The Powerset Proof - REWORDED!

[Graham Cooper]

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

[Graham Cooper]

On Nov 1, 11:03 am, George Greene <gree...@email.unc.edu> wrote:
> Your example has nothing to do with P(N).  N *is*INFINITE*.
> YOUR sets ARE ALL FINITE.  The paradox DOES NOT ARISE


THE POWERSET PROOF REWORDED - REWORDED!

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, ..}

[Rupert]

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).

[Graham Cooper]

It won't hold up to Induction.

IF: k e f(k) then ~k e MISS
~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

[Rupert]

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"?

[Gus Gassmann]

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.