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Exercise 15.1
Solution
We note that in the proof of Theorem 7.13, the cuts $U$ in $B(P)$ are exactly: $$ \begin{align*} U = \sum\lbrace p \in P : p \in U\rbrace \end{align*}$$ We may then reduce the summand to an antichain in $P$. Since $P$ is $\kappa$-cc, every antichain is of length $<\kappa$. For each $\lambda < \kappa$, there are at most $\vert P\vert ^\lambda$ many possible antichains of length $\lambda$. Thus: $$ \begin{align*} \vert B(P)\vert &\leq \mod{\ss{\sum W : W \subseteq P \text{ is an antichain}}} \\ &= \bigcup_{\lambda < \kappa} \mod{\ss{\sum W : W \subseteq P \text{ is an antichain of length $\lambda$}}} \\ &\leq \bigcup_{\lambda < \kappa} \vert P\vert ^\lambda \\ &= \vert P\vert ^{<\kappa} \end{align*}$$
We note that in the proof of Theorem 7.13, the cuts $U$ in $B(P)$ are exactly: $$ \begin{align*} U = \sum\lbrace p \in P : p \in U\rbrace \end{align*}$$ We may then reduce the summand to an antichain in $P$. Since $P$ is $\kappa$-cc, every antichain is of length $<\kappa$. For each $\lambda < \kappa$, there are at most $\vert P\vert ^\lambda$ many possible antichains of length $\lambda$. Thus: $$ \begin{align*} \vert B(P)\vert &\leq \mod{\ss{\sum W : W \subseteq P \text{ is an antichain}}} \\ &= \bigcup_{\lambda < \kappa} \mod{\ss{\sum W : W \subseteq P \text{ is an antichain of length $\lambda$}}} \\ &\leq \bigcup_{\lambda < \kappa} \vert P\vert ^\lambda \\ &= \vert P\vert ^{<\kappa} \end{align*}$$
$\square$
Exercise 15.2
Solution
For any $p \in P$, let $\lambda = \sup\dom(p)$. Since $\kappa$ is regular and $\vert \dom(p)\vert < \kappa$, $\lambda < \kappa$. Define $q : \lambda \to \lbrace 0,1\rbrace$ by: $$ \begin{align*} q(\alpha) = \begin{cases} p(\alpha), &\text{if $\alpha \in \dom(p)$} \\ 0, &\text{otherwise} \end{cases} \end{align*}$$ Clearly $q \leq p$ and $q \in Q$, so $Q$ is dense in $P$.
For any $p \in P$, let $\lambda = \sup\dom(p)$. Since $\kappa$ is regular and $\vert \dom(p)\vert < \kappa$, $\lambda < \kappa$. Define $q : \lambda \to \lbrace 0,1\rbrace$ by: $$ \begin{align*} q(\alpha) = \begin{cases} p(\alpha), &\text{if $\alpha \in \dom(p)$} \\ 0, &\text{otherwise} \end{cases} \end{align*}$$ Clearly $q \leq p$ and $q \in Q$, so $Q$ is dense in $P$.
$\square$
Exercise 15.3
Solution
We modify the hint for singular cardinals in general. Fix a singular cardinal $\kappa$, along with the added subset $X \subseteq \kappa$ (so $\bigcup G$ is the characteristic function of $X$). Also, fix a cofinal sequence of cardinals $\c{\lambda_\gamma : \gamma < \cf(\kappa)}$, and we may assume every $\lambda_\gamma$ is regular (otherwise just take the successor cardinal). Define $g : \kappa \to \cf(\kappa)$ (as sets) by: $$ \begin{align*} g(\alpha) := \min\lbrace \gamma < \cf(\kappa) : \otp(X \cap (\lambda_{\gamma+1} - \lambda_\gamma)) = \lambda_\gamma + \alpha\rbrace \end{align*}$$
We modify the hint for singular cardinals in general. Fix a singular cardinal $\kappa$, along with the added subset $X \subseteq \kappa$ (so $\bigcup G$ is the characteristic function of $X$). Also, fix a cofinal sequence of cardinals $\c{\lambda_\gamma : \gamma < \cf(\kappa)}$, and we may assume every $\lambda_\gamma$ is regular (otherwise just take the successor cardinal). Define $g : \kappa \to \cf(\kappa)$ (as sets) by: $$ \begin{align*} g(\alpha) := \min\lbrace \gamma < \cf(\kappa) : \otp(X \cap (\lambda_{\gamma+1} - \lambda_\gamma)) = \lambda_\gamma + \alpha\rbrace \end{align*}$$
Claim
For all ordinals $\alpha < \kappa$ and $p \in P$, there exists a $q \leq p$ ($q \supseteq p$) such that there is a $\gamma < \cf(\kappa)$ in which $\lambda_{\gamma+1} - \lambda_\gamma \subseteq \dom(q)$, and: $$ \begin{align*} \otp(\lbrace \xi \in \lambda_{\gamma+1} - \lambda_\gamma : q(\xi) = 1\rbrace) = \lambda_\gamma + \alpha \end{align*}$$ In other words, the set: $$ \begin{align*} D_\alpha &:= \lbrace p \in P : (\exists \gamma < \cf(\kappa))(\lambda_{\gamma+1} - \lambda_\gamma \subseteq \dom(p) \\ &\eqbreak \wedge \otp(\lbrace \xi \in \lambda_{\gamma+1} - \lambda_\gamma : q(\xi) = 1\rbrace) = \lambda_\gamma + \alpha)\rbrace \end{align*}$$ is dense in $P$.
Proof
We work in $V$. Fix $p \in P$, and let $p_\gamma := p\restrictedto(\lambda_{\gamma+1} - \lambda_\gamma)$. Then there exists a $\gamma$ such that $\vert \dom(p_\gamma)\vert < \lambda_\gamma$, for otherwise we have: $$ \begin{align*} \vert \dom(p)\vert = \sum_{\gamma < \cf(\kappa)} \vert \dom(p_\gamma)\vert = \sum_{\gamma < \cf(\kappa)} \lambda_\gamma = \kappa \end{align*}$$ contradicting that $\vert \dom(p)\vert < \kappa$. With this $\gamma$, we have that $\dom(p_\gamma) \subseteq \lambda_{\gamma+1} - \lambda_\gamma$ is bounded as $\lambda_{\gamma+1}$ is regular. We may then define $q_\gamma$ (on the domain $\lambda_{\gamma+1} - \lambda_\gamma$) by extending $p_\gamma$ such that $q_\gamma(\xi) = 1$ for enough ordinals above $\sup\dom(p_\gamma)$, followed by $q_\gamma(\xi) = 0$ for the remaining ordinals, such that $\otp(\lbrace \xi \in \lambda_{\gamma+1} - \lambda_\gamma : q(\xi) = 1\rbrace) = \lambda_\gamma + \alpha$. Then let $q := \bigsqcup_{\delta \neq \gamma} p_\delta \sqcup q_\gamma$, and we have that $q \supseteq p$ has the desired property.
This proves the result, as the genericity of $G$ implies that for all $\alpha$ there is a $q \in D_\alpha$ such that $q \subseteq f = \bigcup G$. Thus, for every $\alpha < \kappa$ there is a $\gamma < \cf(\kappa)$ such that $\otp(\lbrace \xi \in \lambda_{\gamma+1} - \lambda_\gamma : f(\xi) = 1\rbrace) = \lambda_\gamma + \alpha$. Then, in $V[G]$, $g$ is a well-defined injective map from $\kappa$ to $\cf(\kappa)$ (as sets).
For all ordinals $\alpha < \kappa$ and $p \in P$, there exists a $q \leq p$ ($q \supseteq p$) such that there is a $\gamma < \cf(\kappa)$ in which $\lambda_{\gamma+1} - \lambda_\gamma \subseteq \dom(q)$, and: $$ \begin{align*} \otp(\lbrace \xi \in \lambda_{\gamma+1} - \lambda_\gamma : q(\xi) = 1\rbrace) = \lambda_\gamma + \alpha \end{align*}$$ In other words, the set: $$ \begin{align*} D_\alpha &:= \lbrace p \in P : (\exists \gamma < \cf(\kappa))(\lambda_{\gamma+1} - \lambda_\gamma \subseteq \dom(p) \\ &\eqbreak \wedge \otp(\lbrace \xi \in \lambda_{\gamma+1} - \lambda_\gamma : q(\xi) = 1\rbrace) = \lambda_\gamma + \alpha)\rbrace \end{align*}$$ is dense in $P$.
Proof
We work in $V$. Fix $p \in P$, and let $p_\gamma := p\restrictedto(\lambda_{\gamma+1} - \lambda_\gamma)$. Then there exists a $\gamma$ such that $\vert \dom(p_\gamma)\vert < \lambda_\gamma$, for otherwise we have: $$ \begin{align*} \vert \dom(p)\vert = \sum_{\gamma < \cf(\kappa)} \vert \dom(p_\gamma)\vert = \sum_{\gamma < \cf(\kappa)} \lambda_\gamma = \kappa \end{align*}$$ contradicting that $\vert \dom(p)\vert < \kappa$. With this $\gamma$, we have that $\dom(p_\gamma) \subseteq \lambda_{\gamma+1} - \lambda_\gamma$ is bounded as $\lambda_{\gamma+1}$ is regular. We may then define $q_\gamma$ (on the domain $\lambda_{\gamma+1} - \lambda_\gamma$) by extending $p_\gamma$ such that $q_\gamma(\xi) = 1$ for enough ordinals above $\sup\dom(p_\gamma)$, followed by $q_\gamma(\xi) = 0$ for the remaining ordinals, such that $\otp(\lbrace \xi \in \lambda_{\gamma+1} - \lambda_\gamma : q(\xi) = 1\rbrace) = \lambda_\gamma + \alpha$. Then let $q := \bigsqcup_{\delta \neq \gamma} p_\delta \sqcup q_\gamma$, and we have that $q \supseteq p$ has the desired property.
$\blacksquare$
$\square$
Exercise 15.4
Solution
The proof of Exercise 15.3 may be repeated. Note that for the proof of the claim, we have furthermore that $p_\gamma = \emptyset$ for some $\gamma$ large enough.
The proof of Exercise 15.3 may be repeated. Note that for the proof of the claim, we have furthermore that $p_\gamma = \emptyset$ for some $\gamma$ large enough.
$\square$
Exercise 15.5
Solution
We will provide a solution that does not use the hint. We shall show that: $$ \begin{align*} W := \ss{\prod_{\alpha < \kappa} f(\alpha) : f \in \prod_{\alpha < \kappa} W_\alpha} \end{align*}$$ is a partition which is a common refinement of $\c{W_\alpha : \alpha < \kappa}$.
Antichain: Let $f, g \in \prod_{\alpha < \kappa} W_\alpha$ such that $f \neq g$. Then $f(\alpha) \neq g(\alpha)$ for some $\alpha$. Since $f(\alpha), g(\alpha) \in W_\alpha$ and $W_\alpha$ is an antichain, $f(\alpha) \cdot g(\alpha) = 0$. Thus $\bb{\prod_{\alpha < \kappa} f(\alpha)} \cdot \bb{\prod_{\alpha < \kappa} g(\alpha)} = 0$.
$\sum W = 1$: Let $w := \sum W$. It suffices to show that for all $u \neq 0$, $u \cdot w \neq 0$. Indeed, fix $u \in B$ and let $G$ be a generic filter such that $u \in G$. Let $W_\alpha' := \lbrace -u\rbrace \cup (\lbrace u \cdot w_\alpha : w \in W_\alpha\rbrace - \lbrace 0\rbrace)$. Then $W_\alpha'$ is a partition, and since $G$ is generic, $G \cap W_\alpha' \neq \emptyset$. Of course $-u \notin G$, so we have that $u_\alpha := u \cdot w_\alpha \in G$ for some $w_\alpha \in W_\alpha$. We let $f : \kappa \to V$ be defined by $f(\alpha) := u_\alpha$. By the premise, $f \in V$. Since $G$ is generic, we have that $v := \prod_{\alpha<\kappa} f(\alpha) \in G$. Then $u \cdot v = v \neq 0$, and clearly $v \in W$.
$W \leq W_\alpha$ for all $\alpha$: For all $f \in \prod_{\alpha<\kappa} W_\alpha$, $\prod_{\alpha<\kappa} f(\alpha) \leq f(\alpha) \in W_\alpha$.
We will provide a solution that does not use the hint. We shall show that: $$ \begin{align*} W := \ss{\prod_{\alpha < \kappa} f(\alpha) : f \in \prod_{\alpha < \kappa} W_\alpha} \end{align*}$$ is a partition which is a common refinement of $\c{W_\alpha : \alpha < \kappa}$.
Antichain: Let $f, g \in \prod_{\alpha < \kappa} W_\alpha$ such that $f \neq g$. Then $f(\alpha) \neq g(\alpha)$ for some $\alpha$. Since $f(\alpha), g(\alpha) \in W_\alpha$ and $W_\alpha$ is an antichain, $f(\alpha) \cdot g(\alpha) = 0$. Thus $\bb{\prod_{\alpha < \kappa} f(\alpha)} \cdot \bb{\prod_{\alpha < \kappa} g(\alpha)} = 0$.
$\sum W = 1$: Let $w := \sum W$. It suffices to show that for all $u \neq 0$, $u \cdot w \neq 0$. Indeed, fix $u \in B$ and let $G$ be a generic filter such that $u \in G$. Let $W_\alpha' := \lbrace -u\rbrace \cup (\lbrace u \cdot w_\alpha : w \in W_\alpha\rbrace - \lbrace 0\rbrace)$. Then $W_\alpha'$ is a partition, and since $G$ is generic, $G \cap W_\alpha' \neq \emptyset$. Of course $-u \notin G$, so we have that $u_\alpha := u \cdot w_\alpha \in G$ for some $w_\alpha \in W_\alpha$. We let $f : \kappa \to V$ be defined by $f(\alpha) := u_\alpha$. By the premise, $f \in V$. Since $G$ is generic, we have that $v := \prod_{\alpha<\kappa} f(\alpha) \in G$. Then $u \cdot v = v \neq 0$, and clearly $v \in W$.
$W \leq W_\alpha$ for all $\alpha$: For all $f \in \prod_{\alpha<\kappa} W_\alpha$, $\prod_{\alpha<\kappa} f(\alpha) \leq f(\alpha) \in W_\alpha$.
$\square$
Exercise 15.8
Solution
Let $G \subseteq P$ be a filter (generic or not). Let: $$ \begin{align*} D := \lbrace (p_1,p_2) \in P \times P : p_1 \perp p_2\rbrace \end{align*}$$ We see that $D$ is dense in $P \times P$ — indeed, let $(p_1,p_2) \in P \times P$. Let $q_i, r_i \leq p_i$ be incompatible. If $q_1 \perp q_2$, then $(q_1,q_2) \leq (p_1,p_2)$ and $(q_1,q_2) \in D$. Otherwise, we must have $r_1 \perp q_2$, so again $(r_1,q_2) \leq (p_1,p_2)$ and $(r_1,q_2) \in D$. But observe that $(G \times G) \cap D = \emptyset$, as if $(p_1,p_2) \in G \times G$ then $p_1$ and $p_2$ must be compatible (by the compatibility condition of filters).
Let $G \subseteq P$ be a filter (generic or not). Let: $$ \begin{align*} D := \lbrace (p_1,p_2) \in P \times P : p_1 \perp p_2\rbrace \end{align*}$$ We see that $D$ is dense in $P \times P$ — indeed, let $(p_1,p_2) \in P \times P$. Let $q_i, r_i \leq p_i$ be incompatible. If $q_1 \perp q_2$, then $(q_1,q_2) \leq (p_1,p_2)$ and $(q_1,q_2) \in D$. Otherwise, we must have $r_1 \perp q_2$, so again $(r_1,q_2) \leq (p_1,p_2)$ and $(r_1,q_2) \in D$. But observe that $(G \times G) \cap D = \emptyset$, as if $(p_1,p_2) \in G \times G$ then $p_1$ and $p_2$ must be compatible (by the compatibility condition of filters).
$\square$
Exercise 15.9
Lemma 15.9.A
If $P = \prod_i P_i$, then $B(P)$ is the completion of the direct sum of the algebras $B(P_i)$.
Proof
It suffices to show that $\prod_i B(P_i)$ embeds into $B(P)$ — indeed, from that since $B(P)$ is a complete algebra and $\prod_i P_i$ embeds densely into $B(P)$, then so does $\prod_i B(P_i)$. By the uniqueness of completion of Boolean algebras, $B\bb{\prod_i B(P_i)} = B(P)$.
We follow the construction of $B(P)$ in Theorem 7.13. Let $U \in \prod_i B(P_i)$, so $U = \prod_i U_i$, where $U_i$ is a regular cut in $P_i$, and $U_i \neq P_i$ on finitely many $i \in I$. We shall show that $U$ is a regular cut in $P$. Indeed, let $p = (p_i)_{i \in I} \notin U$. Then $p_i \notin U_i$ for some $i \in I$. Since $U_i$ is regular, there exists $p_i' \leq p_i$ such that $U_{p_i'} \cap U_i = \emptyset$. For $j \neq i$, let $p_j' := p_j$. Let $p' := (p_i')_{i \in I}$. Then $U_{p'} = \prod_i U_{p_i'}$ and $U_{p'} \cap U = \emptyset$, so $U$ is regular.
If $P = \prod_i P_i$, then $B(P)$ is the completion of the direct sum of the algebras $B(P_i)$.
Proof
It suffices to show that $\prod_i B(P_i)$ embeds into $B(P)$ — indeed, from that since $B(P)$ is a complete algebra and $\prod_i P_i$ embeds densely into $B(P)$, then so does $\prod_i B(P_i)$. By the uniqueness of completion of Boolean algebras, $B\bb{\prod_i B(P_i)} = B(P)$.
We follow the construction of $B(P)$ in Theorem 7.13. Let $U \in \prod_i B(P_i)$, so $U = \prod_i U_i$, where $U_i$ is a regular cut in $P_i$, and $U_i \neq P_i$ on finitely many $i \in I$. We shall show that $U$ is a regular cut in $P$. Indeed, let $p = (p_i)_{i \in I} \notin U$. Then $p_i \notin U_i$ for some $i \in I$. Since $U_i$ is regular, there exists $p_i' \leq p_i$ such that $U_{p_i'} \cap U_i = \emptyset$. For $j \neq i$, let $p_j' := p_j$. Let $p' := (p_i')_{i \in I}$. Then $U_{p'} = \prod_i U_{p_i'}$ and $U_{p'} \cap U = \emptyset$, so $U$ is regular.
$\blacksquare$
Solution
By Lemma 15.9.A, we have: $$ \begin{align*} B(P) = B\bb{\prod_i P_i} = B\bb{\prod_i B(P_i)} = B\bb{\prod_i B(Q_i)} = B\bb{\prod_i Q_i} = B(Q) \end{align*}$$
By Lemma 15.9.A, we have: $$ \begin{align*} B(P) = B\bb{\prod_i P_i} = B\bb{\prod_i B(P_i)} = B\bb{\prod_i B(Q_i)} = B\bb{\prod_i Q_i} = B(Q) \end{align*}$$
$\square$
Exercise 15.10
Solution
For $\alpha < \kappa$ let each $P_\alpha$ be the notion of forcing (14.2) adding a single Cohen real (for $\alpha \neq \beta$, $P_\alpha = P_\beta$). Define $F : \prod_\alpha P_\alpha \to P$ such that for $p_\alpha \in P_\alpha$, we have: $$ \begin{align*} F\bb{(p_\alpha)_{\alpha < \kappa}}(\beta,n) := p_\beta(n) \end{align*}$$ Clearly for all $p \in \prod_\alpha P_\alpha$, $\ran(p) \subseteq \lbrace 0,1\rbrace$. Furthermore, $\dom(p)$ is finite as $\prod_\alpha P_\alpha$ is of finite support, so all but finitely many $\alpha$'s are such that $p_\alpha = \emptyset$.
$F$ is invertible, with the inverse $p \mapsto \prod_\alpha p(\alpha,-)$ (which is well-defined as $\dom(p)$ is finite). Finally, if $p, q \in \prod_\alpha P_\alpha$ and $p \leq q$, then $p_\alpha \supseteq q_\alpha$ for all $\alpha$, so $F(p) \supseteq F(q)$.
For $\alpha < \kappa$ let each $P_\alpha$ be the notion of forcing (14.2) adding a single Cohen real (for $\alpha \neq \beta$, $P_\alpha = P_\beta$). Define $F : \prod_\alpha P_\alpha \to P$ such that for $p_\alpha \in P_\alpha$, we have: $$ \begin{align*} F\bb{(p_\alpha)_{\alpha < \kappa}}(\beta,n) := p_\beta(n) \end{align*}$$ Clearly for all $p \in \prod_\alpha P_\alpha$, $\ran(p) \subseteq \lbrace 0,1\rbrace$. Furthermore, $\dom(p)$ is finite as $\prod_\alpha P_\alpha$ is of finite support, so all but finitely many $\alpha$'s are such that $p_\alpha = \emptyset$.
$F$ is invertible, with the inverse $p \mapsto \prod_\alpha p(\alpha,-)$ (which is well-defined as $\dom(p)$ is finite). Finally, if $p, q \in \prod_\alpha P_\alpha$ and $p \leq q$, then $p_\alpha \supseteq q_\alpha$ for all $\alpha$, so $F(p) \supseteq F(q)$.
$\square$
Exercise 15.11
Solution
Let $A \subseteq P \times Q$ be an antichain. Let $B := \lbrace q \in Q : (\exists p \in P) \, (p,q) \in A\rbrace$. We consider two cases.
Let $A \subseteq P \times Q$ be an antichain. Let $B := \lbrace q \in Q : (\exists p \in P) \, (p,q) \in A\rbrace$. We consider two cases.
- Suppose $B$ is countable. For each $q \in B$, let $A_q := \lbrace p \in P : (p,q) \in A\rbrace$. Since $A$ is an antichain, for two conditions $(p,q)$, $(p',q)$ in $A_q$, $(p,q) \perp (p',q)$ so $p \perp p'$. Thus $A_q \subseteq P$ is an antichain. Since $P$ satisfies c.c.c., $A_q$ must be countable. Thus $A = \bigcup_{q \in B} A_q \times \lbrace q\rbrace$ is countable.
- Suppose $B$ is uncountable. Since $Q$ has property (K), there exists an uncountable $B' \subseteq B$ such that elements in $B'$ are pairwise compatible. Now for $q_1, q_2 \in B'$, if $(p_1,q_1) \in A$ and $(p_2,q_2) \in A$ then we must have $p_1 \perp p_2$, for otherwise $(p_1,q_1) \not\perp (p_2,q_2)$. This implies that $A' := \lbrace p \in P : (\exists q \in B') \, (p,q) \in A\rbrace$ is an uncountable antichain, contradicting that $P$ satisfies c.c.c.
$\square$
Exercise 15.12
Solution
We elaborate on each identity used in the hint. Suppose $V[G] \models 2^{\cf(\kappa)} < \kappa$, i.e. $F(\cf(\kappa)) < \kappa$.
We elaborate on each identity used in the hint. Suppose $V[G] \models 2^{\cf(\kappa)} < \kappa$, i.e. $F(\cf(\kappa)) < \kappa$.
- $(\kappa^{\cf(\kappa)})^{V[G]} = (\kappa^{\cf(\kappa)})^N$ follows from Lemma 15.19.
- $(\kappa^{\cf(\kappa)})^N \leq (2^\kappa)^N$ follows from simple cardinal arithmetic: $\kappa^{\cf(\kappa)} \leq (2^\kappa)^{\cf(\kappa)} = 2^{\kappa \cdot \cf(\kappa)} = 2^\kappa$.
- $(2^\kappa)^N \leq \vert B(P^{\leq\cf(\kappa)})\vert ^\kappa$ follows from Lemma 15.1.
- To show that $\vert B(P^{\leq\cf(\kappa)})\vert ^\kappa = (F(\cf(\kappa)))^\kappa$, we first show that $\vert P^{\leq\cf(\kappa)}\vert = F(\cf(\kappa))$. Given $\alpha < \cf(\kappa)$, we have $((\alpha,0,0),0) \in P^{\leq\cf(\kappa)}$, so we have that $\vert P^{\leq\cf(\kappa)}\vert \geq F(\cf(\kappa))$. On the other hand: $$ \begin{align*} \vert P^{\leq\cf(\kappa)}\vert &\leq \vert \lbrace (\lambda,\alpha,\beta,n) : \lambda \leq \cf(\kappa), \alpha < \lambda, \beta < F(\lambda), n \in \lbrace 0,1\rbrace\rbrace\vert \\ &\leq \sup_{\lambda < \cf(\kappa)} \lambda \cdot \lambda \cdot F(\lambda) \cdot 2 \\ &= \sup_{\lambda < \cf(\kappa)} F(\lambda) \\ &\leq F(\cf(\kappa)) \end{align*}$$ Thus $\vert P^{\leq\cf(\kappa)}\vert = F(\cf(\kappa))$. By Exercise 15.1, since $P^{\leq\cf(\kappa)}$ satisfies the $(\cf(\kappa))^+$-chain condition: $$ \begin{align*} \vert B(P^{\leq\cf(\kappa)})\vert \leq \vert P^{\leq\cf(\kappa)}\vert ^{<(\cf(\kappa))^+} = \vert P^{\leq\cf(\kappa)}\vert ^{\cf(\kappa)} \end{align*}$$ Thus: $$ \begin{align*} \vert P^{\leq\cf(\kappa)}\vert ^\kappa \leq \vert B(P^{\leq\cf(\kappa)})\vert ^\kappa \leq (\vert P^{\leq\cf(\kappa)}\vert ^{\cf(\kappa)})^\kappa = \vert P^{\leq\cf(\kappa)}\vert ^\kappa \end{align*}$$ so $\vert B(P^{\leq\cf(\kappa)})\vert ^\kappa = (F(\cf(\kappa)))^\kappa$.
- Since $F(\cf(\kappa)) < \kappa$, we have that $(F(\cf(\kappa)))^\kappa = 2^\kappa = \kappa^+$, since the ground model satisfies $\GCH$.
- Finally, $\kappa^+ = (\kappa^+)^{V[G]}$ as all cardinals and cofinalities are preserved.
$\square$
Exercise 15.13
Solution
Let $P$ denote the forcing notion (15.18). As explained in the paragraph below (15.18), in $V[G]$ we have that $f$ is a function on $\aleph_1$ onto $\aleph_\omega$. Note that $-$ here denotes set exclusion, so $\alpha + \omega - \alpha = \lbrace \alpha + n : n \in \omega\rbrace$.
Let $P$ denote the forcing notion (15.18). As explained in the paragraph below (15.18), in $V[G]$ we have that $f$ is a function on $\aleph_1$ onto $\aleph_\omega$. Note that $-$ here denotes set exclusion, so $\alpha + \omega - \alpha = \lbrace \alpha + n : n \in \omega\rbrace$.
Claim
There exists an ordinal $\alpha$ such that $f``(\alpha + \omega - \alpha) = X$.
Proof
Let: $$ \begin{align*} D := \lbrace p \in P : (\exists \alpha \in \omega_1)(\alpha + \omega - \alpha \subseteq \dom(p) \wedge p``(\alpha + \omega - \alpha) = X)\rbrace \end{align*}$$ We shall show that $D$ is dense in $P$. Indeed, let $p \in P$. Since $\dom(p)$ is countable and $\aleph_1$ is regular, there exists a limit ordinal $\alpha > \sup\dom(p)$. Since $P$ is $\aleph_0$-closed, no countable subsets of $\aleph_\omega$ are added. Thus, in $V$ there exists a one-to-one $h$ on $\omega$ onto $X$. Then let: $$ \begin{align*} q := p \cup \lbrace (\alpha + n, h(n)) : n \in \omega\rbrace \end{align*}$$ Then clearly $q \leq p$ and $q \in D$, as desired. Since $G$ is generic, $G \cap D \neq \emptyset$. So $f = \bigcup G$ must have some $\alpha$ such that $f(\alpha + \omega - \alpha) = X$.
Thus, the function $g$ in the hint is well-defined in $V[G]$. Clearly it is one-to-one.
There exists an ordinal $\alpha$ such that $f``(\alpha + \omega - \alpha) = X$.
Proof
Let: $$ \begin{align*} D := \lbrace p \in P : (\exists \alpha \in \omega_1)(\alpha + \omega - \alpha \subseteq \dom(p) \wedge p``(\alpha + \omega - \alpha) = X)\rbrace \end{align*}$$ We shall show that $D$ is dense in $P$. Indeed, let $p \in P$. Since $\dom(p)$ is countable and $\aleph_1$ is regular, there exists a limit ordinal $\alpha > \sup\dom(p)$. Since $P$ is $\aleph_0$-closed, no countable subsets of $\aleph_\omega$ are added. Thus, in $V$ there exists a one-to-one $h$ on $\omega$ onto $X$. Then let: $$ \begin{align*} q := p \cup \lbrace (\alpha + n, h(n)) : n \in \omega\rbrace \end{align*}$$ Then clearly $q \leq p$ and $q \in D$, as desired. Since $G$ is generic, $G \cap D \neq \emptyset$. So $f = \bigcup G$ must have some $\alpha$ such that $f(\alpha + \omega - \alpha) = X$.
$\blacksquare$
$\square$
Exercise 15.14
Solution
Let $P$ denote the forcing notion (15.18).
Let $P$ denote the forcing notion (15.18).
Claim
For each $\alpha < \lambda$, there is an $n \in \omega$ such that the function $f\restrictedto(\omega_{n+1} - \omega_n)$ is eventually equal to $\alpha$.
Proof
Fix $\alpha < \lambda$. Let: $$ \begin{align*} D := \lbrace p \in P : \, &(\exists n \in \omega)(\exists \beta \in \omega_{n+1} - \omega_n) \\ &(\omega_{n+1} - \beta \subseteq \dom(p) \wedge \text{$p\restrictedto(\omega_{n+1} - \beta)$ is constantly $\alpha$})\rbrace \end{align*}$$ We shall show that $D$ is dense in $P$. For each $p \in P$ let: $$ \begin{align*} \alpha_n := \sup\dom(p\restrictedto(\omega_{n+1} - \omega_n)) \end{align*}$$ Then $\alpha_n \leq \omega_{n+1}$. We cannot have $\alpha_n = \omega_{n+1}$ on an unbounded subset of $\omega$ — otherwise, since $\aleph_n$ is regular for all $n$, we have that $\vert \dom(p\restrictedto(\omega_{n+1} - \omega_n))\vert = \aleph_{n+1}$ for each $n$ such that $\alpha_n = \omega_{n+1}$. So: $$ \begin{align*} \vert \dom(p)\vert = \sum_{n \in \omega} \mod{\dom\bb{p\restrictedto(\omega_{n+1} - \omega_n)}} = \sum_{n \in \omega} \aleph_n = \aleph_\omega \end{align*}$$ which contradicts that $\vert \dom(p)\vert < \aleph_\omega$. Therefore, fix any $n$ such that $\alpha_n < \omega_{n+1}$. Let: $$ \begin{align*} q := p \cup \lbrace (\beta,\alpha) : \beta \in \omega_{n+1} - \alpha_n\rbrace \end{align*}$$ Clearly $q \leq p$, and $q$ is eventually constant with value $\alpha$. Furthermore: $$ \begin{align*} \vert \dom(q)\vert \leq \vert \dom(p)\vert + \aleph_{n+1} < \aleph_\omega \end{align*}$$ so $q \in P$, and therefore $q \in D$. Hence $D$ is dense.
Thus, the function $g$ in the hint is well-defined. Clearly $g$ is one-to-one.
For each $\alpha < \lambda$, there is an $n \in \omega$ such that the function $f\restrictedto(\omega_{n+1} - \omega_n)$ is eventually equal to $\alpha$.
Proof
Fix $\alpha < \lambda$. Let: $$ \begin{align*} D := \lbrace p \in P : \, &(\exists n \in \omega)(\exists \beta \in \omega_{n+1} - \omega_n) \\ &(\omega_{n+1} - \beta \subseteq \dom(p) \wedge \text{$p\restrictedto(\omega_{n+1} - \beta)$ is constantly $\alpha$})\rbrace \end{align*}$$ We shall show that $D$ is dense in $P$. For each $p \in P$ let: $$ \begin{align*} \alpha_n := \sup\dom(p\restrictedto(\omega_{n+1} - \omega_n)) \end{align*}$$ Then $\alpha_n \leq \omega_{n+1}$. We cannot have $\alpha_n = \omega_{n+1}$ on an unbounded subset of $\omega$ — otherwise, since $\aleph_n$ is regular for all $n$, we have that $\vert \dom(p\restrictedto(\omega_{n+1} - \omega_n))\vert = \aleph_{n+1}$ for each $n$ such that $\alpha_n = \omega_{n+1}$. So: $$ \begin{align*} \vert \dom(p)\vert = \sum_{n \in \omega} \mod{\dom\bb{p\restrictedto(\omega_{n+1} - \omega_n)}} = \sum_{n \in \omega} \aleph_n = \aleph_\omega \end{align*}$$ which contradicts that $\vert \dom(p)\vert < \aleph_\omega$. Therefore, fix any $n$ such that $\alpha_n < \omega_{n+1}$. Let: $$ \begin{align*} q := p \cup \lbrace (\beta,\alpha) : \beta \in \omega_{n+1} - \alpha_n\rbrace \end{align*}$$ Clearly $q \leq p$, and $q$ is eventually constant with value $\alpha$. Furthermore: $$ \begin{align*} \vert \dom(q)\vert \leq \vert \dom(p)\vert + \aleph_{n+1} < \aleph_\omega \end{align*}$$ so $q \in P$, and therefore $q \in D$. Hence $D$ is dense.
$\blacksquare$
$\square$
Exercise 15.15
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 15.16
Solution
The first hint leads to a solution, but the second only explains why $P$ "does the job of $Q$". The proof of the second hint has working very similar to Exercise 15.13, so in this solution we expand on the first hint only.
To see that $Q'$ in the first hint is dense, let $q \in Q$. Define $q'$ such that for $\alpha < \sup\dom(q)$: $$ \begin{align*} q'(\alpha) := \begin{cases} q(\alpha), &\text{if $\alpha \in \dom(q)$} \\ 0, &\text{if $\alpha \notin \dom(q)$} \end{cases} \end{align*}$$ Clearly $q' \in Q$, and $\dom(q') = \sup\dom(q)$, an initial segment of $\omega_1$. Let $P'$ be the set of all $p \in P$ such that $\dom(p)$ is a limit ordinal. By a similar reasoning as above, $P'$ is indeed dense in $P$. We shall recursively construct an isomorphism $\pi : Q' \to P'$, where $P' \subseteq P$ is dense.
Start by fixing a one-to-one and onto enumeration $h : 2^{\aleph_0} \to \lbrace 0,1\rbrace^\omega$. Let $\pi(\emptyset) := \emptyset$. Now suppose $\pi(q)$ is defined for all $\dom(q) = \alpha$. Let $q \in Q'$ such that $\dom(q) = \alpha + 1$. Define: $$ \begin{align*} \pi(q) := \pi(q\restrictedto\alpha) \cup \lbrace (\dom(\pi(q\restrictedto\alpha)) + n, h(q(\alpha))(n)) : n \in \omega\rbrace \end{align*}$$ Then $\dom(\pi(q))$ is a partial function into $\lbrace 0,1\rbrace$ with domain $\dom(\pi(q\restrictedto\alpha)) + \omega$, which is a limit ordinal. Thus we indeed have $\pi(q) \in P'$. Finally, if $\alpha$ is a limit ordinal, we define $\pi(q) := \bigcup_{\beta<\alpha} \pi(q\restrictedto\beta)$. Again, $\dom(\pi(q))$ is clearly a limit ordinal.
We now show that $\pi$ is an isomorphism.
$\pi$ respects $<$: From the construction of $\pi$, it's easy to see that if $q_1 \subseteq q_2$ then $\pi(q_1) \subseteq \pi(q_2)$. Thus $q_1 < q_2 \implies \pi(q_1) < \pi(q_2)$. On the other hand, suppose $q_1 \perp q_2$. Let $\alpha$ be the least ordinal such that $q_1(\alpha) \neq q_2(\alpha)$. Since $h$ is one-to-one, there exists an $n$ such that $h(q_1(\alpha))(n) \neq h(q_2(\alpha))(n)$, so $\pi(q_1\restrictedto(\alpha+1)) \perp \pi(q_2\restrictedto(\alpha+1))$. Since $\pi$ respects $<$, we have that $\pi(q_1) \perp \pi(q_2)$.
$\pi$ is one-to-one: It's easy to see from the construction that $q_1 \subsetneq q_2 \implies \pi(q_1) \subsetneq \pi(q_2)$, so $\pi$ is one-to-one.
$\pi$ is onto: Let $\alpha$ be a limit ordinal below $\omega_1$. Let $p \in P$ such that $\dom(p) = \alpha$. We induct on $\alpha$. The base case $\alpha = 0$ is clear. If $\alpha = \beta + \omega$ for some limit ordinal $\beta$, then by induction hypothesis there exists some $q \in Q'$ such that $\pi(q) = p\restrictedto\beta$. One may check that: $$ \begin{align*} p = \pi\bb{q \cup \lbrace (\dom(q)+1, h^{-1}(\lbrace p(\beta+n) : n \in \omega\rbrace))\rbrace} \end{align*}$$ If $\alpha \neq \beta + \omega$ for all ordinals $\beta$, then: $$ \begin{align*} p = \bigcup_{\substack{\beta < \alpha \\ \beta \text{ is limit}}} p\restrictedto(\beta + \omega) \end{align*}$$ so if $\pi(q_\beta) = p\restrictedto(\beta + \omega)$, then let $q := \bigcup_{\beta<\alpha} q_\beta$ and we have that $\pi(q) = p$.
The first hint leads to a solution, but the second only explains why $P$ "does the job of $Q$". The proof of the second hint has working very similar to Exercise 15.13, so in this solution we expand on the first hint only.
To see that $Q'$ in the first hint is dense, let $q \in Q$. Define $q'$ such that for $\alpha < \sup\dom(q)$: $$ \begin{align*} q'(\alpha) := \begin{cases} q(\alpha), &\text{if $\alpha \in \dom(q)$} \\ 0, &\text{if $\alpha \notin \dom(q)$} \end{cases} \end{align*}$$ Clearly $q' \in Q$, and $\dom(q') = \sup\dom(q)$, an initial segment of $\omega_1$. Let $P'$ be the set of all $p \in P$ such that $\dom(p)$ is a limit ordinal. By a similar reasoning as above, $P'$ is indeed dense in $P$. We shall recursively construct an isomorphism $\pi : Q' \to P'$, where $P' \subseteq P$ is dense.
Start by fixing a one-to-one and onto enumeration $h : 2^{\aleph_0} \to \lbrace 0,1\rbrace^\omega$. Let $\pi(\emptyset) := \emptyset$. Now suppose $\pi(q)$ is defined for all $\dom(q) = \alpha$. Let $q \in Q'$ such that $\dom(q) = \alpha + 1$. Define: $$ \begin{align*} \pi(q) := \pi(q\restrictedto\alpha) \cup \lbrace (\dom(\pi(q\restrictedto\alpha)) + n, h(q(\alpha))(n)) : n \in \omega\rbrace \end{align*}$$ Then $\dom(\pi(q))$ is a partial function into $\lbrace 0,1\rbrace$ with domain $\dom(\pi(q\restrictedto\alpha)) + \omega$, which is a limit ordinal. Thus we indeed have $\pi(q) \in P'$. Finally, if $\alpha$ is a limit ordinal, we define $\pi(q) := \bigcup_{\beta<\alpha} \pi(q\restrictedto\beta)$. Again, $\dom(\pi(q))$ is clearly a limit ordinal.
We now show that $\pi$ is an isomorphism.
$\pi$ respects $<$: From the construction of $\pi$, it's easy to see that if $q_1 \subseteq q_2$ then $\pi(q_1) \subseteq \pi(q_2)$. Thus $q_1 < q_2 \implies \pi(q_1) < \pi(q_2)$. On the other hand, suppose $q_1 \perp q_2$. Let $\alpha$ be the least ordinal such that $q_1(\alpha) \neq q_2(\alpha)$. Since $h$ is one-to-one, there exists an $n$ such that $h(q_1(\alpha))(n) \neq h(q_2(\alpha))(n)$, so $\pi(q_1\restrictedto(\alpha+1)) \perp \pi(q_2\restrictedto(\alpha+1))$. Since $\pi$ respects $<$, we have that $\pi(q_1) \perp \pi(q_2)$.
$\pi$ is one-to-one: It's easy to see from the construction that $q_1 \subsetneq q_2 \implies \pi(q_1) \subsetneq \pi(q_2)$, so $\pi$ is one-to-one.
$\pi$ is onto: Let $\alpha$ be a limit ordinal below $\omega_1$. Let $p \in P$ such that $\dom(p) = \alpha$. We induct on $\alpha$. The base case $\alpha = 0$ is clear. If $\alpha = \beta + \omega$ for some limit ordinal $\beta$, then by induction hypothesis there exists some $q \in Q'$ such that $\pi(q) = p\restrictedto\beta$. One may check that: $$ \begin{align*} p = \pi\bb{q \cup \lbrace (\dom(q)+1, h^{-1}(\lbrace p(\beta+n) : n \in \omega\rbrace))\rbrace} \end{align*}$$ If $\alpha \neq \beta + \omega$ for all ordinals $\beta$, then: $$ \begin{align*} p = \bigcup_{\substack{\beta < \alpha \\ \beta \text{ is limit}}} p\restrictedto(\beta + \omega) \end{align*}$$ so if $\pi(q_\beta) = p\restrictedto(\beta + \omega)$, then let $q := \bigcup_{\beta<\alpha} q_\beta$ and we have that $\pi(q) = p$.
$\square$
Exercise 15.17
Solution
Call a tree $T$ an $\alpha$-tree if it is a countable normal tree (as in (15.20)) such that $\height(T) = \alpha$. Let $Q' := \lbrace T \in Q : \height(T) \text{ is a limit ordinal}\rbrace$, and clearly it is dense in $Q$. Let $R$ be the notion of forcing that collapses $2^{\aleph_0}$ to $\aleph_1$, and let $R' \subseteq R$ be the set of all $r \in R$ such that $\dom(r)$ is an initial segment of $\omega_1$. We shall show that $R'$ and $Q'$ are isomorphic, so by Exercise 15.16 we have that $B(P) = B(R) = B(Q)$.
Call a tree $T$ an $\alpha$-tree if it is a countable normal tree (as in (15.20)) such that $\height(T) = \alpha$. Let $Q' := \lbrace T \in Q : \height(T) \text{ is a limit ordinal}\rbrace$, and clearly it is dense in $Q$. Let $R$ be the notion of forcing that collapses $2^{\aleph_0}$ to $\aleph_1$, and let $R' \subseteq R$ be the set of all $r \in R$ such that $\dom(r)$ is an initial segment of $\omega_1$. We shall show that $R'$ and $Q'$ are isomorphic, so by Exercise 15.16 we have that $B(P) = B(R) = B(Q)$.
Claim
Let $\alpha$ be a limit ordinal, and let $T$ be an $\alpha$-tree. Then there are exactly $2^{\aleph_0}$ many $(\alpha + \omega)$-trees $T'$ such that $T' < T$.
Proof
Fix an $\alpha$-tree $T$. By (iii) of (15.20), if $T'$ is an $(\alpha + \omega)$-tree and $T' < T$, then $T'$ is completely determined by its $\alpha^\text{th}$ level. Thus, it suffices to determine the cardinality of possible $\alpha^\text{th}$ levels of $T'$.
Since $\alpha$ is countable, there are $2^{\aleph_0}$ many functions $t : \alpha \to \omega$. The $(\alpha + 1)^\text{th}$ level is a countable set of functions $t : \alpha \to \omega$, so there are at most $(2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0}$ many such countable sets.
Conversely, by Lemma 9.21 there exists an almost disjoint family $A$ of functions $t : \alpha \to \omega$ (i.e. for $t_1, t_2 \in A$, $\lbrace \beta : t_1(\beta) = t_2(\beta)\rbrace$ is finite) of size $2^{\aleph_0}$. For each $t \in A$, define an $(\alpha + 1)^\text{th}$ level $L_\alpha^{(t)}$ by: $$ \begin{align*} L_\alpha^{(t)} := \lbrace t' \cup t\restrictedto(\alpha - \dom(t')) : t' \in T\rbrace \end{align*}$$ Since $T$ is countable, so is $L_\alpha^{(t)}$. Let $T_t < T$ be the $(\alpha + \omega)$-tree such that the $\alpha^\text{th}$ level of $T_t$ is $L_\alpha^{(t)}$. Then clearly if $t_1, t_2 \in A$ and $t_1 \neq t_2$, then $T_{t_1} \neq T_{t_2}$. Thus we constructed $2^{\aleph_0}$ many different $(\alpha + \omega)$-trees stronger than $T$.
Let $A$ be the almost disjoint family in the claim, and let $h : 2^{\aleph_0} \to A$ be one-to-one and onto. We may define an isomorphism $\pi : R' \to Q'$ recursively, like in Exercise 15.16, as follows: Let $\pi(\emptyset) = \lbrace \emptyset\rbrace$. If $\pi(q) = T$, let $\pi(q \cup \lbrace \alpha\rbrace) := T_{h(\alpha)}$, as defined in the claim. Finally, if $\dom(q)$ is a limit ordinal, let $\pi(q) := \bigcup_{\beta < \dom(q)} \pi(q\restrictedto \beta)$. One may proceed as in Exercise 15.16 to show that $\pi$ is indeed an isomorphism.
Let $\alpha$ be a limit ordinal, and let $T$ be an $\alpha$-tree. Then there are exactly $2^{\aleph_0}$ many $(\alpha + \omega)$-trees $T'$ such that $T' < T$.
Proof
Fix an $\alpha$-tree $T$. By (iii) of (15.20), if $T'$ is an $(\alpha + \omega)$-tree and $T' < T$, then $T'$ is completely determined by its $\alpha^\text{th}$ level. Thus, it suffices to determine the cardinality of possible $\alpha^\text{th}$ levels of $T'$.
Since $\alpha$ is countable, there are $2^{\aleph_0}$ many functions $t : \alpha \to \omega$. The $(\alpha + 1)^\text{th}$ level is a countable set of functions $t : \alpha \to \omega$, so there are at most $(2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0}$ many such countable sets.
Conversely, by Lemma 9.21 there exists an almost disjoint family $A$ of functions $t : \alpha \to \omega$ (i.e. for $t_1, t_2 \in A$, $\lbrace \beta : t_1(\beta) = t_2(\beta)\rbrace$ is finite) of size $2^{\aleph_0}$. For each $t \in A$, define an $(\alpha + 1)^\text{th}$ level $L_\alpha^{(t)}$ by: $$ \begin{align*} L_\alpha^{(t)} := \lbrace t' \cup t\restrictedto(\alpha - \dom(t')) : t' \in T\rbrace \end{align*}$$ Since $T$ is countable, so is $L_\alpha^{(t)}$. Let $T_t < T$ be the $(\alpha + \omega)$-tree such that the $\alpha^\text{th}$ level of $T_t$ is $L_\alpha^{(t)}$. Then clearly if $t_1, t_2 \in A$ and $t_1 \neq t_2$, then $T_{t_1} \neq T_{t_2}$. Thus we constructed $2^{\aleph_0}$ many different $(\alpha + \omega)$-trees stronger than $T$.
$\blacksquare$
$\square$
Exercise 15.18
Solution
As in the hint, if $\pi$ is a non-trivial automorphism of $T$ and $\pi$ is extended to an automorphism of $T' < T$ with $\height(T') = \alpha + 1$, then as $\pi$ preserves $<$, if $b$ is extended then so must $\pi(b)$.
Let $b$ be a branch (i.e. $b : \alpha \to \omega$, and $b\restrictedto\beta \in T$ for all $\beta < \alpha$) in $T$ such that $\pi(b) \neq b$. Let $c : \alpha \to \omega$ be defined as $c(\beta) := \pi(b)(\beta) + 1$. Let: $$ \begin{align*} X := \lbrace t \cup c\restrictedto(\alpha - \dom(t)) : t \in T\rbrace \end{align*}$$ Let $T' := T \cup X \cup \lbrace b\rbrace$. Since $T$ is countable, $X$ is countable so $T'$ is a normal $(\alpha + 1)$-tree. Clearly $b$ is extended in $T'$, but we observe that $\pi(b)$ is not — given any branch $d \in T'$, we have that either $d = b$, or $d(\beta) > \pi(b)(\beta)$ for sufficiently large $\beta < \alpha$, so $d \neq \pi(b)$.
As in the hint, if $\pi$ is a non-trivial automorphism of $T$ and $\pi$ is extended to an automorphism of $T' < T$ with $\height(T') = \alpha + 1$, then as $\pi$ preserves $<$, if $b$ is extended then so must $\pi(b)$.
Let $b$ be a branch (i.e. $b : \alpha \to \omega$, and $b\restrictedto\beta \in T$ for all $\beta < \alpha$) in $T$ such that $\pi(b) \neq b$. Let $c : \alpha \to \omega$ be defined as $c(\beta) := \pi(b)(\beta) + 1$. Let: $$ \begin{align*} X := \lbrace t \cup c\restrictedto(\alpha - \dom(t)) : t \in T\rbrace \end{align*}$$ Let $T' := T \cup X \cup \lbrace b\rbrace$. Since $T$ is countable, $X$ is countable so $T'$ is a normal $(\alpha + 1)$-tree. Clearly $b$ is extended in $T'$, but we observe that $\pi(b)$ is not — given any branch $d \in T'$, we have that either $d = b$, or $d(\beta) > \pi(b)(\beta)$ for sufficiently large $\beta < \alpha$, so $d \neq \pi(b)$.
$\square$
Exercise 15.19
Solution
The hint basically solves the problem: Let $T \in G$ such that $T \forces \dot{\rho}$ is a nontrivial automorphism of $\mathcal{T}$. We may choose $T$ strong enough so that, in $V[G]$, $\pi := \rho\restrictedto T$ is a non-trivial automorphism. Then $T \subseteq \mathcal{T}$, and the set in the hint is dense below $T$ by Exercise 15.18 (call this set $D$). By genericity of $G$, there exists some $T' \leq T$ in $G$ such that $\pi$ cannot be extended to a non-trivial automorphism of $T'$. This is a contradiction, as we have $T' \subseteq \mathcal{T}$ so $\rho\restrictedto T'$ extends $\pi$, and is a non-trivial automorphism of $T'$.
The hint basically solves the problem: Let $T \in G$ such that $T \forces \dot{\rho}$ is a nontrivial automorphism of $\mathcal{T}$. We may choose $T$ strong enough so that, in $V[G]$, $\pi := \rho\restrictedto T$ is a non-trivial automorphism. Then $T \subseteq \mathcal{T}$, and the set in the hint is dense below $T$ by Exercise 15.18 (call this set $D$). By genericity of $G$, there exists some $T' \leq T$ in $G$ such that $\pi$ cannot be extended to a non-trivial automorphism of $T'$. This is a contradiction, as we have $T' \subseteq \mathcal{T}$ so $\rho\restrictedto T'$ extends $\pi$, and is a non-trivial automorphism of $T'$.
$\square$
Exercise 15.20
Solution
Let $(P,<)$ denote the partial order of all normal countable trees with the property in the hint. We first show that there exist such trees of arbitrarily tall heights. Let $T$ be an $\alpha$-tree, where $\alpha$ is a limit ordinal, satisfying (vi) and (vii). For $t \in T$, define: $$ \begin{align*} \ext(t) := t \cup \lbrace (\beta,0) : \beta \in \alpha - \dom(t)\rbrace \end{align*}$$ Let $X := \lbrace \ext(t) : t \in T\rbrace$, and let $T' := T \cup X$. Then $T' < T$ and $T'$ is of height $\alpha + 1$. We now show that $T'$ satisfies (vi) and (vii).
We now show that the Suslin tree $\mathcal{T}$ is homogeneous. Let $x, y \in \mathcal{T}$, so there exists some $T \in G$ such that $x, y \in T$ (say $\dom(x) = \alpha$). $T$ is homogeneous by the above claim, so there exists an automorphism $\pi : T \to T$ such that $\pi(x) = y$. We then extend $\pi$ to $\pi' : \mathcal{T} \to \mathcal{T}$ by stipulating that for $t \in \mathcal{T}$, $\pi'(t) := \pi(t\restrictedto\alpha) \cup (t - t\restrictedto\alpha)$, and clearly $\pi'$ remains an automorphism. Thus $\mathcal{T}$ is homogeneous.
Let $(P,<)$ denote the partial order of all normal countable trees with the property in the hint. We first show that there exist such trees of arbitrarily tall heights. Let $T$ be an $\alpha$-tree, where $\alpha$ is a limit ordinal, satisfying (vi) and (vii). For $t \in T$, define: $$ \begin{align*} \ext(t) := t \cup \lbrace (\beta,0) : \beta \in \alpha - \dom(t)\rbrace \end{align*}$$ Let $X := \lbrace \ext(t) : t \in T\rbrace$, and let $T' := T \cup X$. Then $T' < T$ and $T'$ is of height $\alpha + 1$. We now show that $T'$ satisfies (vi) and (vii).
- (vi) Let $b \in X$, and let $b'$ be such that $b \sim b'$. Let $\beta < \alpha$ be such that for all $\beta \leq \gamma < \alpha$, $b'(\gamma) = 0$ (such a $\beta$ exists as $b \sim b'$, and $b$ is eventually zero). Let $t' := b'\restrictedto\beta$. By (iii) of (15.20), we may choose $\gamma$ and $t \in T$ such that $b = \ext(t)$ and $\dom(t) = \gamma$. Then since $b \sim b'$, $t \sim t'$ so $t' \in T$ as $T$ satisfies (vi). Then $b' = \ext(t') \in X$.
- (vii) Let $b, b' \in X$, so there exist $t, t'$ such that $b = \ext(t)$ and $b' = \ext(t')$. WLOG assume $\dom(t) \geq \dom(t')$. By (iii) of (15.20), there exists a $t'' \geq t'$ such that $\dom(t) = \dom(t'')$ and $\ext(t') = \ext(t'')$. Then $t, t''$ are at the same level, and since $T$ satisfies (vii), $t \sim t''$. Since $t(\beta) = t''(\beta) = 0$ for $\beta \in \alpha - \dom(t)$, $b \sim b'$.
Claim
Every $T \in P$ is homogeneous.
Proof
Let $x, y \in T$ be at the same level. Then $x \sim y$, so let $\alpha_1,\dots,\alpha_n \in \dom(x)$ be all the ordinals such that $x(\alpha_i) \neq y(\alpha_i)$. Define $\pi : T \to T$ such that for $t \in T$ and $\beta \in \dom(t)$: $$ \begin{align*} \pi(t)(\beta) := \begin{cases} y(\beta), &\text{if $\beta = \alpha_i$ and $t(\beta) = x(\beta)$} \\ x(\beta), &\text{if $\beta = \alpha_i$ and $t(\beta) = y(\beta)$} \\ t(\beta), &\text{otherwise} \end{cases} \end{align*}$$ It's easy to see that $t \sim \pi(t)$, so $\pi$ is well-defined. Clearly if $s \sqsubseteq t$ then $\pi(s) \sqsubseteq \pi(t)$. Observing that $\pi^2 = \id$, $\pi$ is one-to-one and onto. Finally, $\pi(x) = y$.
We shall call this construction the eventually-zero construction.Every $T \in P$ is homogeneous.
Proof
Let $x, y \in T$ be at the same level. Then $x \sim y$, so let $\alpha_1,\dots,\alpha_n \in \dom(x)$ be all the ordinals such that $x(\alpha_i) \neq y(\alpha_i)$. Define $\pi : T \to T$ such that for $t \in T$ and $\beta \in \dom(t)$: $$ \begin{align*} \pi(t)(\beta) := \begin{cases} y(\beta), &\text{if $\beta = \alpha_i$ and $t(\beta) = x(\beta)$} \\ x(\beta), &\text{if $\beta = \alpha_i$ and $t(\beta) = y(\beta)$} \\ t(\beta), &\text{otherwise} \end{cases} \end{align*}$$ It's easy to see that $t \sim \pi(t)$, so $\pi$ is well-defined. Clearly if $s \sqsubseteq t$ then $\pi(s) \sqsubseteq \pi(t)$. Observing that $\pi^2 = \id$, $\pi$ is one-to-one and onto. Finally, $\pi(x) = y$.
$\blacksquare$
We now show that the Suslin tree $\mathcal{T}$ is homogeneous. Let $x, y \in \mathcal{T}$, so there exists some $T \in G$ such that $x, y \in T$ (say $\dom(x) = \alpha$). $T$ is homogeneous by the above claim, so there exists an automorphism $\pi : T \to T$ such that $\pi(x) = y$. We then extend $\pi$ to $\pi' : \mathcal{T} \to \mathcal{T}$ by stipulating that for $t \in \mathcal{T}$, $\pi'(t) := \pi(t\restrictedto\alpha) \cup (t - t\restrictedto\alpha)$, and clearly $\pi'$ remains an automorphism. Thus $\mathcal{T}$ is homogeneous.
$\square$
Exercise 15.21
Lemma 15.21.A
Let $(P,<_P)$ and $(Q,<_Q)$ be two partial orders with $P, Q \subseteq \boldsymbol{\mathrm{ORD}}$. Fix $x_0 \in P$ and $y_0 \in Q$. Suppose that:
Proof
Fix $x_0 \in P$ and $y_0 \in Q$. Consider the following relation on $R$: $$ \begin{gather*} S := \, <_P \cup <_Q \cup \, \underbrace{\lbrace (x,y) : x \leq_P x_0 \wedge y \leq_Q y_0 \wedge x < y\rbrace}_{=:A} \cup \underbrace{\lbrace (x,y) : x \leq_Q y_0 \wedge y \leq_P x_0 \wedge x < y\rbrace}_{=:B} \end{gather*}$$ Let $<_R$ denote the transitive closure of $S$. We see that if $x <_R y$, then there exists a sequence $(x,z_0),(z_0,z_1),\dots,(z_n,y)$ all in $S$, which implies $x < z_0$, $z_i < z_{i+1}$ and $z_n < y$, so $x < y$. In particular, $x \not<_R x$. Furthermore, $<_R$ is transitive by definition. Therefore, $(R,<_R)$ defines a partial order in which (a) holds, and clearly $x_0 \leq_R y_0$.
It remains to show properties (ii) and (iii). The proof of each is the same, so we just show (ii). Before that, we make the following observations:
Let $(P,<_P)$ and $(Q,<_Q)$ be two partial orders with $P, Q \subseteq \boldsymbol{\mathrm{ORD}}$. Fix $x_0 \in P$ and $y_0 \in Q$. Suppose that:
- If $x <_P y$ or $x <_Q y$, then $x < y$.
- For all $x \in P \cap Q$, we have $\pred_P(x) = \pred_Q(x)$.
- $(\pred_P(x_0) \cup \lbrace x_0\rbrace) \cap Q = (\pred_Q(y_0) \cup \lbrace y_0\rbrace) \cap P$.
- $R = P \cup Q$.
- $<_P \, = \, <_R \cap \, (P \times P)$.
- $<_Q \, = \, <_R \cap \, (Q \times Q)$.
- $x_0$ and $y_0$ are comparable.
Proof
Fix $x_0 \in P$ and $y_0 \in Q$. Consider the following relation on $R$: $$ \begin{gather*} S := \, <_P \cup <_Q \cup \, \underbrace{\lbrace (x,y) : x \leq_P x_0 \wedge y \leq_Q y_0 \wedge x < y\rbrace}_{=:A} \cup \underbrace{\lbrace (x,y) : x \leq_Q y_0 \wedge y \leq_P x_0 \wedge x < y\rbrace}_{=:B} \end{gather*}$$ Let $<_R$ denote the transitive closure of $S$. We see that if $x <_R y$, then there exists a sequence $(x,z_0),(z_0,z_1),\dots,(z_n,y)$ all in $S$, which implies $x < z_0$, $z_i < z_{i+1}$ and $z_n < y$, so $x < y$. In particular, $x \not<_R x$. Furthermore, $<_R$ is transitive by definition. Therefore, $(R,<_R)$ defines a partial order in which (a) holds, and clearly $x_0 \leq_R y_0$.
It remains to show properties (ii) and (iii). The proof of each is the same, so we just show (ii). Before that, we make the following observations:
- If $x <_P y$ and $y <_P z$, then immediately $x <_P z$.
- If $x <_Q y$ and $y <_P z$, then $y \in P$ and since $\pred_Q(y) = \pred_P(y)$, $x <_P y$, so $x <_P z$.
- If $(x,y) \in A$ and $y <_P z$, then $y \leq_Q y_0$ and $y \in P$. By (c), $y \leq_P x_0$ and $y \in Q$. Since $x \leq_P x_0$ and $x < y$, we have $x <_P y$ (as $\pred_P(x_0)$ is well-ordered). Therefore $x <_P z$.
- By symmetry, the case of $y <_Q z$ with $x <_P y$, $x <_Q y$, or $(x,y) \in B$ reduces to $x <_Q z$.
- If $x <_P y$ and $(y,z) \in A$, then $x <_P y \leq_P x_0$ and $x < z$, so $(x,z) \in A$.
- If $x <_Q y$ and $(y,z) \in A$, then since $y \in P$ and $\pred_Q(y) = \pred_P(y)$, $x <_P y$ so $(x,z) \in A$ by above.
- If $(x,y) \in A$ and $(y,z) \in A$, then $x, y \leq_P x_0$ and $x < y$, implying $x <_P y$, so $(x,z) \in A$ by above.
- If $(x,y) \in B$ and $(y,z) \in A$, then $x \leq_P x_0$ and $z \leq_P x_0$ and $x < z$, so $x <_P z$.
- By symmetry, the case of $(y,z) \in B$ with $x <_R y$ reduces to $x <_Q z$ or $(x,z) \in B$.
- $(x,y) \in A$.
- $(x,y) \in B$.
- $(x,z) \in B$ and $z <_P y$.
$\blacksquare$
Solution
By the hint, we shall show that $(P,<)$ has the (K) property. Let $W \subseteq P$ be an uncountable set of conditions. Since the trees $T \in P$ are finite, by the $\Delta$-Lemma we first obtain an uncountable set $Z \subseteq W$ such that for all $S, T \in Z$, $S \cap T = R$ for some finite set $R$.
Fix $\alpha \in R$. For all $T \in Z$, consider $\pred_T(\alpha) := \lbrace \beta \in T : \beta <_T \alpha\rbrace$. Since this is a finite subset of $\alpha$, there are only countably many possibilities. Thus, there exists an uncountable set $Z' \subseteq Z$ such that for all $S, T \in Z'$, $\pred_S(\alpha) = \pred_T(\alpha)$. We repeat this for all $\alpha \in R$, and since $R$ is finite we obtain a final uncountable set $Z_0 \subseteq W$ such that for all $\alpha \in R$ and for all $S, T \in Z_0$, $\pred_S(\alpha) = \pred_T(\alpha)$. By Lemma 15.21.A, this implies that $(S \cup T, <_S \cup <_T) \in P$ is stronger than both $(S,<_S)$ and $(T,<_T)$, so $S$ and $T$ are compatible.
By the hint, we shall show that $(P,<)$ has the (K) property. Let $W \subseteq P$ be an uncountable set of conditions. Since the trees $T \in P$ are finite, by the $\Delta$-Lemma we first obtain an uncountable set $Z \subseteq W$ such that for all $S, T \in Z$, $S \cap T = R$ for some finite set $R$.
Fix $\alpha \in R$. For all $T \in Z$, consider $\pred_T(\alpha) := \lbrace \beta \in T : \beta <_T \alpha\rbrace$. Since this is a finite subset of $\alpha$, there are only countably many possibilities. Thus, there exists an uncountable set $Z' \subseteq Z$ such that for all $S, T \in Z'$, $\pred_S(\alpha) = \pred_T(\alpha)$. We repeat this for all $\alpha \in R$, and since $R$ is finite we obtain a final uncountable set $Z_0 \subseteq W$ such that for all $\alpha \in R$ and for all $S, T \in Z_0$, $\pred_S(\alpha) = \pred_T(\alpha)$. By Lemma 15.21.A, this implies that $(S \cup T, <_S \cup <_T) \in P$ is stronger than both $(S,<_S)$ and $(T,<_T)$, so $S$ and $T$ are compatible.
$\square$
Exercise 15.22
Solution
Following the hint, suppose $T_0 \forces \dot{A}$ is uncountable. We first construct an uncountable set of pairs $(T_\eta,\alpha_\eta)$, where $\eta < \omega_1$, as follows: Suppose $(T_\xi,\alpha_\xi)$ has been constructed for $\xi < \eta$. Let $T_\eta \leq T$ be such that $T_\eta \forces (\exists \beta > \sup_{\xi<\eta} \alpha_\xi) \, \beta \in \dot{A}$ (this is possible as $\eta$ is countable). Let $\alpha_\eta$ be such $\beta$. We may choose $T_\eta$ such that $T_\eta \neq T_\xi$ for all $\xi < \eta$, and WLOG assume $\alpha_\eta \in T_\eta$.
Now $W := \lbrace T_\eta : \eta < \omega_1\rbrace$ is an uncountable subset of $P$, and as in the proof of Exercise 15.21 we obtain an uncountable $Z \subseteq W$ such that for all $S, T \in Z$, $(S \cup T, <_S \cup <_T)$ is a condition stronger than both $S$ and $T$.
Following the hint, suppose $T_0 \forces \dot{A}$ is uncountable. We first construct an uncountable set of pairs $(T_\eta,\alpha_\eta)$, where $\eta < \omega_1$, as follows: Suppose $(T_\xi,\alpha_\xi)$ has been constructed for $\xi < \eta$. Let $T_\eta \leq T$ be such that $T_\eta \forces (\exists \beta > \sup_{\xi<\eta} \alpha_\xi) \, \beta \in \dot{A}$ (this is possible as $\eta$ is countable). Let $\alpha_\eta$ be such $\beta$. We may choose $T_\eta$ such that $T_\eta \neq T_\xi$ for all $\xi < \eta$, and WLOG assume $\alpha_\eta \in T_\eta$.
Now $W := \lbrace T_\eta : \eta < \omega_1\rbrace$ is an uncountable subset of $P$, and as in the proof of Exercise 15.21 we obtain an uncountable $Z \subseteq W$ such that for all $S, T \in Z$, $(S \cup T, <_S \cup <_T)$ is a condition stronger than both $S$ and $T$.
Claim
There exists an uncountable $Z' \subseteq Z$ such that for all $T_\eta, T_\delta \in Z'$, $\alpha_\delta \notin T_\eta$ and $\alpha_\eta \notin T_\delta$.
Proof
Let $\xi < \omega_1$ be $0$ or a limit ordinal. Let $X_n := \lbrace \eta < \omega_1 : \alpha_{\xi + n} \notin T_\eta\rbrace$. We must have $X_n$ uncountable for some $n$ — otherwise, $\bigcup_n X_n$ is countable, so $\omega_1 - \bigcup_n X_n = \bigcap_n (\omega_1 - X_n)$ is uncountable, meaning for some $\eta$, $\alpha_{\xi + n} \in T_\eta$ for all $n$, contradicting that $T_\eta$ is finite.
For $\zeta < \omega_1$, define $Z_\zeta, Y_\zeta \subseteq \omega_1$ inductively:
Let $Z' := \bigcup_{\zeta<\omega_1} Z_\zeta$. Then $Z'$ is uncountable, and for $T_\eta, T_\delta \in Z'$, $\alpha_\eta \notin T_\delta$ and $\alpha_\delta \notin T_\eta$.
Now for each $\xi \in Z'$, consider the set $\lbrace \pred_{T_\xi}(\alpha_\xi) : \xi < \omega_1\rbrace \cap R$. This is a finite subset of $R$, so there are only finitely many possibilities. Therefore, we may obtain $Z'' \subseteq Z'$ such that $\lbrace \pred_{T_\xi}(\alpha_\xi) : \xi < \omega_1\rbrace \cap R$ is equal for all $T_\xi \in Z''$. Then the full premise of Lemma 15.21.A is satisfied, so for all $T_\eta, T_\delta \in Z''$ there exists some tree $T$ stronger than both. Since $\alpha_\eta$ and $\alpha_\delta$ are compatible in $T$, $T \forces \dot{A}$ is not an antichain.
There exists an uncountable $Z' \subseteq Z$ such that for all $T_\eta, T_\delta \in Z'$, $\alpha_\delta \notin T_\eta$ and $\alpha_\eta \notin T_\delta$.
Proof
Let $\xi < \omega_1$ be $0$ or a limit ordinal. Let $X_n := \lbrace \eta < \omega_1 : \alpha_{\xi + n} \notin T_\eta\rbrace$. We must have $X_n$ uncountable for some $n$ — otherwise, $\bigcup_n X_n$ is countable, so $\omega_1 - \bigcup_n X_n = \bigcap_n (\omega_1 - X_n)$ is uncountable, meaning for some $\eta$, $\alpha_{\xi + n} \in T_\eta$ for all $n$, contradicting that $T_\eta$ is finite.
For $\zeta < \omega_1$, define $Z_\zeta, Y_\zeta \subseteq \omega_1$ inductively:
- Let $Z_0 := \emptyset$, $Y_0 := \omega_1$.
- Suppose $Z_\zeta$, $Y_\zeta$ have been defined with $Y_\zeta$ uncountable. Let $\xi$ be the least limit ordinal such that $\xi > \sup Z_\zeta$, and for all $\eta \in Z_\zeta$, $\alpha_\xi > \sup T_\eta$. By the first paragraph, there exists some $n$ such that the set of $\delta \in Y_\zeta$ in which $\alpha_{\xi+n} \notin T_\delta$ is uncountable. Let $Y_{\zeta+1}$ be this uncountable set, and $Z_{\zeta+1} := Z_\zeta \cup \lbrace \xi+n\rbrace$.
- If $\zeta$ is a limit ordinal, let $Z_\zeta := \bigcup_{\eta<\zeta} Z_\eta$.
Let $Z' := \bigcup_{\zeta<\omega_1} Z_\zeta$. Then $Z'$ is uncountable, and for $T_\eta, T_\delta \in Z'$, $\alpha_\eta \notin T_\delta$ and $\alpha_\delta \notin T_\eta$.
$\blacksquare$
$\square$
Exercise 15.23
Lemma 15.23.A
Let $P$ be an $\omega$-closed partial order, and let $\lbrace p_n : n < \omega\rbrace \subseteq P$. Then there exists a $q \in P$ such that for all $n$, either $q \leq p_n$, or $q$ and $p_n$ are incompatible.
Proof
Define a sequence $\c{q_n : n < \omega}$ as follows: Start with $q_0 := p_0$. Suppose $q_n$ is defined, and consider two cases:
Let $P$ be an $\omega$-closed partial order, and let $\lbrace p_n : n < \omega\rbrace \subseteq P$. Then there exists a $q \in P$ such that for all $n$, either $q \leq p_n$, or $q$ and $p_n$ are incompatible.
Proof
Define a sequence $\c{q_n : n < \omega}$ as follows: Start with $q_0 := p_0$. Suppose $q_n$ is defined, and consider two cases:
- If $q_n$ and $p_{n+1}$ are incompatible, let $q_{n+1} := q_n$.
- If $q_n$ and $p_{n+1}$ are compatible, let $q_{n+1} \in P$ such that $q_{n+1} \leq q_n$ and $q_{n+1} \leq p_{n+1}$.
$\blacksquare$
Lemma 15.23.B
Let $P$ be an $\omega$-closed partial order. Let $\dot{X}$ be a name such that $\forces \dot{X} \subseteq \omega_1$. For any $\alpha < \omega_1$, there exists a $p \in P$ such that $p$ decides $\dot{X} \cap \alpha$ — that is, there exists an $A \subseteq \alpha$ (in the ground model) such that $p \forces \dot{X} \cap \alpha = \check{A}$.
Proof
We may assume WLOG that $\dot{X} = \lbrace (\check{\gamma},p_\gamma) : \gamma < \omega_1\rbrace$, where $p_\gamma \in P$ for all $\gamma$. By Lemma 15.23.A, there exists a $p \in P$ such that for all $\gamma < \alpha$, $p \leq p_\gamma$ or $p$ is incompatible with $p_\gamma$. In other words, if $G, H$ are $P$-generic with $p \in G$ and $p \in H$, then for all $\gamma$, either $p_\gamma \in G$ and $p_\gamma \in H$, or $p_\gamma \notin G$ and $p_\gamma \notin H$. Therefore, $p$ decides the membership of $\gamma$ in $\dot{X}$ for all $\gamma < \alpha$. Since $P$ is $\omega$-closed, no countable subsets are added in the generic extension, so $\dot{X} \cap \alpha$ must be in the ground model.
Let $P$ be an $\omega$-closed partial order. Let $\dot{X}$ be a name such that $\forces \dot{X} \subseteq \omega_1$. For any $\alpha < \omega_1$, there exists a $p \in P$ such that $p$ decides $\dot{X} \cap \alpha$ — that is, there exists an $A \subseteq \alpha$ (in the ground model) such that $p \forces \dot{X} \cap \alpha = \check{A}$.
Proof
We may assume WLOG that $\dot{X} = \lbrace (\check{\gamma},p_\gamma) : \gamma < \omega_1\rbrace$, where $p_\gamma \in P$ for all $\gamma$. By Lemma 15.23.A, there exists a $p \in P$ such that for all $\gamma < \alpha$, $p \leq p_\gamma$ or $p$ is incompatible with $p_\gamma$. In other words, if $G, H$ are $P$-generic with $p \in G$ and $p \in H$, then for all $\gamma$, either $p_\gamma \in G$ and $p_\gamma \in H$, or $p_\gamma \notin G$ and $p_\gamma \notin H$. Therefore, $p$ decides the membership of $\gamma$ in $\dot{X}$ for all $\gamma < \alpha$. Since $P$ is $\omega$-closed, no countable subsets are added in the generic extension, so $\dot{X} \cap \alpha$ must be in the ground model.
$\blacksquare$
Solution
Start with a condition $p \in G$ such that $p \forces$ ($\dot{C}$ is a closed unbounded set and $\dot{X} \subseteq \omega_1$). Let: $$ \begin{align*} D := \lbrace q \leq p : q \forces (\alpha \in \dot{C} \text{ and } \dot{X} \cap \alpha = S_\alpha)\rbrace \end{align*}$$ We shall show that $D$ is dense below $p$. Indeed, let $p' \leq p$, and suppose $p' = \c{S_\xi : \xi < \gamma}$. Define sequences $\c{q_n : n < \omega}$ and $\c{\alpha_n : n < \omega}$ as follows: Let $q_0 = p'$ and $\alpha_0 = \gamma$. Suppose $q_n, \alpha_n$ are defined. Now $q_n \forces (\exists \beta > \check{\alpha}_n) \, \beta \in \dot{C}$, so there exist $\alpha_{n+1} > \dom(q_n)$ and $q_{n+1}'$ with $\dom(q_{n+1}') > \alpha_{n+1}$ such that $q_{n+1}' \forces \check{\alpha}_{n+1} \in \dot{C}$. By Lemma 15.23.B, there exists some $q_{n+1} \leq q_{n+1}'$ and $A_{\alpha_{n+1}} \subseteq \alpha_{n+1}$ in the ground model such that $q_{n+1} \forces \dot{X} \cap \alpha_{n+1} = \check{A}_{\alpha_{n+1}}$. Note that $A_{\alpha_n} \subseteq A_{\alpha_{n+1}}$ necessarily.
Now let $q := \bigcup_n q_n$, and $\alpha := \lim_n \alpha_n$. Then $q \forces (\forall n \in \omega) \, \check{\alpha}_n \in \dot{C}$, and since $q \forces \dot{C}$ is closed, $q \forces \check{\alpha} \in \dot{C}$. Noting that $\alpha_1 < \dom(q_1) < \alpha_2 < \dom(q_2) < \dots$, we have $\dom(q) = \alpha$, say $q = \c{S_\xi : \xi < \alpha}$. Let $q' := \c{S_\xi : \xi \leq \alpha}$ and $S_\alpha := \bigcup_{n<\omega} A_{\alpha_n}$. Then $q' \forces \dot{X} \cap \alpha = S_\alpha$, and $q' \forces (\alpha \in \dot{C} \text{ and } \dot{X} \cap \alpha = S_\alpha)$.
Start with a condition $p \in G$ such that $p \forces$ ($\dot{C}$ is a closed unbounded set and $\dot{X} \subseteq \omega_1$). Let: $$ \begin{align*} D := \lbrace q \leq p : q \forces (\alpha \in \dot{C} \text{ and } \dot{X} \cap \alpha = S_\alpha)\rbrace \end{align*}$$ We shall show that $D$ is dense below $p$. Indeed, let $p' \leq p$, and suppose $p' = \c{S_\xi : \xi < \gamma}$. Define sequences $\c{q_n : n < \omega}$ and $\c{\alpha_n : n < \omega}$ as follows: Let $q_0 = p'$ and $\alpha_0 = \gamma$. Suppose $q_n, \alpha_n$ are defined. Now $q_n \forces (\exists \beta > \check{\alpha}_n) \, \beta \in \dot{C}$, so there exist $\alpha_{n+1} > \dom(q_n)$ and $q_{n+1}'$ with $\dom(q_{n+1}') > \alpha_{n+1}$ such that $q_{n+1}' \forces \check{\alpha}_{n+1} \in \dot{C}$. By Lemma 15.23.B, there exists some $q_{n+1} \leq q_{n+1}'$ and $A_{\alpha_{n+1}} \subseteq \alpha_{n+1}$ in the ground model such that $q_{n+1} \forces \dot{X} \cap \alpha_{n+1} = \check{A}_{\alpha_{n+1}}$. Note that $A_{\alpha_n} \subseteq A_{\alpha_{n+1}}$ necessarily.
Now let $q := \bigcup_n q_n$, and $\alpha := \lim_n \alpha_n$. Then $q \forces (\forall n \in \omega) \, \check{\alpha}_n \in \dot{C}$, and since $q \forces \dot{C}$ is closed, $q \forces \check{\alpha} \in \dot{C}$. Noting that $\alpha_1 < \dom(q_1) < \alpha_2 < \dom(q_2) < \dots$, we have $\dom(q) = \alpha$, say $q = \c{S_\xi : \xi < \alpha}$. Let $q' := \c{S_\xi : \xi \leq \alpha}$ and $S_\alpha := \bigcup_{n<\omega} A_{\alpha_n}$. Then $q' \forces \dot{X} \cap \alpha = S_\alpha$, and $q' \forces (\alpha \in \dot{C} \text{ and } \dot{X} \cap \alpha = S_\alpha)$.
$\square$
Exercise 15.24
Solution
By Exercise 15.16, let $(R,<)$ denote the notion of forcing that collapses $2^{\aleph_0}$ to $\aleph_1$, and it suffices to show that $B(Q) = B(R)$. For each infinite $\alpha < \omega_1$, let $h_\alpha$ be a one-to-one correspondence between the cardinal $2^{\aleph_0}$ and $P(\alpha)$.
$B(R) \subseteq B(Q)$: Define $f : R \to Q$ by stipulating that $f(p) = \c{S_\xi : \xi < \dom(p)}$, where: $$ \begin{align*} S_\xi := \begin{cases} \emptyset, &\text{if $\xi < \omega$} \\ h_\beta(p(\beta)), &\text{if $\xi = \omega + \beta$ for some ordinal $\beta$} \end{cases} \end{align*}$$ Since $h_\beta$ is one-to-one for all $\beta$, $f$ is also one-to-one and $p < q$ iff $f(p) < f(q)$. Thus $R$ is isomorphic to a subset of $Q$, so $B(R) \subseteq B(Q)$.
$B(Q) \subseteq B(R)$: Define $g : Q \to R$ by stipulating that $g(\c{S_\xi : \xi < \alpha}) = p$, where $\dom(p) = \alpha$ and $p(\xi) := h_\xi^{-1}(S_\xi)$ for all $\xi < \alpha$. Then $g$ is one-to-one and respects $<$, so $Q$ is isomorphic to a subset of $R$, hence $B(Q) \subseteq B(R)$.
By Exercise 15.16, let $(R,<)$ denote the notion of forcing that collapses $2^{\aleph_0}$ to $\aleph_1$, and it suffices to show that $B(Q) = B(R)$. For each infinite $\alpha < \omega_1$, let $h_\alpha$ be a one-to-one correspondence between the cardinal $2^{\aleph_0}$ and $P(\alpha)$.
$B(R) \subseteq B(Q)$: Define $f : R \to Q$ by stipulating that $f(p) = \c{S_\xi : \xi < \dom(p)}$, where: $$ \begin{align*} S_\xi := \begin{cases} \emptyset, &\text{if $\xi < \omega$} \\ h_\beta(p(\beta)), &\text{if $\xi = \omega + \beta$ for some ordinal $\beta$} \end{cases} \end{align*}$$ Since $h_\beta$ is one-to-one for all $\beta$, $f$ is also one-to-one and $p < q$ iff $f(p) < f(q)$. Thus $R$ is isomorphic to a subset of $Q$, so $B(R) \subseteq B(Q)$.
$B(Q) \subseteq B(R)$: Define $g : Q \to R$ by stipulating that $g(\c{S_\xi : \xi < \alpha}) = p$, where $\dom(p) = \alpha$ and $p(\xi) := h_\xi^{-1}(S_\xi)$ for all $\xi < \alpha$. Then $g$ is one-to-one and respects $<$, so $Q$ is isomorphic to a subset of $R$, hence $B(Q) \subseteq B(R)$.
$\square$
Exercise 15.25
Lemma 15.25.A
$\diamond$ is equivalent to the following statement ($\diamond'$): There exists a sequence of functions $h_\alpha$, $\alpha < \omega_1$, such that for every $f : \omega_1 \to \omega_1$, the set $\lbrace \alpha < \omega_1 : f\restrictedto\alpha = h_\alpha\rbrace$ is stationary.
Proof
$\diamond \implies \diamond'$: We first establish two claims.
$\diamond' \implies \diamond$: Let $\c{h_\alpha : \alpha < \omega_1}$ be a $\diamond'$-sequence. For each $\alpha$, let $S_\alpha := \lbrace \beta < \alpha : h_\alpha(\beta) = 1\rbrace$. For any $A \subseteq \omega_1$, let $f_A : \omega_1 \to 2$ be such that $f_A(\alpha) = 1$ iff $\alpha \in A$. Then $S := \lbrace \alpha < \omega_1 : f_A\restrictedto\alpha = h_\alpha\rbrace$ is stationary. Since: $$ \begin{align*} f_A\restrictedto\alpha = h_\alpha &\iff (\forall \beta < \alpha)(f_A(\beta) = 1 \leftrightarrow h_\alpha(\beta) = 1) \\ &\iff A \cap \alpha = \lbrace \beta < \alpha : h_\alpha(\beta) = 1\rbrace \\ &\iff A \cap \alpha = S_\alpha \end{align*}$$ so $\c{S_\alpha : \alpha < \omega_1}$ is a $\diamond$-sequence.
$\diamond$ is equivalent to the following statement ($\diamond'$): There exists a sequence of functions $h_\alpha$, $\alpha < \omega_1$, such that for every $f : \omega_1 \to \omega_1$, the set $\lbrace \alpha < \omega_1 : f\restrictedto\alpha = h_\alpha\rbrace$ is stationary.
Proof
$\diamond \implies \diamond'$: We first establish two claims.
Claim
Let $\kappa$ be an uncountable regular cardinal. Let $f : \kappa \to \kappa \times \kappa$ be a one-to-one correspondence. Then the set $C := \lbrace \alpha < \kappa : f``(\alpha) = \alpha \times \alpha\rbrace$ is closed unbounded.
Proof
$C$ is closed: Let $\alpha < \kappa$ such that $C \cap \alpha$ is unbounded. Then: $$ \begin{align*} f``(\alpha) = \bigcup_{\beta \in C \cap \alpha} f``(\beta) = \bigcup_{\beta \in C \cap \alpha} \beta \times \beta = \alpha \times \alpha \end{align*}$$ $C$ is unbounded: Fix any $\alpha < \kappa$. Let $\beta_0 := \alpha$, and choose $\beta_n$'s such that: $$ \begin{align*} f``(\beta_0) \subseteq \beta_1 \times \beta_1 \subseteq f``(\beta_2) \subseteq \beta_3 \times \beta_3 \subseteq \dots \end{align*}$$ Let $\beta := \lim_n \beta_n$, and clearly $f``(\beta) = \beta \times \beta$.
Let $\kappa$ be an uncountable regular cardinal. Let $f : \kappa \to \kappa \times \kappa$ be a one-to-one correspondence. Then the set $C := \lbrace \alpha < \kappa : f``(\alpha) = \alpha \times \alpha\rbrace$ is closed unbounded.
Proof
$C$ is closed: Let $\alpha < \kappa$ such that $C \cap \alpha$ is unbounded. Then: $$ \begin{align*} f``(\alpha) = \bigcup_{\beta \in C \cap \alpha} f``(\beta) = \bigcup_{\beta \in C \cap \alpha} \beta \times \beta = \alpha \times \alpha \end{align*}$$ $C$ is unbounded: Fix any $\alpha < \kappa$. Let $\beta_0 := \alpha$, and choose $\beta_n$'s such that: $$ \begin{align*} f``(\beta_0) \subseteq \beta_1 \times \beta_1 \subseteq f``(\beta_2) \subseteq \beta_3 \times \beta_3 \subseteq \dots \end{align*}$$ Let $\beta := \lim_n \beta_n$, and clearly $f``(\beta) = \beta \times \beta$.
$\blacksquare$
Claim
Assume $\diamond$. There exists a sequence of sets $A_\alpha \subseteq \alpha \times \alpha$, where $\alpha < \omega_1$, such that for all $A \subseteq \omega_1 \times \omega_1$, the set $\lbrace \alpha < \omega_1 : A \cap (\alpha \times \alpha) = A_\alpha\rbrace$ is stationary.
Proof
Fix any one-to-one correspondence $f : \omega_1 \to \omega_1 \times \omega_1$. Let $C := \lbrace \alpha < \omega_1 : f``(\alpha) = \alpha \times \alpha\rbrace$, which is closed unbounded by the previous claim. Let $\c{S_\alpha : \alpha < \omega_1}$ be a $\diamond$-sequence. Define: $$ \begin{align*} A_\alpha := f``(S_\alpha) \cap (\alpha \times \alpha) \end{align*}$$ Fix $A \subseteq \omega_1 \times \omega_1$. Let $T := \lbrace \alpha < \omega_1 : f_{-1}(A) \cap \alpha = S_\alpha\rbrace$, which is stationary, so $T \cap C$ is stationary. For $\alpha \in C$: $$ \begin{align*} \alpha \in T &\iff f_{-1}(A) \cap \alpha = S_\alpha \\ &\iff f``(f_{-1}(A) \cap \alpha) = f``(S_\alpha) \\ &\iff A \cap f``(\alpha) = f``(S_\alpha) \cap (\alpha \times \alpha) \\ &\iff A \cap (\alpha \times \alpha) = A_\alpha \\ &\iff \alpha \in S \end{align*}$$ So $S \cap C = T \cap C$ is stationary.
We now prove the $\implies$ direction. Let $\c{A_\alpha : \alpha < \omega_1}$ be the sequence from the second claim. For each $\alpha$, let $h_\alpha := A_\alpha$ if $A_\alpha \subseteq \alpha \times \alpha$ is a function, and $h_\alpha = \emptyset$ otherwise. Fix a function $f : \omega_1 \to \omega_1$, and let $C := \lbrace \alpha < \omega_1 : f\restrictedto\alpha = f \cap (\alpha \times \alpha)\rbrace$. We shall show that $C$ is closed unbounded. The argument is very similar to the first claim: to see that $C$ is closed, suppose $C \cap \alpha$ is unbounded. Then $f\restrictedto\alpha = \bigcup_{\beta \in C \cap \alpha} f\restrictedto\beta= \bigcup_{\beta \in C \cap \alpha} f \cap (\beta \times \beta) = f \cap (\alpha \times \alpha)$. To see that $C$ is unbounded, fix any $\alpha < \omega_1$ and construct $\beta_n$'s such that:
$$
\begin{align*}
f\restrictedto\beta_0 \subseteq f \cap (\beta_1 \times \beta_1) \subseteq f\restrictedto\beta_2 \subseteq f \cap (\beta_3 \times \beta_3) \subseteq \dots
\end{align*}$$
Then $\beta := \lim_n\beta_n \in C$. Now since $S := \lbrace \alpha < \omega_1 : f \cap (\alpha \times \alpha) = A_\alpha\rbrace$ is stationary, on $S \cap C$ we have $f\restrictedto\alpha = f \cap (\alpha \times \alpha) = h_\alpha$, as desired.Assume $\diamond$. There exists a sequence of sets $A_\alpha \subseteq \alpha \times \alpha$, where $\alpha < \omega_1$, such that for all $A \subseteq \omega_1 \times \omega_1$, the set $\lbrace \alpha < \omega_1 : A \cap (\alpha \times \alpha) = A_\alpha\rbrace$ is stationary.
Proof
Fix any one-to-one correspondence $f : \omega_1 \to \omega_1 \times \omega_1$. Let $C := \lbrace \alpha < \omega_1 : f``(\alpha) = \alpha \times \alpha\rbrace$, which is closed unbounded by the previous claim. Let $\c{S_\alpha : \alpha < \omega_1}$ be a $\diamond$-sequence. Define: $$ \begin{align*} A_\alpha := f``(S_\alpha) \cap (\alpha \times \alpha) \end{align*}$$ Fix $A \subseteq \omega_1 \times \omega_1$. Let $T := \lbrace \alpha < \omega_1 : f_{-1}(A) \cap \alpha = S_\alpha\rbrace$, which is stationary, so $T \cap C$ is stationary. For $\alpha \in C$: $$ \begin{align*} \alpha \in T &\iff f_{-1}(A) \cap \alpha = S_\alpha \\ &\iff f``(f_{-1}(A) \cap \alpha) = f``(S_\alpha) \\ &\iff A \cap f``(\alpha) = f``(S_\alpha) \cap (\alpha \times \alpha) \\ &\iff A \cap (\alpha \times \alpha) = A_\alpha \\ &\iff \alpha \in S \end{align*}$$ So $S \cap C = T \cap C$ is stationary.
$\blacksquare$
$\diamond' \implies \diamond$: Let $\c{h_\alpha : \alpha < \omega_1}$ be a $\diamond'$-sequence. For each $\alpha$, let $S_\alpha := \lbrace \beta < \alpha : h_\alpha(\beta) = 1\rbrace$. For any $A \subseteq \omega_1$, let $f_A : \omega_1 \to 2$ be such that $f_A(\alpha) = 1$ iff $\alpha \in A$. Then $S := \lbrace \alpha < \omega_1 : f_A\restrictedto\alpha = h_\alpha\rbrace$ is stationary. Since: $$ \begin{align*} f_A\restrictedto\alpha = h_\alpha &\iff (\forall \beta < \alpha)(f_A(\beta) = 1 \leftrightarrow h_\alpha(\beta) = 1) \\ &\iff A \cap \alpha = \lbrace \beta < \alpha : h_\alpha(\beta) = 1\rbrace \\ &\iff A \cap \alpha = S_\alpha \end{align*}$$ so $\c{S_\alpha : \alpha < \omega_1}$ is a $\diamond$-sequence.
$\blacksquare$
Solution
It is proved in Theorem 13.21 that $V = L$ implies $\diamond$, so by Lemma 15.25.A, $V = L$ implies $\diamond'$.
It is proved in Theorem 13.21 that $V = L$ implies $\diamond$, so by Lemma 15.25.A, $V = L$ implies $\diamond'$.
$\square$
Exercise 15.26
Lemma 15.26.A
For each regular cardinal $\kappa$, there exists an unbounded set $X \subseteq \kappa$, consisting only of limit ordinals, that is not stationary.
Proof
Let $X := \lbrace \alpha + \omega : \alpha < \kappa\rbrace$. Clearly $X$ and $C - X$ (where $C$ is the set of all limit ordinals below $\kappa$) are both unbounded. Furthermore, $C - X$ is closed — otherwise, suppose $\lbrace \beta_n : n < \omega\rbrace \subseteq C - X$ with $\beta_n \to \alpha + \omega$ for some $\alpha < \kappa$. Then $\alpha + k < \beta_n < \alpha + \omega$ for some $k \in \omega$, so $\beta_n = \alpha + k'$ for some $k' \in \omega$, contradicting that $\beta_n$ is a limit ordinal. Hence $X$ is non-stationary.
For each regular cardinal $\kappa$, there exists an unbounded set $X \subseteq \kappa$, consisting only of limit ordinals, that is not stationary.
Proof
Let $X := \lbrace \alpha + \omega : \alpha < \kappa\rbrace$. Clearly $X$ and $C - X$ (where $C$ is the set of all limit ordinals below $\kappa$) are both unbounded. Furthermore, $C - X$ is closed — otherwise, suppose $\lbrace \beta_n : n < \omega\rbrace \subseteq C - X$ with $\beta_n \to \alpha + \omega$ for some $\alpha < \kappa$. Then $\alpha + k < \beta_n < \alpha + \omega$ for some $k \in \omega$, so $\beta_n = \alpha + k'$ for some $k' \in \omega$, contradicting that $\beta_n$ is a limit ordinal. Hence $X$ is non-stationary.
$\blacksquare$
Solution
Consider the Suslin tree constructed in Theorem 15.26. Let $\vec{S} := \c{S_\alpha : \alpha < \omega_1}$ be a $\diamond$-sequence. Given another sequence $\vec{T} := \c{T_\alpha : \alpha < \omega_1}$ with $T_\alpha \subseteq \alpha$, we say that $\vec{T}$ almost equals $\vec{S}$ if $T_\alpha = S_\alpha$ for all but non-stationarily many $\alpha$'s. We observe that $\vec{T}$ remains a $\diamond$-sequence — given any $A \subseteq \omega_1$, let $S$ be the stationary set such that $A \cap \alpha = S_\alpha$ for all $\alpha \in S$. Let $X$ be the non-stationary set such that $T_\alpha = S_\alpha$ for all $\alpha \notin X$. Then $T := S - X$ remains stationary, so $A \cap \alpha = T_\alpha$ for $\alpha \in T$.
Now let $C$ be the closed unbounded set in Lemma 15.27 with the maximal antichain $A$, and let $S$ be the stationary set such that $A \cap \alpha = S_\alpha$ for all $\alpha \in S$. Let $X$ be the unbounded, non-stationary set from Lemma 15.26.A consisting only of countable limit ordinals. By the first paragraph, we may let $\vec{T} := \c{T_\alpha : \alpha < \omega_1}$ be a $\diamond$-sequence almost equal to $\vec{S}$, with $T_\alpha = \emptyset$ for all $\alpha \in X$. We run the construction of the Suslin tree $\mathcal{T}$ in Theorem 15.26 using $\vec{T}$. Since we only need to build $T_{\alpha+1}$ such that $S_\alpha$ is a maximal antichain for $\alpha \in T \cap C$ and $T \cap C \cap X = \emptyset$, we are free to construct $T_{\alpha+1}$ for all $\alpha \in X$.
We describe the construction of $T_{\alpha+1}$ for $\alpha \in X$. Let: $$ \begin{align*} J := \lbrace (s,t) : s \neq t \wedge (&\exists \beta < \omega_1)(\beta \text{ is limit} \wedge s \in T_\beta \wedge t \in T_\beta \\ &\wedge \exists \pi \,(\pi : T_\beta \to T_\beta \text{ is an automorphism} \wedge \pi(s) = t))\rbrace \end{align*}$$ We see that $J \subseteq \omega_1 \times \omega_1$ so $\vert J\vert \leq \aleph_1$. Let $f : \omega_1 \to J$ be onto. Let $Y = \lbrace \beta_\gamma : \gamma < \omega_1\rbrace \subseteq X$ be unbounded such that for all $\alpha < \omega_1$, if $f(\alpha) = (s,t)$ then $s, t \in T_{\beta_\alpha}$. For each $\beta_\alpha \in Y$, repeat the construction of $T_{\beta_\alpha+1}$ from Exercise 15.18 so that there does not exist an automorphism $\pi : T_{\beta_\alpha+1} \to T_{\beta_\alpha+1}$ with $\pi(s) = t$, where $f(\alpha) = (s,t)$.
We show that $\mathcal{T}$ is rigid. Suppose not, so there exists a non-trivial automorphism $\pi$ of $\mathcal{T}$. Then $\pi(s) = t$ for some $s, t \in \mathcal{T}$. Let $\beta$ be the least limit ordinal such that $s, t \in T_\beta$. Then $f(\alpha) = (s,t)$ for some $\alpha$, so $\pi\restrictedto T_{\beta_\alpha}$ is an automorphism of $T_{\beta_\alpha}$ satisfying $\pi(s) = t$. Then $\pi\restrictedto T_{\beta_\alpha+1}$ is an automorphism of $T_{\beta_\alpha+1}$ with $\pi(s) = t$, contradicting the construction.
Consider the Suslin tree constructed in Theorem 15.26. Let $\vec{S} := \c{S_\alpha : \alpha < \omega_1}$ be a $\diamond$-sequence. Given another sequence $\vec{T} := \c{T_\alpha : \alpha < \omega_1}$ with $T_\alpha \subseteq \alpha$, we say that $\vec{T}$ almost equals $\vec{S}$ if $T_\alpha = S_\alpha$ for all but non-stationarily many $\alpha$'s. We observe that $\vec{T}$ remains a $\diamond$-sequence — given any $A \subseteq \omega_1$, let $S$ be the stationary set such that $A \cap \alpha = S_\alpha$ for all $\alpha \in S$. Let $X$ be the non-stationary set such that $T_\alpha = S_\alpha$ for all $\alpha \notin X$. Then $T := S - X$ remains stationary, so $A \cap \alpha = T_\alpha$ for $\alpha \in T$.
Now let $C$ be the closed unbounded set in Lemma 15.27 with the maximal antichain $A$, and let $S$ be the stationary set such that $A \cap \alpha = S_\alpha$ for all $\alpha \in S$. Let $X$ be the unbounded, non-stationary set from Lemma 15.26.A consisting only of countable limit ordinals. By the first paragraph, we may let $\vec{T} := \c{T_\alpha : \alpha < \omega_1}$ be a $\diamond$-sequence almost equal to $\vec{S}$, with $T_\alpha = \emptyset$ for all $\alpha \in X$. We run the construction of the Suslin tree $\mathcal{T}$ in Theorem 15.26 using $\vec{T}$. Since we only need to build $T_{\alpha+1}$ such that $S_\alpha$ is a maximal antichain for $\alpha \in T \cap C$ and $T \cap C \cap X = \emptyset$, we are free to construct $T_{\alpha+1}$ for all $\alpha \in X$.
We describe the construction of $T_{\alpha+1}$ for $\alpha \in X$. Let: $$ \begin{align*} J := \lbrace (s,t) : s \neq t \wedge (&\exists \beta < \omega_1)(\beta \text{ is limit} \wedge s \in T_\beta \wedge t \in T_\beta \\ &\wedge \exists \pi \,(\pi : T_\beta \to T_\beta \text{ is an automorphism} \wedge \pi(s) = t))\rbrace \end{align*}$$ We see that $J \subseteq \omega_1 \times \omega_1$ so $\vert J\vert \leq \aleph_1$. Let $f : \omega_1 \to J$ be onto. Let $Y = \lbrace \beta_\gamma : \gamma < \omega_1\rbrace \subseteq X$ be unbounded such that for all $\alpha < \omega_1$, if $f(\alpha) = (s,t)$ then $s, t \in T_{\beta_\alpha}$. For each $\beta_\alpha \in Y$, repeat the construction of $T_{\beta_\alpha+1}$ from Exercise 15.18 so that there does not exist an automorphism $\pi : T_{\beta_\alpha+1} \to T_{\beta_\alpha+1}$ with $\pi(s) = t$, where $f(\alpha) = (s,t)$.
We show that $\mathcal{T}$ is rigid. Suppose not, so there exists a non-trivial automorphism $\pi$ of $\mathcal{T}$. Then $\pi(s) = t$ for some $s, t \in \mathcal{T}$. Let $\beta$ be the least limit ordinal such that $s, t \in T_\beta$. Then $f(\alpha) = (s,t)$ for some $\alpha$, so $\pi\restrictedto T_{\beta_\alpha}$ is an automorphism of $T_{\beta_\alpha}$ satisfying $\pi(s) = t$. Then $\pi\restrictedto T_{\beta_\alpha+1}$ is an automorphism of $T_{\beta_\alpha+1}$ with $\pi(s) = t$, contradicting the construction.
$\square$
Exercise 15.27
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 15.28
Solution
Let $A$ be the set in the hint. Clearly $A$ is uncountable, so we shall show that the elements are pairwise incompatible. Let $x \neq y$ be two elements in $T$, and suppose $(p_x,q_x)$ and $(p_y,q_y)$ are compatible. Then there exists some $z \in T$ such that $(p_x,q_x) > (p_z,q_z)$ and $(p_y,q_y) > (p_z,q_z)$. So $p_x, p_y < p_z$ and $q_x, q_y < q_z$. But since the branches below $p_z$ (resp. $q_z$) are well-ordered, we must have: $$ \begin{align*} (p_x < p_y \text{ or } p_x > p_y) \text{ and } (q_x < q_y \text{ or } q_x > q_y) \end{align*}$$ Since $p_x, q_x$ share the same level (and so do $p_y, q_y$), we have $(p_x,q_x) < (p_y,q_y)$ or $(p_x,q_x) > (p_y,q_y)$. WLOG assume the first. Then $p_x \leq y$ and $q_x \leq y$. Since $p_x, q_x$ share the same level and the branches below $y$ are well-ordered, $p_x = q_x$, a contradiction.
Let $A$ be the set in the hint. Clearly $A$ is uncountable, so we shall show that the elements are pairwise incompatible. Let $x \neq y$ be two elements in $T$, and suppose $(p_x,q_x)$ and $(p_y,q_y)$ are compatible. Then there exists some $z \in T$ such that $(p_x,q_x) > (p_z,q_z)$ and $(p_y,q_y) > (p_z,q_z)$. So $p_x, p_y < p_z$ and $q_x, q_y < q_z$. But since the branches below $p_z$ (resp. $q_z$) are well-ordered, we must have: $$ \begin{align*} (p_x < p_y \text{ or } p_x > p_y) \text{ and } (q_x < q_y \text{ or } q_x > q_y) \end{align*}$$ Since $p_x, q_x$ share the same level (and so do $p_y, q_y$), we have $(p_x,q_x) < (p_y,q_y)$ or $(p_x,q_x) > (p_y,q_y)$. WLOG assume the first. Then $p_x \leq y$ and $q_x \leq y$. Since $p_x, q_x$ share the same level and the branches below $y$ are well-ordered, $p_x = q_x$, a contradiction.
$\square$
Exercise 15.29
Solution
Note that $P$ here refers to the forcing notion described in Example 14.2. It suffices to show that $P$ is isomorphic to a dense subset of $P \times P$. Let $P' := P - \lbrace \emptyset\rbrace$, which is clearly dense in $P$ (so $P' \times P'$ is dense in $P \times P$). Let $\Gamma : \omega \times \omega \to \omega$ be the canonical well-ordering restricted to $\omega$. Define $F : P' \times P' \to P'$ as follows: Given $(p,q) \in P' \times P'$, define $r := F(p,q)$ such that:
Thus, $V[x] = V[x_1][x_2]$ as in the hint, which implies that $x$ is not minimal over the ground model.
Note that $P$ here refers to the forcing notion described in Example 14.2. It suffices to show that $P$ is isomorphic to a dense subset of $P \times P$. Let $P' := P - \lbrace \emptyset\rbrace$, which is clearly dense in $P$ (so $P' \times P'$ is dense in $P \times P$). Let $\Gamma : \omega \times \omega \to \omega$ be the canonical well-ordering restricted to $\omega$. Define $F : P' \times P' \to P'$ as follows: Given $(p,q) \in P' \times P'$, define $r := F(p,q)$ such that:
- $\dom(r) = \Gamma(\max\dom(p), \max\dom(q)) + 1$ (note that $\max\dom(p) = \dom(p) - 1$).
- Let $k \in \dom(r)$, and suppose $\Gamma(i,j) = k$. Then $i \in \dom(p)$ and $j \in \dom(q)$, as $k < \dom(r)$ implies $(i,j) < (\max\dom(p),\max\dom(q))$. Define $r(k) := \Gamma(p(i),q(j))$.
- If $\Gamma(m,n) + 1 = \dom(r)$, then $\dom(p) = m + 1$ and $\dom(q) = n + 1$.
- If $i \in \dom(p)$ and $j \in \dom(q)$, let $a, b$ be such that $\Gamma(a,b) = r(\Gamma(i,j))$. Then $p(i) := a$ and $q(j) := b$.
Thus, $V[x] = V[x_1][x_2]$ as in the hint, which implies that $x$ is not minimal over the ground model.
$\square$
Exercise 15.30
Solution
Let $P$ denote the Sacks forcing notion (i.e. the forcing notion of perfect trees). We first define a function $F : P \times \omega \to \omega$ as follows: $$ \begin{align*} F(p,n) := \min\lbrace k \in \omega : (&\forall t \in p)(\dom(t) = n \to \\ &(\exists s \in p)(\dom(s) \leq k \wedge s \supseteq t \wedge s^\frown 0 \in p \wedge s^\frown 1 \in p))\rbrace \end{align*}$$ Intuitively, $F(p,n)$ is the smallest $k$ such that every $t \in p$ with $\dom(t) = n$ splits by the $k^\text{th}$ level. This is well-defined as $p$ is a perfect tree and there are only finitely many such $t$.
Fix $f : \omega \to \omega$ in $V[a]$. Let $p \in P$ be such that $p \forces \dot{f} : \omega \to \omega$. We define a sequence of perfect trees $q_n \in P$, a function $g : \omega \to \omega$ (in $V$), and natural numbers $k_n \in \omega$ inductively: Let $q_0 := p$, $g(0)$ be any natural number, and $k_0 := F(p,0)$. Suppose $q_n, g(n), k_n$ have been defined. Let: $$ \begin{align*} r_{k,n+1} := \bigcup\lbrace r \leq q_n : r \forces \dot{f}(n+1) < k\rbrace \end{align*}$$ The union of perfect trees is a perfect tree, so $r_{k,n+1} \in P$. Clearly $r_{k,n+1} \subseteq r_{k+1,n+1}$ and $\bigcup_{k < \omega} r_{k,n+1} = q_n$. Let $k_{n+1} := F(q_n,k_n) + 1$, and choose $g(n+1)$ large enough so that: $$ \begin{align*} \lbrace t\restrictedto k_{n+1} : t \in r_{g(n+1),n+1}\rbrace = \lbrace t\restrictedto k_{n+1} : t \in q_n\rbrace \end{align*}$$ This is possible as these sets are finite. Let $q_{n+1} := r_{g(n+1),n+1}$.
Let $q := \bigcap_n q_n$. We show that $q$ is a perfect tree. Let $t \in q$ with $\dom(t) = n$. Then $t \in q_n$. Since $k_n \geq n$, by definition of $F(q_n,k_n)$ there exists $s \supseteq t$ with $\dom(s) \leq F(q_n,k_n)$ such that $s^\frown 0, s^\frown 1 \in q_n$. Since $\lbrace t\restrictedto k_{n+1} : t \in q_n\rbrace = \lbrace t\restrictedto k_{n+1} : t \in q_{n+1}\rbrace$, we have $s^\frown 0, s^\frown 1 \in q_{n+1}$, and since $k_m > k_{n+1}$ for all $m > n+1$, we have $s^\frown 0, s^\frown 1 \in q_m$ for all $m \geq n+1$. Therefore $s^\frown 0, s^\frown 1 \in q$, so $q$ is perfect.
We obtain $q \in P$ such that $q \leq q_n$ for all $n$, so $q \forces (\forall n \in \omega)(\dot{f}(n) < \check{g}(n))$. Therefore $g$ dominates $f$ and $g$ is in the ground model.
Let $P$ denote the Sacks forcing notion (i.e. the forcing notion of perfect trees). We first define a function $F : P \times \omega \to \omega$ as follows: $$ \begin{align*} F(p,n) := \min\lbrace k \in \omega : (&\forall t \in p)(\dom(t) = n \to \\ &(\exists s \in p)(\dom(s) \leq k \wedge s \supseteq t \wedge s^\frown 0 \in p \wedge s^\frown 1 \in p))\rbrace \end{align*}$$ Intuitively, $F(p,n)$ is the smallest $k$ such that every $t \in p$ with $\dom(t) = n$ splits by the $k^\text{th}$ level. This is well-defined as $p$ is a perfect tree and there are only finitely many such $t$.
Fix $f : \omega \to \omega$ in $V[a]$. Let $p \in P$ be such that $p \forces \dot{f} : \omega \to \omega$. We define a sequence of perfect trees $q_n \in P$, a function $g : \omega \to \omega$ (in $V$), and natural numbers $k_n \in \omega$ inductively: Let $q_0 := p$, $g(0)$ be any natural number, and $k_0 := F(p,0)$. Suppose $q_n, g(n), k_n$ have been defined. Let: $$ \begin{align*} r_{k,n+1} := \bigcup\lbrace r \leq q_n : r \forces \dot{f}(n+1) < k\rbrace \end{align*}$$ The union of perfect trees is a perfect tree, so $r_{k,n+1} \in P$. Clearly $r_{k,n+1} \subseteq r_{k+1,n+1}$ and $\bigcup_{k < \omega} r_{k,n+1} = q_n$. Let $k_{n+1} := F(q_n,k_n) + 1$, and choose $g(n+1)$ large enough so that: $$ \begin{align*} \lbrace t\restrictedto k_{n+1} : t \in r_{g(n+1),n+1}\rbrace = \lbrace t\restrictedto k_{n+1} : t \in q_n\rbrace \end{align*}$$ This is possible as these sets are finite. Let $q_{n+1} := r_{g(n+1),n+1}$.
Let $q := \bigcap_n q_n$. We show that $q$ is a perfect tree. Let $t \in q$ with $\dom(t) = n$. Then $t \in q_n$. Since $k_n \geq n$, by definition of $F(q_n,k_n)$ there exists $s \supseteq t$ with $\dom(s) \leq F(q_n,k_n)$ such that $s^\frown 0, s^\frown 1 \in q_n$. Since $\lbrace t\restrictedto k_{n+1} : t \in q_n\rbrace = \lbrace t\restrictedto k_{n+1} : t \in q_{n+1}\rbrace$, we have $s^\frown 0, s^\frown 1 \in q_{n+1}$, and since $k_m > k_{n+1}$ for all $m > n+1$, we have $s^\frown 0, s^\frown 1 \in q_m$ for all $m \geq n+1$. Therefore $s^\frown 0, s^\frown 1 \in q$, so $q$ is perfect.
We obtain $q \in P$ such that $q \leq q_n$ for all $n$, so $q \forces (\forall n \in \omega)(\dot{f}(n) < \check{g}(n))$. Therefore $g$ dominates $f$ and $g$ is in the ground model.
$\square$
Exercise 15.31
Solution
By Theorem 15.38, it suffices to show that every $f : \kappa \to P(\kappa)$ in the generic extension is in the ground model. Following the hint, consider the set $X := \lbrace (\alpha,\beta) : \beta \in f(\alpha)\rbrace \in P(\kappa \times \kappa)$. Define in the generic extension $g : \kappa \times \kappa \to 2$ by $g(\alpha,\beta) = 1$ iff $(\alpha,\beta) \in X$. Since $B$ is $(\kappa,2)$-distributive, $g$ is in the ground model, and so $X$ is in the ground model. Then we may define $f$ in the ground model by: $$ \begin{align*} f(\alpha) := \lbrace \beta < \kappa : (\alpha,\beta) \in X\rbrace \end{align*}$$ so $f$ is also in the ground model.
By Theorem 15.38, it suffices to show that every $f : \kappa \to P(\kappa)$ in the generic extension is in the ground model. Following the hint, consider the set $X := \lbrace (\alpha,\beta) : \beta \in f(\alpha)\rbrace \in P(\kappa \times \kappa)$. Define in the generic extension $g : \kappa \times \kappa \to 2$ by $g(\alpha,\beta) = 1$ iff $(\alpha,\beta) \in X$. Since $B$ is $(\kappa,2)$-distributive, $g$ is in the ground model, and so $X$ is in the ground model. Then we may define $f$ in the ground model by: $$ \begin{align*} f(\alpha) := \lbrace \beta < \kappa : (\alpha,\beta) \in X\rbrace \end{align*}$$ so $f$ is also in the ground model.
$\square$
Exercise 15.32
Solution
Let $g : \cf(\kappa) \to V[G]$ be the function $\alpha \mapsto f\restrictedto\kappa_\alpha$, where $\sup_{\alpha < \cf(\kappa)} \kappa_\alpha = \kappa$. Since $\kappa$ is singular, $\cf(\kappa) < \kappa$, so $g \in V$. Then $f := \bigcup_{\alpha < \cf(\kappa)} g(\alpha) \in V$.
Let $g : \cf(\kappa) \to V[G]$ be the function $\alpha \mapsto f\restrictedto\kappa_\alpha$, where $\sup_{\alpha < \cf(\kappa)} \kappa_\alpha = \kappa$. Since $\kappa$ is singular, $\cf(\kappa) < \kappa$, so $g \in V$. Then $f := \bigcup_{\alpha < \cf(\kappa)} g(\alpha) \in V$.
$\square$
Exercise 15.33
Solution
By Lemma 15.39, $B(P)$ is weakly $(\omega,\omega)$-distributive iff every $f : \omega \to \omega$ in $V[G]$ is dominated by some $g : \omega \to \omega$ in $V$, which we know is false by Lemma 15.30(ii).
By Lemma 15.39, $B(P)$ is weakly $(\omega,\omega)$-distributive iff every $f : \omega \to \omega$ in $V[G]$ is dominated by some $g : \omega \to \omega$ in $V$, which we know is false by Lemma 15.30(ii).
$\square$
Exercise 15.34
Solution
$\implies$: Assume $B$ is weakly $(\omega,\omega_1)$-distributive. Suppose for a contradiction that $\omega_1$ ceases to be a cardinal in $V[G]$. Then in $V[G]$ there exists some onto mapping $f : \omega \to \omega_1$. By Lemma 15.39, there exists a $g : \omega \to \omega_1$ in $V$ such that $g(\alpha) > f(\alpha)$ for all $\alpha$. Then $g$ must be cofinal in $\omega_1$, so $\cf^V(\omega_1) = \omega$, a contradiction.
$\impliedby$: Assume $\omega_1$ remains a cardinal in $V[G]$. Let $f : \omega \to \omega_1$ in $V[G]$. Since $\omega_1^{V[G]} = \omega_1$, $f$ must be bounded — there exists some $\alpha < \omega_1$ such that $f(n) < \alpha$ for all $n$. Then the constant map $n \mapsto \alpha$ is in $V$ and dominates $f$.
$\implies$: Assume $B$ is weakly $(\omega,\omega_1)$-distributive. Suppose for a contradiction that $\omega_1$ ceases to be a cardinal in $V[G]$. Then in $V[G]$ there exists some onto mapping $f : \omega \to \omega_1$. By Lemma 15.39, there exists a $g : \omega \to \omega_1$ in $V$ such that $g(\alpha) > f(\alpha)$ for all $\alpha$. Then $g$ must be cofinal in $\omega_1$, so $\cf^V(\omega_1) = \omega$, a contradiction.
$\impliedby$: Assume $\omega_1$ remains a cardinal in $V[G]$. Let $f : \omega \to \omega_1$ in $V[G]$. Since $\omega_1^{V[G]} = \omega_1$, $f$ must be bounded — there exists some $\alpha < \omega_1$ such that $f(n) < \alpha$ for all $n$. Then the constant map $n \mapsto \alpha$ is in $V$ and dominates $f$.
$\square$
Exercise 15.35
Lemma 15.35.A
Let $B$ be a $\lambda$-saturated complete Boolean algebra. For any $X \subseteq B$:
Proof
All sums and products are well-defined as $B$ is complete.
Let $B$ be a $\lambda$-saturated complete Boolean algebra. For any $X \subseteq B$:
- There exists a $Y \subseteq X$, $\vert Y\vert < \lambda$, such that $\sum X = \sum Y$.
- There exists a $Y \subseteq X$, $\vert Y\vert < \lambda$, such that $\prod X = \prod Y$.
Proof
All sums and products are well-defined as $B$ is complete.
- Let $x := \sum X$. Suppose $X = \lbrace x_\alpha : \alpha < \vert X\vert \rbrace$. Let $y_\alpha := x_\alpha - \sum_{\beta<\alpha} x_\beta$. Then $\lbrace y_\alpha : \alpha < \vert X\vert \rbrace$ is an antichain and $\sum_{\alpha<\vert X\vert } y_\alpha = \sum_{\alpha<\vert X\vert } x_\alpha$. Let $Y := \lbrace \alpha < \vert X\vert : y_\alpha \neq 0\rbrace$, so $\lbrace y_\alpha : \alpha < \vert X\vert \rbrace \cup \lbrace 1 - x\rbrace$ is a partition of $B$. Since $B$ is $\lambda$-saturated, $\vert Y\vert < \lambda$. Then $\sum_{\alpha \in Y} y_\alpha = x$, and since $y_\alpha \leq x_\alpha \leq x$, we have $\sum_{\alpha \in Y} x_\alpha = x$.
- Let $-X := \lbrace -x : x \in X\rbrace$. By Exercise 7.27(ii), $\prod X = -\sum(-X)$. By (i), there exists $Y \subseteq X$, $\vert Y\vert < \lambda$, such that $\sum(-X) = \sum(-Y)$. Then $\prod X = -\sum(-X) = -\sum(-Y) = \prod Y$.
$\blacksquare$
Solution
Let $B$ be a complete Boolean algebra that is $\kappa$-generated and $\lambda$-saturated. Let $X \subseteq B$ with $\vert X\vert \leq \kappa$ generating $B$. By working similar to Exercise 7.18, for all $x \in B$ there exist $X_\alpha \subseteq X$ such that: $$ \begin{align*} x = \sum_{\alpha < \mu} \prod X_\alpha \end{align*}$$ By Lemma 15.35.A, we may take $\mu < \lambda$ and $\vert X_\alpha\vert < \lambda$ for each $\alpha < \mu$. Therefore there are $\kappa^{<\lambda}$ many possible $\prod X_\alpha$'s, so $(\kappa^{<\lambda})^\mu = \kappa^{<\lambda}$ many possible expressions $\sum_{\alpha<\mu} \prod X_\alpha$. Therefore $\vert B\vert \leq \kappa^{<\lambda}$.
Let $B$ be a complete Boolean algebra that is $\kappa$-generated and $\lambda$-saturated. Let $X \subseteq B$ with $\vert X\vert \leq \kappa$ generating $B$. By working similar to Exercise 7.18, for all $x \in B$ there exist $X_\alpha \subseteq X$ such that: $$ \begin{align*} x = \sum_{\alpha < \mu} \prod X_\alpha \end{align*}$$ By Lemma 15.35.A, we may take $\mu < \lambda$ and $\vert X_\alpha\vert < \lambda$ for each $\alpha < \mu$. Therefore there are $\kappa^{<\lambda}$ many possible $\prod X_\alpha$'s, so $(\kappa^{<\lambda})^\mu = \kappa^{<\lambda}$ many possible expressions $\sum_{\alpha<\mu} \prod X_\alpha$. Therefore $\vert B\vert \leq \kappa^{<\lambda}$.
$\square$
Exercise 15.36
Solution
By Exercise 15.35, $\vert B\vert \leq \aleph_0^{<\aleph_1} = 2^{\aleph_0}$. On the other hand, by Exercise 7.32 we have $\vert B\vert = \vert B\vert ^{\aleph_0} \geq 2^{\aleph_0}$. Therefore $\vert B\vert = 2^{\aleph_0}$.
By Exercise 15.35, $\vert B\vert \leq \aleph_0^{<\aleph_1} = 2^{\aleph_0}$. On the other hand, by Exercise 7.32 we have $\vert B\vert = \vert B\vert ^{\aleph_0} \geq 2^{\aleph_0}$. Therefore $\vert B\vert = 2^{\aleph_0}$.
$\square$
Exercise 15.37
Solution
Suppose $< \, \in U$ is a linear order on $A$. Let $E$ be the support of $<$, i.e. $\pi(<) = \, <$ for all $\pi \in \func{fix}(E)$.
Example 15.49: Pick $a, b \in A - E$, and assume WLOG that $a < b$ (as $<$ is a linear order). Let $\pi \in \func{fix}(E)$ such that $\pi(a) = b$ and $\pi(b) = a$. Then $\pi(a) \, \pi(<) \, \pi(b) \implies b < a$, a contradiction.
Example 15.50: We define $f : \lbrace P_n : n \in \omega\rbrace \to P$ as follows: $$ \begin{align*} f(P_n) := \min P_n \end{align*}$$ This is well-defined as $a_n, b_n$ are compatible for each $n \in \omega$. Then $f$ is a choice function on $\lbrace P_n : n \in \omega\rbrace$, a contradiction.
Suppose $< \, \in U$ is a linear order on $A$. Let $E$ be the support of $<$, i.e. $\pi(<) = \, <$ for all $\pi \in \func{fix}(E)$.
Example 15.49: Pick $a, b \in A - E$, and assume WLOG that $a < b$ (as $<$ is a linear order). Let $\pi \in \func{fix}(E)$ such that $\pi(a) = b$ and $\pi(b) = a$. Then $\pi(a) \, \pi(<) \, \pi(b) \implies b < a$, a contradiction.
Example 15.50: We define $f : \lbrace P_n : n \in \omega\rbrace \to P$ as follows: $$ \begin{align*} f(P_n) := \min P_n \end{align*}$$ This is well-defined as $a_n, b_n$ are compatible for each $n \in \omega$. Then $f$ is a choice function on $\lbrace P_n : n \in \omega\rbrace$, a contradiction.
$\square$