Why can no one in sci.math understand my simple point?
in [Logic]
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From: Virgil on 23 Jun 2010 02:31 In article <88dhukFesiU1(a)mid.individual.net>, Sylvia Else <sylvia(a)not.here.invalid> wrote: > On 23/06/2010 2:27 PM, Virgil wrote: > > In article<88d6j2Fqq8U1(a)mid.individual.net>, > > Sylvia Else<sylvia(a)not.here.invalid> wrote: > > > >> On 23/06/2010 11:03 AM, Virgil wrote: > >>> In article > >>> <2000e81b-7c5a-41be-b6af-98f96f2fb630(a)w31g2000yqb.googlegroups.com>, > >>> WM<mueckenh(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. > > > > There are a lot of possible lists here. > > Yes. > > > > One starts with, for example, same listing of the rationals indexed by > > the 0-origin naturals: L0 = {q0, q1, q2, ...}. > > > > For that list one finds an antidiagonal, a0, not in L0, and with it > > forms a new list L1 = {a0, q0, q1, q2, ...} with the antidiagonal a0 > > prepended to the list of which it is the antidiagonal. > > > > This process is clearly recursive, allowing us now, for example, to find > > an antidiagonal a1 to L1 which we can prepend a1 to L1 giving a new list > > L2 = {a1, a0, q0, q1, q2, ...}. > > > > The process also clearly may be in theory repeated infinitely often, so > > that one can derive from it new sequence of the antidiagonals taken in > > the order of their derivation, A = {a0,a1, a2, ...}. > > Yes, but at some point VM made it clear that the issue wasn't that the > diagonal wasn't present in A, but in the 'ultimate' L. The "ultimate L", as presented by WM, does not exist as a standard list, but only as a list open at both ends, so cannot have a standard anti-diagonal, though all sorts f equivalent non-members can be constructed merely by rearranging that "ultimate L" into a list, which may be done in many ways. > > I was concerned that as it was then formatulated, VM's proposition could > be attacked on the basis that the diagonal of the 'ultimate' L couldn't > be constructed - you couldn't even start to do so, because the first > element of the list wasn't defined. It was the 'last' element of an > infinite sequence. One can still create equivalents of antidiagonals simply by reordering that "final" list. Nthing about an "antidiagonal" requires taking the elements of the base list in any particular order, as long as one can take ALL of them one after another in in SOME order. > > Rather than argue that VMs proposition fails on that point, I wanted to > address the flaw, in order to find a more substantial objection. The flaw in WM's argument is that he claims that no nonmember can be found for his "ultimate" open at both ends list, but he is wrong. 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.
From: Sylvia Else on 23 Jun 2010 02:39 On 23/06/2010 4:31 PM, Virgil wrote: > The flaw in WM's argument is that he claims that no nonmember can be > found for his "ultimate" open at both ends list, but he is wrong. I thought he was trying a Cantor like argument to the effect that a non-member is found that should by construction by a member, and then arguing that the contradiction therein had to be resolved by concluding that the set couldn't be listed, even though it was clearly countable. He was certainly wrong, and that was always apparent, but I prefer a specific identified flaw over a general "must be wrong" type of argument, even if the latter amounts to a proof. OK, sometimes they're not available, but they're what I prefer. > > 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. I thought that was what I did. Sylvia.
From: Virgil on 23 Jun 2010 02:42 In article <3c04e794-89f9-4438-995f-d818a4a0ce25(a)5g2000yqz.googlegroups.com>, Newberry <newberryxy(a)gmail.com> wrote: > On Jun 22, 1:14�pm, Virgil <Vir...(a)home.esc> wrote: > > In article > > <e6cef47a-8776-48b7-b21a-627a6366d...(a)g19g2000yqc.googlegroups.com>, > > > > > > > > > > > > �Newberry <newberr...(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. > > I do not know how to quantify the amount of information the list of > naturals carries. If it is not 0 then it is a few bits at most. > > You talk nonsense and you know it. It makes perfect sense to me, and perhaps to some of those who read me. The list of naturals is > compressible the diagonal is generally incompressible. Whatever does "compressible" mean? The list of naturals can be generated by iteration of a finite formula. But Cantor's original anti-diagonal argument uses only binary sequences, i.e, functions from N to a two member set. Cantor's two element sets was {m,w}, but using {0,1} simplifies things even more. Given a list of infinite binary sequences with values in {0,1}, b_0, b_1, b_2,..., with b_m_n being the nth term in the mth sequence. the antidiagonal is a where a_n = 1 - b_n_n > > > > > > > > > > > > > > > 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.
From: Virgil on 23 Jun 2010 03:40 In article <88doghFgf9U1(a)mid.individual.net>, Sylvia Else <sylvia(a)not.here.invalid> wrote: > On 23/06/2010 4:31 PM, Virgil wrote: > > > The flaw in WM's argument is that he claims that no nonmember can be > > found for his "ultimate" open at both ends list, but he is wrong. > > I thought he was trying a Cantor like argument to the effect that a > non-member is found that should by construction by a member, and then > arguing that the contradiction therein had to be resolved by concluding > that the set couldn't be listed, even though it was clearly countable. I do not think that WM has any coherent argument. Though many ambiguities and non-sequiturs that his arguments always include make it difficult to be sure. > > He was certainly wrong, and that was always apparent, but I prefer a > specific identified flaw over a general "must be wrong" type of > argument, even if the latter amounts to a proof. OK, sometimes they're > not available, but they're what I prefer. > > > > > 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. > > I thought that was what I did. Perhaps so, but since it was not in this post, I wasn't sure. My newsreader does not make it easy to recover earlier posts in a thread once they have been marked as read. > > Sylvia.
From: WM on 23 Jun 2010 06:30
On 22 Jun., 23:21, "Mike Terry" <news.dead.person.sto...(a)darjeeling.plus.com> wrote: > > 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! Youe welcome. > > 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. a) Does this list contain the anti-diagonal of (..., An, ... A2, A1, A0, L0)? Then Cantor's argument fails as a list contains its antidiagonal. Reason: you list in your listing above only the lines of lists. That is so because the antidiagonal of every list Ln belongs to another list L(n+1)). b) Does it not? Then a list can not list all anti-diagonals that belong to the countabe set constructed in my argument. > > > > > 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? No. See above. Another example ist the set of all definable reals. There is no bijection with |N. But they belong to a countable set of all finite words. > > > 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. That may be a theorem in ZFC. In mathematics a subset cannot have a larger cardinality than its superset. > > > > > > 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) You will understand if you try to answer question a) above. > > > > > >   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?) If you understand enouf´gh German, then you might look here: 1906, 8. Aug. letter Cantor to Hilbert Lieber Freund .... König will zwei Arten von reellen Zahlen unterscheiden; solche, die âendliche Definitionen" zulassen und solche, die âunendliche Definitionen" erfordern. Eine jede Definition ist aber ihrem Wesen nach eine endliche, d. h. sie erklärt den zu bestimmenden Begriff durch eine endliche Anzahl bereits bekannter Begriffe B1, B2, B3, ...,Bn. âUnendliche Definitionen" (die nicht in endlicher Zeit verlaufen) sind Undinge. Wäre Königs Satz, daà alle âendlich definirbaren" reellen Zahlen einen Inbegriff von der Mächtigkeit ï0 ausmachen, richtig, so hieÃe dies, das ganze Zahlencontinuum sei abzählbar, was doch sicherlich falsch ist. .... Essential: Infinite definitions, are nonsense. If there were only alef_0 definable reals, then the continuum was countable. So Cantor clearly states, that undefinable reals are nonsense. And of course he is right. A number must be in trichotmy with others. Otherwise it is not a number. The idea of undefinable numbers was created in order to save set theory, just as some fools now claim uncountably many languages in order to invalidate my list of all words in all languages. Regards, WM |