
Re: Push Down Lemma
Posted:
Feb 8, 2013 10:10 PM


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_lemma
The 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 wellordered 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?

