116c- Lecture 9

April 30, 2008

We proved König’s theorem and results of Hausdorff and Tarski on cardinal exponentiation, indicated some of their consequences (for example, {\mathfrak c}\ne\aleph_\omega), and showed how to compute under {\sf GCH} the function (\kappa,\lambda)\mapsto\kappa^\lambda.

We stated Easton’s result essentially saying that without additional assumptions, in {\sf ZFC} nothing can be said about the exponential function 2^\lambda beyond monotonicity and König’s theorem.

For singular cardinals the situation is much more delicate. We stated as a sample result Shelah’s theorem that if \aleph_\omega is strong limit, then 2^{\aleph_\omega} is regular and smaller than \aleph_{\min(\omega_4,{\mathfrak c}^+)}.

This result is beyond the scope of this course. Instead, we will prove a particular case of an earlier result of Silver, namely, that \aleph_{\omega_1} is not the first counterexample to {\sf GCH}.

In order to prove Silver’s result, we need to develop the theory of club and stationary sets. We defined these notions and proved some of their basic properties.

116c- Homework 4

April 29, 2008

Homework 4.

Due Wednesday, May 7 at 2:30 pm.

Update. In exercise 2, \lambda=\mu, of course. In exercise 3, (\kappa)^\lambda in the right hand side must be (<\kappa)^\lambda.

Hint for exercise 5: Consider the smallest \alpha such that \alpha+\beta>\beta (ordinal addition).

Update. Here is a quick sketch of the solutions:

  • 1.(a). Write \kappa=\sum_{\alpha<{\rm cf}(\kappa)}\kappa_\alpha for some (not necessarily strictly) increasing {\rm cf}(\kappa)-sequence of cardinals \kappa_\alpha<\kappa. (If \kappa is regular, take \kappa_\alpha=1 for all \alpha. If \kappa is singular, let the \kappa_\alpha be increasing and cofinal in \kappa.) Then 2^\kappa=2^{\sum_{\alpha<{\rm cf}(\kappa)}\kappa_\alpha}=\prod_\alpha 2^{\kappa_\alpha}\le\prod_\alpha 2^{<\kappa}=(2^{<\kappa})^{{\rm cf}(\kappa)} but also (2^{<\kappa})^{{\rm cf}(\kappa)}\le (2^\kappa)^\kappa=2^\kappa and the result follows. If \kappa is strong limit, then 2^{<\kappa}=\kappa, and we get 2^\kappa=\kappa^{{\rm cf}(\kappa)}.
  • 1.(b). Under the assumption, we may choose the \kappa_\alpha as in 1.(a) such that {\rm cf}(\kappa)<\kappa_0 and 2^{\kappa_\alpha}= 2^{\kappa_\beta} for all \alpha<\beta<{\rm cf}(\kappa). Then 2^\kappa=(2^{<\kappa})^{{\rm cf}(\kappa)}=(2^{\kappa_0})^{{\rm cf}(\kappa)}=2^{\kappa_0}=2^{<\kappa}.
  • 2. Write \mu as a disjoint union of \mu sets A_\alpha, \alpha<\mu, each of size \mu. This is possible since \mu\times\mu=\mu. Since each A_\alpha has size \mu, it is necessarily cofinal in \mu. We then have \kappa^\mu\ge\prod_{i\in\mu}\kappa_i=\prod_{\alpha\in\mu}\prod_{i\in A_\alpha}\kappa_i\ge\prod_\alpha \kappa=\kappa^\mu.
  • 3. If \kappa\le 2^\lambda, then 2^\lambda\le\kappa^\lambda\le(2^\lambda)^\lambda=2^\lambda.
  • If \lambda<{\rm cf}(\kappa) then any function f:\lambda\to\kappa is bounded, so {}^\lambda\kappa=\bigcup_{\alpha<\kappa}{}^\lambda\alpha, and \kappa^\lambda\le\sum_\alpha|\alpha|^\lambda\le\kappa\cdot(<\kappa)^\lambda; the other inequality is clear.
  • Suppose that {\rm cf}(\kappa)\le\lambda2^\lambda<\kappa, and \rho\mapsto\rho^\lambda is eventually constant as \rho approaches \kappa. Choose a strictly increasing sequence (\kappa_\alpha:\alpha<{\rm cf}(\kappa)) cofinal in \kappa such that \kappa_\alpha^\lambda=\kappa_\beta^\lambda for \alpha<\beta<{\rm cf}(\kappa). Then \kappa^\lambda\le(\prod_\alpha\kappa_\alpha)^\lambda=\prod_\alpha \kappa_\alpha^\lambda=\prod_\alpha \kappa_0^\lambda=\kappa_0^{\lambda\cdot{\rm cf}(\kappa)}=\kappa_0^\lambda=(<\kappa)^\lambda. The other inequality is clear.
  • Finally, suppose that {\rm cf}(\kappa)\le\lambda, 2^\lambda<\kappa, and \rho\mapsto\rho^\lambda is not eventually constant as \rho approaches \kappa. Notice that if \rho<\kappa then \rho^\lambda<\kappa. Otherwise, \tau^\lambda=(\tau^\lambda)^\lambda\ge\kappa^\lambda\ge\tau^\lambda for any \rho\le\tau<\kappa, and the map \rho\mapsto\rho^\lambda would be eventually constant below \kappa after all. Hence, (<\kappa)^\lambda=\kappa. Choose an increasing sequence of cardinals (\kappa_\alpha:\alpha<{\rm cf}(\kappa)) cofinal in \kappa. Then \kappa^\lambda\le(\prod_\alpha \kappa_\alpha)^\lambda=\prod_\alpha\kappa_\alpha^\lambda\le\prod_\alpha (<\kappa)^\lambda=\prod_\alpha\kappa=\kappa^{{\rm cf}(\kappa)}. The other inequality is clear.
  • 5. Suppose that \beta\ge\omega and for all \alpha, 2^{\aleph_\alpha}=\aleph_{\alpha+\beta}. Let \alpha be least such that \beta<\alpha+\beta; \alpha exists since \beta<\beta+\beta. On the other hand, \alpha is a limit ordinal, since 1+\omega=\omega and \omega\le\beta, so (\lambda+n)+\beta=\lambda+(n+\beta)=\lambda+\beta for any \lambda and any n<\omega, and we would have a contradiction to the minimality of \alpha. Let \gamma<\alpha. Then \gamma+\beta=\beta and 2^{\aleph_{\alpha+\gamma}}=\aleph_{\alpha+\beta}. By exercise 1.(b), 2^{\aleph_{\alpha+\alpha}}=\aleph_{\alpha+\beta} since \aleph_{\alpha+\alpha} is singular (it has cofinality {\rm cf}(\alpha)\le\alpha<\alpha+\alpha\le\aleph_{\alpha+\alpha}). This is a contradiction, since by assumption, 2^{\aleph_{\alpha+\alpha}}=\aleph_{\alpha+\alpha+\beta}>\aleph_{\alpha+\beta}.
  • 4. We want to show that \prod_{\xi<\beta}\aleph_\xi=\aleph_\beta^{|\beta|} for all limit ordinals \beta. Write \beta=|\beta|+\alpha where \alpha=0 or a limit ordinal, and proceed by induction on \alpha, noticing that the case \alpha=0 follows from exercise 2.
  • Notice that any limit ordinal \alpha can be written as \alpha=\omega\cdot\delta for some nonzero ordinal \delta. This is easily established by induction. If \delta is a limit ordinal, then any sequence converging to \delta gives rise in a natural way to a sequence of limit ordinals converging to \alpha. On the other hand, if \delta is a successor, then \alpha=\lambda+\omega for some \lambda=0 or limit. In summary: any limit ordinal is either a limit of limit ordinals, or else it has the form \lambda+\omega for \lambda=0 or limit.
  • We divide our induction on \alpha in two cases. Suppose first \alpha=\lambda+\omega, as above. Then \prod_{\xi<\beta} \aleph_\xi =\aleph_{|\beta|+\lambda}^{|\beta|} \prod_n \aleph_{|\beta|+\lambda+n}, by induction. Since |\beta|=|\beta|\aleph_0, we have \aleph_{|\beta|+\lambda}^{|\beta|}\prod_n \aleph_{|\beta|+\lambda+n}= \prod_n \aleph_{|\beta|+\lambda}^{|\beta|}\aleph_{|\beta|+\lambda+n}=\prod_n \aleph_{|\beta|+\lambda+n}^{|\beta|}, where the last equality holds by Hausdorff’s result that (\kappa^+)^\tau=\kappa^+\cdot\kappa^\tau. Finally, \prod_n\aleph_{|\beta|+\lambda+n}^{|\beta|}=(\prod_n \aleph_{|\beta|+\lambda+n})^{|\beta|}=(\aleph_\beta)^{\aleph_0\cdot|\beta|}=\aleph_\beta^{|\beta|} as wanted, where the previous to last equality is by exercise 2.
  • Now suppose that \alpha is a limit of limit ordinals. Choose a strictly increasing sequence (\gamma_\nu:\nu<{\rm cf}(\alpha)) of limit ordinals cofinal in \alpha. We argue according to which of the four possibilities described in exercise 3 holds. If \aleph_\beta\le 2^{|\beta|} then \aleph_\beta^{|\beta|}=2^{|\beta|}\le\prod_{\xi<\beta} \aleph_\xi\le\aleph_\beta^{|\beta|}.
  • Notice that {\rm cf}(\alpha)={\rm cf}(\beta)={\rm cf}(\aleph_\beta) and {\rm cf}(\beta)\le|\beta|, so the second possibility does not occur.
  • Suppose now that we are in the third possibility, i.e., 2^{|\beta|}<\aleph_\beta and \rho\mapsto\rho^{|\beta|} is eventually constant as \rho approaches \aleph_\beta. Then \aleph_\beta^{|\beta|}=(<\aleph_\beta)^{|\beta|}=\sup_{\nu<{\rm cf}(\alpha)} \aleph_{|\beta|+\gamma_\nu}^{|\beta|}=\sup_\nu\prod_{\xi<|\beta|+\gamma_\nu}\aleph_\xi, by induction (it is here that we used the assumption that \alpha is a limit of limit ordinals). And \sup_\nu \prod_{\xi<|\beta|+\gamma_\nu} \aleph_\xi\le\prod_{\xi<\beta}\aleph_\xi\le\aleph_\beta^{|\beta|}.
  • Finally, if 2^{|\beta|}<\aleph_\beta and \rho\mapsto\rho^{|\beta|} is not eventually constant as \rho approaches \aleph_\beta, then \aleph_\beta^{|\beta|}=\aleph_\beta^{{\rm cf}(\alpha)} and \aleph_\beta^{{\rm cf}(\alpha)}=\prod_{\nu<{\rm cf}(\alpha)}\aleph_{|\beta|+\gamma_\nu}, by exercise 2, but \prod_{\nu<{\rm cf}(\alpha)}\aleph_{|\beta|+\gamma_\nu}\le\prod_{\xi<\beta}\aleph_\xi\le\aleph_\beta^{|\beta|}, and we are done.


April 28, 2008

Peter O’Toole is old, and here he plays an old man. Seriously, that is all he does throughout the movie: Look very old. People nominated him for an Oscar based on that. Very disappointing.

At least, it is not O’Toole’s last role. The movie is well written, and all the actors are very good, which stops it from being a complete failure. The story, on the other hand, did not seem believable to me in the least, and was far from being engaging.

The painted veil

April 28, 2008

This movie is a very nice adaptation of  W. Somerset Maugham’s novel. It only ran in Pasadena at the second-run theatre; apparently it only had a limited release in the States.

I’ve seen recently a few movies or theatre plays based on Maugham’s works, and they have all been satisfying. This one is primarily a love story, which surprised me a little at first. The acting is impecable, of course. I imagine even The incredible Hulk might be near tolerable thanks to Edward Norton. The cinematography was gorgeous. The exteriors alone justify watching this film.

Seraphim falls

April 28, 2008

The best western I have seen in recent years is the extremely violent Australian movie The proposition. What made that movie interesting in my mind was the air of freshness that the new surroundings provide to the genre. Seraphim falls starts promising, and for a while I expected it was going to end up as satisfying an experience as watching The proposition was. It did not.

This movie ends in absurdity, which more and more looks like what Hollywood confuses with depth. The problem is that a metaphorical ending in a story of this sort feels like a cop out. For more than an hour and a half the movie has established the storytelling framework in which it occurs, and the end destroy this. More skillfully executed, it would have been an interesting post-modern take on the western genre. The way it is, you feel an unfulfilled promise and disappointment. It reminds me a bit of the end of 3:10 to Yuma, where suddenly the characters stop being characters and decide to serve as moralizing stand ins for… I don’t know… the scriptwriter? 

That being said, the first hour and forty minutes or so is a very good western, it is well acted and very nicely shot.

The big kill

April 28, 2008

I decided a while ago to get a good working understanding of noir fiction, which among other things means to read the classics. The lady in the lake was entertaining. This one… well, this one was not good.

The end was ridiculously, portentously, absurd. It would have made a cute joke in an early Woody Allen movie, I think. The one liners get old rather quickly, although I suppose they may have been a bit of a novelty back when. It is funny how the over the top language that is used throughout Sin City ended up not being a caricature but rather very close to what we have here. 

All the dames fall for Mike Hammer, the hero of Mickey Spillane’s novels. They fall just because that’s how it is, which makes the tale rather silly, a not particularly clever wish fulfillment adolescent fantasy of sorts. And the writing is quite pedestrian.

There is more violence than in other noir stories I had read. I now believe the explicit violence, nothing too unusual by today’s standards, was actually a selling point for Spillane’s pulps. There is also a lot of anger; Mike is so angry all the time that you fear he’ll end up hospitalized with a serious ulcer. He does not. It would have been a better ending. 

In any case, it was entertaining for what it was, but clearly Raymond Chandler is much better writer than Mickey Spillane.

Suicide kings

April 27, 2008

The memory I had of this movie is much better than what it turned out to be. I remember I went to see this movie when it came out, and it seemed decent. I guess I was younger and foolish, because this was just ridiculous; I think it fails in that it is hard to take it seriously. But, of course, maybe it is intended to be a comedy, in which case it fails because it is not particularly funny.

Anyway, it has Christopher Walken in it, so at least it makes you smile every now and then.