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Topic: Push Down Lemma
Replies: 10   Last Post: Feb 11, 2013 4:00 AM

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Butch Malahide

Posts: 894
Registered: 6/29/05
Re: Push Down Lemma
Posted: Feb 7, 2013 7:13 AM
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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 well-ordered 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.



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