My FINAL Cantor thread!
in [Math]
|
Prev: Symmetry of BP's economic blowout & scarcity of their product?
Next: Ha! I have no job. I am divorced though. It figures. I will hangmyself
From: Graham Cooper on 24 Jun 2010 20:28 On Jun 25, 9:10 am, George Greene <gree...(a)email.unc.edu> wrote: > On Jun 24, 6:50 pm, Graham Cooper <grahamcoop...(a)gmail.com> wrote: > > > > Consider the list of computable reals. > > > > > > Let w = the digit width of the largest set > > > of complete permutations > > > > > > assume w is finite > > This is just idiotic. > Every real is infinitely wide BY DEFINITION, > and again, this width is, BY DEFINITION, THE SMALLEST infinity, > so w is KNOWN IN ADVANCE NOT to be finite. Use another letter. I have not defined w to be the width of any real. w is the max width of the set of sets of reals that contain every possible sequence for some digit width. Herc
From: Graham Cooper on 24 Jun 2010 20:49 On Jun 25, 10:18 am, "Mike Terry" <news.dead.person.sto...(a)darjeeling.plus.com> wrote: > "Graham Cooper" <grahamcoop...(a)gmail.com> wrote in message > > news:df261549-3c6e-4f4d-844e-65f14cbfa8b3(a)a30g2000yqn.googlegroups.com... > > > > Consider the list of computable reals. > > > > Let w = the digit width of the largest set > > > of complete permutations > > > > assume w is finite > > > there are 10 computable copies of the > > > complete permutations of width w > > > each ending in each of digits 0..9 (at position w+1) > > > which generates a set larger than width w > > > so finite w cannot be the maximum size > > > > therefore w is infinite > > > ---- > > or there is no largest set of complete permutations. > > E.g. have you considered the possibility that: > 1) all 1 digit permutations are in the list > 2) all 2 digit permutations are in the list > ... > 3) all n digit permutations are in the list > ... > 4) not all (countably) infinite permutations are in the list > > Then your w does not exist. > > Regards, > Mike. Yes but there is unlimited width to the complete sets of possible sequences should imply there is unlimited width of all sequences. Distinguishing the 2 and claiming you construct a new sequence anyway is a narrow viewpoint. Herc
From: Mike Terry on 25 Jun 2010 15:02 "Graham Cooper" <grahamcooper7(a)gmail.com> wrote in message news:d705aded-4c8b-4edb-916b-0cb74c721ea1(a)d37g2000yqm.googlegroups.com... > On Jun 25, 10:18 am, "Mike Terry" > <news.dead.person.sto...(a)darjeeling.plus.com> wrote: > > "Graham Cooper" <grahamcoop...(a)gmail.com> wrote in message > > > > news:df261549-3c6e-4f4d-844e-65f14cbfa8b3(a)a30g2000yqn.googlegroups.com... > > > > > > Consider the list of computable reals. > > > > > > Let w = the digit width of the largest set > > > > of complete permutations > > > > > > assume w is finite > > > > there are 10 computable copies of the > > > > complete permutations of width w > > > > each ending in each of digits 0..9 (at position w+1) > > > > which generates a set larger than width w > > > > so finite w cannot be the maximum size > > > > > > therefore w is infinite > > > > ---- > > > > or there is no largest set of complete permutations. > > > > E.g. have you considered the possibility that: > > 1) all 1 digit permutations are in the list > > 2) all 2 digit permutations are in the list > > ... > > 3) all n digit permutations are in the list > > ... > > 4) not all (countably) infinite permutations are in the list > > > > Then your w does not exist. > > > > Regards, > > Mike. > > Yes but there is unlimited width to the complete sets of possible > sequences Correct. (i.e. we're both agreeing as far as step (3) above.) > should imply there is unlimited width of all sequences. Aaaargh, now you've gone back to unclear mode. What does "unlimited width of all sequences" mean? Whatever it means, it obviously does not imply all (countably) infinite permutations are in the list, because that's obviously false. (Not even 1 infinite digit permuatation is in the list :-) So... it seems I'm right - there is no "largest set of complete permutations", and so your w does not exist. > > Distinguishing the 2 and claiming you construct a new sequence > anyway is a narrow viewpoint. I'm not claiming that yet, just that your "proof" breaks down at its second line: Let w = the digit width of the largest set of complete permutations There is no largest set, so w doesn't exist. You could now redefine w so that it does exist, and present a new argument using this w. Or you could maybe restate your argument without using w. But as things are right now your argument doesn't work... Mike. > > Herc >
From: Mike Terry on 25 Jun 2010 18:11 "Mike Terry" <news.dead.person.stones(a)darjeeling.plus.com> wrote in message news:4pydnUChP7f9YLnRnZ2dnUVZ7sgAAAAA(a)brightview.co.uk... > "Graham Cooper" <grahamcooper7(a)gmail.com> wrote in message > news:d705aded-4c8b-4edb-916b-0cb74c721ea1(a)d37g2000yqm.googlegroups.com... > > On Jun 25, 10:18 am, "Mike Terry" > > <news.dead.person.sto...(a)darjeeling.plus.com> wrote: > > > "Graham Cooper" <grahamcoop...(a)gmail.com> wrote in message > > > > > > > news:df261549-3c6e-4f4d-844e-65f14cbfa8b3(a)a30g2000yqn.googlegroups.com... > > > > > > > > Consider the list of computable reals. > > > > > > > > Let w = the digit width of the largest set > > > > > of complete permutations > > > > > > > > assume w is finite > > > > > there are 10 computable copies of the > > > > > complete permutations of width w > > > > > each ending in each of digits 0..9 (at position w+1) > > > > > which generates a set larger than width w > > > > > so finite w cannot be the maximum size > > > > > > > > therefore w is infinite > > > > > ---- > > > > > > or there is no largest set of complete permutations. > > > > > > E.g. have you considered the possibility that: > > > 1) all 1 digit permutations are in the list > > > 2) all 2 digit permutations are in the list > > > ... > > > 3) all n digit permutations are in the list > > > ... > > > 4) not all (countably) infinite permutations are in the list > > > > > > Then your w does not exist. > > > > > > Regards, > > > Mike. > > > > Yes but there is unlimited width to the complete sets of possible > > sequences > > Correct. (i.e. we're both agreeing as far as step (3) above.) > > > should imply there is unlimited width of all sequences. > > Aaaargh, now you've gone back to unclear mode. What does "unlimited width > of all sequences" mean? > > Whatever it means, it obviously does not imply all (countably) infinite > permutations are in the list, because that's obviously false. (Not even 1 > infinite digit permuatation is in the list :-) Well maybe *some* infinite digit permutations, since you start off "Consider the list of computable reals". So I think you mean for *all* computable reals to be included? This is OK because I agree the computable reals *can* be listed, although in this case you do not have a "computable list". (Also, be aware that there are many possible lists of computable reals, so you'd be better starting "Consider *a* list of all computable reals".) In any case, the list does not have any *uncomputable* reals in it, and these are of "width" omega (first infinite ordinal). So (4) above is still correct, and your w does not exist. OR... maybe you have a secret proof that *all* infinite digit sequences are computable? Regards, Mike. > > > So... it seems I'm right - there is no "largest set of complete > permutations", and so your w does not exist. > > > > > Distinguishing the 2 and claiming you construct a new sequence > > anyway is a narrow viewpoint. > > I'm not claiming that yet, just that your "proof" breaks down at its second > line: > > Let w = the digit width of the largest set of complete permutations > > There is no largest set, so w doesn't exist. > > You could now redefine w so that it does exist, and present a new argument > using this w. Or you could maybe restate your argument without using w. > But as things are right now your argument doesn't work... > > Mike. > > > > > > Herc > > > >
From: |-|ercules on 26 Jun 2010 01:08
"Mike Terry" <news.dead.person.stones(a)darjeeling.plus.com> wrote >> > should imply there is unlimited width of all sequences. >> >> Aaaargh, now you've gone back to unclear mode. What does "unlimited width >> of all sequences" mean? >> >> Whatever it means, it obviously does not imply all (countably) infinite >> permutations are in the list, because that's obviously false. (Not even 1 >> infinite digit permuatation is in the list :-) > the proof is lost on you. I will say again I am not building infinite amount of finite sized blocks of numbers. I am sampling the pattern of digits early in the string of reals, and using induction to show the structure of digits out to infinity wide, i.e. the TOTAL digit strings of all the computable reals. e.g. 006666666.. 016666666.. 106666666.. 116666666.. The structure of the first 2 digits is a complete permutation set. I prove the maximum width of complete permutations is infinity. Think of a CPS as a tall rectangle which is sampled at larger and larger sizes with induction, revealing the digit structure of the entire list (of computable reals). It's a BREADTH and DEPTH wise induction so it reveals the nature of lists more reliably than drawing a square and striking a line down the center saying NO! (that's diagonalisation if you missed the pun) > > In any case, the list does not have any *uncomputable* reals in it, and > these are of "width" omega (first infinite ordinal). So (4) above is still > correct, and your w does not exist. OR... maybe you have a secret proof > that *all* infinite digit sequences are computable? > I'm glad for the open mindedness of your final statement, because the top down attacks on my proof are merely blind belief that ZFC is complete. Herc |