
Re: Push Down Lemma
Posted:
Feb 10, 2013 12:50 AM


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.
Two comments on the chain of inequalities in the last line.
First, your proof of the inequality X <= X.S is unnecessarily elaborate; it follows from the fact that S is nonempty.
Second, you have not proved the strict inequality X.S < X^2; in fact, if X is infinite (and the axiom of choice is assumed), that strict inequality holds only when S is empty.
> 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.

