```Date: Feb 9, 2013 9:16 PM
Author: William Elliot
Subject: Re: 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).> > 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.> > > 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?> Let kappa be a regular infinite cardinal, identified with its initial> ordinal omega_{alpha}> Since kappa is regular, for any function f defined on kappa, there> exists X subset kappa, with |X| = kappa, such that the restriction f|X> is either injective or constant. Since kappa is an initial ordinal, X> is order-isomorphic to kappa.Where, if at all for this part, is the fact that kappa is regular used?Assume X infinite and f:X -> Y.If |f^-1(y)| = |X|, then f|f^-1(y) is constant.Otherwise assume for all y, |f^-1(y)| < |X|.For all y in f(X), there's some a_y in f^-1(y).S = { a_y | y in Y } subset X;  f|S is injective.|S| = |X|.  Otherwise the contradiction:. . |X| = |\/{ f^-1f(a_y) | y in S }| <= |X|.|S| < |X|^2 = |X|.> Now suppose f:kappa --> P, an ordered set, and f is monotone. If f|X> is injective, then f|X is a strictly monotone map from X to P; i.e.,> either P or its dual contains a chain isomorphic to X and therefore to> kappa. On the other hand, if f|X is constant, it follows from the> monotonicity of f and the fact that X is unbounded in kappa that f is> eventually constant.> Hence, if kappa is regular, and if P is an ordered set containing no> chains of order type kappa or kappa^* [in particular, if P = P(S)> where |S| < kappa], then every monotone function f:kappa --> P is> eventually constant.Why does kappa need to be regular?The Push Down Lemma shows that P(S) has no chains of order type kappaor kappa^*.
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