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

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William Elliot

Posts: 1,232
Registered: 1/8/12
Re: Push Down Lemma
Posted: Feb 11, 2013 4:00 AM
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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 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.
>




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