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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: 2,383 Registered: 1/8/12
Re: Push Down Lemma
Posted: Feb 9, 2013 9:16 PM

> > > Let beta = omega_eta, kappa = aleph_eta.
> > > Assume f:beta -> P(S) is descending, ie
> > > . . for all mu,nu < beta, (mu <= nu implies f(nu) subset f(mu)),
> > > f(0) = S and |S| < kappa.

>
> > > Then there's some xi < beta with f(xi) = f(xi + 1).

> > If S is an infinite set, then the power set P(S) (ordered
> > by inclusion) does not contain any well-ordered chain W
> > with |W| > |S|, and neither of course does its dual.

> > > Can the push down lemma be extended to show f is eventually constant?

> > Yes, easily, if kappa is regular; no, if kappa is singular. Suppose,
> > e.g., that eta = omega and S = omega. Your descending function
> > f:omega_{omega} -> P(omega) cannot be injective, but neither does it
> > have to be eventually constant. For examply, you could have f(mu) = S
> > for the first aleph_0 values of mu, f(mu) = S\{0| for the next aleph_1
> > values, f(mu) = S\{0,1} for the next aleph_2 values, and so on.

>
> Easily? ?How so for regular kappa that f is eventually constant?

> Let kappa be a regular infinite cardinal, identified with its initial
> ordinal omega_{alpha}

> Since kappa is regular, for any function f defined on kappa, there
> exists X subset kappa, with |X| = kappa, such that the restriction f|X
> is either injective or constant. Since kappa is an initial ordinal, X
> is order-isomorphic to kappa.

Where, if at all for this part, is the fact that kappa is regular used?

Assume X infinite and f:X -> Y.

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|. Otherwise the contradiction:
. . |X| = |\/{ f^-1f(a_y) | y in S }| <= |X|.|S| < |X|^2 = |X|.

> Now suppose f:kappa --> P, an ordered set, and f is monotone. If f|X
> is injective, then f|X is a strictly monotone map from X to P; i.e.,
> either P or its dual contains a chain isomorphic to X and therefore to
> kappa. On the other hand, if f|X is constant, it follows from the
> monotonicity of f and the fact that X is unbounded in kappa that f is
> eventually constant.

> Hence, if kappa is regular, and if P is an ordered set containing no
> chains of order type kappa or kappa^* [in particular, if P = P(S)
> where |S| < kappa], then every monotone function f:kappa --> P is
> eventually constant.

Why does kappa need to be regular?

The Push Down Lemma shows that P(S) has no chains of order type kappa
or kappa^*.

Date Subject Author
2/7/13 William Elliot
2/7/13 J. Antonio Perez M.
2/7/13 Butch Malahide
2/7/13 David C. Ullrich
2/7/13 David Hartley
2/7/13 David C. Ullrich
2/8/13 William Elliot
2/8/13 Butch Malahide
2/9/13 William Elliot
2/10/13 Butch Malahide
2/11/13 William Elliot