Why can no one in sci.math understand my simple point?
in [Logic]
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From: Mike Terry on 22 Jun 2010 12:27 "WM" <mueckenh(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, ...). 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. WM has now suggested an alternative list (L0, A0, A1,...) but that is not a valid list of numbers. 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. So that's where we are currently I think. Regards, Mike.
From: WM on 22 Jun 2010 13:05 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. > 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. > 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). The contradiction arose when the definable sequences were recognized to be countable. That was the point - not the present discussion. Regards, WM
From: WM on 22 Jun 2010 13:11 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. It is the same with all Cantor-list. All diagonals of all Cantor-lists belong to a countable set. Regards, WM
From: Mike Terry on 22 Jun 2010 15:05 "WM" <mueckenh(a)rz.fh-augsburg.de> wrote in message news:2af854b2-1a9a-423f-8f6a-f831e78f584f(a)t10g2000yqg.googlegroups.com... > 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 ...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. :-) > > > > 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. I'm struggling to follow. Here's what you are actually saying: - Ln is a list, for each n [Correct] - Ln has an antidiag An [Correct] - The "set of lists Ln" is {L0, L1, ...} - The "possible diagonals of this set are? What??? I guess you mean the set of diagonals of the members of the set? That would be {A0, A1, A2,...} - So parsing your sentence gives: {A0, A1, ....} is a countable set... [Correct] - ... but that {A0, A1, ...} cannot be listed. This is wrong. An obvious listing is (A0, A1, ...) > > It is the same with all Cantor-list. All diagonals of all Cantor-lists > belong to a countable set. This is ambiguous as well. I'm nearly giving up, but I'll suggest two possible precise interpretations, and you can say which of them you mean. (Or suggest another, but if it's not precise I'm giving up! :-) Interpretation 1: If L is a Cantor-list, then there exists a countable set M, such that (anti-?)diagonal of L belongs to M. That is so obvious its banal. (So probably not the right interpretation) Interpretation 2: 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. (So probably still not the right interpretation?) Hmmm, I've taken your phrase "Cantor-list" to mean "a list of real numbers". Maybe that's the problem? Regards, Mike.
From: Virgil on 22 Jun 2010 15:34
In article <5b14b681-b796-4ea0-88c4-a1d15928e8ce(a)u7g2000yqm.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 22 Jun., 01:29, "Mike Terry" > <news.dead.person.sto...(a)darjeeling.plus.com> wrote: > > "WM" <mueck...(a)rz.fh-augsburg.de> wrote in message > > > > > > Correct. Therefore the set of antidiagonals appears to be uncountable > > > = unlistable. But it is countable. That depends of WHICH set of "antidiagonals" one is talking about. the set of only those constructed from diagonals will be countable but the set of all numbers not in the list will be uncountable > > > > I don't see why you say it appears to be uncountable. �(See below) > > > > > > The set of antidiagonals that can be constructed (by a given > > > substitution rule) from an infinite sequence of lists. > > > > Ah, OK. > > > > 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 But it HAS been written as a list (A0, A1, A2, ...), dingbat! |