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Topic: Push Down Lemma
Replies: 10   Last Post: Feb 11, 2013 4:00 AM

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Butch Malahide

Posts: 894
Registered: 6/29/05
Re: Push Down Lemma
Posted: Feb 10, 2013 12:50 AM
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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|.


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.



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