CANTOR DISPROOF <<<<<<<<<<<<<<<<
in [Logic]
|
From: Alan Smaill on 28 Jun 2010 17:41 Transfer Principle <lwalke3(a)lausd.net> writes: > Still, I see nothing wrong with having a theory which does > satisfy either Herc's or WM's intuitions. Of course you're right; there's nothing wrong with having such theories. Is there anyone who has said anything different in sci.logic? Do you think that Herc or WM has actually proposed any such theory? -- Alan Smaill
From: |-|ercules on 28 Jun 2010 19:04 "|-|ercules" <radgray123(a)yahoo.com> wrote > > XXXXX... is not a single real. It's shorthand for a Complete Digit Sequence > for all widths. I should explain my ambiguous use of variables. In your counter examples, X is a standard variable for a real or a list. In my demonstration, X is a 1 digit wide matrix, XX is a 2 digit wide matrix Herc
From: Sylvia Else on 28 Jun 2010 20:08 On 28/06/2010 11:15 PM, |-|ercules wrote: > You prove case n, by logical inspection, that the property holds for all > elements in the list of size n. (a whole object) > > I prove case n, by logical inspection, that the property holds for > prefixes of width n. (a subset of digits) > > You prove, by induction, that all (finite) sizes of lists the property > holds. > > I prove, by induction, that the property holds for all digit widths. > (all digits). You only prove that it holds for finite digit widths. Somehow you want to be allowed to extrapolate from finite to infinite when it suits you, but not for me to allowed to do the same thing when it doesn't suit you. >> >> BTW, it occurs to me that if you have a list containing all the >> prefixes obtained by permuting 1, 2, 3, ... n digits, followed in each >> case by infinite suffixes, then we can use the anti-diagonal technique >> to construct an infinite sequence A whose first k digits differ from >> the corresponding diagonal digits of your list. That sequence A isn't >> in your list. >> >> When you move from n digit prefixes to n+1 digit prefixes, you add >> 10^n+1 new sequences, and we can change the next 10^n+1 digits of A so >> that it is still not in the list. >> >> Thus at each induction step, you increase the number of prefixes in >> the list, but you retain a sequence A that is not in the list. This >> remains true for any finite number of steps. >> >> That is, no induction step extinguishes sequence A which is not in the >> list. It is hardly reasonable to argue that extending this to an >> infinite number of steps creates a list with every inifinite sequence, >> but in the process mysteriously extinguishes sequence A. >> >> Sylvia. > > That's just a proof that finite lists are incomplete, and that sequences > are shorter than > the number of sequences to cover every digit sequence. > Similar to the favorite argument of ANTI-Cantorians! > > A list contains all finite prefixes. > > Step 1: a 1 digit anti-diagonal is not a new sequence > Step n: a n digit anti-diagonal is not a new sequence > Step n+1: a n+1 digit anti-diagonal is not a new sequence (ok IF n is > then n+1 is, but no base step!) You're talking about finite length anti-diagonals. I clearly said "construct an *infinite* sequence A whose *first* k digits differ from the corresponding diagonal digits of your list." As I said, A is not in your list, and it never will be no matter how far you take the induction. Sylvia.
From: |-|ercules on 28 Jun 2010 20:42 "Sylvia Else" <sylvia(a)not.here.invalid> wrote > On 28/06/2010 11:15 PM, |-|ercules wrote: > >> You prove case n, by logical inspection, that the property holds for all >> elements in the list of size n. (a whole object) >> >> I prove case n, by logical inspection, that the property holds for >> prefixes of width n. (a subset of digits) >> >> You prove, by induction, that all (finite) sizes of lists the property >> holds. >> >> I prove, by induction, that the property holds for all digit widths. >> (all digits). > > You only prove that it holds for finite digit widths. Somehow you want > to be allowed to extrapolate from finite to infinite when it suits you, > but not for me to allowed to do the same thing when it doesn't suit you. > Nope! Your prove it for increasing different objects. I sample larger and larger sizes of the one object. Different style of proof! >>> >>> BTW, it occurs to me that if you have a list containing all the >>> prefixes obtained by permuting 1, 2, 3, ... n digits, followed in each >>> case by infinite suffixes, then we can use the anti-diagonal technique >>> to construct an infinite sequence A whose first k digits differ from >>> the corresponding diagonal digits of your list. That sequence A isn't >>> in your list. >>> >>> When you move from n digit prefixes to n+1 digit prefixes, you add >>> 10^n+1 new sequences, and we can change the next 10^n+1 digits of A so >>> that it is still not in the list. >>> >>> Thus at each induction step, you increase the number of prefixes in >>> the list, but you retain a sequence A that is not in the list. This >>> remains true for any finite number of steps. >>> >>> That is, no induction step extinguishes sequence A which is not in the >>> list. It is hardly reasonable to argue that extending this to an >>> infinite number of steps creates a list with every inifinite sequence, >>> but in the process mysteriously extinguishes sequence A. >>> >>> Sylvia. >> >> That's just a proof that finite lists are incomplete, and that sequences >> are shorter than >> the number of sequences to cover every digit sequence. >> Similar to the favorite argument of ANTI-Cantorians! >> >> A list contains all finite prefixes. >> >> Step 1: a 1 digit anti-diagonal is not a new sequence >> Step n: a n digit anti-diagonal is not a new sequence >> Step n+1: a n+1 digit anti-diagonal is not a new sequence (ok IF n is >> then n+1 is, but no base step!) > > You're talking about finite length anti-diagonals. I clearly said > "construct an *infinite* sequence A whose *first* k digits differ from > the corresponding diagonal digits of your list." > > As I said, A is not in your list, and it never will be no matter how far > you take the induction. > > Sylvia. As I said, your proof of finite incompleteness and short sequence lengths both have more trivial proofs, and induction on the diagonal has a similar (but different) proof that anti-diagonals contain no unique sequence. Herc
From: Sylvia Else on 28 Jun 2010 21:19
On 29/06/2010 10:42 AM, |-|ercules wrote: > "Sylvia Else" <sylvia(a)not.here.invalid> wrote >> On 28/06/2010 11:15 PM, |-|ercules wrote: >> >>> You prove case n, by logical inspection, that the property holds for all >>> elements in the list of size n. (a whole object) >>> >>> I prove case n, by logical inspection, that the property holds for >>> prefixes of width n. (a subset of digits) >>> >>> You prove, by induction, that all (finite) sizes of lists the property >>> holds. >>> >>> I prove, by induction, that the property holds for all digit widths. >>> (all digits). >> >> You only prove that it holds for finite digit widths. Somehow you want >> to be allowed to extrapolate from finite to infinite when it suits >> you, but not for me to allowed to do the same thing when it doesn't >> suit you. >> > > Nope! Your prove it for increasing different objects. > > I sample larger and larger sizes of the one object. Different style of > proof! > <sigh> So we're sampling now, not constructing. Fine. By sampling, I suppose you mean that you look through the list of computables to find sequences with the required prefixes. I question whether your inductive reasoning even works with that approach. It is certainly possible to find in the list of computables a sequence that starts with any finite prefix, but that doesn't mean that the inductive step is valid, merely that the result is true. An inductive step would need to take the form: --- If a sequence with a specific prefix p is in the list of computables, then prefixes constituting p followed by 0, p followed by 1, p followed by 2, etc. are also in the list. Then prove (by inspection presumably) that the prefixes 0 thru 9 exist. --- If the proof merely consists of observing that since prefixes one digit longer than p are also of finite length, and therefore exist in the list of computables, then that is not an inductive step (except in the very limited sense that if n is finite, then so is n+1). That aside, since the list of computables contains all finite prefixes, then when you've sampled the list to find sequences starting with all prefixes of length n digits, with the sample constituting a sublist[*] of length 10^n, I can sample the list of computables to find a sequence with a prefix of length 10^n which is not in your sublist. This remains true for all finite n. [*] I use the term loosely. It's really a listing of the subset. Sylvia. |