## 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 $Scapsup(X)$ or ${mathbb Q}capsup(X)$, 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 $Ssubseteqomega_1$ be a given stationary set, and argue by induction on $alpha 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, notice that $Ssetminus(gamma+1)$ 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 $Csubseteqomega_1$ with increasing enumeration ${gamma_beta:beta as follows: Let $(alpha_n:n 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 $Asubset S$ 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. 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 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 $Scap[gamma_{beta_n},gamma_{beta_{n+1}})$. ${sf QED}$

For 3, let $kappa$ be regular and $Ssubseteqkappa$ be stationary, and set $T={alphain S:{rm cf}(alpha)=omega$ or $({rm cf}(alpha)>omega$ and $Scapalpha$ 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 $alphain C'$, it is a limit of points in $C$, so $Ccapalpha$ is club in $alpha$, so $(Ccapalpha)'=C'capalpha$ is also club in $alpha$. Were $Scapalpha$ stationary in $alpha$, it would meet $C'capalpha$, and this would contradict the minimality of $alpha$. It follows that $alphain Tcap C$, 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 $Scapalpha$ is not stationary in $alpha$. Since $Tsubseteq S$, it follows that $Tcapalpha$ 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 $alphain T$, we now claim that there is some $xi such that  for all $eta, the set $T_eta={alphain T:xiin{rm dom}(f_alpha)$ and $f_alpha(xi)geeta}$ is stationary. The proof is by contradiction, assuming that no $xi is as wanted.

It follows then that for all $xi there is some $eta=eta(xi) 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; here, $eta:kappatokappa$ is the function $ximapstoeta_xi$. Notice that $E$ is club, and so is $D$. We claim that $|Dcap T|le 1$. This contradicts that $T$ is stationary, and therefore there must be a $xi as claimed.

Suppose then that $gamma are points in $Dcap T$. Since $alphain D$, then $alphain C$, so $alphain C_xi$ (hence, $alphanotin T^xi_{eta_xi}$) for all $xi. We claim that $gammain{rm dom}(f_alpha)$. To see this, let $xiingammacap{rm dom}(f_alpha)$. Then (by definition of $T^xi_{eta_xi}$), $f_alpha(xi). Since $gammain D$, then $gammain E$ and,  since $xi, then $eta_xi. It follows that $sup{rm ran}(f_alphaupharpoonrightgamma)legamma. Since $f_alpha$ is cofinal in $alpha$, we must necessarily have $gammain{rm dom}(f_alpha)$.

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

Now let $f:Ttokappa$ be the map $f(alpha)=left{begin{array}{cl}f_alpha(xi)&mbox{ if }xiin{rm dom}(f_alpha),\ 0&mbox{ otherwise.}end{array}right.$ Clearly, $f$ is regressive. Also, from the definition of $f$, it follows that ${alphain T:f(alpha)geeta}=T_eta$ for all $eta, so $f$ is unbounded in $kappa$, since each $T_eta$ is in fact stationary, as we showed above. Given any $eta, since $fupharpoonright T_eta$ is regressive, there is some $gamma$ (necessarily, $gammageeta$) such that ${alphain T_eta:f(alpha)=gamma}$ is stationary, by Fodor’s lemma. A simple induction allows us to define a strictly increasing sequence $(gamma_eta:eta such that ${alphain T_{gamma_eta+1}:f(alpha)=gamma_{eta+1}}$ 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 $S$.