
Re: Push Down Lemma
Posted:
Feb 7, 2013 6:34 AM


On Thursday, February 7, 2013 11:43:11 AM UTC+2, William Elliot wrote: > 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?
I strongly urge to write down your maths in a LaTeX supported site and add a link to that site or PDF file, otherwise it is almost impossible to understand what you wrote, and it takes too long...
Tonio

