Why can no one in sci.math understand my simple point?
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From: Virgil on 23 Jun 2010 16:49 In article <8cd5db0d-46a5-4deb-8fe5-9a28acad56c3(a)k39g2000yqb.googlegroups.com>, Newberry <newberryxy(a)gmail.com> wrote: > Cantor's proof starts with the assumption that a bijection EXISTS, not > that it is effective. Actually, a careful reading shows Cantor's proof merely assumes an arbitrary INJECTION from N to R (originally from N to the set of all binary sequences, B) which is NOT presumed initially to be surjective, and then directly proves it not to be surjective by constructing the "antidiagonal"as a member of the codomain not in the image. Thus proving that ANY injection from N to R (or B) fails to be surjective. For some unknown reason, the DIRECT "anti-diagonal" proofs given by Cantor, and his followers, are often misrepresented as being proofs by contradiction, but they never were.
From: WM on 23 Jun 2010 16:55 On 23 Jun., 15:42, 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 that does not help. It is as clear that at any later line you have yet seen less lines than you will have to see. Therefore, without finished infinity, you cannot see any line. Most of them, nearly all of them (infinitely many compared to finitely many) will never be seen. > > > > >> 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. How should that be possible? How many there ever may have been constructed. The majority of infinitely many will remain to be constructed. And even if infinitely many diagonals would have been added, then the resulting list would yield another diagonal. Why do you think that would be a complete list withot its own diagonal? > 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 It is very naive to believe that there would result a list without a first line. In my construction every list has a first line and every diagonal will be placed in front of it. How should a list without first line occur? - any element you might claim is the first is in fact preceded > by infinitely many other elements. That is ridiculous! How should a list without first element come to existence? > > Your proof falls apart if you cannot construct the anti-diagonal which > you claim should have been in the list. If it is in the list, I can construct it and place it in front of the list. Why should that situation change? > > As I observed earlier, this problem can be obviated What problem? There is no problem. There is an infinite sequence of lists and of antidiagonals. If the sequence is interrupted by a list without containing its antidiagonal, then Cantor's argument fails. If the sequence is not interrupted by a list containing its antidiagonal, then the countable set of antidiagonals is not listable. > >> 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. No, you are completely in error. How should a missing first line come into existence? If it occured in the process of construction, I would stop and boast that the preceding list has no antidiagonal. Why should I continue? And, first of all, *how* should I continue? To create another blank line from a list without first line??? Can you tell me that? Can you suppose that you could or would like to produce such nonsense-list? Regards, WM >
From: Mike Terry on 23 Jun 2010 19:02 "WM" <mueckenh(a)rz.fh-augsburg.de> wrote in message news:ccf88754-1d64-46f3-ad9f-372ef6fe91c9(a)i28g2000yqa.googlegroups.com... > 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)? This is not a list of numbers. L0 is not a number, it is a list. Therefore (..., An, ... A2, A1, A0, L0) does not have an anti-diagonal. > 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 > „endliche Definitionen" zulassen und solche, die „unendliche > 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. > „Unendliche Definitionen" (die nicht in endlicher Zeit verlaufen) sind > Undinge. > Wäre Königs Satz, daß alle „endlich definirbaren" reellen Zahlen einen > Inbegriff von der Mächtigkeit 0 ausmachen, richtig, so hieße dies, > das ganze Zahlencontinuum sei abzählbar, was doch sicherlich falsch > ist. > ... I will have a go at that, but I only did "O-level" German 30 or so years ago, so I'm not sure how I'll get on... (Thanks anyway) > > 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 >
From: Mike Terry on 23 Jun 2010 19:38 "Virgil" <Virgil(a)home.esc> wrote in message news:Virgil-12E0AD.00311823062010(a)bignews.usenetmonster.com... > 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. Yes, WM's "list" is simply not a list, and I think all this "trying to turn a clearly wrong statement into something meaningful on WM's behalf" ultimately just makes the thread go on and on without getting anywhere. *WM* has to say what he actually meant if he can, and if he can't then we can leave and go on to another thread where we can maybe achieve something (not sure what! hehe). It seems to me that although WM makes numerous mistakes in his wording, with persistence he actually does end up clarify what he's saying so that things can move forward. But I had my own idea which doesn't really help WM, but neatly avoids the main problem at the moment... We start with the list L0 = (Q0, Q1, Q2, ...) [list of rationals as before] and form it's antidiagonal A0 as before. BUT... we create our new list L1 by inserting A0 at position 2: L1 = (Q0, A0, Q1, Q2, Q3, Q4, ...) And call L1's anti-diagonaly A1, which we stuff into L1 at position 4 to get L2: L2 = (Q0, A0, Q1, A1, Q2, Q3, Q4, ...) .... and so on. (An is the antidiag for Ln, and is inserted at posiiton (2n+2) to form list L_(n+1). The advantage is that there is now a well defined limit list containing ALL the antidiagonals, which I'll name Loo: Loo = (Q0, A0, Q1, A1, Q2, A2, Q3, A3, Q4, A4, Q5, ...) This seems to be as close to what WM is after as I can come up with. But it still doesn't work, as Loo has an antidiagonal which is not in Loo, and there is still no contradiction to be had. Regards, Mike. > > > > 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: Virgil on 23 Jun 2010 19:52
In article <b7fa9621-55d5-4d36-bd93-b2324bb9485e(a)z10g2000yqb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 23 Jun., 15:42, 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 that does not help. It is as clear that at any later line you have > yet seen less lines than you will have to see. Therefore, without > finished infinity, you cannot see any line. Most of them, nearly all > of them (infinitely many compared to finitely many) will never be > seen. Seeing them is not required. Nothing in, for example, FOL+ZFC requires sets or their members to be visible. > > > > > > > > >> 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. > > How should that be possible? How many there ever may have been > constructed. The majority of infinitely many will remain to be > constructed. Any of them need for the construction of a new antidiagonal can be constructed as needed. > And even if infinitely many diagonals would have been added, then the > resulting list would yield another diagonal. Thus no list can ever be "complete", and the set of elements of a list is never the whole, > Why do you think that > would be a complete list withot its own diagonal? There is no way to list or count every real as Cantor and others have proved in several ways and WM has yet to disprove successfully. Given any list (counting) of reals (or binary seqeuences), there are at least as many unlisted (uncounted) as in the list. > > > > 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 > > It is very naive to believe that there would result a list without a > first line. > In my construction every list has a first line and every diagonal will > be placed in front of it. How should a list without first line occur? When one has as many new first lines as entries in the original list. > > > - any element you might claim is the first is in fact preceded > > by infinitely many other elements. > > That is ridiculous! How should a list without first element come to > existence? > > > > > Your proof falls apart if you cannot construct the anti-diagonal which > > you claim should have been in the list. > > If it is in the list The point is that anti-diagonals are NOT in the list from which they are constructed. > > > > As I observed earlier, this problem can be obviated > > What problem? There is no problem. There is with WM's world being imposed on anyone else's, as it won't fit. > > > > >> 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. > > No, you are completely in error. How should a missing first line come > into existence? By straightforward construction. What WM really means to ask is how to get a pseudo-sequence without a first elemnt. One can easily do it as an infinite process completed. Note that in the real world, there are myriads of infinite processes being completed all the time. For example all motions are infinite processes in a very real sense. |