```Date: Feb 8, 2013 10:10 PM
Author: William Elliot
Subject: Re: Push Down Lemma

On Thu, 7 Feb 2013, Butch Malahide wrote:> 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? No, it was taken from sci.math or some other such source.> 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_lemmaThe same source also gave the pressing down lemma for omega_1 > > 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 well-ordered chain W> with |W| > |S|, and neither of course does its dual. Right. Yes, that's a nice description.  On the other hand P(S) can have a nest of length |P(S)|.> 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.Easily?  How so for regular kappa that f is eventually constant?
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