```Date: Feb 11, 2013 4:00 AM
Author: William Elliot
Subject: Re:  Push Down Lemma

On Sat, 9 Feb 2013, Butch Malahide wrote:> On Feb 9, 8:16 pm, William Elliot <ma...@panix.com> wrote:> >> > Where, if at all for this part, is the fact that kappa is regular used?> > Assume X infinite and f:X -> Y.> > X = omega_{omega}, Y = omega, f(x) = min{n in omega: x < omega_n}.> > > 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|.> > {I wonder how he's going to prove this.}> > > Otherwise the contradiction:> > . . |X| = |\/{ f^-1f(a_y) | y in S }| <= |X|.|S| < |X|^2 = |X|.Tilt:  |X|.|S| = max{ |X|,|S| } = |X|> Two comments on the chain of inequalities in the last line.Shucks and Yuck.Let S be a set with a regular infinite cardinality kappa, C a collection of subsets of S, each with cardinality < kappa. If |C| < kappa, then |\/C| < kappa.Proof.Well order S to the order type eta, the initial cardinal correspondingto kappa.  For all A in C, A (embedded in eta) is bounded above by some b_A < eta.  Let B = { b_A | A in C }.  If |C| < kappa, then |B| <= |C| < kappa = cof kappa.  Since |B| < cof kappa, B (as embedded in eta) is bounded above by say b.  Thus since \/C is bounded above by b, \/C can't be cofinal with kappa.  Consequently |\/C| < cof kappa = kappa.  QED.How could this be better formalized?  By using an order isomorphism h, from well ordered S onto eta?> > Why does kappa need to be regular?> > If kappa is regular, then a set of cardinality kappa cannot be> partitioned into fewer than kappa sets each of cardinality less than> kappa; in other words, there is no function defined on kappa which> takes fewer than kappa values and takes each value fewer than kappa> times. That's what it *means* be be a regular cardinal.>
```