## 116c- Homework 5

Homework 5

Due Wednesday, May 14 at 2:30 pm.

Update. In both parts of exercise 2, “closed” should actually be “closed in its supremum;” i.e., the subset $X$of $S$ or of ${\mathbb Q}$ is closed in $S\cap\sup(X)$ or ${\mathbb Q}\cap\sup(X)$, respectively. Or, if you rather, replace the order type $\alpha$ of $X$ with $\alpha+1$. Sorry for the confusion; thanks to Fedor Manin for noticing this.

In exercise 1.(a)iii, “$m” should be “$n.” Thanks to Michael Conley for pointing this out.

Update. Here are sketches of the solutions of exercises 2.(a) and 3:

For 2.(a), let $S\subseteq\omega_1$ be a given stationary set, and argue by induction on $\alpha<\omega_1$ that $S$ contains closed copies $t$ of $\alpha+1$ with $\min(t)$ arbitrarily large. (Of course, if the result of the exercise holds, this must be the case: Given any $\gamma<\omega+1$, notice that $S\setminus(\gamma+1)$ is stationary, so it must contain a closed copy $t$ of $\alpha+1$, and $\min(t)>\gamma$.)

This strengthened version holds trivially for $\alpha$ finite or successor, by induction. So it suffices to show it for $\alpha$ limit, assuming it holds for all smaller ordinals. Define a club $C\subseteq\omega_1$ with increasing enumeration $\{\gamma_\beta:\beta<\omega_1\}$ as follows: Let $(\alpha_n:n<\omega)$ be strictly increasing and cofinal in $\alpha$. Since $S$ contains closed copies $A_n$ of $\alpha_n+1$ for all $n$, with their minima arbitrarily large, by choosing such copies $A_n$ with $\min(A_{n+1})>\sup(A_n)$ and taking their union, we see that $S$ must contain copies of $\alpha$, closed in their supremum, with arbitrarily large minimum element. (I am not claiming that $A=\bigcup_n A_n$ built this way has order type $\alpha$. For example, if $\alpha=\omega+\omega$ and $\alpha_n=\omega+n$, then $A$ would have order type $\omega^2$; but for sure $A\subset S$ is closed in its supremum and has order type at least $\alpha$. So a suitable initial segment of $A$ is as wanted.)

Let $\gamma_0$ be the supremum of such a copy of $\alpha$. At limit ordinals $\beta$, let $\gamma_\beta=\sup_{\delta<\beta}\gamma_\delta$. Once $\gamma_\beta$ is defined, find such a copy of $\alpha$ inside $S$ with minimum larger than $\gamma_\beta$, and let $\gamma_{\beta+1}$ be its supremum.

The set $C$ so constructed is club, so it meets $S$. If they meet in $\gamma_0$ or in a $\gamma_{\beta+1}$, this immediately gives us a closed copy of $\alpha+1$ inside $S$. If they meet in a $\gamma_\beta$ with $\beta$ limit, let $(\beta_n:n<\omega)$ be strictly increasing and cofinal in $\beta$, and consider an appropriate initial segment of $A=(\bigcup_n A_n)\cup\{\gamma_\beta\}$, where $A_n$ is a closed copy of $\alpha_n+1$ in $S\cap[\gamma_{\beta_n},\gamma_{\beta_{n+1}})$. ${\sf QED}$

For 3, let $\kappa$ be regular and $S\subseteq\kappa$ be stationary, and set $T=\{\alpha\in S:{\rm cf}(\alpha)=\omega$ or $({\rm cf}(\alpha)>\omega$ and $S\cap\alpha$ is not stationary in $\alpha)\}$. We claim that $T$ is stationary. To see this, let $C$ be an arbitrary club subset of $\kappa$. Then the set $C'$ of limit points of $C$ is also club (and a subset of $C$), so it meets $S$, since $S$ is stationary. Let $\alpha=\min(C'\cap S)$. Then either $\alpha$ has cofinality $\omega$, so it is in $T$, or else it has uncountable cofinality. In that case, notice that since $\alpha\in C'$, it is a limit of points in $C$, so $C\cap\alpha$ is club in $\alpha$, so $(C\cap\alpha)'=C'\cap\alpha$ is also club in $\alpha$. Were $S\cap\alpha$ stationary in $\alpha$, it would meet $C'\cap\alpha$, and this would contradict the minimality of $\alpha$. It follows that $\alpha\in T\cap C$, and therefore $T$ is stationary, as wanted.

Let now $\alpha$ be an arbitrary point of $T$. If $\alpha$ has cofinality $\omega$, it is the limit of an $\omega$-sequence of successor ordinals. Let $f_\alpha$ be the increasing enumeration of this sequence, and notice that ${\rm ran}(f_\alpha)\cap T=\emptyset$, since all ordinals in $T$ are limit ordinals. Suppose now that $\alpha$ has uncountable cofinality, so $S\cap\alpha$ is not stationary in $\alpha$. Since $T\subseteq S$, it follows that $T\cap\alpha$ is not stationary either, so there is a club subset of $\alpha$ disjoint from $T$, and let $f_{\alpha}$ be the increasing enumeration of this club set.

With the sequences $f_\alpha$ defined as above for all $\alpha\in T$, we now claim that there is some $\xi<\kappa$ such that  for all $\eta<\kappa$, the set $T_\eta=\{\alpha\in T:\xi\in{\rm dom}(f_\alpha)$ and $f_\alpha(\xi)\ge\eta\}$ is stationary. The proof is by contradiction, assuming that no $\xi<\kappa$ is as wanted.

It follows then that for all $\xi<\kappa$ there is some $\eta=\eta(\xi)<\kappa$ such that the set $T_\eta=T_\eta^\xi$ as above is non-stationary. Fix a club $C_\xi$ disjoint from it, and let $C$ be the club $C=\bigtriangleup_\xi C_\xi$. Let $D=C'\cap E$, where $E=\{\alpha<\kappa:\eta[\alpha]\subseteq\alpha\}$; here, $\eta:\kappa\to\kappa$ is the function $\xi\mapsto\eta_\xi$. Notice that $E$ is club, and so is $D$. We claim that $|D\cap T|\le 1$. This contradicts that $T$ is stationary, and therefore there must be a $\xi<\kappa$ as claimed.

Suppose then that $\gamma<\alpha$ are points in $D\cap T$. Since $\alpha\in D$, then $\alpha\in C$, so $\alpha\in C_\xi$ (hence, $\alpha\notin T^\xi_{\eta_\xi}$) for all $\xi<\alpha$. We claim that $\gamma\in{\rm dom}(f_\alpha)$. To see this, let $\xi\in\gamma\cap{\rm dom}(f_\alpha)$. Then (by definition of $T^\xi_{\eta_\xi}$), $f_\alpha(\xi)<\eta_\xi$. Since $\gamma\in D$, then $\gamma\in E$ and,  since $\xi<\gamma$, then $\eta_\xi<\gamma$. It follows that $\sup{\rm ran}(f_\alpha\upharpoonright\gamma)\le\gamma<\alpha$. Since $f_\alpha$ is cofinal in $\alpha$, we must necessarily have $\gamma\in{\rm dom}(f_\alpha)$.

Since $f_\alpha$ is continuous and $\gamma$ is a limit ordinal (since it is in $C'$), it follows that $f_\alpha(\gamma)=\sup_{\xi<\gamma}f_\alpha(\xi)\le\gamma$. But, since $f_\alpha$ is increasing, then also $f_\alpha(\gamma)\ge\gamma$. Hence, $f_\alpha(\gamma)=\gamma$. We have finally reached a contradiction, because $\gamma\in T$, but the sequence $f_\alpha$ was chosen so its range is disjoint from $T$. This proves that $|D\cap T|\le 1$, which of course is a contradiction since $T$ is stationary. It follows that indeed there is some $\xi<\kappa$ such that all the sets $T_\eta=\{\alpha\in T:\xi\in{\rm dom}(f_\alpha)$ and $f_\alpha(\xi)\ge\eta\}$ are stationary for $\eta<\kappa$.

Now let $f:T\to\kappa$ be the map $f(\alpha)=\left\{\begin{array}{cl}f_\alpha(\xi)&\mbox{ if }\xi\in{\rm dom}(f_\alpha),\\ 0&\mbox{ otherwise.}\end{array}\right.$ Clearly, $f$ is regressive. Also, from the definition of $f$, it follows that $\{\alpha\in T:f(\alpha)\ge\eta\}=T_\eta$ for all $\eta<\kappa$, so $f$ is unbounded in $\kappa$, since each $T_\eta$ is in fact stationary, as we showed above. Given any $\eta<\kappa$, since $f\upharpoonright T_\eta$ is regressive, there is some $\gamma$ (necessarily, $\gamma\ge\eta$) such that $\{\alpha\in T_\eta:f(\alpha)=\gamma\}$ is stationary, by Fodor’s lemma. A simple induction allows us to define a strictly increasing sequence $(\gamma_\eta:\eta<\kappa)$ such that $\{\alpha\in T_{\gamma_\eta+1}:f(\alpha)=\gamma_{\eta+1}\}$ is stationary for all $\eta$. Notice that these $\kappa$ many subsets of $T$ (hence, of $S$) so defined are all disjoint. By adding to one of them whatever (if anything) remains of $S$ after removing all these sets, we obtain a partition of  $S$ into $\kappa$ many disjoint stationary subsets, as wanted. ${\sf QED}$

There seems to be a problem with 2(a). Say we take $S=\omega_1$ and $\alpha=\omega$. Then any subset of $S$ of order type $\alpha$ is bounded, but $\bigcup \alpha \notin \alpha$, so it is not closed in $S$.