Date: Feb 7, 2013 4:43 AM Author: William Elliot Subject: Push Down Lemma 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?