CANTOR'S PROOF: An AD(n) =/= L(n,n) -> An AD(n) =/= L(n,n)
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From: |-|ercules on 17 Jun 2010 21:53 CONSTRUCT a *new sequence of digits* (somehow?) like so An AD(n) =/= L(n,n) Then we PROVE that it's a new sequence of digits An AD(n) =/= L(n,n) VOILA! Superinfinity! Just ignore the fact that it never comes up with a new sequence of digits. Consider this LOGIC LOOP as an infinite sequence of propositions. AD = 0 D = 1 WHILE TRUE AD = AD + a different digit than the Dth digit of the Dth real) / 10^D YOU HAVEN'T PRODUCED A NEW DIGIT SEQUENCE YET! D++ WEND How does this make a *new digit sequence* that is not present on the list of computable reals? Herc -- If you ever rob someone, even to get your own stuff back, don't use the phrase "Nobody leave the room!" ~ OJ Simpson
From: |-|ercules on 19 Jun 2010 05:34 CANTOR'S PROOF of modifying the diagonal does not create any new sequence at all. All it does is this CONSTRUCT a digit sequence like so An AD(n) =/= L(n,n) And then you say, it's different to each number like so PROOF An AD(n) =/= L(n,n) But you have not demonstrated a NEW SEQUENCE OF DIGITS. All you've done is this [ An AD(n) =/= L(n,n) -> An AD(n) =/= L(n,n) ] -> Higher Infinities Your actual 'proof' is a specific example of the above 'proof'! [ 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
From: George Greene on 19 Jun 2010 14:54 On Jun 17, 9:53 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > CONSTRUCT a *new sequence of digits* (somehow?) like so > > An AD(n) =/= L(n,n) That is NOT a construction. That is just an ASSERTION ABOUT each individual element of the anti-diagonal. And it is mis-stated in any case. AD DOES NOT TAKE (n) as an argument! AD(.) is the anti-diagonal OF A LIST! It takes L as an argument. To CONSTRUCT this, you would have to say what the nth element of the anti-diagonal WAS EQUAL to,NOT what it was UNequal to! What you ACTUALLY MEANT to say here was An[ AD(L,n) = 9 - L(n,n) ] THAT is a construction. The statement you made will follow FROM this as a consequence, since Ane{0..9}[ 9 - n =/= n ] > Then we PROVE that it's a new sequence of digits > > An AD(n) =/= L(n,n) Well, yes, this does prove that the anti-diagonal of L is NOT ON the list L. But that actually does NOT mean that ANYthing is NEW! It just means that the list DID NOT ALREADY HAVE this element on it! It means that L WAS NOT "the list" (or even A list) of EVERY real! However, since we didn't make any assumptions at all about L here, other than that it was square and denumerable (in both directions), this proves THAT ANY AND EVERY list WITH THAT SHAPE is NOT "the list of all" reals! In other words, THERE CANNOT BE ANY "list of ALL" reals! > VOILA! Superinfinity! This is not where the superinfinity part comes from. It is possible that the class of all reals either a) doesn't exist, or b) even if it does, does not HAVE a SIZE. Actually getting to a superinfinity number REQUIRES MORE than this, but since you already had a list that was both infinitely wide and infinitely long, the axiom of infinity has sort of been assumed already. You are going to need to invoke choice and power as well, to get an actual new number, but this is of course all above your head. > Just ignore the fact that it never comes up with a new sequence of digits. AD(L) *IS* a new sequence, DUMBASS. YOU JUST PROVED it was NEW! It WAS NOT ON the ORIGINAL list L! > > Consider this LOGIC LOOP as an infinite sequence of propositions. > > AD = 0 > D = 1 > WHILE TRUE > AD = AD + a different digit than the Dth digit of the Dth real) / 10^D > YOU HAVEN'T PRODUCED A NEW DIGIT SEQUENCE YET! > D++ > WEND > > How does this make a *new digit sequence* that is not present on the > list of computable reals? > > Herc > -- > If you ever rob someone, even to get your own stuff back, don't use the phrase > "Nobody leave the room!" ~ OJ Simpson
From: |-|ercules on 19 Jun 2010 16:21 "George Greene" <greeneg(a)email.unc.edu> wrote > On Jun 19, 5:34 am, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> CANTOR'S PROOF of modifying the diagonal does not create any new sequence at all. >> >> All it does is this >> >> CONSTRUCT a digit sequence like so >> An AD(n) =/= L(n,n) >> >> And then you say, it's different to each number like so >> >> PROOF >> An AD(n) =/= L(n,n) > > THAT'S RIGHT, DUMBASS, IT IS DIFFERENT from EVERY number on the list! > >> >> But you have not demonstrated a NEW SEQUENCE OF DIGITS. > > YES, WE HAVE! THIS NUMBER IS NOT ON the OLD list! > THEREFORE, IT IS *NEW*!! YOU HAD SAID that the OLD list was > the list of ALL the numbers! IT WASN'T! THIS NUMBER was NOT ON it! > THIS NUMBER IS THEREFORE *NEW*!! Rubbish! You don't show any new pattern of digits, you just SAY it's new. If all digits of a single infinite expansion can be contained with increasing finite prefixes, and the computable set of reals has EVERY finite prefix, then all digits of EVERY infinite expansion are contained (in increasing segments). Find some suitable substitution for the word "contained" since this bothers you, and that proposition disproves that modifying the diagonal produces a new sequence of digits. Yes I know An AD(n) = (L(n,n) + 1) mod 9 -> An AD(n) =/= L(n,n) and millions of other zero entropy formula. Herc
From: George Greene on 20 Jun 2010 02:36
On Jun 19, 4:21 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > Yes I know > > An AD(n) = (L(n,n) + 1) mod 9 No, you don't. ACTUALLY, AD(n) = 9 - L(n,n) |