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Exercise 20.1.
Solution. We first prove an intermediate claim:Claim.. $\kappa$ is regular.
Proof. Suppose $\kappa$ is singular. Then, by taking a cofinal subsequence, the intersection of $\kappa$-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 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.
$\blacksquare$
Let $\Sigma = \lbrace \sigma_\alpha : \alpha < \lambda\rbrace$ be a $\kappa$-satisfiable collection of $\L_{\kappa,\kappa}$ 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 that $\bigcup_{\beta < \delta} y_\beta \in P_\kappa(\lambda)$, so $\bigcap_{\beta < \delta} [P_\kappa(\lambda)]^{\supseteq y_\beta} \in F$.
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 $\L_{\kappa,\kappa}$ language, which one may easily prove using the same technique as that of the original theorem.
$\square$
Exercise 20.2.
Solution. The hint solves the problem.$\square$
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: (iii) Suppose $\Sigma$ is a set of sentences in the language of $\L_{\kappa,\omega}$, 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 mimic-ed completely.
(iii) $\implies$ (ii): Let $F$ be a $\kappa$-complete filter on $S$ generated $E \subseteq P(S)$, with $\vert E\vert \leq \lambda$. We consider the $\L_{\kappa,\omega}$-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 $\L_{\kappa,\omega}$ consisting of:
- All sentences true in $(S,X)_{X \subseteq S}$.
- $\dot{X}(c)$ for all $X \in E$.
$$ \begin{align*}
X \in U \iff \A \models \dot{X}(c)
\end{align*}$$ One may verify similarly that $U$ is a $\kappa$-complete ultrafilter. To see that $F \subseteq U$, we simply observe that $X \in U$ for all $X \in E$, and since $F$ is generated by $E$, $F \subseteq U$.
(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 $C \in U$.
$\square$
Exercise 20.5.
Solution. The problems 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 statement is simply an application of Łoś Theorem. Indeed, for all $f : P_\kappa(\lambda) \to V$, we have that:
$$ \begin{align*}
&\iffbreak [f] \in \Ult_U\lbrace (V_{\lambda_x},\in) : x \in P_\kappa(\lambda)\rbrace \\
&\iff \lbrace x \in P_\kappa(\lambda) : f(x) \in V_{\lambda_x}\rbrace \in U \\
&\iff \lbrace x \in P_\kappa(\lambda) : \rank^V(f(x)) < \lambda_x\rbrace \in U \\
&\iff \rank^M([f]) < \lambda \\
&\iff [f] \in V_\lambda^M
\end{align*}$$
$\square$
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 $V \models \exists y \, \phi(x,y) \implies V_\kappa \models \exists x \, \phi(x)$. By 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). My Mostowski's Collapsing Theorem, we have that its 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 that $\pi(M) \in V_\kappa$. 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 that $V \models \varphi(x,y)$, so $V \models \exists y \, \varphi(x,y)$.On the other hand, suppose we have $V \models \varphi(x,y)$ 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, we have that $M \cap V_{j(\kappa)} \models \varphi(x,y)$, so $M \cap V_{j(\kappa)} \models \exists y \, \varphi(x,y)$. But since $j(V_\kappa) = M \cap V_{j(\kappa)}$, by elementarity again this implies that $V_\kappa \models \exists y \, \varphi(x,y)$.
$\square$
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 this implies that:
$$ \begin{align*}
P_\kappa(\lambda) - Z = \lbrace y : (\exists x \subsetneq y) \, x \in X \wedge y \notin Z_x\rbrace \in D
\end{align*}$$ Define $f : P_\kappa(\lambda) \to X$ by stipulating that:
$$ \begin{align*}
f(y) :=
\begin{cases}
x, &\text{if for some $x \subsetneq y$, $x \in X$ and $y \notin Z_x$} \\
\emptyset, &\text{otherwise}
\end{cases}
\end{align*}$$ By definition we see that $f(x) \subsetneq x$ for all $x \in X$. Since $X$ witnesses the strong normality of $D$, we have that 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, we 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 $x,y \in X \cap Z$, and suppose that $x \subsetneq y$. 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 $Z_x$ (by choice of $X$).
(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 stipulating that:
$$ \begin{align*}
F(x,y) :=
\begin{cases}
1, &\text{if $x \subsetneq y$ and $\lambda_x < \kappa_y$, or $y \subsetneq x$ and $\lambda_y < \kappa_x$} \\
0, &\text{otherwise}
\end{cases}
\end{align*}$$ 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 $y \subsetneq x\rbrace$. But 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 $y \in \hat{z} \cap D$. 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 $f$, we have that $x \subsetneq j"(\lambda)$ and $x \in j(X)$. By (iv), this implies that:
$$ \begin{align*}
\vert x\vert \leq \lambda_x < \kappa_{j"(\lambda)} < j"(\lambda) \cap \kappa = \kappa
\end{align*}$$ Therefore we must have $x = j(y)$ for some $y$. This means that $(j(f))(j"(\lambda)) = j (y)$, so $f(x) = y$ for almost all $x$ by the remark proceeding (20.21).
$\square$
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, $\crit(j) < \delta$. 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 $\alpha < \xi < \gamma$ and all $\alpha > \kappa$, there exists a $j : V_\alpha \to V_\beta$ such that $j(\kappa) > \xi$. Suppose not, so we have $\gamma = \xi + 1$ 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 $j(\kappa) = \xi$. Thus, we fix any $j : V_\alpha \to V_\beta$, $k : V_\beta \to V_\delta$ with $j(\kappa) = k(\kappa) = \xi$. Since $\kappa < \xi < \beta$, we have that $k(\xi) > \xi$. Thus, $k \circ j : V_\alpha \to V_\delta$ is an elementary embedding with $(k \circ j)(\kappa) > \xi$, a contradiction. In particular, we have that $\gamma$ 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 $\beta$, and hence all $\beta > \xi$, such that there exists a $j : V_\beta \to V_\delta$ with $j(\xi) \geq \gamma$. Choose $\alpha$ sufficiently large and such that for all $\ell : V_\alpha \to V_\beta$, we have $\ell(\alpha) < \gamma$. By the previous claim, we may also choose $\alpha$ large enough so we can 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 $j(\gamma) = \gamma$ - indeed, 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 $j"(\gamma) \subseteq \gamma$, we have that $\lbrace j(\alpha_n) : n \in \omega\rbrace \subseteq \gamma$, so we must have $j(\gamma) = \gamma$.
In that case, we see that $j\restrictedto V_{\gamma+2} : V_{\gamma+2} \to V_{\gamma+2}$ 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 $\cf(\gamma) > \omega$ we have that $\lambda < \gamma$. 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 $z \in V_\lambda$, we have that $V \models \psi(x,y,z)$, so by Exercise 20.6 we have that $V_\lambda \models \psi(x,y,z)$. Hence $V_\lambda \models \varphi(x,y)$. This implies that $V_\lambda \models \exists y \, \varphi(x,y)$, and since $V_\kappa \prec V_\lambda$ we have that $V_\kappa \models \varphi(x,y)$, as desired.
It remains to prove the hint. But this follows from 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 $P_\kappa^\ast (\lambda) := \lbrace X \subseteq \lambda : \text{order-type}(X) = \kappa\rbrace$. Here, 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, but we illustrate the proof for completeness. 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 shall show that $U$ 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.
$\impliedby$: Let $U$ be a normal measure on $P_\kappa^\ast (\lambda)$, and let $M := \Ult_U(V)$ and let $j : V \to M$ be the canonical elementary embedding. The normality of $U$ gives us the $P_\kappa^\ast (\lambda)$ analog of Lemma 20.13, and that give $[f],[g] \in M$ we have:
$$ \begin{align*}
[f] = [g] &\iff (j(f))(j"(\lambda)) = (j(g))(j"(\lambda)) \\
[f] \in [g] &\iff (j(f))(j"(\lambda)) \in (j(g))(j"(\lambda))
\end{align*}$$ and therefore $[f] = (j(f))(j"(\lambda))$. We also have the analog of (20.23), where $\lambda$ is represented by the function $x \mapsto \lambda_x$. Here however, we instead have $\lambda_x = \kappa$ be all $x \in P_\kappa^\ast (\lambda)$, and therefore $j(\kappa) = \lambda$. We also have $j(\gamma) = \gamma$ for all $\gamma < \kappa$ by $\kappa$-completeness of $U$, and since $\lambda > \kappa$ we have $\crit(j) = \kappa$. Finally, the proof of $M^\lambda \subseteq M$ is verbatim to that of Lemma 20.14.
$\square$
Exercise 20.12.
Solution. We expand on the hint. Suppose $\kappa = \mu$. By Lemma 20.27, $V_\mu$ is a model of Vop\v{e}nka's Principle. The proof of Lemma 20.25 shows that there exists an $\alpha < \mu$ such that for all $\lambda < \alpha$, $V_\alpha \models (\text{$\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 $\kappa < i(\mu)$ we have that:
$$ \begin{align*}
N \models (\text{$\exists$ huge cardinal below $i(\mu)$})
\end{align*}$$ By elementarity of $i$, this gives:
$$ \begin{align*}
V \models (\text{$\exists$ huge cardinal below $\mu$})
\end{align*}$$ 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 exist 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 inaccessibility of $\kappa$ in $M$. Since $\kappa < j(\kappa)$, we have that:
$$ \begin{align*}
M \models (\text{$\exists$ measurable cardinal below $j(\kappa)$})
\end{align*}$$ by elementarity this implies that there exist 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, we have that:
$$ \begin{align*}
X = \lbrace \alpha < \kappa : \exists M \, (\exists k : V \to M) \, \crit(k) = \alpha \wedge M^{k^n(\alpha)} \subseteq M\rbrace
\end{align*}$$ 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 we have that $j(M)^{j^{n+1}(\kappa)} \subseteq j(M)$. 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:
- $\kappa$ is $n$-huge.
- There is a normal $\kappa$-complete measure $U$ over some $P(\lambda)$ (where fineness normality are defined analogously to paragraph preceding Lemma 20.2, and Definition 20.12), where $\kappa = \lambda_0 < \dots < \lambda_n = \lambda$, 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 $\lambda_i := j^i(\kappa)$, and $\lambda := \lambda_n$. 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 that $M^\lambda \subseteq M$, and therefore $j"(\lambda) \in M$. The proof that $U$ is indeed a normal measure follows very 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 it's easy to see that $j"(\lambda) \cap j(\lambda_{i+1}) = j"(\lambda_{i+1})$, which has order-type $\lambda_{i+1}$. So we indeed have that $\lbrace x \in P(\lambda) : \otp(x \cap \lambda_{i+1}) = \lambda_i\rbrace \in U$ for all $i < n$.
$\impliedby$: Let $U$ be a normal $\kappa$-complete measure on $P(\lambda)$, let $M := \Ult_U(V)$ and let $j : V \to M$ be the canonical elementary embedding. The normality of $U$ gives us the $P(\lambda)$ analog of Lemma 20.13. This gives us:
$$ \begin{align*}
\lambda_{i+1} = \otp(j"(\lambda) \cap j(\lambda_{i+1})) = [\c{\otp(x \cap \lambda_{i+1}) : x \in P(\lambda)}]_U
\end{align*}$$ By (ii), we have that 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, we have that 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 that $L[A] \subseteq L[j(A)]$ 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 Corollary 20.20(iii). Then 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. Note that 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 of Theorem 17.7 shows that if $j : V \to M$ then $\lbrace \kappa_n : n < \omega\rbrace \notin M$, and therefore $M \neq V$).
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 for $n = 1$ has been done by the choice of $j$. Now suppose $V_\lambda \subseteq M_n$. By considering $j\restrictedto M_n : M_n \to M_{n+1}$, we have that $V_\lambda^{M_n} \subseteq M_{n+1}$. But $V_\lambda^{M_n} = V_\lambda \cap M_n = V_\lambda$, so $V_\lambda \subseteq M_{n+1}$ as desired.
$\blacksquare$
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 that $V \models \exists y \, \varphi(x,y)$. 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 that $V_\kappa \models \exists y \, \varphi(x,y)$.
$\square$
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 $\lambda_f$'s, and let $j : V \to M$ witness the $\nu$-strongess 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 $\delta \geq \alpha$ 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 we have that $M \models \varphi(g,\lambda_g)$ 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 above. Now define $g : V_\kappa \to V_\kappa$ inductively as follows: Let:
$$ \begin{align*}
g(\alpha) :=
\begin{cases}
x_\alpha, &\text{if $\alpha \in A$ and $x_\alpha$ witnesses $\varphi(g\restrictedto\alpha,\lambda_{g\restrictedto\alpha})$} \\
\emptyset, &\text{otherwise}
\end{cases}
\end{align*}$$ Let $x := j(g)(\kappa)$. By elementarity of $j$, $x$ witnesses $\varphi(g,\lambda_g)$ in $M$, and hence in $V$ by absoluteness elaborated above. 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), we have that $k(x) = x$ as $\vert \TC(x)\vert \leq \lambda_g$ and $x$ is finite (so $\max(x) < \lambda_g$). Thus:
$$ \begin{align*}
(j_E(f))(\kappa) = k^{-1}((j(f))(\kappa)) = k^{-1}(x) = x
\end{align*}$$ contradicting that $x$ witnesses $\varphi(g,\lambda_g)$.
$\square$