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_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?