Why can no one in sci.math understand my simple point?
in [Logic]
|
From: WM on 23 Jun 2010 09:35 On 23 Jun., 15:25, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > WM <mueck...(a)rz.fh-augsburg.de> writes: > > Therefore, there is not every subset of an infinite set. > > What on Earth does this mean? It is Fraenkels text. I thought that you could understand German? In 1928 Fraenkel had realized Skolems proof (if a first-order-logic theory is consistent, then it has a countable model). In order to circumvent this problem, Fraenkel (and the later interpreters of the axiom of power set) must keep open the possibility that only countably many subsets of omega may exist. Hence, not all subsets exist "automatically". Reagrds, WM
From: Sylvia Else on 23 Jun 2010 09:42 On 23/06/2010 11:10 PM, WM wrote: > On 23 Jun., 14:51, Sylvia Else<syl...(a)not.here.invalid> wrote: > >> Cantor doesn't rely on being able to identify a last digit. He's just >> saying that no matter how far down the list you look, you'll find that >> the element at that point doesn't match the anti-diagonal. > > > That is wrong. Cantor uses the alleged "fact", that infinity can be > completed, i.e., that that the infinite list can be finished. > If he would only assume what you say, then the anti-diagonal could > remain in the unknown part of the list. You know and appreciate that > after *any* line number n there are infinitely many more lines? Yet it's clear that if you look at any later line m, you will still find that it doesn't match the anti-diagonal. > >> But you can't >> even begin to formulate his proof if you can't identify the first >> element of the list (and hence first digit of the anti-diagonal) either. > > Isn't it enough, also in my case, to know that every antidiagonal has > a first digit? In fact it has. Every antidiagonal is constructed from > a list with a first line (that is the previous antidiagonal) and the > remaining list. Every line has a finite number n. I agree that every anti-diagonal you added is well defined. But for your proof to work you also have to look at the anti-diagonal of the list after you've added all of the (infinitely many) constructed anti-diagonals. To construct the first digit of that anti-diagonal you have to look at the first element in the list. But it has no first element - any element you might claim is the first is in fact preceded by infinitely many other elements. So you have no way of deciding what the first digit of the anti-diagonal should be, let alone any of the other digits. Your proof falls apart if you cannot construct the anti-diagonal which you claim should have been in the list. As I observed earlier, this problem can be obviated by changing how the list is constructed. I can't see why you're opposing that, unless you consider it more than a cosmetic change for reasons that escape me. >> >> First and last are interchangeable, of course, but with your >> construction above, you can't specify either the first or the last. > > As I told you, my notation is only an abbreviation for the following > definition: > 1) Take a list L0 of all rational numbers. > 2) Construct its antidiagonal A0. > 3) Add it at position 0 to get (A0,L0) > 4) Construct the antidiagonal A1. > 5) and so on. > With a resulting 'list' which is infinite at both ends. > There occurs never a problem, because we know Hilberts hotel, don't > we? I'd have thought the Grand Hotel Cigar Mystery was the one to watch. Sylvia.
From: Newberry on 23 Jun 2010 10:31 On Jun 23, 6:42 am, Sylvia Else <syl...(a)not.here.invalid> wrote: > On 23/06/2010 11:10 PM, WM wrote: > > > On 23 Jun., 14:51, Sylvia Else<syl...(a)not.here.invalid> wrote: > > >> Cantor doesn't rely on being able to identify a last digit. He's just > >> saying that no matter how far down the list you look, you'll find that > >> the element at that point doesn't match the anti-diagonal. > > > That is wrong. Cantor uses the alleged "fact", that infinity can be > > completed, i.e., that that the infinite list can be finished. > > If he would only assume what you say, then the anti-diagonal could > > remain in the unknown part of the list. You know and appreciate that > > after *any* line number n there are infinitely many more lines? > > Yet it's clear that if you look at any later line m, you will still find > that it doesn't match the anti-diagonal. > > > > >> But you can't > >> even begin to formulate his proof if you can't identify the first > >> element of the list (and hence first digit of the anti-diagonal) either. > > > Isn't it enough, also in my case, to know that every antidiagonal has > > a first digit? In fact it has. Every antidiagonal is constructed from > > a list with a first line (that is the previous antidiagonal) and the > > remaining list. Every line has a finite number n. > > I agree that every anti-diagonal you added is well defined. But for your > proof to work you also have to look at the anti-diagonal of the list > after you've added all of the (infinitely many) constructed > anti-diagonals. To construct the first digit of that anti-diagonal you > have to look at the first element in the list. But it has no first > element - any element you might claim is the first is in fact preceded > by infinitely many other elements. So you have no way of deciding what > the first digit of the anti-diagonal should be, let alone any of the > other digits. > > Your proof falls apart if you cannot construct the anti-diagonal which > you claim should have been in the list. Why does WM have to construct the anti-diagonal and Cantor does not? > > As I observed earlier, this problem can be obviated by changing how the > list is constructed. I can't see why you're opposing that, unless you > consider it more than a cosmetic change for reasons that escape me. > > > > >> First and last are interchangeable, of course, but with your > >> construction above, you can't specify either the first or the last. > > > As I told you, my notation is only an abbreviation for the following > > definition: > > 1) Take a list L0 of all rational numbers. > > 2) Construct its antidiagonal A0. > > 3) Add it at position 0 to get (A0,L0) > > 4) Construct the antidiagonal A1. > > 5) and so on. > > With a resulting 'list' which is infinite at both ends. > > > There occurs never a problem, because we know Hilberts hotel, don't > > we? > > I'd have thought the Grand Hotel Cigar Mystery was the one to watch. > > Sylvia.
From: Newberry on 23 Jun 2010 10:37 On Jun 22, 11:42 pm, Virgil <Vir...(a)home.esc> wrote: > In article > <3c04e794-89f9-4438-995f-d818a4a0c...(a)5g2000yqz.googlegroups.com>, > > > > > > Newberry <newberr...(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 Cantor's proof starts with the assumption that a bijection EXISTS, not that it is effective. From this assumtion alone you cannot construct the anti-diagonal as you and some other people strangely claim. Then it is reasonable to conclude that the anti-diagonal is a contradiction in terms just like R = {x | ~(x in x}. > > > > > > > > > > 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 -- Hide quoted text - > > - Show quoted text -
From: Virgil on 23 Jun 2010 15:19
In article <ccf88754-1d64-46f3-ad9f-372ef6fe91c9(a)i28g2000yqa.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > 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)? If L0 is a list then the question is irrelevant. If you mean {..., An, ...A0, L0(0), L0(1), ...L0(n),...}, then there are all sorts of "anti-diagonals" possible depending on how you reorder that set into a list. > Then Cantor's argument fails as a list contains its antidiagonal. But you list is not a list of reals, since one of its members is a list of reals rather than a real, and it is only for list of reals that antidiagonals are to be constructed. > > b) Does it not? Then a list can not list all anti-diagonals that > belong to the countabe set constructed in my argument. That depends on the rule or rules by which those antidiagonals are to be constructed. > > > > > > > > > > If we use Cantor's definiton of "countable", then the set > > > {..., An, ...A0, L0(0), L0(1), ...L0(n),...} > > > is uncountable. I can count it by arranging it as the list {L0(0), A0, L0(1),A1, LO(2),A2, ...} > > > > 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. If Cantor's definition is not equivalent to the one you just rejected, what is Cantor's definition? WM continues to moan about what is and is not countable but refuses to give his definition of countable. If WM's definition of countable for an infinite set is not euivalnt to there existing a bijection between it and the naturals, then WM's is wrong. > 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. Irrelevant, as it cannot hold withing any standard set theory.. Every countable set, and every subset of a countable set injects to the set of naturals and the set of naturals surjects onto it. > > > > > > 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. Which is entirely consistent with having every subset of a countable set countable. And that is a theorem in FOL+ZFC and, as far as I know, all other standard set theories having countable sets. > > > > > > > > > > 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. Since one of the elements of your list (..., An, ... A2, A1, A0, L0), is not a real at all but a list of reals, and your listing is not in standard order, the issue of an antidiagonal is irrelevant. However, it is trivial, as shown above, to list the reals in L0 along with all the anti-diagonals listed and find an antidiagonal to that. |