Why can no one in sci.math understand my simple point?
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From: Sylvia Else on 23 Jun 2010 20:26 On 24/06/2010 6:08 AM, Virgil wrote: > 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. I was just considering it a matter of perspective. If you have a single ended list, you can consider the end to be the first element or the last element. I can't see that the choice makes any difference mathematically, other than to invert whatever ordering rule applies to the list. Sylvia.
From: Sylvia Else on 23 Jun 2010 20:39 On 23/06/2010 8:44 PM, WM wrote: > On 23 Jun., 06:47, Sylvia Else<syl...(a)not.here.invalid> wrote: > >> >> Rather than argue that VMs proposition fails on that point, I wanted to >> address the flaw, in order to find a more substantial objection. > > My initials ar WM. > > There is no flaw in the argument: Your bijection either contains all > constructed antidiagonals. Then a list contains also its antidiagonal, > because every list has an antidiagonal and your bijection (that is > only a permutation of my list) contains all lines and antidiagonals. > I.e., there is no missing antidiagonal of a "limit" list outside of > the bijection. > > Or there is a last diagonal of the limit list that does not belong to > your bijection (and to my list). Then there is a countable set (the > set that includes this last diagonal) that is not listable. > > Think over that only a little bit. It is not difficult to understand. > > Regrads, WM Thing is, I just cannot see that you've constructed a countable set that should contain its own anti-diagonal when expressed as a list. All you've done is construct a countable set of countable sets, each of which contains the anti-diagonals of its predecessors, but none of which by construction can be expected to contain its own anti-diagonal. You've then claimed, but without any kind of formal argument, that if you take this to the limit, the result is a countable set that should contain its own anti-diagonal when expressed as a list. As proofs go, it's rather underwhelming. Sylvia.
From: Virgil on 23 Jun 2010 21:47 In article <5LSdnYvXabOVBr_RnZ2dnUVZ7tKdnZ2d(a)brightview.co.uk>, "Mike Terry" <news.dead.person.stones(a)darjeeling.plus.com> wrote: > "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 thought of something vaguely similar, and saw that it also did not support WM's argument. As far as I can see, there is no countable set (by definition admitting a bijection from N) for which an antidiagonal does not exist, thus EVERY COUNTABLE SET fails to contain all reals. This convinces me that the set of reals is NOT one of those many countable sets of reals, i.e., is not countable.
From: Virgil on 23 Jun 2010 21:52 In article <88fn10FqspU3(a)mid.individual.net>, Sylvia Else <sylvia(a)not.here.invalid> wrote: > On 24/06/2010 6:08 AM, Virgil wrote: > > 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. > > I was just considering it a matter of perspective. If you have a single > ended list, you can consider the end to be the first element or the last > element. I can't see that the choice makes any difference > mathematically, other than to invert whatever ordering rule applies to > the list. > > Sylvia. The elements of a list are standardly indexed by the naturals and so standardly inherit the ordering of the naturals. In this sense, at least, a non-empty list always has a first and an infinite list never has a last. But that standard order is merely a matter of custom, not necessity.
From: Newberry on 23 Jun 2010 23:38
On Jun 23, 1:49 pm, Virgil <Vir...(a)home.esc> wrote: > In article > <8cd5db0d-46a5-4deb-8fe5-9a28acad5...(a)k39g2000yqb.googlegroups.com>, > > Newberry <newberr...(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 OK, so construct it assuming injection. > 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. |