Why can no one in sci.math understand my simple point?
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From: Virgil on 23 Jun 2010 16:04 In article <71a50fc4-4bc5-46b5-9ef0-8dc08a2437d1(a)r27g2000yqb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 23 Jun., 14:24, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > > "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> writes: > > > Well, the algorithm is clearly computable. If you give me a purported > > > list of all computable numbers, I can compute a missing number. > > > > Give how? We can't in any literal sense be given infinitary objects such > > as an infinite list of computable reals. > > > > That is wrong. WM trying to correct Aatu? C'est rire!
From: Virgil on 23 Jun 2010 16:08 In article <88eea0Fh36U1(a)mid.individual.net>, Sylvia Else <sylvia(a)not.here.invalid> wrote: > On 23/06/2010 10:40 PM, WM wrote: > > On 23 Jun., 13:54, Sylvia Else<syl...(a)not.here.invalid> wrote: > >> On 23/06/2010 8:34 PM, WM wrote: > >> > >> > >> > >> > >> > >>> On 23 Jun., 04:18, Sylvia Else<syl...(a)not.here.invalid> wrote: > >>>> On 23/06/2010 11:33 AM, Sylvia Else wrote: > >> > >>>>> On 23/06/2010 11:03 AM, Virgil wrote: > >>>>>> In article > >>>>>> <2000e81b-7c5a-41be-b6af-98f96f2fb...(a)w31g2000yqb.googlegroups.com>, > >>>>>> WM<mueck...(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. > >> > >>>> Hmm... > >> > >>>> A1 is the antidiagonal of L1 L2 L3... > >> > >>>> A2 is the antidiagonal of L1 A1 L2 L3... > >> > >>>> A3 is the antidiagonal of L1 A1 L2 A2 L3 L4... > >> > >>>> Each An is thus constructed from a list that is different from the list > >>>> into which it is inserted. So the construction does not lead to a list > >>>> that should contain its own anti-diagonal, and it doesn't. > >> > >>> Ln = > >> > >>> An > >>> ... > >>> A2 > >>> A1 > >>> A0 > >>> L0 > >> > >>> Does your bijection contain the anti-diagonal of > >>> (..., An, ... A2, A1, A0, L0)? > >> > >> I don't understand why you've recast it back to that form. > > > > That is a an abbreviation of the construction I proposed. Of course > > the "..." stand only for an infinite sequence of well defined digits > > at finite places. > > > >> You can't > >> even form the anti-diagonal of that - what would the first digit of the > >> antidiagonal be? > > > > What would the last digit of a normal Cantor-diagonal? Why should the > > first digit be more important than the last one? An infinite sequence > > of digits (that is not converging and not defined by a finite formula, > > like Cantor's diagonal sequence) is as undefined when the last digit > > is missing as it is when the first digit is missing. > > 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. 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. > > First and last are interchangeable, of course Not for infinite sequences when one of them clearly exists and the other clearly doesn't. The first ordinal number is 0. If first and last are so interchaneable, lets see WM interchange it with "the last ordinal number".
From: Virgil on 23 Jun 2010 16:15 In article <46cd79fe-ca18-4b2c-bcc4-5d2032fb3769(a)w31g2000yqb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> 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? > > > 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? No. You have to be able to tell how that, and any other digit of that "antidiagonal" was determined. 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. > > > > 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. > > There occurs never a problem, because we know Hilberts hotel, don't > we? There is also no problem in finding AN antidiagonal for the doubly open ended list (...,a2,a1,a0, LO.1, L0.2, L0.2, ...) once a surjection from N to it has been chosen.
From: Virgil on 23 Jun 2010 16:26 In article <e8d6cf87-4adf-492a-be49-04a4106e2dd5(a)z10g2000yqb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 23 Jun., 14:57, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > > Tim Little <t...(a)little-possums.net> writes: > > > In fact it would be a lot easier if instead of reals, we talked about > > > sets of natural numbers, and instead of computable reals, we talked > > > about recursively enumerable sets of natural numbers. > > > > It would be even more easy if instead of the uncountability of the reals > > we talked about the standard proof of Cantor's theorem: > > > > �Let f be a function taking elements of a set A to subsets of A. There > > �is then a subset of A not in the range of f. For consider the set > > > > � �D = {x in A | x not in f(x)}. > > > > �There is no a in A such that f(a) = D: if there were, we'd have a in > > �f(a) iff a not in f(a), a contradiction. > > That could be a good proof, if we knew that all subsets of an infinite > set A would exist. > But already Fraenkel wrote in the third edition of > his famous book (1928, p. 279f) with respect to the axiom of power > set, that: der Begriff "Teilmenge� eine andere, wesentlich engere > Bedeutung hat als in der CANTORschen Mengenlehre. In dieser konnten > wir bei der Bildung der Potenzmenge Um eine beliebige Gesamtheit von > Elementen aus m zu einer Teilmenge von m zusammenfassen und waren dann > sicher, da� diese sich unter den Elementen von Um findet. Jetzt ist > uns eine derartige, weitgehende Freiheit gew�hrende "Bildung� einer > Teilmenge von m nicht gestattet, also auch ihr Auftreten unter den > Elementen von Um keineswegs gesichert. (Contrary to the second edition > of 1923, Fraenkel now knew Skolem's proof of the same year and had to > explain how it could be circumvented.) > > Therefore, there is not every subset of an infinite set. Why then > should exist the subset of A that contains its pre-image if it does > not contain it, and does not contain ist, if it contains it? Why should it not exist? My own take on set theory is that everything should be allowed that does not allow proofs of statements of the form "P and not P". WM is a minimalist, but offers nothing but his own prejudices as to why he should be able to impose his views on everyone. It is not as if there were anything like Russell's paradox that results from, say, FOL+ZFC. If there ever were, that might justify WM's paranoia re infiniteness, but it certainly hasn't happened yet.
From: Virgil on 23 Jun 2010 16:34
In article <dc0a1be6-1934-42d1-aac5-30c2f4a577e0(a)z8g2000yqz.googlegroups.com>, Newberry <newberryxy(a)gmail.com> wrote: > 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? Because WM claims it can't be constructed, but it can, by reordering the doubly infinite list, in infinitely many ways. > > > > 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. |