
Re: Push Down Lemma
Posted:
Feb 7, 2013 7:13 AM


On Feb 7, 3:43 am, William Elliot <ma...@panix.com> wrote: > The push down lemma:
Can you cite an authoritative source that uses the overblown and inappropriate name "push down lemma" for this triviality, or was that your own idea? When I read your subject line, I thought you were referring to the celebrated "Pressing Down Lemma" aka Fodor's Lemma:
http://en.wikipedia.org/wiki/Fodor's_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).
If I have correctly deciphered your godawful notation, what you're saying amounts to this: If S is an infinite set, then the power set P(S) (ordered by inclusion) does not contain any wellordered chain W with W > S, and neither of course does its dual. Right. The most familiar special case is that P(omega), although it contains uncountable chains ordered like the reals, contains no chain of type omega_1 or omega_1^*.
> Can the push down lemma be extended to show f is eventually constant?
Yes, easily, if kappa is regular; no, if kappa is singular. Suppose, e.g., that eta = omega and S = omega. Your descending function f:omega_{omega} > P(omega) cannot be injective, but neither does it have to be eventually constant. For examply, you could have f(mu) = S for the first aleph_0 values of mu, f(mu) = S\{0 for the next aleph_1 values, f(mu) = S\{0,1} for the next aleph_2 values, and so on.

