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

 Messages: [ Previous | Next ]
 J. Antonio Perez M. Posts: 2,736 Registered: 12/13/04
Re: Push Down Lemma
Posted: Feb 7, 2013 6:34 AM

On Thursday, February 7, 2013 11:43:11 AM UTC+2, William Elliot wrote:
> The push down lemma:
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>
>
> Let beta = omega_eta, kappa = aleph_eta.
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> Assume f:beta -> P(S) is descending, ie
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> for all mu,nu < beta, (mu <= nu implies f(nu) subset f(mu)),
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> f(0) = S and |S| < kappa.
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>
>
> Then there's some xi < beta with f(xi) = f(xi + 1).
>
>
>
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> 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

Date Subject Author
2/7/13 William Elliot
2/7/13 J. Antonio Perez M.
2/7/13 Butch Malahide
2/7/13 David C. Ullrich
2/7/13 David Hartley
2/7/13 David C. Ullrich
2/8/13 William Elliot
2/8/13 Butch Malahide
2/9/13 William Elliot
2/10/13 Butch Malahide
2/11/13 William Elliot