
Re: Push Down Lemma
Posted:
Feb 11, 2013 4:00 AM


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 ff^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; fS 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 corresponding to 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. >

