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From: hagman on 19 Jun 2010 13:09 On 19 Jun., 11:38, "|-|ercules" <radgray...(a)yahoo.com> wrote: > [ An AD(n) = (L(n,n) + 1) mod 9 -> An AD(n) =/= L(n,n) ] -> Higher Infinities > > Does anyone agree with the above version of Cantor's proof? > > Herc > -- > "And God posted an angel with a flaming sword at the gates of Cantor's > paradise, that the slow-witted and the deliberately obtuse might not > glimpse the wonders therein." ~ Barb Knox What you post can hardly be called a proof and hence cannot in this form be easily agreed with. For those in the know, it is clear how to grow a proof out of this germ. (Namely a proof of the statement that the set of all maps N -> {0, ..., 9} is not countable as evidenced by the fact that for each map L: N x N -> {0,...,9} there exists a map AD: N -> {0,...,9} such that none of the maps L_k : N-> {0, ..., 9}, n |-> L(k,n) equals AD) For someone like you it may be possible to fill the ambigous gaps in a way producing nonsense. hagman
From: Mike Terry on 19 Jun 2010 20:33 "|-|ercules" <radgray123(a)yahoo.com> wrote in message news:883hfpFd77U1(a)mid.individual.net... > [ An AD(n) = (L(n,n) + 1) mod 9 -> An AD(n) =/= L(n,n) ] -> Higher Infinities An AD(n) = (L(n,n) + 1) mod 9 -> An AD(n) =/= L(n,n) Agree so far. You are missing the key next step: ... -> An AD =/= L(n) which is saying that : AD =/= L(1) [i.e. AD is not 1st digit sequence in list] AD =/= L(2) [...or 2nd digit sequence in list] AD =/= L(3) [...or 3rd digit sequence in list] ... [...and so on...] i.e. IOW, AD is not in the list. It is a new digit sequence :-) Whether this implies anything about Higher Infinities is too vague a question to try and answer. (If you say precisely what you're asking I'll try and answer...) > > Does anyone agree with the above version of Cantor's proof? It's a bit "sketchy" :-) But if you add in my line above, and chop out the metaphysical bit on the end, it could be recognised as a reasonable sketch of the basic result. The sort of thing you might write on the palm of your hand to take in to an exam, intending to pad it out with the missing details to fill in the many gaps... Mike
From: George Greene on 21 Jun 2010 23:02
On Jun 19, 5:38 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: > [ An AD(n) = (L(n,n) + 1) mod 9 -> An AD(n) =/= L(n,n) ] -> Higher Infinities > > Does anyone agree with the above version of Cantor's proof? THIS IS NOT a version of Cantor's proof. Cantor's Theorem says that no set is as big as its powerset. It does NOT MENTION "9" or any other number. The closest it comes to mentioning numbers has to do with xef(x) being TRUE or FALSE. That gets complemented with "not", NOT with 1-x. THOUGH OF COURSE you could treat xef(x) being true as 1, and xef(x) being false (or ~xef(x) being true) as 0, and THEN you could complement with 1-x. If the underlying set that x was taken from WAS DENUMERABLE (which the positions on a list ALWAYS ARE, whether it is a list of numbers or the list of digits in the expansion of a real number), then you could interpret the membership or non-membership of EACH AND EVERY one of these x's, in any subset of the infinite set, as a 1 or a 0 IN THE BINARY expansion of a real. THAT IS THE CONNECTION between Cantor's theorem and real numbers. What YOU are talking about simply IS NOT a version of Cantor's Theorem, WHICH IS NOT ABOUT infinity in general; THE EXACT SAME proof holds FOR FINITE sets. |