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Exercise 24.1.

Solution. We first note that under this exercise's hypothesis, $\aleph_{\omega_1}$ is a strong limit cardinal (in particular, $2^{\aleph_1} < \aleph_{\omega_1}$) - indeed, let $\alpha < \omega_1$, and since stationary sets are unbounded there exists an $\alpha < \gamma < \omega_1$ such that $2^{\aleph_\alpha} \leq 2^{\aleph_\gamma} \leq \aleph_{\gamma+\beta} < \aleph_{\omega_1}$. Hence, $\aleph_{\omega_1}^{\aleph_1} = 2^{\aleph_{\omega_1}}$.

We note that if $\alpha < \omega_1$ is a limit ordinal, then $\cf(\alpha) = \omega$. Thus, for all limit ordinals such that $2^{\aleph_\alpha} \leq \aleph_{\alpha+\beta}$, we have that $\aleph_\alpha^{\aleph_0} \leq (2^{\aleph_\alpha})^{\aleph_0} \leq 2^{\aleph_\alpha} \leq \aleph_{\alpha+\beta}$. By Exercise 24.2, $2^{\aleph_{\omega_1}} = \aleph_{\omega_1}^{\aleph_1} \leq \aleph_{\omega_1+\beta}$, as desired.

$\square$

Exercise 24.2.

Solution. We note that in Lemma 24.3, if $\varphi$ takes constant value $\beta$, then clearly $\Vert \varphi\Vert \geq \beta$. Thus, the ranks of the functions $\varphi : \omega_1 \to \omega_1$ is unbounded in $\omega_1$, and hence is the set $\omega_1$. In particular, for every $\beta < \omega_1$ there exists a function $\varphi : \omega_1 \to \omega_1$ such that $\Vert \varphi\Vert = \beta$.

Assume that $\aleph_\alpha^{\aleph_0} = \aleph_{\alpha+\beta}$ for a stationary set of $\alpha < \omega_1$ (note that if $\alpha < \omega_1$ is limit, then $\cf(\alpha) = \omega$). For each $h : \omega_1 \to \aleph_{\omega_1}$, let $f_h$ be defined as in the proof of Lemma 8.14 from Lemma 8.15. Let $F := \lbrace f_h : h \in {\aleph_{\omega_1}}^{\omega_1}\rbrace$. Then $F$ is an almost disjoint family of functions, with $F \subseteq \prod_{\alpha<\omega_1} {\aleph_\alpha}^{\omega_1}$. By assumption, $\vert {\aleph_\alpha}^{\omega_1}\vert \leq \aleph_{\alpha+\beta}$.

By Exercise 5.19, for all $\alpha < \omega_1$ we have that $\aleph_\alpha^{\aleph_1} = \aleph_\alpha^{\aleph_0} \cdot 2^{\aleph_1}$. We have $2^{\aleph_1} < \aleph_{\omega_1}$ by hypothesis, and by the unboundedness of stationary set there exists $\alpha < \gamma < \omega_1$ such that $\aleph_\alpha^{\aleph_0} \leq \aleph_\gamma^{\aleph_0} \leq \aleph_{\gamma+\beta} < \aleph_{\omega_1}$. Thus, we may apply Lemma 24.3 with $\Vert \varphi\Vert = \beta$ to get that that $\vert F\vert \leq \aleph_{\omega_1 + \beta}$. Hence $\aleph_{\omega_1}^{\aleph_1} = \aleph_{\omega_1 + \beta}$, as desired.

$\square$

Exercise 24.3.

Solution. We note that Exercise 24.1 can be easily generalised to all singular cardinals of uncountable cofinality (replace $\omega_1$ with $\kappa$). If $\kappa$ is singular, then $E^\kappa_\omega$ is a stationary subset of $\kappa$, so the statement follows directly from Exercise 24.1.

$\square$

Exercise 24.4.

Solution. The proof of this is verbatim to that of Exercise 24.1. In fact, we only require that $2^{\aleph_\alpha} \leq \aleph_{\alpha+\beta}$ for a stationary many $\alpha < \eta$.

$\square$

Exercise 24.5.

Solution. This is an immediate consequence of Exercise 24.7, as the hypothesis of this exercise implies the hypothesis of Exercise 24.7 for stationarily many $\alpha < \omega_1$, which is sufficient for the statement to hold. Note that the hypothesis ensures that $\aleph_{\omega_1}$ is a strong limit cardinal, so that $2^{\aleph_{\omega_1}} = \aleph_{\omega_1}^{\aleph_1}$.

$\square$

Exercise 24.6.

Solution. Define a map $f : \omega_1 \to \omega_1$ by:
$$ \begin{align*}
f(\alpha) = \min\lbrace \beta < \omega_1 : 2^{\aleph_{\omega_1+\alpha}} \leq \aleph_{\omega_1+\alpha+\beta}\rbrace
\end{align*}$$ By the hypothesis of this exercise, $f$ is regressive on $\omega_1$. Thus, there exists a stationary $S \subseteq \omega_1$ such that $f(\alpha) = \gamma$ for all $\alpha \in S$, where $\gamma < \omega_1$ is fixed. Let $S' := \lbrace \omega_1 + \alpha : \alpha \in S\rbrace$. Clearly $S' \subseteq \omega_1 + \omega_1$ is stationary, and $2^{\aleph_\alpha} \leq \aleph_{\alpha+\gamma}$ for all $\alpha \in S'$. By Exercise 24.4, this implies that $2^{\aleph_{\omega_1+\omega_1}} \leq \aleph_{\omega_1+\omega_1+\gamma} < \aleph_{\omega_1+\omega_1+\omega_1}$.

$\square$

Exercise 24.7.

Lemma 24.7.A. Fix $\beta < \omega_1$. If $\varphi : \omega_1 \to \omega_1$ is such that $\varphi(\alpha) \leq \beta$ on a stationary subset of $\omega_1$, then $\Vert \varphi\Vert = \beta$.

Proof. Suppose the statement holds for $\gamma < \beta$. Let $\varphi$ be such that $\varphi(\alpha) \leq \beta$ on a stationary subset $S$. Let $\psi < \varphi$, so there exists a closed unbounded subset $C \subseteq \omega_1$ such that $\psi(\alpha) < \varphi(\alpha)$ for all $\alpha \in C$. Then $\psi(\alpha) < \beta$ on the stationary subset $S' := C \cap S \cap (\omega_1 - \beta)$. Furthermore, $\psi$ is regressive on $S'$, so by Fodor's Theorem there exists a stationary $S" \subseteq S'$ such that $\psi(\alpha) = \gamma < \beta$ for all $\alpha \in S"$. By induction hypothesis, $\Vert \psi\Vert \leq \gamma < \beta$, so:
$$ \begin{align*}
\Vert \varphi\Vert = \sup\lbrace \Vert \psi\Vert + 1 : \psi < \varphi\rbrace \leq \beta
\end{align*}$$

$\blacksquare$

Lemma 24.7.B. Let $\varphi : \omega_1 \to \omega_1$ be such that $\varphi(\alpha) = \alpha$ for all $\alpha$. Then $\Vert \varphi\Vert = \omega_1$.

Proof. Let $\varphi_\beta : \omega_1 \to \omega_1$ be the function taking the constant value $\beta$. Clearly $\Vert \varphi_\beta\Vert \geq \beta$ (in fact, $\Vert \varphi_\beta\Vert = \beta$) and $\varphi_\beta < \varphi$ (as $\varphi_\beta(\alpha) < \varphi(\alpha)$ on a tail segment of $\omega_1$). Thus $\Vert \varphi\Vert \geq \omega_1$. Conversely, suppose $\psi < \varphi$, so $\psi(\alpha) < \varphi(\alpha) = \alpha$ on a closed unbounded (in particular, stationary) subset of $\omega_1$. Then $\psi$ is regressive on a stationary subset of $\omega_1$, so by Fodor's Theorem it takes a constant value $\gamma < \alpha$ on a stationary subset of $\omega_1$. By Lemma 24.7.A again $\Vert \psi\Vert \leq \gamma$, so $\Vert \varphi\Vert \leq \omega_1$.

$\blacksquare$

Solution. By repeating the proof Exercise 24.2, in which we instead apply Lemma 24.3 with $\varphi(\alpha) = \alpha$. We have $\Vert \varphi\Vert = \omega_1$ by Lemma 24.7.B. Note that by the hypothesis, $\aleph_\alpha^{\aleph_1} = \aleph_\alpha^{\aleph_0} \cdot 2^{\aleph_1} < \aleph_{\omega_1}$. Thus, by Lemma 24.3 we have that if $\aleph_\alpha^{\aleph_0} \leq \aleph_{\alpha+\alpha}$ for all $\alpha < \omega_1$, then $\aleph_{\omega_1}^{\aleph_1} \leq \aleph_{\omega_1+\omega_1}$. Then repeat the proof of Lemma 8.15 to conclude that the assertion in the exercise holds.

$\square$

Exercise 24.8.

Solution. Note that we have to further assume that $\kappa$ is singular, as otherwise this would contradict Easton's Theorem 15.18. Under this additional assumption, the hypothesis of Exercise 24.9 is fulfilled and so $2^\kappa = \kappa^{\cf(\kappa)} < \aleph_\gamma$.

$\square$

Exercise 24.9.

Solution. Using the same proof, Lemma 24.3 may be generalised to singular cardinals in the following manner:

Theorem. {\itshape Let $\aleph_\eta$ be a singular cardinal, and assume $\aleph_\alpha^{\cf(\eta)} < \aleph_\eta$ for all $\alpha < \eta$. Let $\varphi : \cf(\eta) \to \eta$ and let $F$ be an almost disjoint family of functions:
$$ \begin{align*}
F \subseteq \prod_{\alpha < \cf(\eta)} A_\alpha
\end{align*}$$ such that:
$$ \begin{align*}
\vert A_\alpha\vert \leq \aleph_{\alpha + \varphi(\alpha)}
\end{align*}$$ for every $\alpha < \cf(\eta)$. Then $\vert F\vert \leq \aleph_{\eta + \Vert \varphi\Vert }$.}

But since $\vert \lbrace \varphi : \cf(\eta) \to \eta\rbrace\vert = \vert \eta\vert ^{\cf(\eta)}$, we clearly have that $\Vert \varphi\Vert \leq \vert \eta\vert ^{\cf(\eta)}$. Thus $\vert \eta + \vert \eta\vert ^{\cf(\eta)}\vert = \vert \eta\vert ^{\cf(\eta)}$, so $\eta + \vert \eta\vert ^{\cf(\eta)} < (\vert \eta\vert ^{\cf(\eta)})^+ =: \gamma$, as desired.

$\square$

Remark. The sets $S_\eta$ are the canonical stationary sets (of order $\eta$).

Note. This definition coincides with the same definition presented in Exercise 8.13.

Exercise 24.11.

Solution. In fact, given an arbitrary cardinal $\kappa$ it is easy to find a partial order of cofinality $\kappa$. We consider the partial order $(\kappa,<^\ast )$, where under $<^\ast $ every two elements in $\kappa$ are incomparable. Then the only cofinal subset of $\kappa$ under $<^\ast $ is $\kappa$ itself, so the cofinality of $(\kappa,<^\ast )$ is $\kappa$.

$\square$

Exercise 24.12.

Solution. The statement in the exercise is false as stated. This is because the lexicographical ordering on $\omega \times \omega_1$ is a linear ordering, and every linear order has true cofinality (as the whole set is itself a cofinal chain).

Instead, we shall show that if we define $<$ on $\omega \times \omega_1$ by:
$$ \begin{align*}
(\alpha,\beta) < (\gamma,\eta) \iff \alpha < \gamma \wedge \beta < \eta
\end{align*}$$ then $(\omega \times \omega_1,<)$ has no true cofinality (note that $<$ is clearly a partial order). To see this, let $C \subseteq \omega \times \omega_1$ be a chain. Suppose $C = \lbrace (\alpha_i,\beta_i) : i < \vert C\vert \rbrace$ such that $i < j \implies (\alpha_i,\beta_i) < (\alpha_j,\beta_j)$. Then $\lbrace \alpha_i : i < \vert C\vert \rbrace$ is an infinite subset of $\omega$ of size $\vert C\vert $, so we must have $\vert C\vert = \aleph_0$. Let $\beta := \sup_{i<\omega} \beta_i$. Then $(0,\beta+1) \not\leq (\alpha_i,\beta_i)$ for all $i < \omega$, as $\beta + 1 > \beta_i$. Hence every chain of $\omega \times \omega_1$ is not cofinal.

$\square$

Exercise 24.13.

Solution. Clearly $d$ is an upper bound, as $c_\gamma(\alpha) < d(\alpha)$ on a tail segment of $\omega_1$, and all bounded subsets of $\omega_1$ are non-stationary. Let $f$ be an upper bound of $\lbrace c_\gamma : \gamma < \omega_1\rbrace$, and suppose for a contradiction that $\lbrace \alpha < \omega_1 : f(\alpha) < d(\alpha)\rbrace \notin I$, i.e. it is stationary. Then $f$ is regressive on a stationary set, so by Fodor's theorem there exists some stationary subset $S$ such that $f$ takes a constant value $\gamma$ for all $\alpha \in S$. This means that $\lbrace \alpha < \omega_1 : c_\gamma(\alpha) \geq f(\alpha)\rbrace \notin I$, so $c \not<_I f$, contradicting that $f$ is an upper bound. Thus $d \leq_I f$ necessarily.

We now show that $d$ is not an exact upper bound. By Solovay's Theorem 8.10, there exists a partition $\lbrace S_\gamma : \gamma < \omega_1\rbrace$ of stationary subsets of $\omega_1$. Let $T_\gamma := S_\gamma - (\gamma + 1)$. Clearly $T_\gamma$ remains stationary. Define $g : \omega_1 \to \omega_1$ by stipulating that:
$$ \begin{align*}
g(\alpha) =
\begin{cases}
\gamma, &\text{if $\alpha \in T_\gamma$} \\
0, &\text{if $\alpha \notin T_\gamma$ for all $\gamma$}
\end{cases}
\end{align*}$$ Since the $T_\gamma$'s are disjoint, $g$ is well-defined. We see that for any $\gamma$, we we have that $T_{\gamma+1} \subseteq \lbrace \alpha < \omega_1 : g(\alpha) > c_\gamma(\alpha)\rbrace$, so $g \not\leq_I c_\gamma$. To see that $g <_I d$, we have that:
$$ \begin{align*}
\lbrace \alpha < \omega_1 : g(\alpha) \geq \alpha\rbrace &= \lbrace \alpha < \omega_1 : (\exists \gamma \geq \alpha) \, \alpha \in T_\gamma\rbrace \\
&= \ss{\alpha < \omega_1 : \alpha \in \bigcup_{\gamma \geq \alpha} T_\gamma} \\
&= \bigcup_{\gamma < \omega_1} (\gamma + 1) \cap S_\gamma \\
&= \emptyset \in I
\end{align*}$$ as desired.

$\square$