The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » sci.math.* » sci.math

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

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

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
William Elliot

Posts: 2,637
Registered: 1/8/12
Push Down Lemma
Posted: Feb 7, 2013 4:43 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

The push down 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).
Proof is by contradiction.

Can the push down lemma be extended to show f is eventually constant?

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.