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

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William Elliot

Posts: 1,669
Registered: 1/8/12
Push Down Lemma
Posted: Feb 7, 2013 4:43 AM
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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?




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