Adriana Xenides Dead
in [Logic]
|
From: |-|ercules on 8 Jun 2010 22:46 "Daryl McCullough" <stevendaryl3016(a)yahoo.com> wrote ... > |-|ercules says... >> >>> "Jesse F. Hughes" <jesse(a)phiwumbda.org> wrote >>>> But I don't see how this analogy is supposed to make Cantor's theorem >>>> appear dubious. >> >>What about this statement? >> >>All possible digit sequences are computable to all, as in an infinite amount >>>of, finite lengths. > > The correct statement is this: > > I. For every real number r, for every natural number n, there > exists a computable real r' such that r agrees with r' in > the first n decimal places. > > Note the logical form of this statement: > > forall r, forall n, exists r' ... > > The order of quantifiers makes a difference! If the change > the order of quantifiers we get a similar-looking but false > statement: > > II. For every real number r, there exists a computable real r', > for every natural number n: > r agrees with r' in the first n decimal places. > > This has the logical form: > > forall r, exists r', forall n, ... > > It differs from the first statement in that the order > of the quantifiers has been changed. > > Statement I is true. Statement II is false. The claim that > "not all reals are computable" is equivalent to the claim > "Statement II is false". The diagonal argument proves that > Statement II is false, not that Statement I is false. > Right, but statement 1 is sufficient to disprove that modifying the diagonal produces a new sequence of digits. This is what a new sequence of digits looks like: 123 456 789 DIAG = 159 ANTIDIAG = 260 260 is a NEW SEQUENCE OF DIGITS. WHERE is the new sequence of digits coming from when all possible digit sequences are computable to all, as in an infinite amount of, finite lengths? Herc
From: |-|ercules on 8 Jun 2010 22:59 "Jesse F. Hughes" <jesse(a)phiwumbda.org> wrote > "|-|ercules" <radgray123(a)yahoo.com> writes: > >> "Jesse F. Hughes" <jesse(a)phiwumbda.org> wrote >>> "|-|ercules" <radgray123(a)yahoo.com> writes: >>> >>>> "Jesse F. Hughes" <jesse(a)phiwumbda.org> wrote >>>>> "|-|ercules" <radgray123(a)yahoo.com> writes: >>>>> >>>>>>>> Given a set of labeled boxes containing numbers inside them, >>>>>>>> can you possibly find a box containing all the label numbers of boxes >>>>>>>> that don't contain their own label number? >>>>>> >>>>>> Have a go mate! >>>>> >>>>> The answer is no, near as I can figure. >>>>> >>>>> Now, if you also knew that, for each set of numbers, there is a box >>>>> containing that set, then you'd have a paradox. Near as I can figure, >>>>> you *don't* know that. >>>>> >>>>> In set theory, on the other hand, we *do* know the analogous claim. >>>> >>>> So, no box ever containing the numbers of boxes not containing their own numbers >>>> means higher infinities exist? >>> >>> *Given* that every set of numbers is contained in some box, I guess >>> so. >>> >>> But I don't see how this analogy is supposed to make Cantor's theorem >>> appear dubious. >> >> >> So, as many have put it, the holy grail of mathematics, the infinite >> paradise is based on no box containing the numbers of boxes that >> don't contain their own number? > > I wouldn't call Cantor's theorem a holy grail, nor claim that it is > *based on* your odd analogy, but in any case this discussion is pretty > pointless. > > Your analogy does not refute the simple fact: Cantor's theorem is a > theorem of ZF. What's the deduction step in ZF that says "higher infinities exist". Herc
From: Tim Little on 8 Jun 2010 23:55 On 2010-06-09, |-|ercules <radgray123(a)yahoo.com> wrote: > Right, but statement 1 is sufficient to disprove that modifying the > diagonal produces a new sequence of digits. As predicted, Herc can't understand the difference nor why it makes nonsense of his claims. > WHERE is the new sequence of digits coming from when all possible > digit sequences are computable to all, as in an infinite amount of, > finite lengths? For *all* N, the sequence differs from the Nth entry in the list at the Nth digit (and possibly other positions as well). It is new because for *every* sequence in the list, the question "is it the same as this sequence" is answered "no". Unfortunately Herc really is dumb enough to think that with enough "no" answers, it really means "yes". - Tim
From: Tim Little on 8 Jun 2010 23:57 On 2010-06-09, |-|ercules <radgray123(a)yahoo.com> wrote: > What's the deduction step in ZF that says "higher infinities exist". Herc is dumb enough to think that every proof in ZF consists of a single deduction step, *the* step. - Tim
From: |-|ercules on 9 Jun 2010 00:50
"Tim Little" <tim(a)little-possums.net> wrote > On 2010-06-09, |-|ercules <radgray123(a)yahoo.com> wrote: >> Right, but statement 1 is sufficient to disprove that modifying the >> diagonal produces a new sequence of digits. > > As predicted, Herc can't understand the difference nor why it makes > nonsense of his claims. > > >> WHERE is the new sequence of digits coming from when all possible >> digit sequences are computable to all, as in an infinite amount of, >> finite lengths? > > For *all* N, the sequence differs from the Nth entry in the list at > the Nth digit (and possibly other positions as well). It is new > because for *every* sequence in the list, the question "is it the same > as this sequence" is answered "no". > So you think the antidiagonal comes up with an actual NEW SEQUENCE OF DIGITS and this does not contradict that ALL sequences of digits are on the computable list of reals up to all (an infinite amount of) digit positions? Herc |