Why can no one in sci.math understand my simple point?
in [Math]
|
Prev: power of a matrix limit A^n
Next: best way of testing Dirac's new radioactivities additive creation Chapt 14 #163; ATOM TOTALITY
From: Virgil on 23 Jun 2010 00:27 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. 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, ...}. But as listability = countability everywhere except in WM's world, WM's arguments claiming that tone can have ne without the other fails.
From: Virgil on 23 Jun 2010 00:28 In article <88d97pF74kU1(a)mid.individual.net>, Sylvia Else <sylvia(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-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. > > 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. > > Sylvia. "I think she's got it!"
From: Sylvia Else on 23 Jun 2010 00:47 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. 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. Rather than argue that VMs proposition fails on that point, I wanted to address the flaw, in order to find a more substantial objection. Sylvia.
From: Newberry on 23 Jun 2010 00:53 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. The list of naturals is compressible the diagonal is generally incompressible. > > > > > > > > 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 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. |