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Exercise 8.1.

Solution. Let $f$ be a normal function.

To see that it's closed, suppose $\c{\alpha_\gamma : \gamma < \beta}$ is a strictly increasing sequence of fixed points of $f$ for some $\beta < \kappa$. Let $\alpha = \lim_{\gamma \to \beta} \alpha_\gamma$. Then, by continuity:
$$ \begin{align*}
f(\alpha) = f\bb{\lim_{\gamma \to \beta} \alpha_\gamma} = \lim_{\gamma \to \beta} f(\alpha_\gamma) = \lim_{\gamma < \beta} \alpha_\gamma = \alpha
\end{align*}$$ To see that it's unbounded, we first note that by Lemma 2.4, $f(\alpha) \geq \alpha$ for all $\alpha$.

To see that it's unbounded, fix any $\alpha_0 < \kappa$. Define inductively $\alpha_{n+1} := f(\alpha_n)$. Since $f$ is normal, this is an increasing sequence. Let $\alpha = \sup_{n < \omega} \alpha_n$. Then:
$$ \begin{align*}
f(\alpha) = f\bb{\lim_{n \to \omega} \alpha_n} = \lim_{n \to \omega} f(\alpha_n) = \lim_{n \to \omega} \alpha_{n+1} = \alpha
\end{align*}$$ so $\alpha$ is a fixed point that is $>\alpha_0$.

$\square$

Exercise 8.2.

Solution. Call the set $A$. Suppose $\c{\alpha_\gamma : \gamma < \beta}$ is a strictly increasing sequence in the set $A$ for some $\beta < \gamma$. Let $\alpha = \lim_{\gamma \to \beta} \alpha_\gamma$. Then for any $\xi < \alpha$, we have $\xi < \alpha_\gamma$ for some $\gamma < \beta$. Since $\alpha_\gamma \in A$, $f(\xi) < \alpha_\gamma \leq \alpha$, so $A$ is closed.

To see that it's unbounded, suppose otherwise. Then there exists $\lambda < \kappa$ such that for all $\alpha > \lambda$, there exists $\alpha' < \alpha$ such that $f(\alpha') \geq \alpha$. Fix any $\alpha_0 > \lambda$, and let:
$$ \begin{align*}
\alpha_{n+1} := \min\lbrace \delta \geq \alpha_n : f(\xi) < \delta \text{ for all } \xi < \alpha_n\rbrace
\end{align*}$$ Let $\alpha = \lim_{n \to \infty} \alpha_n$. We claim that $\alpha \in A$. Let $\xi < \alpha$, so $\xi < \alpha_n$ for some $n < \omega$. By construction, $f(\xi) < \alpha_{n+1} \leq \alpha$.

$\square$

Exercise 8.3.

Solution. By Lemma 8.23, we have in particular that intersection of two closed unbounded sets are closed unbounded. If $\Lim(C)$ is another closed unbounded set, then $(S \cap C) \cap \Lim(C) = S \cap (C \cap \Lim(C)) \neq \emptyset$, so $S \cap C$ is also stationary.

$\square$

Exercise 8.4.

Solution. Since $X \subseteq \kappa$ is nonstationary, there exists a closed unbounded $C \subseteq \kappa$ such that $C \cap X = \emptyset$. Let $f : X \to \kappa$ be defined as:
$$ \begin{align*}
f(\alpha) := \sup(C \cap \alpha)
\end{align*}$$ We show that $f$ is regressive. Otherwise, there exists some $\alpha \in X$ such that $f(\alpha) \geq \alpha$. Clearly $f(\alpha) \leq \alpha$ always, so we have that $\sup(C \cap \alpha) = \alpha$. This implies, in particular, that $\alpha$ is a limit ordinal, so by closedness of $C$ we have that $\alpha \in C$. This contradicts that $C \cap X = \emptyset$.

Now let $Y_\gamma := \lbrace \alpha : f(\alpha) \leq \gamma\rbrace$ with $\gamma < \kappa$. Since $C$ is unbounded, there exists $\beta \in C$ such that $\beta > \gamma$. This implies that $\sup(C \cap \delta) \geq \beta$ for all $\delta > \beta$, so $\sup Y_\gamma \leq \beta$. Thus $Y_\gamma$ is bounded.

$\square$

Exercise 8.5.

Lemma 8.5.A. Let $\alpha$ be a limit ordinal, and let $\beta \geq \mathrm{cf}(\alpha)$. Then there exists a sequence of ordinals $\left\langle \alpha_\gamma : \gamma < \beta\right\rangle$ such that $\alpha_\gamma < \alpha$ for all $\gamma < \beta$, and $\alpha = \sum_{\gamma < \beta} \alpha_\gamma$.

Proof. Let $\c{\alpha_\gamma : \gamma < \beta}$ be an increasing sequence of ordinals such that $\alpha_\gamma < \alpha$ for all $\gamma < \beta$, and $\lim_{\gamma < \beta} \alpha_\gamma = \alpha$. For $\gamma < \beta$ define $\alpha_\gamma'$ inductively as follows:

1. $\alpha_0' := \alpha_0$.
2. Suppose $\alpha_\gamma'$ is defined and $\sum_{\delta < \gamma} \alpha_\delta' = \alpha_\gamma$. Now $\alpha_\gamma < \alpha_{\gamma+1}$, so by Lemma 2.25(ii) let $\alpha_{\gamma+1}'$ be the unique ordinal such that $\alpha_\gamma + \alpha_{\gamma+1}' = \alpha_{\gamma+1}$. Then $\sum_{\delta<\gamma+1} \alpha_\delta' = \alpha_{\gamma+1}$.
3. If $\gamma$ is a limit ordinal, $\alpha_\gamma' := \sup_{\delta<\gamma} \alpha_\delta'$.

Then we have:
$$ \begin{align*}
\alpha = \sup_{\gamma < \beta} \alpha_\gamma = \sup_{\gamma < \beta} \sum_{\delta < \gamma}\alpha_\delta' = \sum_{\delta < \beta} \alpha_\delta'
\end{align*}$$

$\blacksquare$

Solution. We induct on the statement in the hint that $\forall\gamma$ $\exists$closed $A \subseteq S$ of length $\alpha$ such that $\gamma < \min{A}$. Since $S$ is unbounded, the base case $\alpha = 1$ is trivial (the exercise holds trivially for $\alpha = 0$). If $\alpha = \beta + 2$, then by induction hypothesis let $A_{\beta+1}$ be a closed subset of length $\beta+1$ with $\min{A_{\beta+1}} > \gamma$. Then $A_{\beta+1}$ is bounded, as $A_{\beta+1}$ is countable but $\omega_1$ is regular uncountable. Let $A_\alpha = A_{\beta+1} \cup \lbrace \delta\rbrace$ where $\delta \in S$ is such that $\delta > \sup A_{\beta+1}$. Then $A_\alpha$ is closed, as for any $\lbrace \xi_n : n \in \omega\rbrace \subseteq A_\alpha$, we have that:

1. If $\xi_n < \sup{A_{\beta+1}}$ for all $n$, then $\lim_n \xi_n \in A_\beta \subseteq A_\alpha$ by closedness of $A_{\beta+1}$ (and that $\sup A_{\beta+1} \in A_{\beta+1}$).
2. If $\xi_n = \sup{A_{\beta+1}}$ for some $n$, then $\lim_n \xi_n = \sup{A_{\beta+1}} \in A_\alpha$.

If $\alpha$ is a limit ordinal, then by Lemma 8.5.A write $\alpha = \sum_n \alpha_n$, where $\alpha_n < \alpha$ for all $n$. Let $A_n$ be a closed set of ordinals such that $\sup \bigcup_{i < n} A_i < \min A_n$ (by induction hypothesis). Let $A_\alpha := \bigcup_{n=0}^\infty A_n$. Clearly $A_\alpha$ is of length $\alpha$, and $A_\alpha$ is closed as if $\delta < \sup A_\alpha$ is a limit ordinal, then $\delta < \sup A_n$ for some $n \in \omega$, and $\delta \in \bigcup_{i \leq n} A_i$ by closure of all $A_i$'s (finite disjoint union of closed sets of ordinals is still closed. This is also why the exercise doesn't immediately generalise to stationary sets $S \subseteq \omega_2$).

It remains to show the case where $\alpha = \beta + 1$, where $\beta$ is a limit ordinal. The reason why we can't simply mimic the proof for $\beta + 2$ is because if $A_\beta$ is a closed set of ordinals of length $\beta$, we need not have $\sup A_\beta \in S$. Thus, we use the hint to help us resolve this case.

For $\xi < \omega_1$ let $A_\xi$ be closed sets of ordinals $A$ of length $\alpha$ each such that $\lambda_\xi := \sup\bigcup_{\nu<\xi} A_\nu < \min A_\xi$.

Claim.. The set $\lbrace \lambda_\xi : \xi < \omega_1\rbrace$ is a closed unbounded subset of $\omega_1$.

Proof. Unboundedness is immediate. For closure, fix a limit ordinal $\delta < \omega_1$. Then:
$$ \begin{align*}
\sup\lbrace \lambda_\xi : \xi < \delta\rbrace = \sup_{\xi < \delta} \sup \bigcup_{\nu < \xi} A_\nu = \bigcup_{\xi < \delta} \bigcup \bigcup_{\nu < \xi} A_\nu = \bigcup \bigcup_{\nu < \delta} A_\nu = \lambda_\delta
\end{align*}$$ Since $\lambda_\delta$ is clearly in the set, closedness follows.

$\blacksquare$

Since $S$ is stationary, $S \cap \lbrace \lambda_\xi : \xi < \omega_1\rbrace \neq \emptyset$, so $\lambda_\xi \in S$ for some $S$. Let $\xi = \lim_n \xi_n$, and by Lemma 8.5.A we may write $\beta = \sum_n \beta_n$, where $\beta_n < \beta$ for all $n$. Pick initial segments $B_{\xi_n} \subseteq A_{\xi_n}$ of length $\beta_n + 1$ (successor length is picked to ensure closure). Then $\sum_n (\beta_n + 1) = \beta$, so let $A = \bigcup_{n=0}^\infty B_{\xi_n}$. Then $\sup A = \lambda_\xi \in S$, so $A \cup \lbrace \lambda_\xi\rbrace$ is a closed set of ordinals of length $\alpha + 1$.

$\square$

Exercise 8.6.

Solution. Suppose $\kappa$ is Mahlo. Let $X \subseteq \kappa$ be the set of all inaccessible cardinals below $\kappa$. Since $\kappa$ is Mahlo, $X$ is stationary. Define $f : X \to \kappa$ by stipulating that $f(\lambda) = \alpha$, in which $\lambda$ is the $\alpha^\text{th}$ inaccessible. by Fodor's Theorem, $f$ is constant on a stationary set. This is clearly not possible.

$\square$

Exercise 8.7.

Solution. First suppose $\kappa$ is a limit cardinal in which the set of all strong limit cardinals below $\kappa$ is unbounded. Let $\lambda < \kappa$ be a cardinal, and by unboundedness there exists $\lambda < \eta < \kappa$ such that $\eta$ is a strong limit cardinal. This implies that $2^\lambda < \eta < \kappa$. Thus, $\kappa$ is a strong limit.

The second and third claims follow from (by definition) that weakly inaccessible $+$ strong limit gives inaccessible, and weakly Mahlo $+$ strong limit gives Mahlo.

$\square$

Exercise 8.8.

Lemma 8.8.A. Let $I$ be an ideal on a set $X$, and let $F$ be its dual filter. Let $Y \subseteq X$. Then $Y \in I$ iff $Z \cap Y = \emptyset$ for some $Z \in F$.

Proof. If $Y \in I$, then $X - Y \in F$ and $Y \cap (X - Y) = \emptyset$. On the other hand, if $Z \cap Y = \emptyset$ and $Z \in F$, then $Z \subseteq X - Y$ and by upward closure $X - Y \in F$. Thus $Y \in I$.

$\blacksquare$

Solution. Let $F$ be the dual filter of $I$.

$\implies$: Suppose $I$ is normal. We may mimic the proof of Fodor's Theorem: Let $f$ be a regressive function on $S \notin I$. For $\gamma < \kappa$ consider the set $S_\gamma := \lbrace \alpha \in S : f(\alpha) = \gamma\rbrace$, and suppose for a contradiction that $S_\gamma \in I$ for all $\gamma < \kappa$. By Lemma 8.8.A, let $C_\gamma \in F$ such that $C_\gamma \subseteq \kappa - S_\gamma$. By normality of $F$, let $C := \bigtriangle_{\gamma<\kappa} C_\gamma \in F$. Since $S \notin I$, $S \cap C \neq \emptyset$, so let $\alpha \in S \cap C$. This implies that $\alpha \in \bigcap_{\gamma < \alpha} C_\gamma \subseteq \bigcap_{\gamma < \alpha} \kappa - S_\gamma$, i.e. $\alpha \notin S_\gamma$ for all $\gamma < \alpha$. Thus, $f(\alpha) \neq \gamma$ for all $\gamma < \alpha$, so $f(\alpha) \geq \alpha$, contradicting that $f$ is regressive.

$\impliedby$: Let $X_\alpha,S_0,f$ be defined as in the hint where we suppose $\bigtriangle X_\alpha \notin F$. Note that $f$ is well-defined, as for $\alpha < \kappa$:
$$ \begin{align*}
\alpha \in S_0 \iff \alpha \notin \bigtriangle_{\alpha < \kappa} X_\alpha \iff \alpha \notin \bigcap_{\xi < \alpha} X_\alpha \iff (\exists \xi < \alpha) \, \alpha \notin X_\xi
\end{align*}$$ $f$ is regressive on $S_0 \notin I$, so by the hypothesis there is some $S \subseteq S_0$, $S \notin I$, such that $f$ is constant on $S$. Suppose $f(\alpha) = \gamma$ for all $\alpha \in S$. Then $X_\gamma \cap S = \emptyset$, as $\alpha \in S \implies f(\alpha) = \gamma \implies \alpha \notin X_\gamma$. Since $X_\gamma \in F$, by Lemma 8.8.A we have that $S \in I$, a contradiction.

$\square$

Exercise 8.9.

Solution. Suppose $F$ is a normal nonprincipal filter on $\omega$. Let $f : \omega \to \omega$ be a function such that $f(n + 1) = n$ for $n \in \omega$. Then $f$ is regressive, so by Exercise 8.8 we have that $f$ is constant on some $S \in F$. But $f_{-1}(n)$ is finite for all $n$, so $F$ contains a finite set, contradicting its nonprincipal property.

$\square$

Exercise 8.10.

Solution. Suppose there exists a such an ideal $I$. Let $F$ be its dual filter. Let $\lambda := \cf(\kappa)$. Let $\c{X_\alpha : \alpha < \lambda}$ be a sequence of subsets of $\kappa$ such that $\kappa - X_\alpha$ is bounded for all $\alpha < \lambda$, and $\bigcap_{\alpha < \lambda} X_\alpha = \emptyset$. Since $F$ contains all subsets of $\kappa$ which complements are bounded, $X_\alpha \in F$ for all $\alpha < \lambda$.

For $\lambda \leq \alpha < \kappa$, let $X_\alpha := \kappa$. Since $F$ is normal, $X := \bigtriangle_{\alpha < \kappa} X_\alpha \in F$. Now observe that if $\alpha \geq \lambda$, then:
$$ \begin{align*}
\alpha \in \bigtriangle_{\alpha < \kappa} X_\alpha \iff \alpha \in \bigcap_{\xi < \alpha} X_\xi \subseteq \bigcap_{\xi < \lambda} X_\xi = \emptyset
\end{align*}$$ Thus, $\bigtriangle_{\alpha < \kappa} X_\alpha \subseteq \lambda$ is bounded. This implies that $\bigtriangle_{\alpha < \kappa} X_\alpha \in I$, a contradiction.

$\square$

Exercise 8.11.

Exercise 8.11(i).

Solution. This follows from that if $S \cap \alpha$ is stationary, then $T \cap \alpha$ is also stationary.

$\square$

Exercise 8.11(ii).

Solution. By Exercise 8.11(i), we have $T(S \cup T) \supseteq \Tr(S) \cup \Tr(T)$. Conversely, let $\alpha \in \Tr(S \cup T)$. Then $\cf(\alpha) > \omega$ and $(S \cup T) \cap \alpha = (S \cap \alpha) \cup (T \cap \alpha)$ is stationary. If neither $S \cap \alpha$ nor $T \cap \alpha$ is stationary, then for some closed unbounded sets $C_S,C_T \subseteq \kappa$ we have $(S \cap \alpha) \cap C_S = \emptyset$ and $(T \cap \alpha) \cap C_T = \emptyset$. Then $C_S \cap C_T$ is still closed unbounded, and $(S \cup T) \cap \alpha \cap (C_S \cap C_T) = \emptyset$, a contradiction.

$\square$

Exercise 8.11(iii).

Solution. Let $\alpha \in \Tr(\Tr(S))$, so $\cf(\alpha) > \omega$ and $\Tr(S) \cap \alpha$ is stationary. Now:
$$ \begin{align*}
\Tr(S) \cap \alpha = \lbrace \beta < \alpha : \cf(\beta) > \omega \text{ and } S \cap \beta \text{ is stationary}\rbrace
\end{align*}$$ Let $C$ be a closed unbounded subset of $\alpha$. Note that $\Lim(C) \subseteq C$ is closed unbounded by Lemma 8.11(iii).A. Then $\Lim(C) \cap (\Tr(S) \cap \alpha) \neq \emptyset$, so fix $\beta \in \Lim(C) \cap (\Tr(S) \cap \alpha)$. We have $\cf(\beta) > \omega$ and $S \cap \beta$ is stationary. But since $\beta \in \Lim(C)$ as well, it is a limit point, $C \cap \beta$ is closed unbounded. Therefore $(C \cap \beta) \cap (S \cap \beta) \neq \emptyset$. Thus $(S \cap \alpha) \cap C \neq \emptyset$ and so $S \cap \alpha$ is stationary. Hence $\alpha \in \Tr(S)$.

$\square$

Exercise 8.11(iv).

Solution. We only deal with ordinals of cofinality $>\omega$.

We note that $S \simeq T$ mod $I_\text{NS}$ iff there exists some closed unbounded set $C$ such that $S \cap C = T \cap C$. Note that $S \cap \Lim(C) = T \cap \Lim(C)$.

Claim.. $\Tr(S \cap \Lim(C)) = \Tr(S) \cap \Lim(C)$.

Proof. We have $\Tr(S \cap \Lim(C)) \subseteq \Tr(S)$ by Exercise 8.11(i). Furthermore, if $\alpha \in \Tr(S \cap \Lim(C))$, then $(S \cap \Lim(C)) \cap \alpha$ is a stationary subset of $\alpha$. It is in particular unbounded, and since $\Lim(C)$ is closed we have $\alpha \in \Lim(C)$. Thus, we have $\Tr(S \cap \Lim(C)) \subseteq \Tr(S) \cap \Lim(C)$.

Now suppose $\alpha \in \Tr(S) \cap \Lim(C)$. Then $S \cap \alpha$ is stationary. By Lemma 8.11(iv).A, $\alpha$ is a limit point of $\Lim(C)$, thus $C \cap \alpha$ is closed unbounded. Therefore $(S \cap \alpha) \cap (\Lim(C) \cap \alpha) = (S \cap \Lim(C)) \cap \alpha$ is stationary (by Exercise 8.3), so $\alpha \in \Tr(S \cap \Lim(C))$.

$\blacksquare$

Since $\Lim(C)$ is closed unbounded and:
$$ \begin{align*}
\Tr(S) \cap \Lim(C) = \Tr(S \cap \Lim(C)) = \Tr(T \cap \Lim(C)) = \Tr(T) \cap \Lim(C)
\end{align*}$$ we have $\Tr(S) \simeq \Tr(T)$ mod $I_\text{NS}$.

$\square$

Exercise 8.12.

Solution. Fix $\alpha < \kappa$ such that $\cf(\alpha) > \omega$.

$\Tr(E_\lambda^\kappa) \supseteq \lbrace \alpha < \kappa : \cf(\alpha) \geq \lambda^+\rbrace$: Suppose $\cf(\alpha) \geq \lambda^+$. Let $C$ be a closed unbounded subset of $\alpha$. Fix an increasing sequence $\lbrace \beta_\xi : \xi < \lambda\rbrace$ in $C$, which is possible as $C$ is unbounded and $\lambda < \cf(\alpha)$. Since $C$ is closed, $\beta := \lim_{\xi\to\lambda} \beta_\xi \in C$. Then $\cf(\beta) = \cf(\lambda) = \lambda$ by Lemma 3.7(ii), so $\beta \in C \cap (E_\lambda^\kappa \cap \alpha)$. Hence $\alpha \in \Tr(E_\lambda^\kappa)$.

$\Tr(E_\lambda^\kappa) \subseteq \lbrace \alpha < \kappa : \cf(\alpha) \geq \lambda^+\rbrace$: Suppose $\cf(\alpha) \leq \lambda$. Let $\c{\alpha_\xi : \xi < \cf(\alpha)}$ be a sequence of ordinals cofinal in $\alpha$, such that if $\xi$ is a limit ordinal then $\alpha_\xi = \lim_{\gamma \to \xi} \alpha_\gamma$. Let $C := \lbrace \alpha_\xi : \xi < \cf(\alpha)\rbrace$. Clearly $C$ is closed unbounded in $\alpha$, but $C \cap (E_\lambda^\kappa \cap \alpha) = \emptyset$, for if $\alpha_\xi \in C$ then:
$$ \begin{align*}
\cf(\alpha_\xi) \leq \cf(\xi) \leq \xi < \cf(\alpha) \leq \lambda
\end{align*}$$

$\square$

Exercise 8.13.

Lemma 8.13.A. If $\lambda < \mu$ are regular, then $E_\lambda^\kappa < E_\mu^\kappa$.

Proof. We note that if $\alpha$ is a limit ordinal with $\cf(\alpha) > \lambda$, then $E_\lambda^\alpha$ is still stationary (in the sense where $\alpha$ need not be an uncountable regular cardinal). Thus if $\alpha \in E_\mu^\kappa$ and $\cf(\alpha) = \mu > \lambda$, then $E_\lambda^\alpha = E_\lambda^\kappa \cap \alpha$ is stationary. Thus $E_\lambda^\kappa < E_\mu^\kappa$.

$\blacksquare$

Lemma 8.13.B. For stationary sets $S,T$, the following are equivalent:

1. $S < T$.
2. There exists a closed unbounded set $C$ such that $T \cap C \subseteq \Tr(S)$.
3. $T - \Tr(S)$ is non-stationary.

Therefore, if $S < T$ and $X$ is a stationary subset of $T$, then $S < X$.

Proof. (i) $\iff$ (ii): We have:
$$ \begin{align*}
S < T &\iff S \cap \alpha \text{ is stationary for almost all } \alpha \in T \\
&\iff \lbrace \alpha \in T : S \cap \alpha \text{ is not stationary}\rbrace \text{ is non-stationary} \\
&\iff \text{There is a closed unbounded $C$ such that } T \cap C \subseteq \Tr(S)
\end{align*}$$ (ii) $\iff$ (iii): It suffices to show that for two stationary sets $S,T$, we have $S \cap C \subseteq T$ for some closed unbounded set iff $S - T$ is non-stationary. For $\implies$, let $C$ be closed unbounded such that $S \cap C \subseteq T$. If $S - T$ is stationary, then $C \cap (S - T) \neq \emptyset$, which contradicts that $C \cap (S - T) \subseteq T$. For $\impliedby$, suppose $S - T$ is non-stationary. Then $\kappa - (S - T)$ is closed unbounded. We see that for $\alpha \in \kappa$, we have:
$$ \begin{align*}
\alpha \in S \cap (\kappa - (S - T)) &\implies \alpha \in S \wedge \alpha \notin S - T \\
&\implies \alpha \in S \wedge (\alpha \notin S \vee \alpha \in T) \\
&\implies \alpha \in S \wedge \alpha \in T
\end{align*}$$ so $S \cap (\kappa - (S - T)) \subseteq T$.

Therefore, if $X \subseteq T$ is stationary, then $X = T \cap D$ for some closed unbounded $D$, so if $S < T$ then $X \cap C = (T \cap D) \cap C \subseteq \Tr(S)$, so $S < X$.

$\blacksquare$

Lemma 8.13.C. If $\Tr(S)$ is stationary, then $S$ is stationary.

Proof. Let $C$ be a closed unbounded set. By Lemma 8.11(iii).A $\Lim(C)$ is also closed unbounded. Since $\Tr(S)$ is stationary, let $\alpha \in \Tr(S) \cap \Lim(C)$. Then $C \cap \alpha$ is closed unbounded in $\alpha$ and $S \cap \alpha$ is stationary, so $S \cap C \cap \alpha \neq \emptyset$. Thus $S \cap C \neq \emptyset$.

$\blacksquare$

Lemma 8.13.D. If $E$ is the canonical stationary set of order $\alpha$ and $S$ is a stationary set such that $o(S) > \alpha$, then $S \cap \Tr(E) \neq \emptyset$.

Proof. We begin by remarking that all statements in Exercise 8.11 (except Exercise 8.11(iv)) do not assume that $S,T$ are stationary, so we may apply these results on sets which are not necessarily stationary.

Suppose the assertion is false, so we let $S$ be a $<$-minimal counterexample (this is possible as $<$ is a well-founded relation). Then $S \cap \Tr(E) = \emptyset$. Let $T < S$ be stationary of order $\alpha$, so by Lemma 8.13.C there exists a closed unbounded $F$ such that $S \cap F \subseteq \Tr(T)$. Let $A := T \cap E$, $B := T \cap \Tr(E)$ and $C := T - (A \cup B)$. Since $A \subseteq E$, by Exercise 8.11(i) we have that $\Tr(A) \subseteq \Tr(E)$, so $S \cap \Tr(A) = \emptyset$. We also have that $B \subseteq \Tr(E)$, so by Exercise 8.11(i) and Exercise 8.11(iii) we have that $\Tr(B) \subseteq \Tr(\Tr(E)) \subseteq \Tr(E)$. Thus $S \cap \Tr(B)$ as well. This implies that:
$$ \begin{align*}
S \cap F &= S \cap F \cap \Tr(T) \\
&= S \cap F \cap \Tr(A \cup B \cup C) \\
&= S \cap F \cap (\Tr(A) \cup \Tr(B) \cup \Tr(C)) && \text{by Exercise 8.11(ii)} \\
&= S \cap F \cap \Tr(C)
\end{align*}$$ where the last equality hold as, as shown, $\Tr(A)$ and $\Tr(B)$ have empty intersection with $S$. Therefore $S \cap F \subseteq \Tr(C)$. Furthermore, if $D$ is any closed unbounded set, then $\emptyset \neq S \cap F \cap D \subseteq \Tr(C) \cap D$, so we have that $C$ is stationary by Lemma 8.13.C. By Lemma 8.13.B, this implies that $C < S$.

Now since $C \subseteq T$, by Lemma 8.13.B if every stationary $X$ such that $X < T$, we have $X < C$. This implies that $o(C) \geq o(T) = \alpha$. Since $E$ is canonical of order $\alpha$, if $o(C) = \alpha$ then we must have $C \cap E = \emptyset$, a contradiction. Thus $o(C) > \alpha$. But $C < S$, which contradicts the minimality of $S$.

$\blacksquare$

Lemma 8.13.E. If $E$ is the canonical stationary set of order $\alpha$, and $S$ is a stationary set such that $o(S) > \alpha$, then $E < S$.

Proof. Suppose $E \not< S$. By Lemma 8.13.B, this implies that $S - \Tr(E)$ is stationary. But $o(S - \Tr(E)) \geq o(S) > \alpha$ and $(S - \Tr(E)) \cap \Tr(E) = \emptyset$, contradicting Lemma 8.13.D.

$\blacksquare$

Solution. We shall induct on $\alpha$ that if $\lambda$ is the $\alpha^\text{th}$ regular cardinal, then $E_\lambda^\kappa$ is the canonical stationary set of order $\alpha$. We first prove that $E_\lambda^\kappa$ satisfies (ii).

(ii) Base Case - $\alpha = 0$: For $\alpha = 0$, if $S$ is a stationary set such that $S < E_\omega^\kappa$, the in particular almost all ordinals in $E_\omega^\kappa$ have uncountable cofinality, which is false. Thus $o(E_\omega^\kappa) = \sup\emptyset = 0$.

(ii) Successor Case - $\lambda$ regular: Let $\lambda$ be the $\alpha^\text{th}$ regular cardinal, and suppose $E_\lambda^\kappa$ is the canonical stationary set of order $\alpha$. Then $\lambda^+$ is the $(\alpha + 1)^\text{th}$ regular cardinal.

$o(E_{\lambda^+}^\kappa) = \alpha + 1$: Since $E_\lambda^\kappa < E_{\lambda^+}^\kappa$, by Lemma 8.13.A we have immediately that $o(E_{\lambda^+}^\kappa) \geq \alpha + 1$. Now suppose for a contradiction that $o(E_{\lambda^+}^\kappa) > \alpha + 1$. Let $S < E_{\lambda^+}^\kappa$ such that $o(S) \geq \alpha + 1$, and by Lemma 8.13.D we have that $ E_\lambda^\kappa < S$. Thus $E_\lambda^\kappa < S < E_{\lambda^+}^\kappa$, so by Lemma 8.13.B $E_{\lambda^+}^\kappa - \Tr(S)$ and $S - \Tr(E_\lambda^\kappa)$ are non-stationary.

Let $C$ be a closed unbounded set that is disjoint from both these sets. Then $\Lim(C)$ is closed unbounded, so let $\alpha \in \Lim(C) \cap E_{\lambda^+}^\kappa$. Since $\Lim(C) \subseteq C$ and $C \cap (E_{\lambda^+}^\kappa - \Tr(S)) = \emptyset$, $S \cap \alpha$ is stationary. Now $\cf(\alpha) = \lambda^+$, so $D := \lbrace \alpha_\gamma : \gamma < \lambda^+\rbrace$ be a normal sequence cofinal in $\alpha$ such that $\alpha_\gamma$ is limit for all $\gamma$. Then $D$ is a closed unbounded subset of $\alpha$ of order type $\lambda^+$, so $\Lim(C) \cap D$ remains closed unbounded in $\alpha$. Fix $\beta \in (S \cap \alpha) \cap \Lim(C) \cap D$, and since $\beta = \alpha_\gamma$ for some $\gamma < \lambda^+$, $\cf(\beta) \leq \lambda$. At the same time, since $\beta \in S \cap \Lim(C)$ and $C \cap (S - \Tr(E_{\lambda^\kappa})) = \emptyset$, we have that $E_\lambda^\kappa \cap \beta$ is stationary. Let $F \subseteq \beta$ be closed unbounded in $\beta$ of order-type $\cf(\beta)$ (constructed in a similar way to $D$). Then again, we may let $\delta \in E_\lambda^\kappa \cap \beta \cap F$, and we have that $\cf(\delta) < \lambda$ as $\delta \in F$. Yet $\cf(\delta) = \lambda$ as $\delta \in E_\lambda^\kappa$, a contradiction.

$E_{\lambda^+}^\kappa$ is canonical: Let $S$ be a stationary set of order $\alpha + 1$. By Lemma 8.13.D, $E_\lambda^\kappa < S$, so by Lemma 8.13.B we have that $S \cap C \subseteq \Tr(E_\lambda^\kappa)$ for some closed unbounded set $C$. By Exercise 8.12, $S \cap C \subseteq \lbrace \alpha \in \kappa : \cf(\alpha) \geq \lambda^+\rbrace$. Now observe that if $S \cap E_{\lambda^+}^\kappa = \emptyset$ (so $S \cap C \subseteq \lbrace \alpha \in \kappa : \cf(\alpha) > \lambda^+\rbrace$), then by Exercise 8.12 again we have that $S \cap C \subseteq \Tr(E_{\lambda^+}^\kappa)$. By Lemma 8.13.B this implies that $E_{\lambda^+}^\kappa < S$. This contradicts that $o(E_{\lambda^+}^\kappa) = \alpha + 1$, as we just showed.

(ii) Successor Case - $\lambda$ singular: Suppose $\lambda$ is singular, and $\lambda^+$ is the $\alpha^\text{th}$ regular cardinal. Assume that for each $\mu < \lambda$, if $\mu$ is the $\eta^\text{th}$ regular cardinal then $E_\mu^\kappa$ is the canonical stationary set of order $\eta$.

$o(E_{\lambda^+}^\kappa) = \alpha$: We first observe that if $S$ is a stationary set of order $\alpha$, then $S \cap C \subseteq \lbrace \alpha < \kappa : \cf(\alpha) \geq \lambda^+\rbrace$ for some closed unboudned $C$. Indeed, we have that $o(S) > o(E_\mu^\kappa)$ so $E_\mu^\kappa < S$ by Lemma 8.13.E. This implies that for all regular $\mu < \lambda$, there exists a closed unbounded set $C_\mu$ such that $S \cap C_\mu \subseteq \Tr(E_\mu^\kappa) = \lbrace \alpha < \kappa : \cf(\alpha) \geq \mu^+\rbrace$. Let $C := \bigcap_{\mu<\lambda} C_\mu$, so $C$ is still closed unbounded ($\lambda < \kappa$), and we have that $S \cap C \subseteq \lbrace \alpha < \kappa : \cf(\alpha) \geq \lambda\rbrace$. Since $\lambda$ is singular, we in fact have $S \cap C \subseteq \lbrace \alpha < \kappa : \cf(\alpha) \geq \lambda^+\rbrace$

We now proceed to show that $o(E_{\lambda^+}^\kappa) = \alpha$. Once again it's easy to see that $o(E_{\lambda^+}^\kappa) \geq \alpha$, so suppose for a contradiction that $o(E_{\lambda^+}^\kappa) > \alpha$. Let $S < E_{\lambda^+}^\kappa$ be a stationary set of order $\alpha$. By above and Lemma 8.13.B, there exists a closed unbounded $C$ such that $S \cap C \subseteq \lbrace \alpha < \kappa : \cf(\alpha) \geq \lambda^+\rbrace$ and $E_{\lambda^+}^\mu \cap C \subseteq \Tr(S)$. Let $\alpha \in E_{\lambda^+}^\kappa D$ (possible as $E_{\lambda^+}^\kappa$ is stationary). Since $\cf(\alpha) = \lambda^+$, there exists a closed unbounded subset $D \subseteq \alpha$ of order-type $\lambda^+$ consisting only of limit ordinals. We may thus take $\beta \in D \cap \Lim(C) \cap S$. Then $\cf(\beta) < \lambda$ as $\beta \in D$, but $\cf(\beta) \geq \lambda^+$ as $S \cap \Lim(C) \subseteq \lbrace \alpha < \kappa : \cf(\alpha) \geq \lambda^+\rbrace$, a contradiction.

$E_{\lambda^+}^\kappa$ is canonical: Let $S$ be a stationary set of order $\alpha$. As we proved above, there exists a closed unbounded set $C$ such taht $S \cap C \subseteq \lbrace \alpha < \kappa : \cf(\alpha) \geq \lambda^+\rbrace$. If $\cf(\alpha) > \lambda^+.$ for all $\alpha \in S \cap C$, then $S \cap C \subseteq \Tr(E_{\lambda^+}^\kappa)$, so $E_{\lambda^+} < X$ by Lemma 8.13.B. This implies that $o(X) \geq \alpha + 1$, a contradiction.

(ii) Limit Case: It is almost verbatim to the working of the successor singular cardinal case (replace $\lambda^+$ with just regular limit $\lambda$).

(i): We conclude by showing that$E_\lambda^\kappa$ satisfies (i) of the definition of canonical stationary set of order $\alpha$, where $\lambda$ is the $\alpha^\text{th}$ regular cardinal. Let $S \subseteq E_\lambda^\kappa$ be stationary. By Lemma 8.13.B, we see that $o(S) \geq o(E_\lambda^\kappa) = \alpha$. If $o(S) = \beta > \alpha$, then let $\mu$ be the $\beta^\text{th}$ regular cardinal. Since $E_\mu^\kappa$ satisfies (ii) of the definition of stationary set of order $\alpha$, $S \cap E_\mu^\kappa \neq \emptyset$. This is not possible, as $\mu > \lambda$ so $E_\lambda^\kappa \cap E_\mu^\kappa = \emptyset$. Hence $o(S) = \alpha$ necessarily.

$\square$

Exercise 8.14.

Solution.

1. $\implies$: Suppose $\kappa$ is not weakly inaccessible. Then $\kappa = \mu^+$ for some $\mu$ (we only define $o(\kappa)$ on regular cardinals). Let $S \subseteq \kappa$ be stationary such that $o(S) > \mu$. For each regular $\lambda < \kappa$, we have $E_\lambda^\kappa < S$ by Exercise 8.13 and Lemma 8.13.E, so by Lemma 8.13.B there exists a closed unbounded set $C_\lambda$ such that $S \cap C_\lambda \subseteq \Tr(E_\lambda^\kappa)$. Let $C := \bigcap_{\lambda\leq\mu} C_\lambda$, so $C$ remains closed unbounded as $\mu < \kappa$. Since $\Tr(E_\lambda^\kappa) = \lbrace \alpha < \kappa : \cf(\lambda) \geq \lambda^+\rbrace$, this implies that $S \cap C \subseteq \lbrace \alpha < \kappa : \cf(\lambda) \geq \mu^+\rbrace = \emptyset$, a contradiction.
   
   $\impliedby$: Suppose $\kappa$ is weakly inaccessible. Then $\kappa = \sup\lbrace \lambda < \kappa : \lambda \text{ is regular}\rbrace$. Since $\kappa$ is regular, $\vert \lbrace \lambda < \kappa : \lambda \text{ is regular}\rbrace\vert = \kappa$, i.e. $\kappa$ is the $\kappa^\text{th}$ regular cardinal. Thus for each $\alpha < \kappa$, there if $\lambda$ is the $\alpha^\text{th}$ regular cardinal then $o(\kappa) \geq E_\lambda^\kappa = \alpha$, so $o(\kappa) \geq \kappa$.
2. $\implies$: Suppose $o(\kappa) \geq \kappa + 1$. By (i), we have that $\kappa$ is weakly inaccessible. Let $S \subseteq \kappa$ be stationary such that $o(S) = \kappa$. Let $\c{\lambda_\xi : \xi < \kappa}$ be an increasing enumeration of regular cardinals below $\kappa$. Then $E_{\lambda_\xi}^\kappa < S$ for all $\xi$, so there exists a closed unbounded $C_\xi$, consisting only of cardinals (the set of all cardinals below $\kappa$ is closed unbounded) such that $S \cap C_\xi \subseteq \Tr(E_{\mu_\xi}^\kappa)$. Let $C := \bigtriangle_{\xi<\kappa} C_\xi$, so $C$ is closed unbounded. We shall show that $S \cap C$ is a stationary set of regular cardinals, which implies that $\kappa$ is weakly Mahlo. Indeed, if $\lambda \in S \cap C$ then $\lambda \in S \cap C_{\lambda_\xi} \subseteq \Tr(E_{\lambda_\xi}^\kappa)$ for all $\xi < \lambda$, so $\cf(\lambda) > \lambda_\xi$. Thus we must have $\cf(\lambda) = \lambda$, for otherwise we would have $\cf(\lambda) > \lambda_{\cf(\lambda)} \geq \cf(\lambda)$, a contradiction.
   
   $\impliedby$: Suppose $\kappa$ is weakly Mahlo. Then $S := \lbrace \lambda < \kappa : \lambda \text{ is regular}\rbrace$ is stationary. Since for each $\lambda < \kappa$ regular, we have $S \cap (\kappa - \lambda^+) \subseteq \Tr(E_\lambda^\kappa)$ ($\kappa - \lambda^+$ is an end-segment of $\kappa$, hence closed unbounded). Thus $E_\lambda^\kappa < S$ for all regular $\lambda < \kappa$, so $o(S) \geq \kappa$. Therefore $o(\kappa) \geq \kappa + 1$.

$\square$

Exercise 8.15.

Solution. Closedness: Let $\beta < \kappa$ and let $\c{x_\xi : \xi < \beta}$ be a sequence of sets in $\lbrace x \in P_\kappa(A) : x \supseteq \alpha\rbrace$. Let $x := \bigcup_{\xi<\beta} x_\xi$. Clearly $\alpha \subseteq x$. Furthermore, since $\vert x_\xi\vert < \kappa$ for all $\xi < \alpha$ and $\kappa$ is regular, we have:
$$ \begin{align*}
\vert x\vert = \sup\lbrace \vert x_\xi\vert : \xi < \alpha\rbrace < \kappa
\end{align*}$$ so indeed $x \in P_\kappa(A)$.

Unboundedness: Let $y \in P_\kappa(A)$. Then $\vert y\vert < \kappa$, and since $\vert \alpha\vert < \kappa$ we have $\vert \alpha \cup y\vert \leq \vert \alpha\vert + \vert y\vert < \kappa$. Then $\alpha \cup y \in \lbrace x \in P_\kappa(A) : x \supseteq \alpha\rbrace$.

$\square$

Exercise 8.16.

Solution. Let $g$ be a function on $X \subseteq P_\kappa(A)$ such that $A - X \notin F$. For each $Y \subseteq F$ and $a \in A$, let:
$$ \begin{align*}
S_a^Y := \lbrace x \in X : a \in g(x) - Y\rbrace
\end{align*}$$ Suppose $S_a^\emptyset$ is not $F$-positive for all $a \in A$. Let $C_a^Y := P_\kappa(A) - S_a^Y$ for all $a$, so $C_a^\emptyset \in F$. Since $F$ is normal, we have that $C^\emptyset := \bigtriangle_{a \in A} C_a^\emptyset \in F$ (and $C_a^Y$ defined accordingly). By Lemma 8.8.A, we have that $C^\emptyset \cap X$ is $F$-positive (by letting $I$ be the dual filter of $F$). Let $x \in C^\emptyset \cap X$. Then $x \in \bigcap_{a \in x} C_a^\emptyset = \bigcap_{a \in x} (P_\kappa(A) - S_a^\emptyset)$, i.e. $x \notin S_a^\emptyset$ for all $a \in x$. This implies that $g(x) \cap x = \emptyset$, but since $g(x) \in [x]^{<\omega}$ we have that $g(x) = \emptyset$. Thus, $g$ takes the constant value $\emptyset$ at an $F$-positive set $C^\emptyset \cap X$.

Now suppose $S_{a_n}^{\lbrace a_0,\dots,a_{n-1}\rbrace}$ is $F$-positive for some $a_0,\dots,a_{n-1} \in A$ (where if $n = 0$ then the superscript is the empty set). We can then repeat the argument above with $g$ restricted to the $F$-positive domain $S_{a_n}^{\lbrace a_0,\dots,a_{n-1}\rbrace}$. If we fix any $x \in S_{a_0}^X$, then this argument will terminate in $\vert g(x)\vert $ steps. Say $S_{a_n}^{\lbrace a_0,\dots,a_{n-1}\rbrace}$ is not $F$-positive for all $a_n$ (for some $n$). By repeating the argument we have that for all $x \in C^{\lbrace a_0,\dots,a_{n-1}\rbrace} \cap S_{a_{n-1}}^{\lbrace a_0,\dots,a_{n-2}\rbrace}$ (which is $F$-positive), we have that $a_i \in g(x)$ for all $i$ but $g(x) - \lbrace a_0,\dots,a_{n-1}\rbrace = \emptyset$. Then $g(x) = \lbrace a_0,\dots,a_{n-1}\rbrace$ on an $F$-positive set.

$\square$

Exercise 8.17.

Solution. By Lemma 8.26, it suffices to show that $C_f \in F$ for all $f : [A]^{<\omega} \to P_\kappa(A)$. Suppose not, so there exists some $f$ such that $C(f) \notin F$, i.e. $X := P_\kappa(A) - C(f)$ is $F$-positive. We define $g : X \to P_\kappa(A)$ by:
$$ \begin{align*}
g(x) = \text{any $e \in [x]^{<\omega}$ such that $f(e) \not\subseteq x$}
\end{align*}$$ This is well-defined as $x \notin C(f)$. By Exercise 8.16, there exists an $F$-positive set $S$ such that $g$ takes a constant value $e_0 \in [A]^{<\omega}$ on $S$. That is, $f(e_0) \not\subseteq x$ for all $x \in S$. This is a contradiction, as $S$ being $F$-positive implies that it is unbounded, so there must exist some $y \in S$ such that $f(e_0) \subseteq y$.

$\square$

Exercise 8.18.

Solution. The proof is very similar to that of Exercise 8.15.

Closedness: Let $\alpha < \kappa$ and let $\c{x_\xi : \xi < \alpha}$ be a sequence of sets in $\lbrace x \in P_\kappa(A) : \vert x\vert \geq \aleph_1\rbrace$. Let $x := \bigcup_{\xi<\alpha} x_\xi$. Clearly $\vert x\vert \geq \aleph_1$. Furthermore, since $\vert x_\xi\vert < \kappa$ for all $\xi < \alpha$ and $\kappa$ is regular, we have:
$$ \begin{align*}
\vert x\vert = \sup\lbrace \vert x_\xi\vert : \xi < \alpha\rbrace < \kappa
\end{align*}$$ so indeed $x \in P_\kappa(A)$.

Unboundedness: Let $y \in P_\kappa(A)$. Fix any $z \subseteq X$ such that $\vert z\vert = \aleph_1$ (this is alwyas possible under $\AC$). Then $\vert y\vert < \kappa$, and since $\vert z\vert = \aleph_1 < \kappa$ we have $\vert z \cup y\vert \leq \vert z\vert + \vert y\vert < \kappa$. Then $z \cup y \in \lbrace x \in P_\kappa(A) : \vert x\vert \geq \aleph_1\rbrace$.

$\square$

Exercise 8.19.

Solution. Closedness: Let $\beta < \kappa$ and let $\c{x_\xi : \xi < \beta}$ be a sequence of sets in $\lbrace x \in P_\kappa(\lambda) : x \cap \kappa \in \kappa\rbrace$. Let $x := \bigcup_{\xi<\beta}$. Once again as in the proof of Exercise 8.18, $\vert x\vert < \kappa$. Furthermore, $x \cap \kappa = \sup_{\xi<\alpha} x_\xi \cap \kappa$ is the supremum of a set of ordinals, so $x \cap \kappa$ is an ordinal and $\vert x \cap \kappa\vert \leq \vert x\vert < \kappa$. Thus $x \in \kappa$.

Unboundedness: Let $y \in X$. Let $\alpha = \sup{y}$. Since $\vert y\vert < \kappa \leq \lambda$ and $\kappa$ is regular, $\alpha < \kappa$. Then $y \cap \kappa \subseteq (\alpha + 1) \cap \kappa \in \kappa$.

$\square$