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Topic: The Powerset Proof - REWORDED!
Replies: 14   Last Post: Nov 8, 2012 4:11 AM

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rt servo

Posts: 19
Registered: 2/1/05
Re: The Powerset Proof - REWORDED!
Posted: Nov 1, 2012 2:35 PM
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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:
>>
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>>
>>
>>
>>
>>
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>>

>>> 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

>




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