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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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>>> On Oct 31, 8:33 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
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>>>> P(N)
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>>>> You have some SETS of NUMBERS
>>>> like {8,11,12,77}
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>>>> Each set HAS a number
>>>> 1 - {1,2,3}
>>>> 2 - {8,11,12,77}
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>>>> What# is the SET of all SETS that don't contain themselves??
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>>>> Herc
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>>> 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.
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>>> Yes. That's the proof. It shows that there cannot be a surjective
>>> function f from N onto P(N).
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>> It won't hold up to Induction.
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>> IF: k e f(k) then ~k e MISS
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> 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
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>> 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
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>> THAT MUCH IS TRUE!
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>> THAT IS HOW YOU PROVE A PROPERTY HOLDS FOR ALL N.
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>> But the Base Step does not hold for ANY SINGLE value of k.
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>> Herc
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