A BLATENT FLAW in Cantor's diag proof
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From: |-|ercules on 8 Jun 2010 16:33 "William Hughes" <wpihughes(a)hotmail.com> wrote .. > On Jun 8, 5:19 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> "George Greene" <gree...(a)email.unc.edu> wrote >> >> > On Jun 7, 11:17 pm, Tim Little <t...(a)little-possums.net> wrote: >> >> On 2010-06-08, |-|ercules <radgray...(a)yahoo.com> wrote: >> >> Delete "(infinite)", and your statement is correct. There is no such >> >> thing as an infinite finite length though, so inserting the word >> >> "infinite" there makes no sense. >> >> > It makes a LITTLE bit of sense. What he MEANT was that >> > there are infiniteLY MANY of these different finite lengths and >> > that there is NO finite upper bound on them. >> >> Partial credit for George. >> >> All possible digit sequences are computable to all, as in an infinite amount of, finite lengths >> >> TRUE or NOT? >> >> Herc > > True. However, this does not mean what you think it > means. The fact that for a number y, > > every subsequence of the digit sequence of y > that has a last digit > > is computable, > > does not mean that the > digit sequence of y (which does not have a last digit) is > computable. > > There is no way to combine all the computable subsequences > to get a computable sequence. > > - William Hughes Is a *new digit sequence* missing from the list? Here is the claim again: >> All possible digit sequences are computable to all, as in an infinite amount of, finite lengths Herc
From: William Hughes on 8 Jun 2010 16:34 On Jun 8, 5:29 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > > The infinitely many long sequences of all possible digit sequences DOESN'T MISS A SEQUENCE OF DIGITS. Since every sequence has a last digit, any sequence of digits that does not have a last digit is missed. - William Hughes
From: |-|ercules on 8 Jun 2010 16:39 "William Hughes" <wpihughes(a)hotmail.com> wrote > On Jun 8, 5:29 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> >> The infinitely many long sequences of all possible digit sequences DOESN'T MISS A SEQUENCE OF DIGITS. > > Since every sequence has a last digit, any sequence > of digits that does not have a last digit is missed. You're as confused as when you said there is no algorithm to produce an infinite list. Herc
From: William Hughes on 8 Jun 2010 16:45 On Jun 8, 5:39 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > "William Hughes" <wpihug...(a)hotmail.com> wrote > > > On Jun 8, 5:29 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: > > >> The infinitely many long sequences of all possible digit sequences DOESN'T MISS A SEQUENCE OF DIGITS. > > > Since every sequence has a last digit, any sequence > > of digits that does not have a last digit is missed. > > You're as confused as when you said there is no algorithm to produce an infinite list. > > Herc Is this digit sequence (which does not have a last 3) 33333... in this list 1 3 2 33 3 333 .... of sequences (all of which have a last 3). Yes or No. - William Hughes
From: |-|ercules on 8 Jun 2010 16:52
"William Hughes" <wpihughes(a)hotmail.com> wrote > On Jun 8, 5:39 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> "William Hughes" <wpihug...(a)hotmail.com> wrote >> >> > On Jun 8, 5:29 pm, "|-|ercules" <radgray...(a)yahoo.com> wrote: >> >> >> The infinitely many long sequences of all possible digit sequences DOESN'T MISS A SEQUENCE OF DIGITS. >> >> > Since every sequence has a last digit, any sequence >> > of digits that does not have a last digit is missed. >> >> You're as confused as when you said there is no algorithm to produce an infinite list. >> >> Herc > > Is this digit sequence (which does not have a last 3) > > 33333... > > in this list > > 1 3 > 2 33 > 3 333 > ... > > of sequences (all of which have a last 3). > > Yes or No. > > - William Hughes No. For the standard interpretation, in functional programming that's how you represent an infinite stream. Herc |