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Topic: Push Down Lemma
Replies: 10   Last Post: Feb 11, 2013 4:00 AM

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J. Antonio Perez M.

Posts: 2,736
Registered: 12/13/04
Re: Push Down Lemma
Posted: Feb 7, 2013 6:34 AM
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On Thursday, February 7, 2013 11:43:11 AM UTC+2, William Elliot wrote:
> The push down lemma:
>
>
>
> Let beta = omega_eta, kappa = aleph_eta.
>
> Assume f:beta -> P(S) is descending, ie
>
> for all mu,nu < beta, (mu <= nu implies f(nu) subset f(mu)),
>
> f(0) = S and |S| < kappa.
>
>
>
> Then there's some xi < beta with f(xi) = f(xi + 1).
>
> Proof is by contradiction.
>
>
>
> Can the push down lemma be extended to show f is eventually constant?



I strongly urge to write down your maths in a LaTeX supported site and add a link to that site or PDF file, otherwise it is almost impossible to understand what you wrote, and it takes too long...

Tonio



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