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Exercise 22.1.

Solution. Let $\lbrace Z_\alpha : \alpha < \lambda\rbrace$ be the sets defined in the hint. Clearly they are pairwise disjoint. To see that they are of positive $I$-measure, we observe that:
$$ \begin{align*}
Z_\alpha = X_\alpha - \bigcup_{\beta<\alpha} X_\beta = X_\alpha - \bigcup_{\beta<\alpha} (X_\alpha \cap X_\beta)
\end{align*}$$ Since $\alpha < \lambda \leq \kappa$ and $I$ is $\kappa$-complete, $\bigcup_{\beta<\alpha} (X_\alpha \cap X_\beta) \in I$. If $Z_\alpha \in I$, then $X_\alpha = Z_\alpha \cup \bigcup_{\beta < \alpha} (X_\alpha \cap X_\beta) \in I$, a contradiction.

$\square$

Exercise 22.2.

Solution. The proof is precisely the last two paragraphs of the proof of Lemma 22.3.

$\square$

Exercise 22.3.

Solution. Note that proving the hint solves the problem - if $x_\alpha \in M$, then $x_\alpha = a$ for some $a \in M$. Then $x_\alpha \cap k = a \cap k$ for all $k$, so $\Vert \dot{x}_\alpha \cap k = \check{a} \cap k\Vert = 1$ for all $k$. This contradicts the hint.

Fix $a \in M$. We see that:
$$ \begin{align*}
\Vert \dot{x}_\alpha \cap k = \check{a} \cap k\Vert &= \Vert (\forall n \in k)(n \in \dot{x}_\alpha \leftrightarrow n \in \check{a})\Vert \\
&= \prod_{n < k} \Vert n \in \dot{x}_\alpha \leftrightarrow n \in \check{a}\Vert \\
&= \bb{\prod_{\substack{n < k \\ n \in a}} \Vert n \in \dot{x}_\alpha\Vert } \cdot \bb{\prod_{\substack{n < k \\ n \notin a}} -\Vert n \in \dot{x}_\alpha\Vert } \\
&= \bb{\prod_{\substack{n < k \\ n \in a}} u_{\alpha,n}} \cdot \bb{\prod_{\substack{n < k \\ n \notin a}} -u_{\alpha,n}} \\
&= [D_k]
\end{align*}$$ where $D_k \in \F$ such that for $t \in \lbrace 0,1\rbrace^{\lambda \times \omega}$:
$$ \begin{align*}
t \in D_k \iff \text{for all $n < k$, we have } t(\alpha,n) =
\begin{cases}
1, &\text{if $n \in a$} \\
0, &\text{if $n \notin a$}
\end{cases}
\end{align*}$$ Clearly $\mu(D_k) = \frac{1}{2^k}$ under the product measure, as desired.

$\square$

Exercise 22.4.

Solution. As in the hint suppose $\lbrace Y_\xi : \xi < \omega_1\rbrace$ is a counterexample, and let $Z_\nu$ be as in the hint. Clearly $Z_\alpha \subseteq Z_\beta$ for $\alpha < \beta$. To see that $\bigcup_{\nu<\omega_1} Z_\nu = \kappa$, we have:
$$ \begin{align*}
\kappa - \bigcup_{\nu<\omega_1} Z_\nu &= \kappa - \bigcup_{\nu<\omega_1} \bb{\bigcap_{\xi \geq \nu} (\kappa - Y_\xi)} \\
&= \bigcap_{\nu < \omega_1}\bb{\kappa - \bigcap_{\xi\geq\nu}(\kappa - Y_\xi)} \\
&= \bigcap_{\nu < \omega_1} \bigcup_{\xi\geq\nu}Y_\xi
\end{align*}$$ Thus, we have $\alpha \notin \bigcup_{\nu<\omega_1} Z_\nu$ for some $\alpha$ iff there exists an unbounded (hence uncountable) subset $X \subseteq \omega_1$ such that $\alpha \in Y_\xi$ for all $\xi \in X$. Then $\alpha \in \bigcap_{\xi \in X} Y_\xi$, contradicting that $\lbrace Y_\xi : \xi < \omega_1\rbrace$ is a counterexample.

Since $I$ is $\sigma$-saturated, there exists some $\gamma < \omega_1$ such that $\kappa - Z \in I$, where $Z := \bigcup_{\nu<\gamma} Z_\nu$. Then $Y_\gamma \subseteq \kappa - Z$ so $Y_\gamma \in I$, another contradiction.

$\square$

Exercise 22.5.

Solution. Let $J$ denote the ideal generated by $I$ in $V[G]$. The proof that $J$ is $\kappa$-complete is verbatim to the first two paragraphs of the proof of Lemma 22.11 ($\sigma$-saturated and countable chain condition are the same thing).

For the proof that $J$ is $\sigma$-saturated, we may also follow the proof of Lemma 22.11, the following statement may not hold anymore as $I$ need not be prime:

By the $\kappa$-completeness of $I$, we have $Y = \bigcap_{\xi<\omega_1} Y_\xi \notin I$. Thus $Y \neq \emptyset$,\dots

However, we may instead use Exercise 22.4 to obtain an uncountable subset $W \subseteq \omega_1$ such that $\bigcap_{\xi \in W} Y_\xi$ is non-empty. Take $\alpha \in \bigcap_{\xi \in W} Y_\xi$, and the rest of the proof follows.

$\square$

Exercise 22.6.

Solution. Let $P$ be the forcing defined in (15.1), with $\dom(p) \subseteq \lambda \times \omega$ finite, which adjoins $\lambda$ many Cohen reals. If $f = \bigcup G$ where $G$ is generic over $P$, then $f_\alpha := f(\alpha,-)$ is each a Cohen real. Under this definition, we have $V[G] = V[\c{f_\alpha : \alpha < \lambda}]$.

For each $n \in \omega$, let $W_n$ be the maximal antichain of $P$ such that each element of $W_n$ decides $g(n)$ (i.e. there exists some $m \in \omega$ such that $p_n \forces g(n) = m$). $P$ satisfies the countable chain condition, so $W_n$ is countable. Thus, $W := \bigcup W_n$ is also countable. Let $A \subseteq \lambda$ be the set of ordinals such that $\alpha \in A$ iff $(\alpha,n) \in \dom(p)$ for some $n \in \omega$ and $p \in W$. Clearly $A$ is countable, and $g \in V[\c{f_\alpha : \alpha \in A}]$.

Now suppose $\beta \notin A$ (so $f_\beta$ is generic over $V[\c{f_\alpha : \alpha \in A}]$). Let $P_\beta \subseteq P$ be the set of all conditions $p$ such that $\dom(p) \subseteq \lbrace \beta\rbrace \times \omega$ (so it is the Cohen forcing which adjoins the real $f_\beta$). Define:
$$ \begin{align*}
D := \lbrace p \in P_\beta : (\exists n \in \omega) \, p(n) > g(n)\rbrace
\end{align*}$$ Clearly $\dom(p)$ is finite for all $p \in P_\beta$, clearly $D$ is dense in $P_\beta$. If $G_\beta$ is the generic filter over $P_\beta$ with $f_\beta = \bigcup G_\beta$, then $G_\beta \cap D \neq \emptyset$ so $f_\beta(n) > g(n)$ for some $n$.

$\square$

Exercise 22.7.

Solution. Let $\c{f_\alpha : n \in \omega}$ be the functions in Exercise 22.6. Assuming on the contrary that $\kappa$ is real-valued measurable with a nontrivial $\kappa$-additive measure $\mu$, one may repeat the proof of Lemma 10.16 to define the $(\aleph_0,\aleph_0)$-matrices $A_{n,k}$, followed by the set $B$ such that $\mu(B) \geq \frac{1}{2}$. We also let $g : \omega \to \omega$ such that $g(n) = k_n$.

Now if $\alpha \in B$, then we have $f_\alpha(n) \leq g(n)$ for all $n$. By Exercise 22.6, $B$ must be countable. This contradicts that $\mu(B) > 0$.

$\square$

Exercise 22.8.

Lemma 22.8.A. Let $B$ be a $\kappa$-complete Boolean algebra, and let $X \subseteq B$ such that $\vert X\vert = \kappa$ and $\sum X$ exists. Then there exists a $Y \subseteq B$ such that elements in $Y$ are pairwise disjoint, and $\sum Y$ exists with $\sum Y = \sum X$.

Proof. Suppose $X = \lbrace x_\alpha : \alpha < \kappa\rbrace$. Define $y_\alpha$:
$$ \begin{align*}
y_\alpha := x_\alpha \cdot \bb{-\sum_{\beta < \alpha} x_\beta}
\end{align*}$$ $y_\alpha$ is well-defined as $\beta < \kappa$ and $B$ is $\kappa$-complete. Clearly if $\beta < \gamma$ then $y_\beta \cdot y_\gamma = 0$, so $Y := \lbrace y_\alpha : \alpha < \kappa\rbrace$ has pairwise disjoint elements. We now show that $\sum Y = \sum X$. Clearly $y_\alpha \leq x_\alpha$ for all $\alpha < \kappa$, so $\sum Y \leq \sum X$. Conversely, we inductively show that $x_\alpha \leq \sum_{\beta \leq \alpha} y_\alpha$ for $\alpha$. Suppose this holds for $\beta < \alpha$. Then $x_\alpha \cdot \sum_{\beta < \alpha} x_\beta \leq \sum_{\beta \leq \alpha} y_\alpha$. But $x_\alpha \cdot \bb{-\sum_{\beta < \alpha} x_\beta} = y_\alpha \leq \sum_{\beta \leq \alpha} y_\alpha$, so $x_\alpha \leq \sum_{\beta \leq \alpha} y_\alpha$, as desired. Therefore, we must have $\sum Y$ exists, with $\sum Y = \sum X$.

$\blacksquare$

Solution. Let $W \subseteq B := P(\kappa)/I$ with $\vert W\vert = \kappa$. By Lemma 22.8.A, we may assume WLOG that elements in $W$ are pairwise disjoint. Let $Z \supseteq W$ be such that $Z$ is a partition of $B$ (which exists by Zorn's Lemma), so by maximality $\sum Z = 1$. Write $Z := \lbrace z_\alpha : \alpha < \kappa\rbrace$ ($\vert Z\vert = \kappa$ by $\kappa^+$-saturatedness). Let $\mathcal{A} = \lbrace A_\alpha : \alpha < \kappa\rbrace \subseteq P(\kappa)$ be such that $z_\alpha = [A_\alpha]$ for all $\alpha < \kappa$. Let $A_\alpha' := A_\alpha - \bigcup_{\beta < \alpha} A_\beta$, so $\mathcal{A}' := \lbrace A_\alpha : \alpha < \kappa\rbrace$ have pairwise disjoint elements.

Let:
$$ \begin{gather*}
C := \bigcup_{\substack{\beta < \kappa \\ z_\beta \in W}} A_\alpha', \;\; D := \bigcup_{\substack{\beta < \kappa \\ z_\beta \notin W}} A_\alpha'
\end{gather*}$$ Clearly $C \cap D = \emptyset$. Now fix $\alpha < \kappa$ such that $z_\alpha \in W$. Observe that $A_\alpha = C \cup D$, so $A_\alpha - C = A_\alpha \cap D$. Therefore:
$$ \begin{align*}
A_\alpha \cap D &= A_\alpha \cap \bigcup_{\substack{\beta < \kappa \\ z_\beta \notin W}} A_\beta' \\
&= \bigcup_{\substack{\beta < \kappa \\ z_\beta \notin W}} A_\alpha \cap A_\beta' \\
&= \bigcup_{\substack{\beta < \alpha \\ z_\beta \notin W}} A_\alpha \cap A_\beta' \\
&\subseteq \bigcup_{\substack{\beta < \alpha \\ z_\beta \notin W}} A_\alpha \cap A_\beta
\end{align*}$$ Since $Z$ consists of pairwise disjoint elements, $A_\alpha \cap A_\beta \in I$ for all $\beta < \alpha$. By $\kappa$-completeness of $I$ (implicitly assumed in the exercise), $A_\alpha - C = A_\alpha \cap D \in I$. Thus, $z_\alpha = [A_\alpha] \leq [C]$. By a similar reasoning, if $z_\alpha \notin W$, we have that $z_\alpha \leq [D]$. Thus, we have:
$$ \begin{align*}
1 = \sum_{\alpha < \kappa} z_\alpha \leq [C] + [D]
\end{align*}$$ Therefore:
$$ \begin{align*}
\sum W = \sum_{\substack{\alpha < \kappa \\ z_\alpha \in W}} = \bb{\sum_{\alpha < \kappa} z_\alpha} \cdot [C] = [C]
\end{align*}$$ so $P(\kappa)/I$ is $\kappa^+$-complete.

$\square$

Exercise 22.9.

Solution. The hint solves the problem. Note that $\GCH$ is used to ensure that $2^{\kappa^+} > 2^\kappa$, as in general this need not be true.

$\square$

Exercise 22.10.

Solution. Let $Y$ be as in the hint. Note that for the last line in the hint, we mat let $Z \supseteq S \notin I$ be the set such that $f$ is constant on $S$. If $f(x) \in X_\alpha$ for all $x \in S$ (where $\alpha$ is fixed), then $S \subseteq Y_\alpha$ so $S \subseteq Z \cap Y_\alpha \notin I$.

It remains to show that $Y$ is the least upper bound. Let $Z \subsetneq Y$ such that $Y - Z \notin I$ (so $[Z] < [Y]$). By the hint/above paragraph, there exists some $\alpha$ such that $(Y - Z) \cap X_\alpha \notin I$. Since $(Y - Z) \cap X_\alpha \subseteq X_\alpha - Z$, this implies that $[Z] \not\geq [X_\alpha]$, as desired.

$\square$

Exercise 22.11.

Solution. We expand on the hint. Note that by almost disjoint we mean that $Y_i \cap Y_j$ is a bounded subset of $\kappa$ for all $i \neq j$.

To see that $X_i$ is of positive measure, we note that since $I$ is normal, $\bigtriangle_{\alpha<\kappa} (\kappa - Z_\alpha)$ is of measure one, so $\bigtriangle_{\alpha<\kappa} (X_i - Z_\alpha) = X_i \cap \bigtriangle_{\alpha<\kappa} (\kappa - Z_\alpha)$ is of positive measure.

To see that $Y_i$ contains almost all elements of $X_i$, we note that $Y_i \subseteq X_i$, and for all $\alpha \in X_i$:
$$ \begin{align*}
\alpha \notin Y_i &\iff \alpha \notin \bigtriangle_{\xi<\kappa} (X_i - Z_\xi) \\
&\iff \alpha \notin \bigcap_{\xi < \alpha} (X_i - Z_\xi) \\
&\iff \alpha \in X_i - \bigcap_{\xi < \alpha} (X_i - Z_\xi) \\
&\iff \alpha \in \bigcup_{\xi < \alpha} X_i \cap Z_\xi
\end{align*}$$ Now $X_i \cap Z_{\xi}$ is measure zero for all $\xi < \alpha$, and since $I$ is $\kappa$-complete we have that $\bigcup_{\xi<\alpha} X_i \cap Z_\xi$ is measure zero. Thus $X_i - Y_i$ is measure zero.

This above workings also imply that $Y_i \cap Z_\alpha \subseteq \alpha + 1$, as if $\beta \in X_i \cap Z_\alpha$ and $\beta > \alpha$, then $\beta \in \bigcup_{\xi<\beta+1} X_i \cap Z_\xi$ so $\beta \notin Y_i$.

Finally, if $i \neq j$ then WLOG assume $i < j$. Then $Y_i = Z_\alpha$ for some $\alpha < \kappa$ (w.r.t. $X_i$), so $Y_j \cap Y_i = Y_j \cap Z_\alpha \subseteq \alpha + 1$.

$\square$

Exercise 22.12.

Solution. The hint shows that $I$ must be $\kappa$-saturated. Note that the fact that $f$ is only regressive almost everywhere and not regressive everywhere does not matter, as we can repeat the hint with a regressive function $f'$ such that $f' = f$ almost everywhere, and if $f'$ is bounded on a measure one set $S'$, then $f$ is bounded on a measure one set $S$ where $S' - S$ is measure zero.

This also implies that $I$ is normal. We may use Exercise 8.8. Let $f$ be a function that is regressive on $S_0 \notin I$. Let $g \supseteq f$ such that if $\alpha \notin S_0$ then $g(\alpha) := 0$. Then $g$ is regressive, so there is a measure one set $X$ such that $g(\alpha) < \gamma$ for all $\alpha \in X$. Let $S \subseteq S_0$, then $f$ is bounded by $\gamma$ on $S$. Then $S = \lbrace \alpha \in S_0 : f(\alpha) < \gamma\rbrace = \bigcup_{\beta<\gamma} \lbrace \alpha \in S_0 : f(\alpha) = \beta\rbrace$. Since $I$ is $\kappa$-complete, $\lbrace \alpha \in S_0 : f(\alpha) = \beta\rbrace \notin I$ for some $\beta$, and $f$ is constant on this set.

$\square$

Exercise 22.13.

Solution. Let $(T,\supseteq)$ be the tree in the hint. We elaborate on that for $\alpha < \kappa$, $T(\alpha)$ is non-empty. We induct on $\alpha < \kappa$. $T(0) = \lbrace \kappa\rbrace \neq \emptyset$, and since $I$ is atomless every set in $T(\alpha)$ can be split into two disjoint subsets of positive measure, so $T(\alpha+1) \neq \emptyset$. If $\alpha$ is a limit ordinal, then by induction hypothesis we have $0 < \vert T(\beta)\vert < \kappa$ for all $\beta < \alpha$, and we have that the set of intersections of all branches before level $\alpha$ is of size at most $2^{\sup_{\beta<\alpha} \vert T(\beta)\vert } < \kappa$ by inaccessibility of $\kappa$. Thus, $T(\alpha)$ must be non-empty, for otherwise the set of intersections form a partition of $\kappa$ into measure zero sets of size $<\kappa$, contradicting $\kappa$-completeness of $I$.

We now show that $T$ has no branch of length $\kappa$. Otherwise, let $\lbrace X_\alpha : \alpha < \kappa\rbrace \subseteq T$ be a branch (so $\alpha < \beta \implies X_\alpha \supseteq X_\beta$). Let $Y_\alpha := X_\alpha - X_{\alpha+1}$. By the way that $T$ is constructed, $Y_\alpha$ is of positive measure for all $\alpha$. Furthermore, if $\alpha < \beta$ then:
$$ \begin{align*}
Y_\alpha \cap Y_\beta = (X_\alpha - X_{\alpha+1}) \cap (X_\beta - X_{\beta+1}) \subseteq (X_\alpha - X_{\alpha+1}) \cap (X_{\alpha+1} - X_{\beta+1}) = \emptyset
\end{align*}$$ Thus $\lbrace Y_\alpha : \alpha < \kappa\rbrace$ is a partition of $\kappa$ into $\kappa$ many sets of positive measure. This contradicts that $\sat(I) \leq \kappa$.

$\square$

Exercise 22.14.

Solution. By there exists a minimal. Note that we are proving the statement in $M[G]$.

Let $f : \kappa \to M$ be a function in $M[G]$ such that $[f]_G = \kappa$ in $\Ult_G(M)$ (we require $I$ to be precipitous in order to make sense of the cardinal $\kappa$ in $\Ult_G(M)$). Let $X \in P(\kappa)/I$ such that $X \forces \text{$\dot{f}$ represents $\kappa$ in $\Ult_G(M)$}$. We shall show that $f : X \to M$ is a minimal unbounded function.

We first show that $f$ is unbounded. Suppose there exists a $Y \subseteq X$ of positive measure such that $f(\alpha) < \gamma$ for all $\alpha \in Y$ and some $\gamma < \kappa$. Then:
$$ \begin{align*}
Y \subseteq \lbrace \alpha < \kappa : f(\alpha) < \gamma\rbrace
\end{align*}$$ Let $G$ be a generic filter containing $Y$, which implies that $X \in G$ as well. Since $\gamma < \kappa$ and $\kappa$ is the least ordinal moved, we have that $\Ult_G(M) \models [f]_G \in j_G(\gamma) = \gamma \in \kappa$. This contradicts the well-foundedness of $\Ult_G(M)$.

We now show that $f$ is minimal. Suppose there exists $Y \subseteq X$ of positive measure and am unbounded function $g : Y \to M$ such that $g(\alpha) < f(\alpha)$ for all $\alpha \in Y$. Again, let $G$ be a generic filter containing $Y$. This implies that $[g]_G \in [f]_G = \kappa$, so $[g]_G = \alpha$ for some $\alpha < \kappa$. Since $\kappa$ is the least ordinal moved, $g$ is constant on $Y$, contradicting its unboundedness.

$\square$