
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 wellordered 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 fX > is either injective or constant. Since kappa is an initial ordinal, X > is orderisomorphic 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 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. 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 fX > is injective, then fX 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 fX 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^*.

