
Re: The Powerset Proof  REWORDED!
Posted:
Nov 1, 2012 2:35 PM


On 01/11/2012 11:06 AM, Rupert wrote: > On 1 Nov., 10:44, Graham Cooper <grahamcoop...@gmail.com> wrote: >> On Nov 1, 7:19 pm, Rupert <rupertmccal...@yahoo.com> wrote: >> >> >> >> >> >> >> >> >> >>> On Oct 31, 8:33 pm, Graham Cooper <grahamcoop...@gmail.com> wrote: >> >>>> P(N) >> >>>> You have some SETS of NUMBERS >>>> like {8,11,12,77} >> >>>> Each set HAS a number >>>> 1  {1,2,3} >>>> 2  {8,11,12,77} >> >>>> What# is the SET of all SETS that don't contain themselves?? >> >>>> Herc >> >>> To put it another way if you have a function f from N into P(N), then >>> you can form the set S of all natural numbers n such that n is not a >>> member of f(n), and this set S will not be in the range of the >>> function f. Because if we had S=f(k) for some natural number k, then >>> we would have that k is a member of f(k) if and only if it is not, >>> which is a contradiction. >> >>> Yes. That's the proof. It shows that there cannot be a surjective >>> function f from N onto P(N). >> >> It won't hold up to Induction. >> >> IF: k e f(k) then ~k e MISS > > What's the meaning of the notation "MISS"?
I think you are missing the bigger picture. Somehow Cooper seems to think the an induction proof is the *only* way to show that a property holds for every n in N. Of course he also seems to think that a property holding for N must therefore also hold for every n in N (and vice versa). Unless you can convince him of the error of his ways (and you *can't*), any further dialog with him is futile.
>> ~k e f(k) then k e MISS >> ergo ~MISS e f >> >> THEN: k+1 e f(k+1) then ~k+1 e MISS >> ~k+1 e f(k+1) then k+1 e MISS >> ergo ~MISS e f >> >> THAT MUCH IS TRUE! >> >> THAT IS HOW YOU PROVE A PROPERTY HOLDS FOR ALL N. >> >> But the Base Step does not hold for ANY SINGLE value of k. >> >> Herc >

