its anti-diagonal is a defined real too. Then exists a > countable set M, namely the set of all defined reals, that is > countable. Nevertheless it cannot be listed. I don't believe that was Cantor's definition of list. (But I'm prepared to be persuaded if you've got some references for this?) Regards, Mike. > > Regards, WM
From: Sylvia Else on 22 Jun 2010 20:03
On 23/06/2010 6:38 AM, WM 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)? I can't see that you've proved that it doesn't. All you've proved is that it doesn't contain the anti-diagonal of (A0, A2, A2, ...). But that means nothing of significance, because your construction doesn't require or imply that it would so contain. Sylvia. |