Why can no one in sci.math understand my simple point?
in [Logic]
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From: WM on 22 Jun 2010 15:34 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. If we use Cantor's definiton of "countable", then the set {..., An, ...A0, L0(0), L0(1), ...L0(n),...} is uncountable. 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. > > > > This is wrong. An obvious listing is (A0, A1, ...) The set {..., An, ...A0, L0(0), L0(1), ...L0(n),...} cannot be listed. > 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. Regards, WM
From: Virgil on 22 Jun 2010 15:51 In article <163b7c66-426c-4665-acdb-81f92c2b4e64(a)5g2000yqz.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 22 Jun., 01:32, Virgil <Vir...(a)home.esc> wrote: > > In article > > <e0b8cbfa-70a9-4be3-a08f-117e0af7f...(a)q12g2000yqj.googlegroups.com>, > > > > > > > > > > > > �WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > On 21 Jun., 22:17, Virgil <Vir...(a)home.esc> wrote: > > > > > > > But the set of all diagonals constructed according to the scheme given > > > > > above is countable but cannot be listed. > > > > > > Whatever do yu mean by "the set of all diagonals"? > > > > If you mean one "diagonal" for each possible list, that set of diagonals > > > > is not countable. > > > > If you mean the set of all "diagonals" for a single list, that is not > > > > countable either, since each of uncountably many permutations of the > > > > list produces a different "diagonal". > > > > > Take a list of all rationals. Construct, accordingto a given > > > substitution rule the antidiagonal. Add it at first position to the > > > list. Construct, according to the given rule the antidiagonal, add > > > it ... and so on. > > > > You are saying given a list, create its anti-diagonal and prepend it to > > that list. Repeat with the new list ad infinitum. > > > > > > > > > The number of antidiagonal is countable. Nevertheless it cannot be > > > listed. > > > > I see no problem in listing the set of such anti-diagonals. > > Including all rational numbers! The set of all rational numbers has been "listed" many times. I have done it myself a number of times (established a surjection or bijection from N to Q). > > > They can > > even be listed in the order in which each anti-diagonal is to be created. > > > > Note that the list of anti-diagonals so far created is finite so long as > > the number of iterations is finite, and only becomes infinite when the > > process achieves infinitely many iterations, and thus it is countable. > > > > Which such anti-diagonals does WM claim remain unlisted? > > That of the intended list of all those antidiagonals and all rational > numbers. If one has separate listings, one of each, it is easy to construct a joint listing by merely alternating entries. > > > > > > The complete infinite binary tree contains, by definition, all real > > > numbers between 0 and 1 as infinite paths, i. e., as infinite > > > sequences { 0, 1 }^N of bits. And infinitely many of them twice under the usual correspondence. > > > The complete infinite binary tree (including all those infinite paths > > > which consist of nodes and edges only) is constructed by a countable > > > number of steps. With each new level doubling the number of finite paths so steps-to-paths is as n-to-2^n. Then in a completed tree one has steps(aleph_0) to paths (2^aaleph_0), but 2^aleph_0 is the cardinality of an uncountalbe set. > > > In no step more than one infinite paths is extended. > > > Hence there are not more than countably many infinite paths. False! Countable, by definition, implies listable, and Cantor�s diagonal proof for binary sequences shows that the set of paths (each path being essentially an infinite binary sequence) is not listable, thus not countable. > > > > I have no idea what sort of definition of "countability" of infinite > > sets that WM is using, > > I know that you cannot read the above text. > But perhaps somewhat else can. Since WM claims a set which is countable need not be listable, or, equivalently a set which is not listable need not be countable, both of which are false according to any standard definitions of 'countability' and 'listability', WM owes us his, obviusly nonstandard, definitions of those two terms. If WM claims the standard definitions, then his other claims re countability and listability are merely, and clearly, false.
From: Virgil on 22 Jun 2010 16:00 In article <dd5850eb-6401-43e5-a342-df3166644cab(a)a30g2000yqn.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > 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 However the finite sublists of so far constructed antidiagonals does have a "limit", it being a list of all the antidiagonals buildable by this iteration. WM's hang up is that in his world no infinite process is ever allowed to be completed so that he is, like Xeno, unable to move past what the rest of us move past easily.
From: Virgil on 22 Jun 2010 16:10 In article <b71db24f-d637-4afd-a717-d2b5055f4fbf(a)a30g2000yqn.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> 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. There is certainly a list of successive anti-diagonals that is an endless list, but infinite sequences do not have "limits" without some sort of convergence rules, which is clearly not the case here. > > The list containing (L0, A0, A1, A2,...) would only then be a list Ln, > if its antidiagonal is the antidiagonal of some Ln. Assuming that L_(n+1) is the antidiagonal to endless list (L_n, ..., L_0, A_0, A_1, A_2,...), there clearly is always, for each n in N, a finite list (L_0, L_1, L_2, ..., L_n). And the union of all such finite lists forma an infinite list. > > 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. WM does not have a valid proof here.
From: Virgil on 22 Jun 2010 16:14
In article <e6cef47a-8776-48b7-b21a-627a6366d00d(a)g19g2000yqc.googlegroups.com>, Newberry <newberryxy(a)gmail.com> wrote: > 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? The list of naturals has an "infinite amount of information" in many senses, but a finite rule of construction contains it all. > > > 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. |