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 <> 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:

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 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?