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Exercise 20.1
Solution
We first prove an intermediate claim:
By hypothesis, say $U$ is the $\kappa$-complete ultrafilter extending $F$. Consider the ultraproduct: $$ \begin{align*} M := \prod_{x \in P_\kappa(\lambda)} \Ult_U(M_x) \end{align*}$$ Then $M$ is a model for the whole $\Sigma$, as for all $\sigma_\alpha \in \Sigma$: $$ \begin{align*} M \models \sigma_\alpha &\iff \lbrace x \in P_\kappa(\lambda) : M_x \models \sigma_\alpha\rbrace \in U \\ &\impliedby \lbrace x \in P_\kappa(\lambda) : \alpha \in x\rbrace \in U \\ &\iff [[\lambda]^{\kappa}]^{\supseteq\alpha+1} \in U \end{align*}$$ Note that we used Łoś' Theorem for the $\mathcal{L}_{\kappa,\kappa}$ language, which one may easily prove using the same technique as that of the original theorem.
We first prove an intermediate claim:
Claim
$\kappa$ is regular.
Proof
Suppose $\kappa$ is singular. Then, by taking a cofinal subsequence, the intersection of a $\kappa$-sequence of sets in a filter $F$ also belongs to $F$. Thus, $\kappa$-completeness of a filter implies $\kappa^+$-completeness. Consider: $$ \begin{align*} F := \lbrace X \subseteq \kappa^+ : \vert \kappa^+ - X\vert < \kappa^+\rbrace \end{align*}$$ It's easy to see that $F$ is a $\kappa$-complete filter on $\kappa^+$, and hence by hypothesis can be extended to a $\kappa$-complete, hence $\kappa^+$-complete, ultrafilter on $\kappa^+$. This implies that $\kappa^+$ is measurable, contradicting that every measurable cardinal is a (strong) limit cardinal.
Let $\Sigma = \lbrace \sigma_\alpha : \alpha < \lambda\rbrace$ be a $\kappa$-satisfiable collection of $\mathcal{L}_{\kappa,\kappa}$ sentences. For each $x \in P_\kappa(\lambda)$, let $M_x$ be a model for all sentences in $\lbrace \sigma_\alpha : \alpha \in x\rbrace$, i.e. $M_x \models \bigwedge_{\alpha \in x} \sigma_\alpha$. Define the filter $F$ on $P_\kappa(\lambda)$ as follows:
$$
\begin{align*}
F := \ss{[P_\kappa(\lambda)]^{\supseteq y} : y \in P_\kappa(\lambda)}, \;\; \text{where $[A]^{\supseteq X} := \lbrace Y \in A : X \subseteq Y\rbrace$}
\end{align*}$$
We see that $F$ is $\kappa$-complete — if $\c{[P_\kappa(\lambda)]^{\supseteq y_\beta} : \beta < \delta}$ is a $\delta < \kappa$ collection of sets in $F$ (where $y_\beta \in P_\kappa(\lambda)$), then $\bigcap_{\beta < \delta} [P_\kappa(\lambda)]^{\supseteq y_\beta} = [P_\kappa(\lambda)]^{\supseteq \bigcup_{\beta < \delta} y_\beta}$, and by regularity of $\kappa$ we have $\bigcup_{\beta < \delta} y_\beta \in P_\kappa(\lambda)$, so $\bigcap_{\beta < \delta} [P_\kappa(\lambda)]^{\supseteq y_\beta} \in F$.$\kappa$ is regular.
Proof
Suppose $\kappa$ is singular. Then, by taking a cofinal subsequence, the intersection of a $\kappa$-sequence of sets in a filter $F$ also belongs to $F$. Thus, $\kappa$-completeness of a filter implies $\kappa^+$-completeness. Consider: $$ \begin{align*} F := \lbrace X \subseteq \kappa^+ : \vert \kappa^+ - X\vert < \kappa^+\rbrace \end{align*}$$ It's easy to see that $F$ is a $\kappa$-complete filter on $\kappa^+$, and hence by hypothesis can be extended to a $\kappa$-complete, hence $\kappa^+$-complete, ultrafilter on $\kappa^+$. This implies that $\kappa^+$ is measurable, contradicting that every measurable cardinal is a (strong) limit cardinal.
$\blacksquare$
By hypothesis, say $U$ is the $\kappa$-complete ultrafilter extending $F$. Consider the ultraproduct: $$ \begin{align*} M := \prod_{x \in P_\kappa(\lambda)} \Ult_U(M_x) \end{align*}$$ Then $M$ is a model for the whole $\Sigma$, as for all $\sigma_\alpha \in \Sigma$: $$ \begin{align*} M \models \sigma_\alpha &\iff \lbrace x \in P_\kappa(\lambda) : M_x \models \sigma_\alpha\rbrace \in U \\ &\impliedby \lbrace x \in P_\kappa(\lambda) : \alpha \in x\rbrace \in U \\ &\iff [[\lambda]^{\kappa}]^{\supseteq\alpha+1} \in U \end{align*}$$ Note that we used Łoś' Theorem for the $\mathcal{L}_{\kappa,\kappa}$ language, which one may easily prove using the same technique as that of the original theorem.
$\square$
Exercise 20.3
Solution
Note that we assume implicitly that $\kappa, \lambda$ are both regular cardinals. We may mimic the proof of Lemma 20.2 to obtain the solution here. Let:
(i) $\implies$ (iii): By a bijection we of course have that there exists a fine measure on $P_\kappa(A)$ for any $\vert A\vert = \lambda$. The (ii) $\implies$ (iii) part of the proof of Lemma 20.2 can be mimicked completely.
(iii) $\implies$ (ii): Let $F$ be a $\kappa$-complete filter on $S$ generated by $E \subseteq P(S)$, with $\vert E\vert \leq \lambda$. We consider the $\mathcal{L}_{\kappa,\omega}$-language which has a unary predicate symbol $\dot{X}$ for each $X \subseteq S$, and a constant symbol $c$. Let $\Sigma$ be the set of $\mathcal{L}_{\kappa,\omega}$ sentences consisting of:
(ii) $\implies$ (i): Let $F$ be the filter defined in (20.1). Similar to Exercise 7.2, one can show that $F$ is the $\kappa$-complete filter generated by $\lbrace \lbrace \alpha\rbrace^\wedge : \alpha < \lambda\rbrace$. Thus $F$ is generated by at most $\lambda$ many sets, so it may be extended to a fine measure by (ii).
Note that we assume implicitly that $\kappa, \lambda$ are both regular cardinals. We may mimic the proof of Lemma 20.2 to obtain the solution here. Let:
(iii) Suppose $\Sigma$ is a set of sentences in the language of $\mathcal{L}_{\kappa,\omega}$, and $\vert \Sigma\vert \leq \lambda$. If every subset of $\Sigma$ of size less than $\kappa$ has a model, then $\Sigma$ has a model.
We shall show that (i) $\implies$ (iii) $\implies$ (ii) $\implies$ (i).(i) $\implies$ (iii): By a bijection we of course have that there exists a fine measure on $P_\kappa(A)$ for any $\vert A\vert = \lambda$. The (ii) $\implies$ (iii) part of the proof of Lemma 20.2 can be mimicked completely.
(iii) $\implies$ (ii): Let $F$ be a $\kappa$-complete filter on $S$ generated by $E \subseteq P(S)$, with $\vert E\vert \leq \lambda$. We consider the $\mathcal{L}_{\kappa,\omega}$-language which has a unary predicate symbol $\dot{X}$ for each $X \subseteq S$, and a constant symbol $c$. Let $\Sigma$ be the set of $\mathcal{L}_{\kappa,\omega}$ sentences consisting of:
- All sentences true in $(S,X)_{X \subseteq S}$.
- $\dot{X}(c)$ for all $X \in E$.
(ii) $\implies$ (i): Let $F$ be the filter defined in (20.1). Similar to Exercise 7.2, one can show that $F$ is the $\kappa$-complete filter generated by $\lbrace \lbrace \alpha\rbrace^\wedge : \alpha < \lambda\rbrace$. Thus $F$ is generated by at most $\lambda$ many sets, so it may be extended to a fine measure by (ii).
$\square$
Exercise 20.4
Solution
Let $C \subseteq P_\kappa(\lambda)$ be closed unbounded, and let $j : V \to M$ be the elementary embedding induced by ultrapower by $U$, with $j(\kappa) > \lambda$. Since $C$ is closed unbounded (in particular unbounded), $D = j(C)$ is a directed subset of $j(C)$, and $\vert D\vert \leq \vert P_\kappa(\lambda)\vert = \lambda^{<\kappa}$. Since $j(\kappa)$ is inaccessible in $M$ and $\kappa < \lambda < j(\kappa)$, $j(\kappa) > \lambda^{<\kappa}$. Thus, as $j(C)$ is closed unbounded in $M$, $\bigcup D \in j(C)$ (as $D$ is directed). Now observe that: $$ \begin{align*} y \in \bigcup D \iff (\exists x \in C) \, y \in j(x) \end{align*}$$ Now $x \in P_\kappa(\lambda)$, $\vert x\vert < \kappa$ so $j(x) = j``(x)$. Therefore, $y \in j(x)$ iff $y = j(\alpha)$ for some $\alpha \in x$. Since $C$ is closed unbounded, clearly $\bigcup C = \lambda$. Therefore $\bigcup D = j(\lambda)$, so $j(\lambda) \in j(C)$, so by Lemma 20.13 $C \in U$.
Let $C \subseteq P_\kappa(\lambda)$ be closed unbounded, and let $j : V \to M$ be the elementary embedding induced by ultrapower by $U$, with $j(\kappa) > \lambda$. Since $C$ is closed unbounded (in particular unbounded), $D = j(C)$ is a directed subset of $j(C)$, and $\vert D\vert \leq \vert P_\kappa(\lambda)\vert = \lambda^{<\kappa}$. Since $j(\kappa)$ is inaccessible in $M$ and $\kappa < \lambda < j(\kappa)$, $j(\kappa) > \lambda^{<\kappa}$. Thus, as $j(C)$ is closed unbounded in $M$, $\bigcup D \in j(C)$ (as $D$ is directed). Now observe that: $$ \begin{align*} y \in \bigcup D \iff (\exists x \in C) \, y \in j(x) \end{align*}$$ Now $x \in P_\kappa(\lambda)$, $\vert x\vert < \kappa$ so $j(x) = j``(x)$. Therefore, $y \in j(x)$ iff $y = j(\alpha)$ for some $\alpha \in x$. Since $C$ is closed unbounded, clearly $\bigcup C = \lambda$. Therefore $\bigcup D = j(\lambda)$, so $j(\lambda) \in j(C)$, so by Lemma 20.13 $C \in U$.
$\square$
Exercise 20.5
Solution
The problem statement is false as it stands. Let $M = \Ult_U(V)$. We shall instead show that the transitive collapse of the ultraproduct is $V_\lambda^M$ (therefore the ultraproduct is isomorphic to $V_\lambda^M$). Note that as in (20.23), $\lambda_x$ is the order-type of $x$, and $\lambda$ is represented by the function $x \mapsto \lambda_x$ in $M$.
This is simply an application of Łoś' Theorem. Indeed, for all $f : P_\kappa(\lambda) \to V$: $$ \begin{align*} &\iffbreak [f] \in \Ult_U\lbrace (V_{\lambda_x},\in) : x \in P_\kappa(\lambda)\rbrace \\ &\iff \lbrace x \in P_\kappa(\lambda) : f(x) \in V_{\lambda_x}\rbrace \in U \\ &\iff \lbrace x \in P_\kappa(\lambda) : \rank^V(f(x)) < \lambda_x\rbrace \in U \\ &\iff \rank^M([f]) < \lambda \\ &\iff [f] \in V_\lambda^M \end{align*}$$
The problem statement is false as it stands. Let $M = \Ult_U(V)$. We shall instead show that the transitive collapse of the ultraproduct is $V_\lambda^M$ (therefore the ultraproduct is isomorphic to $V_\lambda^M$). Note that as in (20.23), $\lambda_x$ is the order-type of $x$, and $\lambda$ is represented by the function $x \mapsto \lambda_x$ in $M$.
This is simply an application of Łoś' Theorem. Indeed, for all $f : P_\kappa(\lambda) \to V$: $$ \begin{align*} &\iffbreak [f] \in \Ult_U\lbrace (V_{\lambda_x},\in) : x \in P_\kappa(\lambda)\rbrace \\ &\iff \lbrace x \in P_\kappa(\lambda) : f(x) \in V_{\lambda_x}\rbrace \in U \\ &\iff \lbrace x \in P_\kappa(\lambda) : \rank^V(f(x)) < \lambda_x\rbrace \in U \\ &\iff \rank^M([f]) < \lambda \\ &\iff [f] \in V_\lambda^M \end{align*}$$
$\square$
Exercise 20.6
Solution
Consider the $\Sigma_1$ formula $\exists y \, \phi(x,y)$, where $\phi$ is quantifier-free with $x \in V_\kappa$. Since $\Sigma_1$ formulas are upward absolute, it suffices to show that $V \models \exists y \, \phi(x,y) \implies V_\kappa \models \exists y \, \phi(x,y)$. By the Löwenheim-Skolem Theorem, let $M \prec V$ such that $\vert M\vert < \kappa$ and $\TC(\lbrace x\rbrace) \subseteq M$ (as $V_\kappa = H_\kappa$ by Lemma 13.23.A). By Mostowski's Collapsing Theorem, the transitive collapse $\pi(M)$ is elementarily equivalent to $M$, so $\pi(M) \models \exists y \, \phi(x,y)$. Since $\pi(M)$ is transitive and $\vert \pi(M)\vert < \kappa$, by Lemma 13.23.A we have $\pi(M) \in V_\kappa$. By upward absoluteness, $V_\kappa \models \exists y \, \phi(x,y)$.
Consider the $\Sigma_1$ formula $\exists y \, \phi(x,y)$, where $\phi$ is quantifier-free with $x \in V_\kappa$. Since $\Sigma_1$ formulas are upward absolute, it suffices to show that $V \models \exists y \, \phi(x,y) \implies V_\kappa \models \exists y \, \phi(x,y)$. By the Löwenheim-Skolem Theorem, let $M \prec V$ such that $\vert M\vert < \kappa$ and $\TC(\lbrace x\rbrace) \subseteq M$ (as $V_\kappa = H_\kappa$ by Lemma 13.23.A). By Mostowski's Collapsing Theorem, the transitive collapse $\pi(M)$ is elementarily equivalent to $M$, so $\pi(M) \models \exists y \, \phi(x,y)$. Since $\pi(M)$ is transitive and $\vert \pi(M)\vert < \kappa$, by Lemma 13.23.A we have $\pi(M) \in V_\kappa$. By upward absoluteness, $V_\kappa \models \exists y \, \phi(x,y)$.
$\square$
Exercise 20.7
Solution
Let $\exists y \, \varphi(x,y)$ be a formula with $\varphi$ being $\Pi_1$, and $x \in V_\kappa$. If $V_\kappa \models \exists y \, \varphi(x,y)$, then $V_\kappa \models \varphi(x,y)$ for some $y \in V_\kappa$. Since $\kappa$ is inaccessible, by Exercise 20.6 we have $V \models \varphi(x,y)$, so $V \models \exists y \, \varphi(x,y)$.
On the other hand, suppose $V \models \varphi(x,y)$ for some $y \in V_\lambda$. Since $\kappa$ is supercompact, there exists an elementary embedding $j : V \to M$ such that $j(\kappa) > \lambda$. Then $j(x) = x$, and $y \in M \cap V_{j(\kappa)}$. Since $\varphi$ is $\Pi_1$ and hence downward absolute, $M \cap V_{j(\kappa)} \models \varphi(x,y)$, so $M \cap V_{j(\kappa)} \models \exists y \, \varphi(x,y)$. But since $j(V_\kappa) = M \cap V_{j(\kappa)}$, by elementarity this implies $V_\kappa \models \exists y \, \varphi(x,y)$.
Let $\exists y \, \varphi(x,y)$ be a formula with $\varphi$ being $\Pi_1$, and $x \in V_\kappa$. If $V_\kappa \models \exists y \, \varphi(x,y)$, then $V_\kappa \models \varphi(x,y)$ for some $y \in V_\kappa$. Since $\kappa$ is inaccessible, by Exercise 20.6 we have $V \models \varphi(x,y)$, so $V \models \exists y \, \varphi(x,y)$.
On the other hand, suppose $V \models \varphi(x,y)$ for some $y \in V_\lambda$. Since $\kappa$ is supercompact, there exists an elementary embedding $j : V \to M$ such that $j(\kappa) > \lambda$. Then $j(x) = x$, and $y \in M \cap V_{j(\kappa)}$. Since $\varphi$ is $\Pi_1$ and hence downward absolute, $M \cap V_{j(\kappa)} \models \varphi(x,y)$, so $M \cap V_{j(\kappa)} \models \exists y \, \varphi(x,y)$. But since $j(V_\kappa) = M \cap V_{j(\kappa)}$, by elementarity this implies $V_\kappa \models \exists y \, \varphi(x,y)$.
$\square$
Exercise 20.8
Solution
Note that we are assuming $D$ is normal when proving the equivalences here.
(i) $\implies$ (ii): Suppose $X \in D$ witnesses the strong normality of $D$, and assume WLOG that $\emptyset \in X$. Let $\lbrace Z_x : x \in X\rbrace \subseteq D$, and suppose for a contradiction that $Z := \bigtriangle_{x \in X} Z_x \notin D$. Then: $$ \begin{align*} P_\kappa(\lambda) - Z = \lbrace y : (\exists x \subsetneq y) \, x \in X \wedge y \notin Z_x\rbrace \in D \end{align*}$$ Define $f : P_\kappa(\lambda) \to X$ by: $$ \begin{align*} f(y) := \begin{cases} x, &\text{if for some $x \subsetneq y$, $x \in X$ and $y \notin Z_x$} \\ \emptyset, &\text{otherwise} \end{cases} \end{align*}$$ By definition $f(x) \subsetneq x$ for all $x \in X$. Since $X$ witnesses the strong normality of $D$, there exists a $Y \in D$, $Y \subseteq X$, such that $f$ is constant on $Y$. We note that $f$ can't take the constant value $\emptyset$ on $Y$, as $Y$ and $P_\kappa(\lambda) - Z$ are disjoint. Thus $f$ takes the value of some non-empty $x$ such that for all $y \in Y$, $y \notin Z_x$. In other words, $Y$ and $Z_x$ are disjoint, which is also not possible.
(ii) $\implies$ (iii): Following the hint, let $Z := \bigtriangle_{x \in X} Z_x$ and show that $X \cap Z$ is homogeneous for $F$ in the sense of (iii). Assume WLOG that on $X$, $F_x$ takes the constant value $0$ on all $x \in X$. Let $x, y \in X \cap Z$ with $x \subsetneq y$. Since $y \in Z$, $y \in Z_x$ as $x \subsetneq y$ and $x \in X$. Thus $F(x,y) = F_x(y) = 0$ as $F_x$ takes the constant value $0$ on $Z_x$.
(iii) $\implies$ (iv): We note that (iv) follows from the hint. Indeed, suppose every $X \in D$ contains $x, y \in X$ such that $x \subsetneq y$ and $\lambda_x < \kappa_y$. Define a partition $F : [P_\kappa(\lambda)]^2 \to \lbrace 0,1\rbrace$ by: $$ \begin{align*} F(x,y) := \begin{cases} 1, &\text{if $x \subsetneq y$ and $\lambda_x < \kappa_y$, or $y \subsetneq x$ and $\lambda_y < \kappa_x$} \\ 0, &\text{otherwise} \end{cases} \end{align*}$$ By (iii), there exists $X \in D$ such that $F$ is constant on $\lbrace \lbrace x,y\rbrace \in [X]^2 : x \subsetneq y$ or $y \subsetneq x\rbrace$. By the hint, there must exist $\lbrace x,y\rbrace \in [X]^2$ such that $x \subsetneq y$ and $\lambda_x < \kappa_y$, so $F(x,y) = 1$. By homogeneity, $F$ takes constant value $1$ on $\lbrace \lbrace x,y\rbrace \in [X]^2 : x \subsetneq y$ or $y \subsetneq x\rbrace$, which gives the desired set for (iv).
It remains to prove the hint. This is rather straightforward: Let $X \in D$ and fix any $x \in X$, so $\lambda_x < \kappa$ as $\vert x\vert < \kappa$. Let $z := x \cup \lbrace \lambda_x\rbrace$. Then $\hat{z} \in D$, so $\hat{z} \cap X \neq \emptyset$. Let $y \in \hat{z} \cap X$. Then clearly $\lambda_x \in y \cap \kappa = \kappa_y$, so $\lambda_x < \kappa_y$.
(iv) $\implies$ (i): We follow the hint. Let $f : X \to X$ be as in the hint, so $j(f) : j(X) \to j(X)$ is a function in $\Ult_D$. Note that since $X \in D$, $j``(\lambda) \in j(X)$, so $(j(f))(j``(\lambda)) \in j(X)$. Let $x := (j(f))(j``(\lambda))$. By choice of $f$, $x \subsetneq j``(\lambda)$ and $x \in j(X)$. By (iv): $$ \begin{align*} \vert x\vert \leq \lambda_x < \kappa_{j``(\lambda)} < j``(\lambda) \cap \kappa = \kappa \end{align*}$$ Therefore we must have $x = j(y)$ for some $y$. This means $(j(f))(j``(\lambda)) = j(y)$, so $f(x) = y$ for almost all $x$ by the remark preceding (20.21).
Note that we are assuming $D$ is normal when proving the equivalences here.
(i) $\implies$ (ii): Suppose $X \in D$ witnesses the strong normality of $D$, and assume WLOG that $\emptyset \in X$. Let $\lbrace Z_x : x \in X\rbrace \subseteq D$, and suppose for a contradiction that $Z := \bigtriangle_{x \in X} Z_x \notin D$. Then: $$ \begin{align*} P_\kappa(\lambda) - Z = \lbrace y : (\exists x \subsetneq y) \, x \in X \wedge y \notin Z_x\rbrace \in D \end{align*}$$ Define $f : P_\kappa(\lambda) \to X$ by: $$ \begin{align*} f(y) := \begin{cases} x, &\text{if for some $x \subsetneq y$, $x \in X$ and $y \notin Z_x$} \\ \emptyset, &\text{otherwise} \end{cases} \end{align*}$$ By definition $f(x) \subsetneq x$ for all $x \in X$. Since $X$ witnesses the strong normality of $D$, there exists a $Y \in D$, $Y \subseteq X$, such that $f$ is constant on $Y$. We note that $f$ can't take the constant value $\emptyset$ on $Y$, as $Y$ and $P_\kappa(\lambda) - Z$ are disjoint. Thus $f$ takes the value of some non-empty $x$ such that for all $y \in Y$, $y \notin Z_x$. In other words, $Y$ and $Z_x$ are disjoint, which is also not possible.
(ii) $\implies$ (iii): Following the hint, let $Z := \bigtriangle_{x \in X} Z_x$ and show that $X \cap Z$ is homogeneous for $F$ in the sense of (iii). Assume WLOG that on $X$, $F_x$ takes the constant value $0$ on all $x \in X$. Let $x, y \in X \cap Z$ with $x \subsetneq y$. Since $y \in Z$, $y \in Z_x$ as $x \subsetneq y$ and $x \in X$. Thus $F(x,y) = F_x(y) = 0$ as $F_x$ takes the constant value $0$ on $Z_x$.
(iii) $\implies$ (iv): We note that (iv) follows from the hint. Indeed, suppose every $X \in D$ contains $x, y \in X$ such that $x \subsetneq y$ and $\lambda_x < \kappa_y$. Define a partition $F : [P_\kappa(\lambda)]^2 \to \lbrace 0,1\rbrace$ by: $$ \begin{align*} F(x,y) := \begin{cases} 1, &\text{if $x \subsetneq y$ and $\lambda_x < \kappa_y$, or $y \subsetneq x$ and $\lambda_y < \kappa_x$} \\ 0, &\text{otherwise} \end{cases} \end{align*}$$ By (iii), there exists $X \in D$ such that $F$ is constant on $\lbrace \lbrace x,y\rbrace \in [X]^2 : x \subsetneq y$ or $y \subsetneq x\rbrace$. By the hint, there must exist $\lbrace x,y\rbrace \in [X]^2$ such that $x \subsetneq y$ and $\lambda_x < \kappa_y$, so $F(x,y) = 1$. By homogeneity, $F$ takes constant value $1$ on $\lbrace \lbrace x,y\rbrace \in [X]^2 : x \subsetneq y$ or $y \subsetneq x\rbrace$, which gives the desired set for (iv).
It remains to prove the hint. This is rather straightforward: Let $X \in D$ and fix any $x \in X$, so $\lambda_x < \kappa$ as $\vert x\vert < \kappa$. Let $z := x \cup \lbrace \lambda_x\rbrace$. Then $\hat{z} \in D$, so $\hat{z} \cap X \neq \emptyset$. Let $y \in \hat{z} \cap X$. Then clearly $\lambda_x \in y \cap \kappa = \kappa_y$, so $\lambda_x < \kappa_y$.
(iv) $\implies$ (i): We follow the hint. Let $f : X \to X$ be as in the hint, so $j(f) : j(X) \to j(X)$ is a function in $\Ult_D$. Note that since $X \in D$, $j``(\lambda) \in j(X)$, so $(j(f))(j``(\lambda)) \in j(X)$. Let $x := (j(f))(j``(\lambda))$. By choice of $f$, $x \subsetneq j``(\lambda)$ and $x \in j(X)$. By (iv): $$ \begin{align*} \vert x\vert \leq \lambda_x < \kappa_{j``(\lambda)} < j``(\lambda) \cap \kappa = \kappa \end{align*}$$ Therefore we must have $x = j(y)$ for some $y$. This means $(j(f))(j``(\lambda)) = j(y)$, so $f(x) = y$ for almost all $x$ by the remark preceding (20.21).
$\square$
Exercise 20.9
Unsolved
If you would like to suggest a solution, feel free to drop me an email and I'll be happy to discuss it.
If you would like to suggest a solution, feel free to drop me an email and I'll be happy to discuss it.
Exercise 20.10
Lemma 20.10.A
For any ordinal $\lambda$, there does not exist an elementary embedding $j : V_{\lambda+2} \to V_{\lambda+2}$.
Proof
Suppose we have $j : V_{\lambda+2} \to V_{\lambda+2}$. We see that $j(\lambda) \geq \lambda$, and since $j(\lambda) < j(\lambda+1) = \lambda+1$, $j(\lambda) = \lambda$. Thus $\crit(j) < \lambda$. As in the proof of Kunen's Theorem 17.7, let $\kappa_n := j^n(\kappa)$, and let $\delta := \lim_{n \to \infty} \kappa_n$. Then $\delta \leq \lambda$ and $\delta$ satisfies the premise of Lemma 17.8, so there exists $F : \delta^\omega \to \omega$ such that whenever $A$ is a subset of $\delta$ of size $\delta$ and $\gamma < \delta$, there exists some $s \in A^\omega$ such that $F(s) = \gamma$. Then (see Exercise 6.4): $$ \begin{align*} \rank(F) + 1 \leq \rank(\delta^{\delta^\omega}) = \max\lbrace \rank(\delta),\rank(\delta^\omega)\rbrace + 1 = \max\lbrace \delta,\delta + 1\rbrace + 1 = \delta + 2 \end{align*}$$ Thus $\rank(F) \leq \delta + 1 \leq \lambda + 1$, so $F \in V_{\lambda+2}$. A similar contradiction to that in Lemma 17.8 can be derived.
For any ordinal $\lambda$, there does not exist an elementary embedding $j : V_{\lambda+2} \to V_{\lambda+2}$.
Proof
Suppose we have $j : V_{\lambda+2} \to V_{\lambda+2}$. We see that $j(\lambda) \geq \lambda$, and since $j(\lambda) < j(\lambda+1) = \lambda+1$, $j(\lambda) = \lambda$. Thus $\crit(j) < \lambda$. As in the proof of Kunen's Theorem 17.7, let $\kappa_n := j^n(\kappa)$, and let $\delta := \lim_{n \to \infty} \kappa_n$. Then $\delta \leq \lambda$ and $\delta$ satisfies the premise of Lemma 17.8, so there exists $F : \delta^\omega \to \omega$ such that whenever $A$ is a subset of $\delta$ of size $\delta$ and $\gamma < \delta$, there exists some $s \in A^\omega$ such that $F(s) = \gamma$. Then (see Exercise 6.4): $$ \begin{align*} \rank(F) + 1 \leq \rank(\delta^{\delta^\omega}) = \max\lbrace \rank(\delta),\rank(\delta^\omega)\rbrace + 1 = \max\lbrace \delta,\delta + 1\rbrace + 1 = \delta + 2 \end{align*}$$ Thus $\rank(F) \leq \delta + 1 \leq \lambda + 1$, so $F \in V_{\lambda+2}$. A similar contradiction to that in Lemma 17.8 can be derived.
$\blacksquare$
Lemma 20.10.B
If $\kappa$ is extendible, then for every $\alpha > \kappa$ and $\gamma > \kappa$, there exists an ordinal $\beta$ and an elementary embedding $j : V_\alpha \to V_\beta$ with $\crit(j) = \kappa$ and $j(\kappa) > \gamma$. In other words, elementary embeddings of arbitrarily large $j(\kappa)$ may be chosen.
Proof
Suppose not, so there exists some $\gamma > \kappa$ such that for sufficiently large $\alpha$, if $j : V_\alpha \to V_\beta$ then $j(\kappa) < \gamma$. We may choose $\gamma$ to be the least such ordinal.
We first show that for all $\kappa < \xi < \gamma$ and all $\alpha > \kappa$, there exists a $j : V_\alpha \to V_\beta$ such that $j(\kappa) > \xi$. Suppose not, so $\gamma = \xi + 1$ for some $\xi$. Then there exists an unbounded class of $\alpha$ such that there exists $j : V_\alpha \to V_\beta$ with $j(\kappa) = \xi$. Since for any $\zeta < \alpha$, $j\restrictedto V_\zeta : V_\zeta \to V_{j(\zeta)}$ remains an elementary embedding, we have that for all $\alpha > \kappa$, there exists $j : V_\alpha \to V_\beta$ with $j(\kappa) = \xi$. Fix any $j : V_\alpha \to V_\beta$, $k : V_\beta \to V_\delta$ with $j(\kappa) = k(\kappa) = \xi$. Since $\kappa < \xi < \beta$, $k(\xi) > \xi$. Thus $k \circ j : V_\alpha \to V_\delta$ is an elementary embedding with $(k \circ j)(\kappa) > \xi$, a contradiction. In particular, $\gamma$ is a limit ordinal.
We next show that if $\beta$ is sufficiently large and $j : V_\beta \to V_\delta$ is an elementary embedding, then $j``(\gamma) \subseteq \gamma$. Otherwise for some $\xi < \gamma$, there exists an unbounded class of $\beta$, hence all $\beta > \xi$, such that there exists a $j : V_\beta \to V_\delta$ with $j(\xi) \geq \gamma$. Choose $\alpha$ sufficiently large so that for all $\ell : V_\alpha \to V_\beta$, $\ell(\kappa) < \gamma$. By the previous claim, we may also choose $\alpha$ large enough to find $k : V_\alpha \to V_\beta$ with $k(\kappa) > \xi$. Then $j \circ k : V_\alpha \to V_\delta$ and $(j \circ k)(\kappa) \geq \gamma$, a contradiction.
We now prove the lemma. Let $\alpha \geq \gamma + 2$ be sufficiently large and $j : V_\alpha \to V_\beta$ such that $j``(\gamma) \subseteq \gamma$. We consider two cases:
If $\kappa$ is extendible, then for every $\alpha > \kappa$ and $\gamma > \kappa$, there exists an ordinal $\beta$ and an elementary embedding $j : V_\alpha \to V_\beta$ with $\crit(j) = \kappa$ and $j(\kappa) > \gamma$. In other words, elementary embeddings of arbitrarily large $j(\kappa)$ may be chosen.
Proof
Suppose not, so there exists some $\gamma > \kappa$ such that for sufficiently large $\alpha$, if $j : V_\alpha \to V_\beta$ then $j(\kappa) < \gamma$. We may choose $\gamma$ to be the least such ordinal.
We first show that for all $\kappa < \xi < \gamma$ and all $\alpha > \kappa$, there exists a $j : V_\alpha \to V_\beta$ such that $j(\kappa) > \xi$. Suppose not, so $\gamma = \xi + 1$ for some $\xi$. Then there exists an unbounded class of $\alpha$ such that there exists $j : V_\alpha \to V_\beta$ with $j(\kappa) = \xi$. Since for any $\zeta < \alpha$, $j\restrictedto V_\zeta : V_\zeta \to V_{j(\zeta)}$ remains an elementary embedding, we have that for all $\alpha > \kappa$, there exists $j : V_\alpha \to V_\beta$ with $j(\kappa) = \xi$. Fix any $j : V_\alpha \to V_\beta$, $k : V_\beta \to V_\delta$ with $j(\kappa) = k(\kappa) = \xi$. Since $\kappa < \xi < \beta$, $k(\xi) > \xi$. Thus $k \circ j : V_\alpha \to V_\delta$ is an elementary embedding with $(k \circ j)(\kappa) > \xi$, a contradiction. In particular, $\gamma$ is a limit ordinal.
We next show that if $\beta$ is sufficiently large and $j : V_\beta \to V_\delta$ is an elementary embedding, then $j``(\gamma) \subseteq \gamma$. Otherwise for some $\xi < \gamma$, there exists an unbounded class of $\beta$, hence all $\beta > \xi$, such that there exists a $j : V_\beta \to V_\delta$ with $j(\xi) \geq \gamma$. Choose $\alpha$ sufficiently large so that for all $\ell : V_\alpha \to V_\beta$, $\ell(\kappa) < \gamma$. By the previous claim, we may also choose $\alpha$ large enough to find $k : V_\alpha \to V_\beta$ with $k(\kappa) > \xi$. Then $j \circ k : V_\alpha \to V_\delta$ and $(j \circ k)(\kappa) \geq \gamma$, a contradiction.
We now prove the lemma. Let $\alpha \geq \gamma + 2$ be sufficiently large and $j : V_\alpha \to V_\beta$ such that $j``(\gamma) \subseteq \gamma$. We consider two cases:
- Suppose $\cf(\gamma) = \omega$. We can show that $j(\gamma) = \gamma$ — if $\lbrace \alpha_n : n \in \omega\rbrace$ is cofinal in $\gamma$ (in $V$), then $\lbrace j(\alpha_n) : n \in \omega\rbrace$ is cofinal in $j(\gamma)$ (in $M$). Since $j``(\gamma) \subseteq \gamma$, $\lbrace j(\alpha_n) : n \in \omega\rbrace \subseteq \gamma$, so $j(\gamma) = \gamma$. Then $j\restrictedto V_{\gamma+2} : V_{\gamma+2} \to V_{\gamma+2}$ is an elementary embedding, contradicting Lemma 20.10.A.
- Suppose $\cf(\gamma) > \omega$. Let $\kappa_n := j^n(\kappa)$, and let $\lambda := \lim_{n\to\infty} \kappa_n$. Then $j(\lambda) = \lambda$ (see the proof of Theorem 17.7), and since $\cf(\gamma) > \omega$, $\lambda < \gamma$. Then $j\restrictedto V_{\lambda+2} : V_{\lambda+2} \to V_{\lambda+2}$ is an elementary embedding, contradicting Lemma 20.10.A.
$\blacksquare$
Solution
Let $\exists y \, \varphi(x,y)$ be a formula with $\varphi$ being $\Pi_2$, and $x \in V_\kappa$. By Exercise 20.7 and that every extendible cardinal is supercompact (Theorem 20.4), if $V_\kappa \models \exists y \, \varphi(x,y)$ then $V \models \exists y \, \varphi(x,y)$.
On the other hand, suppose $V \models \exists y \, \varphi(x,y)$. We write $\varphi(x,y) = \forall z \, \psi(x,y,z)$, where $\psi$ is $\Pi_1$. Let $y \in V_\gamma$ such that $V \models \forall z \, \psi(x,y,z)$. Assuming the hint has been shown, fix any inaccessible $\lambda > \max\lbrace \gamma,\kappa\rbrace$ such that $V_\kappa \prec V_\lambda$. For any $z \in V_\lambda$, $V \models \psi(x,y,z)$, so by Exercise 20.6 $V_\lambda \models \psi(x,y,z)$. Hence $V_\lambda \models \varphi(x,y)$, so $V_\lambda \models \exists y \, \varphi(x,y)$, and since $V_\kappa \prec V_\lambda$ we have $V_\kappa \models \exists y \, \varphi(x,y)$.
It remains to prove the hint. This follows from the fact that if $j : V_\alpha \to V_\beta$ with $\crit(j) = \kappa$, then $j(\kappa)$ is inaccessible (as $\kappa$ is inaccessible), and by Lemma 20.10.B $j(\kappa)$ may be arbitrarily large.
Let $\exists y \, \varphi(x,y)$ be a formula with $\varphi$ being $\Pi_2$, and $x \in V_\kappa$. By Exercise 20.7 and that every extendible cardinal is supercompact (Theorem 20.4), if $V_\kappa \models \exists y \, \varphi(x,y)$ then $V \models \exists y \, \varphi(x,y)$.
On the other hand, suppose $V \models \exists y \, \varphi(x,y)$. We write $\varphi(x,y) = \forall z \, \psi(x,y,z)$, where $\psi$ is $\Pi_1$. Let $y \in V_\gamma$ such that $V \models \forall z \, \psi(x,y,z)$. Assuming the hint has been shown, fix any inaccessible $\lambda > \max\lbrace \gamma,\kappa\rbrace$ such that $V_\kappa \prec V_\lambda$. For any $z \in V_\lambda$, $V \models \psi(x,y,z)$, so by Exercise 20.6 $V_\lambda \models \psi(x,y,z)$. Hence $V_\lambda \models \varphi(x,y)$, so $V_\lambda \models \exists y \, \varphi(x,y)$, and since $V_\kappa \prec V_\lambda$ we have $V_\kappa \models \exists y \, \varphi(x,y)$.
It remains to prove the hint. This follows from the fact that if $j : V_\alpha \to V_\beta$ with $\crit(j) = \kappa$, then $j(\kappa)$ is inaccessible (as $\kappa$ is inaccessible), and by Lemma 20.10.B $j(\kappa)$ may be arbitrarily large.
$\square$
Exercise 20.11
Solution
Let $P_\kappa^\ast (\lambda) := \lbrace X \subseteq \lambda : \text{order-type}(X) = \kappa\rbrace$. A normal ultrafilter over $P_\kappa^\ast (\lambda)$ takes the same definition as that preceding Lemma 8.23.
$\implies$: The proof in this direction is almost the same as that of Lemma 20.14. Suppose $j : V \to M$ and $j(\kappa) = \lambda$, with $M^{j(\kappa)} \subseteq M$. For $X \subseteq \lambda$ such that $\otp(X) = \kappa$, let $U$ be as defined in the hint, which is possible as $M^\lambda \subseteq M$ (so that $\lbrace j(\gamma) : \gamma < \lambda\rbrace \in M$). We show $U$ is a normal $\kappa$-complete ultrafilter.
Let $P_\kappa^\ast (\lambda) := \lbrace X \subseteq \lambda : \text{order-type}(X) = \kappa\rbrace$. A normal ultrafilter over $P_\kappa^\ast (\lambda)$ takes the same definition as that preceding Lemma 8.23.
$\implies$: The proof in this direction is almost the same as that of Lemma 20.14. Suppose $j : V \to M$ and $j(\kappa) = \lambda$, with $M^{j(\kappa)} \subseteq M$. For $X \subseteq \lambda$ such that $\otp(X) = \kappa$, let $U$ be as defined in the hint, which is possible as $M^\lambda \subseteq M$ (so that $\lbrace j(\gamma) : \gamma < \lambda\rbrace \in M$). We show $U$ is a normal $\kappa$-complete ultrafilter.
- Obviously $\emptyset \notin U$. We have $j(P_\kappa^\ast (\lambda)) = P_{j(\kappa)}^\ast (j(\lambda))$, and $j``(\lambda) \subseteq j(\lambda)$, with $\otp(j``(\lambda)) = \otp(\lambda) = \lambda < j(\kappa)$. Thus $j``(\lambda) \in P_\kappa^\ast (\lambda)$.
- Upward Closure: Immediate.
- $\kappa$-Directedness: Let $\lbrace X_\beta : \beta < \mu\rbrace \subseteq U$ where $\mu < \kappa$. Then: $$ \begin{align*} \bigcap_{\beta < \mu} X_\beta \in U &\iff j``(\lambda) \in j\bb{\bigcap_{\beta < \mu} X_\beta} \\ &\iff j``(\lambda) \in \bigcap_{\beta < \mu} j(X_\beta) \\ &\iff (\forall \beta < \mu) \, j``(\lambda) \in j(X_\beta) \\ &\iff (\forall \beta < \mu) \, X_\beta \in U \end{align*}$$ Since the last statement holds, $\bigcap_{\beta < \mu} X_\beta \in U$.
- Normality: Let $f : P_\kappa^\ast (\lambda) \to \lambda$ and $X \in U$ be such that $f(x) \in x$ for all non-empty $x \in X$, i.e. $f$ is regressive on $X$. By elementarity, $j(f)$ is regressive on $j(X)$, and since $X \in U$, $j``(\lambda) \in j(X)$. Thus $(j(f))(j``(\lambda)) \in j``(\lambda)$, so $(j(f))(j``(\lambda)) = j(\gamma)$ for some $\gamma < \lambda$. Thus $f(x) = \gamma$ for almost all $x$, as desired.
$\square$
Exercise 20.12
Solution
We expand on the hint. Suppose $\kappa = \mu$. By Lemma 20.27, $V_\mu$ is a model of Vopěnka's Principle. The proof of Lemma 20.25 shows that there exists an $\alpha < \mu$ such that for all $\lambda < \alpha$, $V_\alpha \models$ ($\kappa$ is extendible) implies $\kappa$ is extendible, and $V_\alpha$ contains an extendible cardinal $\lambda < \alpha$. Thus $\lambda$ is extendible, hence supercompact, in $V$. Yet $\lambda < \mu$, contradicting minimality of $\mu$.
Suppose $\mu < \kappa$. Let $j : V \to M$ be elementary such that $j(\kappa) = \lambda$ and $M^\lambda \subseteq M$. Let $i : V \to N$ be also elementary such that $i(\mu) > \lambda + 2$ (so $V_{\lambda+2} \subseteq N$ as $N^{\lambda+2} \subseteq N$). If $U$ is a normal measure witnessing the hugeness of $\kappa$ in $V$, then since $\kappa, U \in N$ and $\kappa < i(\mu)$: $$ \begin{align*} N \models (\text{$\exists$ huge cardinal below $i(\mu)$}) \end{align*}$$ By elementarity of $i$: $$ \begin{align*} V \models (\text{$\exists$ huge cardinal below $\mu$}) \end{align*}$$ which contradicts the minimality of $\kappa$.
We expand on the hint. Suppose $\kappa = \mu$. By Lemma 20.27, $V_\mu$ is a model of Vopěnka's Principle. The proof of Lemma 20.25 shows that there exists an $\alpha < \mu$ such that for all $\lambda < \alpha$, $V_\alpha \models$ ($\kappa$ is extendible) implies $\kappa$ is extendible, and $V_\alpha$ contains an extendible cardinal $\lambda < \alpha$. Thus $\lambda$ is extendible, hence supercompact, in $V$. Yet $\lambda < \mu$, contradicting minimality of $\mu$.
Suppose $\mu < \kappa$. Let $j : V \to M$ be elementary such that $j(\kappa) = \lambda$ and $M^\lambda \subseteq M$. Let $i : V \to N$ be also elementary such that $i(\mu) > \lambda + 2$ (so $V_{\lambda+2} \subseteq N$ as $N^{\lambda+2} \subseteq N$). If $U$ is a normal measure witnessing the hugeness of $\kappa$ in $V$, then since $\kappa, U \in N$ and $\kappa < i(\mu)$: $$ \begin{align*} N \models (\text{$\exists$ huge cardinal below $i(\mu)$}) \end{align*}$$ By elementarity of $i$: $$ \begin{align*} V \models (\text{$\exists$ huge cardinal below $\mu$}) \end{align*}$$ which contradicts the minimality of $\kappa$.
$\square$
Exercise 20.13
Solution
We note that since every huge cardinal is measurable (Theorem 17.3), the least huge cardinal is at least the least measurable cardinal. It thus suffices to show that there exists a measurable cardinal below any huge cardinal.
Let $\kappa$ be a huge cardinal. Let $j : V \to M$ be such that $\crit(j) = \kappa$ and $M^{j(\kappa)} \subseteq M$. Now $j(\kappa)$ is inaccessible, and $U \in P(P(\kappa))$, so $U \in M$. Thus $U$ witnesses the measurability of $\kappa$ in $M$. Since $\kappa < j(\kappa)$: $$ \begin{align*} M \models (\text{$\exists$ measurable cardinal below $j(\kappa)$}) \end{align*}$$ By elementarity, there exists a measurable cardinal below $\kappa$ in $V$.
We note that since every huge cardinal is measurable (Theorem 17.3), the least huge cardinal is at least the least measurable cardinal. It thus suffices to show that there exists a measurable cardinal below any huge cardinal.
Let $\kappa$ be a huge cardinal. Let $j : V \to M$ be such that $\crit(j) = \kappa$ and $M^{j(\kappa)} \subseteq M$. Now $j(\kappa)$ is inaccessible, and $U \in P(P(\kappa))$, so $U \in M$. Thus $U$ witnesses the measurability of $\kappa$ in $M$. Since $\kappa < j(\kappa)$: $$ \begin{align*} M \models (\text{$\exists$ measurable cardinal below $j(\kappa)$}) \end{align*}$$ By elementarity, there exists a measurable cardinal below $\kappa$ in $V$.
$\square$
Exercise 20.14
Solution
Define a measure $D$ on $\kappa$ by stipulating that for $X \subseteq \kappa$: $$ \begin{align*} X \in D \iff \kappa \in j(X) \end{align*}$$ As proven in Lemma 17.3 and the remark preceding Lemma 17.5, $D$ is a normal measure. We shall show that $X := \lbrace \alpha < \kappa : \alpha \text{ is $n$-huge}\rbrace \in D$. Indeed: $$ \begin{align*} X = \lbrace \alpha < \kappa : \exists M \, (\exists k : V \to M) \, \crit(k) = \alpha \wedge M^{k^n(\alpha)} \subseteq M\rbrace \end{align*}$$ Thus: $$ \begin{align*} j(X) = \lbrace \alpha < j(\kappa) : \exists N \, (\exists k : M \to N) \, \crit(k) = \alpha \wedge N^{j(k^n(\alpha))} \subseteq N\rbrace \end{align*}$$ But we note that since $\kappa$ is, in particular, $n$-huge, we have $M^{j^n(\kappa)} \subseteq M$, so by elementarity $j(M)^{j^{n+1}(\kappa)} \subseteq j(M)$. Since $j\restrictedto M$ yields an elementary embedding with domain $M$, we have that $\kappa \in j(X)$, so $X \in D$.
Define a measure $D$ on $\kappa$ by stipulating that for $X \subseteq \kappa$: $$ \begin{align*} X \in D \iff \kappa \in j(X) \end{align*}$$ As proven in Lemma 17.3 and the remark preceding Lemma 17.5, $D$ is a normal measure. We shall show that $X := \lbrace \alpha < \kappa : \alpha \text{ is $n$-huge}\rbrace \in D$. Indeed: $$ \begin{align*} X = \lbrace \alpha < \kappa : \exists M \, (\exists k : V \to M) \, \crit(k) = \alpha \wedge M^{k^n(\alpha)} \subseteq M\rbrace \end{align*}$$ Thus: $$ \begin{align*} j(X) = \lbrace \alpha < j(\kappa) : \exists N \, (\exists k : M \to N) \, \crit(k) = \alpha \wedge N^{j(k^n(\alpha))} \subseteq N\rbrace \end{align*}$$ But we note that since $\kappa$ is, in particular, $n$-huge, we have $M^{j^n(\kappa)} \subseteq M$, so by elementarity $j(M)^{j^{n+1}(\kappa)} \subseteq j(M)$. Since $j\restrictedto M$ yields an elementary embedding with domain $M$, we have that $\kappa \in j(X)$, so $X \in D$.
$\square$
Exercise 20.15
Lemma 20.15.A
Let $\kappa$ be an uncountable cardinal. The following are equivalent:
Proof
(i) $\implies$ (ii): Let $j : V \to M$ witness the $n$-hugeness of $\kappa$, and let $\lambda_i := j^i(\kappa)$, $\lambda := \lambda_n$. Define $U$ by stipulating that for all $X \subseteq P(\lambda)$: $$ \begin{align*} X \in U \iff j``(\lambda) \in j(X) \end{align*}$$ Note that such definition is possible, as the $n$-hugeness of $\kappa$ implies $M^\lambda \subseteq M$, and therefore $j``(\lambda) \in M$. The proof that $U$ is indeed a normal measure follows closely to the $\implies$ direction of Exercise 20.9. We show that for all $i < n$, $\lbrace x \in P(\lambda) : \otp(x \cap \lambda_{i+1}) = \lambda_i\rbrace \in U$.
We have: $$ \begin{align*} &\iffbreak \lbrace x \in P(\lambda) : \otp(x \cap \lambda_{i+1}) = \lambda_i\rbrace \in U \\ &\iff j``(\lambda) \in j\bb{\lbrace x \in P(\lambda) : \otp(x \cap \lambda_{i+1}) = \lambda_i\rbrace} \\ &\iff j``(\lambda) \in \lbrace x \in P(j(\lambda)) : \otp(x \cap j(\lambda_{i+1})) = j(\lambda_i)\rbrace \\ &\iff \otp(j``(\lambda) \cap j(\lambda_{i+1})) = \lambda_{i+1} \end{align*}$$ But $j``(\lambda) \cap j(\lambda_{i+1}) = j``(\lambda_{i+1})$, which has order-type $\lambda_{i+1}$. So indeed $\lbrace x \in P(\lambda) : \otp(x \cap \lambda_{i+1}) = \lambda_i\rbrace \in U$ for all $i < n$.
(ii) $\implies$ (i): Let $U$ be a normal $\kappa$-complete measure on $P(\lambda)$, $M := \Ult_U(V)$ and $j : V \to M$ the canonical elementary embedding. The normality of $U$ gives us the $P(\lambda)$ analog of Lemma 20.13. This gives: $$ \begin{align*} \lambda_{i+1} = \otp(j``(\lambda) \cap j(\lambda_{i+1})) = [\c{\otp(x \cap \lambda_{i+1}) : x \in P(\lambda)}]_U \end{align*}$$ By (ii), for almost all $x \in P(\lambda)$, $\otp(x \cap \lambda_{i+1}) = \lambda_i$. Thus: $$ \begin{align*} [\c{\otp(x \cap \lambda_{i+1}) : x \in P(\lambda)}]_U = [\c{\lambda_i : x \in P(\lambda)}]_U = j(\lambda_i) \end{align*}$$ Therefore $\lambda = j^n(\kappa)$. Finally, the proof of $M^\lambda \subseteq M$ is verbatim to that of Lemma 20.14.
Let $\kappa$ be an uncountable cardinal. The following are equivalent:
- $\kappa$ is $n$-huge.
- There is a normal $\kappa$-complete measure $U$ over some $P(\lambda)$ (where fineness and normality are defined analogously to the paragraph preceding Lemma 20.2 and Definition 20.12), where $\kappa = \lambda_0 < \cdots < \lambda_n = \lambda$, such that for all $i < n$: $$ \begin{align*} \lbrace x \in P(\lambda) : \otp(x \cap \lambda_{i+1}) = \lambda_i\rbrace \in U \end{align*}$$
Proof
(i) $\implies$ (ii): Let $j : V \to M$ witness the $n$-hugeness of $\kappa$, and let $\lambda_i := j^i(\kappa)$, $\lambda := \lambda_n$. Define $U$ by stipulating that for all $X \subseteq P(\lambda)$: $$ \begin{align*} X \in U \iff j``(\lambda) \in j(X) \end{align*}$$ Note that such definition is possible, as the $n$-hugeness of $\kappa$ implies $M^\lambda \subseteq M$, and therefore $j``(\lambda) \in M$. The proof that $U$ is indeed a normal measure follows closely to the $\implies$ direction of Exercise 20.9. We show that for all $i < n$, $\lbrace x \in P(\lambda) : \otp(x \cap \lambda_{i+1}) = \lambda_i\rbrace \in U$.
We have: $$ \begin{align*} &\iffbreak \lbrace x \in P(\lambda) : \otp(x \cap \lambda_{i+1}) = \lambda_i\rbrace \in U \\ &\iff j``(\lambda) \in j\bb{\lbrace x \in P(\lambda) : \otp(x \cap \lambda_{i+1}) = \lambda_i\rbrace} \\ &\iff j``(\lambda) \in \lbrace x \in P(j(\lambda)) : \otp(x \cap j(\lambda_{i+1})) = j(\lambda_i)\rbrace \\ &\iff \otp(j``(\lambda) \cap j(\lambda_{i+1})) = \lambda_{i+1} \end{align*}$$ But $j``(\lambda) \cap j(\lambda_{i+1}) = j``(\lambda_{i+1})$, which has order-type $\lambda_{i+1}$. So indeed $\lbrace x \in P(\lambda) : \otp(x \cap \lambda_{i+1}) = \lambda_i\rbrace \in U$ for all $i < n$.
(ii) $\implies$ (i): Let $U$ be a normal $\kappa$-complete measure on $P(\lambda)$, $M := \Ult_U(V)$ and $j : V \to M$ the canonical elementary embedding. The normality of $U$ gives us the $P(\lambda)$ analog of Lemma 20.13. This gives: $$ \begin{align*} \lambda_{i+1} = \otp(j``(\lambda) \cap j(\lambda_{i+1})) = [\c{\otp(x \cap \lambda_{i+1}) : x \in P(\lambda)}]_U \end{align*}$$ By (ii), for almost all $x \in P(\lambda)$, $\otp(x \cap \lambda_{i+1}) = \lambda_i$. Thus: $$ \begin{align*} [\c{\otp(x \cap \lambda_{i+1}) : x \in P(\lambda)}]_U = [\c{\lambda_i : x \in P(\lambda)}]_U = j(\lambda_i) \end{align*}$$ Therefore $\lambda = j^n(\kappa)$. Finally, the proof of $M^\lambda \subseteq M$ is verbatim to that of Lemma 20.14.
$\blacksquare$
Solution
We note that by the formulation of (ii) in Lemma 20.15.A, the statement "$\kappa$ is $n$-huge" is absolute between $V_\lambda$ and $V$, where $\lambda$ is large enough (to contain the normal measure in (ii)). Let $\kappa_n := j^n(\kappa)$, and fix some $n < \omega$. Define a normal $\kappa$-complete measure $U$ on $P(\kappa_n)$ by stipulating that $X \in U$ iff $j``(\kappa_n) \in j(X)$. This is well-defined, as $\kappa_{n+1} \in V_\lambda$ and $j``(\kappa_n) \subseteq \kappa_{n+1}$, therefore $j(``\kappa_n) \in V_\lambda$. We may proceed with the proof in Lemma 20.15.A with $\lambda_i = \kappa_i$, and we have that $U$ is a measure satisfying (ii) of Lemma 20.15.A, so $\kappa$ is $n$-huge. By elementarity, $\kappa_i$ is $n$-huge for all $i < \omega$.
We note that by the formulation of (ii) in Lemma 20.15.A, the statement "$\kappa$ is $n$-huge" is absolute between $V_\lambda$ and $V$, where $\lambda$ is large enough (to contain the normal measure in (ii)). Let $\kappa_n := j^n(\kappa)$, and fix some $n < \omega$. Define a normal $\kappa$-complete measure $U$ on $P(\kappa_n)$ by stipulating that $X \in U$ iff $j``(\kappa_n) \in j(X)$. This is well-defined, as $\kappa_{n+1} \in V_\lambda$ and $j``(\kappa_n) \subseteq \kappa_{n+1}$, therefore $j(``\kappa_n) \in V_\lambda$. We may proceed with the proof in Lemma 20.15.A with $\lambda_i = \kappa_i$, and we have that $U$ is a measure satisfying (ii) of Lemma 20.15.A, so $\kappa$ is $n$-huge. By elementarity, $\kappa_i$ is $n$-huge for all $i < \omega$.
$\square$
Exercise 20.16
Solution
Suppose $V = L[A]$. Let $j : V \to M$, with critical point $\kappa$, such that $A \in M$. Then $M = L[j(A)] \subseteq L[A] = V$. On the other hand, since $A \in L[j(A)]$, we have $L[A] \subseteq L[j(A)]$ by Lemma 13.27.A. Thus $L[A] = L[j(A)]$, so there exists a non-trivial elementary embedding $j : L[A] \to L[A]$, contradicting Kunen's Theorem 17.7.
Suppose $V = L[A]$. Let $j : V \to M$, with critical point $\kappa$, such that $A \in M$. Then $M = L[j(A)] \subseteq L[A] = V$. On the other hand, since $A \in L[j(A)]$, we have $L[A] \subseteq L[j(A)]$ by Lemma 13.27.A. Thus $L[A] = L[j(A)]$, so there exists a non-trivial elementary embedding $j : L[A] \to L[A]$, contradicting Kunen's Theorem 17.7.
$\square$
Exercise 20.17
Solution
We may repeat the proof of Lemma 20.19 and Corollary 20.20(iii). The main difference is that if $j : V \to M$, we require $M$ to compute the ultrapowers by the normal measures on $P_\kappa(\lambda)$ correctly. But for this to hold true, it suffices to choose $j$ such that $V^\mu \subseteq M$ for $\mu$ large enough.
We may repeat the proof of Lemma 20.19 and Corollary 20.20(iii). The main difference is that if $j : V \to M$, we require $M$ to compute the ultrapowers by the normal measures on $P_\kappa(\lambda)$ correctly. But for this to hold true, it suffices to choose $j$ such that $V^\mu \subseteq M$ for $\mu$ large enough.
$\square$
Exercise 20.18
Lemma 20.18.A
If $\kappa$ is a strong cardinal, then for all $\lambda \geq \kappa$ and any set $x$, there exists an elementary embedding $j : V \to M$ such that $j(\kappa) > \lambda$ and $x \in M$. Consequently, $\kappa$ is strong iff it is $\lambda$-strong for all $\lambda \geq \kappa$.
Proof
It suffices to replace "$x \in M$" with "$V_\lambda \subseteq M$". Fix $\lambda \geq \kappa$, and let $j : V \to M$ be such that $V_{\lambda+1} \subseteq M$. We first note that $\lambda < \sup\lbrace j^n(\kappa) : n < \omega\rbrace$ — otherwise, since $j^n(\kappa) < \lambda$ for all $n$, $\lbrace j^n(\kappa) : n < \omega\rbrace \in V_{\lambda + 1}$, contradicting Kunen's Theorem 17.7 (the proof shows that if $j : V \to M$ then $\lbrace \kappa_n : n < \omega\rbrace \notin M$, so $M \neq V$).
Let $n < \omega$ such that $\lambda < j^n(\kappa)$. If we can show that $V_\lambda \subseteq M_n := j^n(V)$, then we are done. We do this by induction — the case $n = 1$ is done by choice of $j$. Suppose $V_\lambda \subseteq M_n$. By considering $j\restrictedto M_n : M_n \to M_{n+1}$, $V_\lambda^{M_n} \subseteq M_{n+1}$. But $V_\lambda^{M_n} = V_\lambda \cap M_n = V_\lambda$, so $V_\lambda \subseteq M_{n+1}$.
If $\kappa$ is a strong cardinal, then for all $\lambda \geq \kappa$ and any set $x$, there exists an elementary embedding $j : V \to M$ such that $j(\kappa) > \lambda$ and $x \in M$. Consequently, $\kappa$ is strong iff it is $\lambda$-strong for all $\lambda \geq \kappa$.
Proof
It suffices to replace "$x \in M$" with "$V_\lambda \subseteq M$". Fix $\lambda \geq \kappa$, and let $j : V \to M$ be such that $V_{\lambda+1} \subseteq M$. We first note that $\lambda < \sup\lbrace j^n(\kappa) : n < \omega\rbrace$ — otherwise, since $j^n(\kappa) < \lambda$ for all $n$, $\lbrace j^n(\kappa) : n < \omega\rbrace \in V_{\lambda + 1}$, contradicting Kunen's Theorem 17.7 (the proof shows that if $j : V \to M$ then $\lbrace \kappa_n : n < \omega\rbrace \notin M$, so $M \neq V$).
Let $n < \omega$ such that $\lambda < j^n(\kappa)$. If we can show that $V_\lambda \subseteq M_n := j^n(V)$, then we are done. We do this by induction — the case $n = 1$ is done by choice of $j$. Suppose $V_\lambda \subseteq M_n$. By considering $j\restrictedto M_n : M_n \to M_{n+1}$, $V_\lambda^{M_n} \subseteq M_{n+1}$. But $V_\lambda^{M_n} = V_\lambda \cap M_n = V_\lambda$, so $V_\lambda \subseteq M_{n+1}$.
$\blacksquare$
Solution
Let $\varphi(x,y)$ be a $\Pi_1$ formula, with $x \in V_\kappa$. As in Exercise 20.6 and Exercise 20.7, if $V_\kappa \models \exists y \, \varphi(x,y)$ then the inaccessibility of $\kappa$ implies $V \models \exists y \, \varphi(x,y)$. Conversely, suppose $V \models \exists y \, \varphi(x,y)$, so $V \models \varphi(x,y)$ for some $y \in V_\lambda$, where $\lambda \geq \kappa$. Let $j : V \to M$, with $\crit(j) = \kappa$, such that $j(\kappa) > \lambda$ (possible by Lemma 20.18.A) and $V_\lambda \subseteq M$. Then $y \in M \cap V_{j(\kappa)}$, so we may repeat the proof of Exercise 20.7 to conclude $V_\kappa \models \exists y \, \varphi(x,y)$.
Let $\varphi(x,y)$ be a $\Pi_1$ formula, with $x \in V_\kappa$. As in Exercise 20.6 and Exercise 20.7, if $V_\kappa \models \exists y \, \varphi(x,y)$ then the inaccessibility of $\kappa$ implies $V \models \exists y \, \varphi(x,y)$. Conversely, suppose $V \models \exists y \, \varphi(x,y)$, so $V \models \varphi(x,y)$ for some $y \in V_\lambda$, where $\lambda \geq \kappa$. Let $j : V \to M$, with $\crit(j) = \kappa$, such that $j(\kappa) > \lambda$ (possible by Lemma 20.18.A) and $V_\lambda \subseteq M$. Then $y \in M \cap V_{j(\kappa)}$, so we may repeat the proof of Exercise 20.7 to conclude $V_\kappa \models \exists y \, \varphi(x,y)$.
$\square$
Exercise 20.19
Solution
We mimic the proof of Laver's Theorem 20.21. Suppose the assertion is false. For each $g : \kappa \to V_\kappa$, let $\lambda_g \geq \kappa$ be the least cardinal such that for some $x$ with $\vert \TC(x)\vert \leq \lambda_g$, $j_E(g)(\kappa) \neq x$ for all $(\kappa,\lambda)$-extenders $E$. Let $\nu$ be greater than all the $\lambda_g$'s, and let $j : V \to M$ witness the $\nu$-strongness of $\kappa$ (so $j(\kappa) > \nu$ and $V_\nu \subseteq M$).
Let $\varphi(h,\delta)$ be the statement that $h : \alpha \to V_\alpha$ for some cardinal $\alpha$, and $\delta \geq \alpha$ is the least cardinal such that for some $x$ with $\vert \TC(x)\vert \leq \delta$, there exists no $(\kappa,\lambda)$-extender $E$ such that $j_E(h)(\kappa) = x$. Given any $g : \kappa \to V_\kappa$, by our assumption in the first paragraph we have such a counterexample $x$ with $\vert \TC(x)\vert \leq \lambda_g < \nu$, so $x \in V_\nu \in M$. Furthermore, $\nu$ is also chosen to be large enough so that $M$ contains all $(\kappa,\lambda_g)$-extenders for all $g$. Therefore, the statement "there is no $(\kappa,\lambda_g)$-extender $E$ such that $j_E(g)(\kappa) = x$" is absolute between $V$ and $M$, and $M \models \varphi(g,\lambda_g)$ for all $g : \kappa \to V_\kappa$.
Let $A$ be the set of all $\alpha < \kappa$ such that $\varphi(h,\lambda_h)$ for all $h : \alpha \to V_\alpha$. Then $j(A)$ is the set of all $\alpha < j(\kappa)$ such that $\varphi(h,\lambda_h)$ for all $h : \alpha \to V_\alpha$, so $\kappa \in j(A)$ by the above. Now define $g : V_\kappa \to V_\kappa$ inductively as follows: $$ \begin{align*} g(\alpha) := \begin{cases} x_\alpha, &\text{if $\alpha \in A$ and $x_\alpha$ witnesses $\varphi(g\restrictedto\alpha,\lambda_{g\restrictedto\alpha})$} \\ \emptyset, &\text{otherwise} \end{cases} \end{align*}$$ Let $x := j(g)(\kappa)$. By elementarity of $j$, $x$ witnesses $\varphi(g,\lambda_g)$ in $M$, and hence in $V$ by absoluteness. Let $E$ be the $(\kappa,\lambda_g)$-extender constructed from (20.31) to (20.37). Let $k : \Ult_E \to M$ be elementary such that $k \circ j_E = j$. By Lemma 20.29(i), $k(x) = x$ as $\vert \TC(x)\vert \leq \lambda_g$ and $\max(x) < \lambda_g$. Thus: $$ \begin{align*} (j_E(f))(\kappa) = k^{-1}((j(f))(\kappa)) = k^{-1}(x) = x \end{align*}$$ contradicting that $x$ witnesses $\varphi(g,\lambda_g)$.
We mimic the proof of Laver's Theorem 20.21. Suppose the assertion is false. For each $g : \kappa \to V_\kappa$, let $\lambda_g \geq \kappa$ be the least cardinal such that for some $x$ with $\vert \TC(x)\vert \leq \lambda_g$, $j_E(g)(\kappa) \neq x$ for all $(\kappa,\lambda)$-extenders $E$. Let $\nu$ be greater than all the $\lambda_g$'s, and let $j : V \to M$ witness the $\nu$-strongness of $\kappa$ (so $j(\kappa) > \nu$ and $V_\nu \subseteq M$).
Let $\varphi(h,\delta)$ be the statement that $h : \alpha \to V_\alpha$ for some cardinal $\alpha$, and $\delta \geq \alpha$ is the least cardinal such that for some $x$ with $\vert \TC(x)\vert \leq \delta$, there exists no $(\kappa,\lambda)$-extender $E$ such that $j_E(h)(\kappa) = x$. Given any $g : \kappa \to V_\kappa$, by our assumption in the first paragraph we have such a counterexample $x$ with $\vert \TC(x)\vert \leq \lambda_g < \nu$, so $x \in V_\nu \in M$. Furthermore, $\nu$ is also chosen to be large enough so that $M$ contains all $(\kappa,\lambda_g)$-extenders for all $g$. Therefore, the statement "there is no $(\kappa,\lambda_g)$-extender $E$ such that $j_E(g)(\kappa) = x$" is absolute between $V$ and $M$, and $M \models \varphi(g,\lambda_g)$ for all $g : \kappa \to V_\kappa$.
Let $A$ be the set of all $\alpha < \kappa$ such that $\varphi(h,\lambda_h)$ for all $h : \alpha \to V_\alpha$. Then $j(A)$ is the set of all $\alpha < j(\kappa)$ such that $\varphi(h,\lambda_h)$ for all $h : \alpha \to V_\alpha$, so $\kappa \in j(A)$ by the above. Now define $g : V_\kappa \to V_\kappa$ inductively as follows: $$ \begin{align*} g(\alpha) := \begin{cases} x_\alpha, &\text{if $\alpha \in A$ and $x_\alpha$ witnesses $\varphi(g\restrictedto\alpha,\lambda_{g\restrictedto\alpha})$} \\ \emptyset, &\text{otherwise} \end{cases} \end{align*}$$ Let $x := j(g)(\kappa)$. By elementarity of $j$, $x$ witnesses $\varphi(g,\lambda_g)$ in $M$, and hence in $V$ by absoluteness. Let $E$ be the $(\kappa,\lambda_g)$-extender constructed from (20.31) to (20.37). Let $k : \Ult_E \to M$ be elementary such that $k \circ j_E = j$. By Lemma 20.29(i), $k(x) = x$ as $\vert \TC(x)\vert \leq \lambda_g$ and $\max(x) < \lambda_g$. Thus: $$ \begin{align*} (j_E(f))(\kappa) = k^{-1}((j(f))(\kappa)) = k^{-1}(x) = x \end{align*}$$ contradicting that $x$ witnesses $\varphi(g,\lambda_g)$.
$\square$