## 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

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\lambda$$2^\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.

## Venus

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.

## The last kiss

April 27, 2008

I do not really have much to say about this movie; I found it kind of mediocre. It presents itself as more than it ends up being, and rather than having characters that face their problems in interesting or challenging ways, it felt dissapointing, a bit cliched. It is not a bad movie, though. I suppose there are worst ways of spending two hours.

## God grew tired of us

April 27, 2008

This is an excellent documentary about a horrible ongoing tragedy. But there is a lot of hope in the story; John Bul Dau, one of the “lost boys of Sudan” the movie is about, is inspirational, a great leader. In the midst of all their suffering, I could not believe how much energy and optimism he displayed. He is truly an admirable person.

Part of the documentary follows several kids that are relocated to the States (thanks to Catholic Charities International). I found particularly interesting to see the culture clash that the group suffers, arriving to and having to survive in the States with what looks like very little assistance.

Although they are very grateful, we learn that the older ones have to hold one or two jobs in order to pay back the cost of their move. Of course, the jobs they find are not particularly appealing or well paid, plus they have to face discrimination and ignorance. The younger ones, on the other hand, get to go to school and several of them try very quickly to absorb the American life style, leaving behind their roots and traditions, which leads to an interesting clash with people like John Bul Dau, who makes every effort to keep their memory and connections alive.

I highly recommend this moving and sobering documentary.

## 116c- Lecture 8

April 24, 2008

We defined infinite sums and products and showed that if $\max{2,\kappa_i}\le\lambda_i$ for all $i\in I$, then $\sum_{i\in I}\kappa_i\le\prod_{i\in I}\lambda_i$.

We also showed that if $(\kappa_i:i<\mbox{cf}(\kappa))$ is an increasing sequence of cardinals cofinal in $\kappa$, then $\prod_{i\in\mbox{\small cf}(\kappa)}\kappa_i=\kappa^{\mbox{\small cf}(\kappa)}$. In particular, $\prod_n\aleph_n=\aleph_\omega^{\aleph_0}$.

We defined singular cardinals and showed that (with choice) all successor cardinals are regular and all limit cardinals $\aleph_\alpha$ are singular unless $\alpha=\aleph_\alpha$. We showed that, indeed, there are fixed points of the aleph function, as a particular case of a result about normal functions. We defined (weakly) inaccessible cardinals as the regular limit cardinals (thus, regular fixed points of the aleph function).

Correction. I believe during lecture I mixed two arguments by mistake, making one of the proofs come out unnecessarily confusing, so I will present the correct argument here, for clarity.

In lecture we showed that if $F$ is a normal function, then it has a proper class of fixed points. Thus, we can enumerate them in increasing order. Let $G$ be this enumeration.

Claim. $G$ was also normal.

Proof. We need to check that $G$ is continuous. Let $\gamma$ be a limit ordinal and suppose that $\tau=\sup_{\beta<\gamma}G(\beta)$. We need to show that $G(\gamma)=\tau$.

By definition, this means that:

1. $\tau$ is a fixed point of $F$, and
2. $\tau$ is the $\gamma$-th fixed point of $F$.

But, clearly, if $\tau$ is a fixed point, then it must be the $\gamma$-th one, since we have already enumerated $\gamma$ fixed points below $\tau$, and any fixed point below $\tau$ is below some $G(\beta)$ with $\beta<\gamma$, so it is not even the $\beta$-th one.

So we only need to check that $F(\tau)=\tau$. But $\tau=\sup_{\beta<\gamma}G(\beta)$ and each $G(\beta)$ is a fixed point of $F$ (again, by definition of $G$), so $\tau=\sup_{\beta<\gamma}F(G(\beta))=F(\sup_{\beta<\gamma}G(\beta))=F(\tau)$, where the previous to last equality is by continuity of $F$. ${\sf QED}$

It follows that $G$ itself has a proper class of fixed points. It is also the case that there is a proper class of fixed points of $F$ that are limits of fixed points of $F$: Simply notice that the argument above shows that any limit of fixed points of $F$ is itself a fixed point. Thus, we have:

Corollary. The function $H:{\sf ORD}\to{\sf ORD}$ enumerating the limit points of $G$ (i.e., the fixed points of $F$ that are themselves limit of fixed points) is normal.

I believe during lecture I mixed at some point $H$ and $G$ (although I never explicitly mentioned $H$). Hopefully the above clarifies the argument. For the particular case of $F(\alpha)=\aleph_\alpha$, we have that $G$ enumerates the ordinals $\alpha$ such that $\alpha=\aleph_\alpha$, so $G(0)$ is the first such cardinal. The function $H$ enumerates the limit points of $G$, so $H(0)=G(\omega)$. Notice that $\mbox{cf}(H(0))=\mbox{cf}(G(\omega))=\omega$. One can easily see that if $\kappa$ is a weakly inaccessible cardinal, then $\kappa$ is a fixed point of $F$, $G$ and $H$.

In fact, define $F_0(\alpha)=\aleph_\alpha$ for all $\alpha$, let $F_{\beta+1}$ be the enumeration of the fixed points of $F_\beta$, and let $F_\gamma$ (for $\gamma$ limit) enumerate the ordinals $\kappa$ that are simultaneously fixed points of all the $F_\beta$ for $\beta<\gamma$. Then, if $\kappa$ is weakly inaccessible, then $F_\alpha(\kappa)=\kappa$ for all $\alpha<\kappa$.

Remark.

1.  We did not prove that weakly inaccessible cardinals exist. The examples given in lecture of fixed points of the aleph function have cofinality $\omega$ and, similarly, we can produce fixed points of arbitrarily large cofinality, but the argument falls short of finding regular fixed points (in fact, we can show that each $F_\alpha$ as defined above is normal, but the argument does not show that we can “diagonalize” to obtain a $\kappa$ fixed for all $F_\alpha$ with $\alpha<\kappa$). In fact, it is consistent with ${\sf ZFC}$ that all limit cardinals are singular. However, it is the general consensus among set theorists that the existence of inaccessible cardinals is one of the axioms of set theory that the original list ${\sf ZFC}$ somehow missed.
2. We defined normal functions as proper classes; however, we can as well define for any ordinal $\alpha$ a function $F:\alpha\to{\sf ORD}$ to be normal iff it is strictly increasing and continuous. The same argument as in lecture (or above) then shows that if $\mbox{cf}(\kappa)>\omega$ and $F:\kappa\to\kappa$ is normal, then there is a closed and unbounded subset of $\kappa$ consisting of fixed points of $F$. It turns out that closed unbounded sets are very important in infinitary combinatorics, and we will study them in more detail in subsequent lectures.