Proof. To show that homogeneity holds over some $A \subseteq \mathscr {A}$ , it suffices to show that every infinite $D \subseteq \mathscr {A}$ disjoint from A has a duplicate $D'$ that is also disjoint from A. To see this, suppose that $D = B \cup C$ is disjoint from A and $B \sim C$ . Then fix some duplicate $D'$ of D that is disjoint from A, within which there is a duplicate $B'$ of B. We can then let $\pi _1$ be an automorphism that swaps B and $B'$ while point-wise fixing A and $\pi _2$ be an automorphism that swaps $B'$ and C while point-wise fixing A. The composition of $\pi _1$ and $\pi _2$ is then a desired automorphism.

(1) Suppose Tail holds. Let A be a set of urelements with the least tail cardinal $\kappa $ . Suppose that D is disjoint from A, which has size at most $\kappa $ . Let E be a tail of $D \cup A$ . By the minimality of $\kappa $ , E must have size $\kappa $ ; so for some $E'\subseteq E$ , $E' \sim D$ . $E'$ is then a duplicate of D that is disjoint from A. If Tail does not hold, then $\mathscr {A}$ is a proper class and we can fix an $A \subseteq \mathscr {A}$ without tails. Suppose that $D \subseteq \mathscr {A}$ is disjoint from A. We may assume D is infinite and hence has size $\kappa $ for some infinite cardinal $\kappa $ . Then there must be some $E \subseteq \mathscr {A}$ of size $\kappa ^+$ that is disjoint from A, which means $E - D$ has size $\kappa ^+$ . So E has a subset which duplicates D.

(2) Suppose that $A \subseteq A' \subseteq \mathscr {A}$ and homogeneity holds over A. Let D be an infinite set of urelements disjoint from $A'$ . Let $E = (A'- A) \cup D$ . Since E is infinite, by AC it can be partitioned into a pair of duplicates $E_1$ and $E_2$ such that $E \sim E_1 \sim E_2$ . By homogeneity over A, there is an automorphism $\pi $ such that $\pi E_1 = E$ and $\pi $ point-wise fixes A. Since $E_1$ has a duplicate disjoint from A, by applying $\pi $ it follows that E has a duplicate disjoint from A. Let $E'$ be such a duplicate of E, which is also disjoint from $A'$ . Since $D \subseteq E$ , $E'$ has a subset that duplicates D. Hence, homogeneity holds over $A'$ .

(3) is an immediate consequence of (1) and (2).

Proof. By the remark in the proof of Lemma 2.4, Plenitude implies that homogeneity holds over every set of urelements. Fix some $\langle \kappa , \alpha \rangle $ and some $A \subseteq \mathscr {A}$ extending $ker(u)$ with $\kappa $ -many urelements. For each $x\in V_\alpha (A)$ , define $\theta _x$ to be the least cardinal such that there is some y with $\varphi (x, y, u)$ and $ker(y) \sim \theta _x$ , and let $\kappa ' = Sup\{ \theta _x : x \in V_\alpha (A)\}$ . Fix some infinite B of size $\kappa '$ that is disjoint from A, which exists by Plenitude. Then for every $ x \in V_\alpha (A)$ , fix some $y'$ such that $\varphi (x, y', u)$ and $ker(y') \sim \theta _x$ . $ker(y') - A$ is equinumerous to a subset of B, so by homogeneity over A, there is an automorphism $\pi $ that moves $ker(y')$ into B and point-wise fixes A. It follows that $\varphi (x, \pi y', u)$ and $\pi y' \in V(A \cup B)$ . Thus, each $x \in V_\alpha (A)$ has some $y \in V(A\cup B)$ with $\varphi (x, y, u)$ , so there is some large enough $\alpha '$ such that every $x \in V_\alpha (A)$ has some $y \in V_{\alpha '}(A\cup B)$ with $\varphi (x, y, u)$ . Furthermore, for every $A'$ extending $ker(u)$ by $\kappa $ -many urelements, by homogeneity over $ker(u)$ , there is an automorphism $\pi $ with $\pi A = A'$ that point-wise fixes $\ker (u)$ ; so $\pi B$ will be such that every $x \in V_\alpha (A')$ has some $y \in V_{\alpha '}(A' \cup \pi B)$ with $\varphi (x, y, u)$ . Therefore, ${\left \langle \kappa ', \alpha '\right \rangle }$ is indeed a $(\varphi , ker(u))$ -extension of ${\left \langle \kappa , \alpha \right \rangle }$ .

Proof. Suppose that Plenitude holds. It suffices to show that the DC $_\kappa $ -scheme holds for every regular cardinal $\kappa $ since by a standard argument as in [Reference Jech10, Theorem 8.1], this will imply the DC $_\lambda $ -scheme for every singular $\lambda $ as well. So suppose that $\kappa $ is regular and $\forall x \exists y \varphi (x, y, u)$ with some parameter u. We will construct a set $\bar {x}$ that is closed under $<\kappa $ -sequences (i.e., $\bar {x}^{<\kappa } \subseteq \bar {x}$ ) and the $\varphi $ -relation (i.e., $\forall x \in \bar {x} \exists y \in \bar {x} \varphi (x, y, u)$ ). This will suffice for the DC $_\kappa $ -scheme because we can apply DC $_\kappa $ to $\bar {x}$ to get a desired $\kappa $ -sequence.

We first construct a $\kappa $ -sequence of pairs of ordinals $\langle \langle \lambda _\alpha , \gamma _\alpha \rangle : \alpha < \kappa \rangle $ by recursion as follows. Let $A_0$ be a set of urelements that extends $\ker (u)$ by $\lambda _0$ -many urelements and $\gamma _0$ be an ordinal with cf $(\gamma _0) = \kappa $ . For each ordinal $\alpha < \kappa $ , we let $\langle \lambda _{\alpha +1} , \gamma _{\alpha +1} \rangle $ be the lexicographical-least $(\varphi , ker(u))$ -extension of ${\left \langle \lambda _\alpha ,\gamma _\alpha \right \rangle }$ with cf $(\gamma _\alpha ) = \kappa $ , which exists by Lemma 2.6; we take the union at the limit stage. By Plenitude, there exists a $\kappa $ -sequence of sets of urelements $\langle A_\alpha : \alpha < \kappa \rangle $ , where $A_\alpha $ extends $\bigcup _{\beta < \alpha } A_\beta \cup ker(u)$ by $\lambda _\alpha $ -many urelements. Let $\bar {x} = \bigcup _{\alpha < \kappa } V_{\gamma _\alpha } (A_\alpha )$ . For any $x \in V_{\gamma _\alpha } (A_\alpha )$ , there is some B disjoint from $A_\alpha $ witnessing the fact that ${\left \langle \lambda _{\alpha +1}, \gamma _{\alpha +1}\right \rangle }$ is a $(\varphi , ker(u))$ -extension of ${\left \langle \lambda _\alpha ,\gamma _\alpha \right \rangle }$ . By homogeneity over $A_\alpha $ , it follows that $A_{\alpha +1} - A_\alpha $ is a witness as well, so there is some $y \in V_{\gamma _{\alpha +1}}(A_{\alpha +1})$ with $\varphi (x, y, u)$ . This shows that $\bar {x}$ is closed under the $\varphi $ -relation. Finally, $\bar {x}^{<\kappa } \subseteq \bar {x}$ because $\text {cf} (\gamma _\alpha ) = \kappa $ for each $\alpha < \kappa $ and $\kappa $ is regular. This completes the proof.

Proof. (1) Assume Closure and Duplication. Suppose that $\forall x \in w \exists y \varphi (x, y, u)$ , where w is a set and u is a parameter. For every $x \in w$ , let $\theta _x$ be the least $\theta $ realized by the kernel of some y such that $\varphi (x, y, u)$ , and define $\theta $ as the supremum of all such $\theta _x$ . Let $A \subseteq \mathscr {A}$ be such that $ker(w) \cup \ker (u) \subseteq A$ and homogeneity holds over A, which exists by Lemma 2.4(3). By Closure and Duplication, there is a $B \subseteq \mathscr {A}$ of size $\theta $ that is disjoint from A. Then for every $x \in w$ , fix a $y'$ such that $\varphi (x, y', u)$ and $ker(y') \sim \theta _x$ . By homogeneity over A, there is an automorphism that moves $ker(y')$ into $A \cup B$ while point-wise fixing A. Thus, every $x \in w$ has a $y \in V(A \cup B)$ such that $\varphi (x, y, u)$ . Therefore, Collection holds by applying Proposition 1.2.

(2) Assume Plenitude. Closure immediately follows. By (1), it remains to show Duplication holds. Given an infinite $A \subseteq \mathscr {A}$ , $A \sim \kappa $ for some infinite cardinal $\kappa $ and we can fix some B that realizes $\kappa ^+$ . So B contains a subset that duplicates A.

Proof. Assume that every set of urelements has a tail and suppose that $\forall x \in w \exists y \varphi (x, y, u)$ . There is an $A \subseteq \mathscr {A}$ such that $ker(w) \cup ker(u) \subseteq A$ and homogeneity holds over A. Let B be a tail of A. For every $x \in w$ and $y'$ such that $\varphi (x, y', u)$ , $ker(y') - A$ can be mapped injectively into B. So by homogeneity over A, there is an automorphism that moves $ker(y')$ into $A \cup B$ while point-wise fixing A. Therefore, every $x \in w$ has some $y \in V(A \cup B)$ such that $\varphi (x, y, u)$ and hence Collection holds by Proposition 1.2.

Proof. Since $\mathscr {A}$ is a proper class, for every x, there is some y with $ker(x) \subsetneq ker(y)$ . It follows from the DC $_\kappa $ -scheme that there exists a $\kappa $ -sequence f with $ker(f\mathord {\upharpoonright } \alpha ) \subsetneq ker(f(\alpha ))$ for every $\alpha < \kappa $ . Fix a well-ordering $\prec $ of $ker(f)$ . Then the map

$$ \begin{align*}\alpha \mapsto \text{the}\prec\text{-least element of }ker(f(\alpha +1)) - ker(f\mathord{\upharpoonright} (\alpha + 1))\end{align*} $$

is an injection from $\kappa $ to $ker(f)$ . Therefore, $\kappa $ is realized.

Proof. Again, we may assume $\kappa $ is regular. Suppose that $\forall x \exists y \varphi (x, y, u)$ with some parameter u. Let A be a set of urelements extending $ker(u)$ over which homogeneity holds and B be a tail of A. Since B has size at least $\kappa $ , B can be partitioned into $\kappa $ -many pieces $\{B_\eta : \eta < \kappa \}$ , where $B_\eta \sim B$ for each $\eta $ . We define a $\kappa $ -sequence of ordinals ${\left \langle \gamma _\alpha : \alpha < \kappa \right \rangle }$ by recursion, where $\gamma _\alpha $ is the least ordinal such that:

1. (i) $\gamma _\alpha> \bigcup _{\eta < \alpha } \gamma _\eta $ and cf $(\gamma _\alpha ) = \kappa $ ;

2. (ii) for every x in $ \bigcup _{\eta < \alpha } V_{\gamma _\eta }(\bigcup _{\eta < \alpha }B_\eta \cup A)$ , there is a $y \in V_{\gamma _\alpha }(\bigcup _{\eta \leq \alpha } B_\eta \cup A)$ with $\varphi (x, y, u)$ .

Such $\gamma _\alpha $ exists, because homogeneity holds over $\bigcup _{\eta < \alpha }B_\eta \cup A$ by Lemma 2.4 (2) and each $B_\eta $ is a tail of A, which allows us to find a sufficiently large $\gamma _\alpha $ . Let $\bar {x} = \bigcup _{\alpha < \kappa } V_{\gamma _\alpha }(\bigcup _{\eta \leq \alpha }B_\eta \cup A)$ . It follows that $\bar {x}$ is closed under the $\varphi $ -relation and $\bar {x}^{<\kappa } \subseteq \bar {x}$ . We can then apply DC $_\kappa $ to $\bar {x}$ to get a desired $\kappa $ -sequence.

Proof. (1) Assume $\mathscr {A}$ is a set and $\forall x \exists y \varphi (x, y, u)$ . Fix some regular $\kappa $ . Define $\langle \gamma _\alpha : \alpha < \kappa \rangle $ by recursion by letting $\gamma _\alpha $ be the least ordinal such that cf $(\alpha ) = \kappa $ and $\forall x \in \bigcup _{\eta < \alpha } V_{\gamma _\eta } (\mathscr {A}) \exists y \in V_{\gamma _\alpha }(\mathscr {A}) \varphi (x, y, u)$ . Let $\bar {x} = \bigcup _{\alpha < \kappa } V_{\gamma _\alpha }(\mathscr {A})$ . As before, we can then apply DC $_\kappa $ to $\bar {x}$ to get a desired $\kappa $ -sequence.

(2) By Lemma 2.10, DC $_{<Ord}$ implies that either Plenitude holds or $\mathscr {A}$ is a set. But (Plenitude $\lor $ $\mathscr {A}$ is a set) implies Collection by Proposition 1.2 and Theorem 2.8, so DC $_{<Ord}$ implies Collection.

(3) Suppose that RP $^-$ holds. Again, by Theorem 2.8 we may assume that Plenitude fails. Now we show that Tail holds, which suffices by Theorem 2.9. So fix an $A \subseteq \mathscr {A}$ and let $\kappa $ be the least cardinal that is not realized by any set of urelements disjoint from A. $\kappa $ exists because Plenitude fails. Then by RP $^-$ there is a transitive set t extending $\{\kappa , A\}$ such that t reflects $\forall \lambda < \kappa \exists B (B \sim \lambda \land B \cap A = \emptyset )$ . It is not hard to check $C = \bigcup \{B \in t : B \subseteq \mathscr {A} \land B \cap A = \emptyset \}$ is a tail of A.

Now assume Collection.

(4) Let x be a set of realized cardinals. Then there is a set y such that for every $\kappa \in x$ , there is some $A \in y$ such that $A \sim \kappa $ . Let $B = \bigcup \{A: A \in y\}$ . Then the cardinality of B is at least the supremum of x and hence Closure holds.

(5) Suppose that Plenitude fails. Given a set A of urelements, let w be the set of cardinals realized by some $B \subseteq \mathscr {A}$ disjoint from A. By Collection there is some set v such that for every $\lambda \in w$ , there is some $B \in v$ such that $B \sim \lambda $ and $B \cap A = \emptyset $ . $C = \bigcup \{B \in v: B \subseteq \mathscr {A} \land B \cap A = \emptyset \}$ is a tail of A.

(6) To show the DC $_{\omega }$ -scheme holds, we may assume that Plenitude fails by Theorem 2.7 and that $\mathscr {A}$ is a proper class by (1). Then by Collection and (5), every set of urelements must have an infinite tail, so the DC $_{\omega }$ -scheme holds by Lemma 2.11.

(7) RP holds by (6) and Theorem 2.1.

Proof. (1) $\to $ (2) is immediate. (2) $\to $ (3) and (3) $\to $ (4) are proved in Theorem 2.12. (4) $\rightarrow $ (1) by Theorems 2.8, 2.9, and 2.12 (7).

Open Question 2.13. Does $\text {ZFU}_{\text {R}}$ prove any of the following?

1. (1) Tail $\rightarrow $ Collection.

2. (2) Collection $\rightarrow $ RP $^-$ .

3. (3) RP $^- \rightarrow $ Collection.

Proof. Fix some $a' \in \mathscr {A} - A$ . Let $\pi $ be a permutation that only swaps a and $a'$ . To see that $\pi \mathscr {I} = \mathscr {I}$ , let $B \in \mathscr {I}$ . We may assume $a \in B$ and $a' \notin B$ . Then $\pi B = (B - \{a\}) \cup \{a'\}$ , which is in $\mathscr {I}$ . Also, $B = \pi ((B - \{a\}) \cup \{a'\})$ . Therefore, $\pi \mathscr {I} = \mathscr {I}$ .

Proof. It is clear that $U^{\mathscr {I}}$ is transitive and contains all the urelements and pure sets. Thus, $U^{\mathscr {I}}$ satisfies Foundation, Extensionality, Infinity, and $\mathscr {A}$ is a proper class in $U^{\mathscr {I}}$ . It is also immediate that $U^{\mathscr {I}}$ satisfies Pairing, Union, Powerset, and Separation. $U^{\mathscr {I}} \models $ AC because for any set x in $U^{\mathscr {I}}$ , any bijection in U between x and an ordinal has the same kernel as x and hence also lives in $U^{\mathscr {I}}$ . It remains to show that Replacement holds in $U^{\mathscr {I}}$ .

Suppose that $U^{\mathscr {I}} \models \forall x \in w \exists ! y \varphi (x, y, u)$ for some $w, u \in U^{\mathscr {I}}$ . Let $v = \{ y \in U^{\mathscr {I}} : \exists x \in w\ \varphi ^{U^{\mathscr {I}}}(x, y, u) \}$ , which is a set in U. It suffices to show that $ker(v) \subseteq ker(w) \cup ker(u)$ . Suppose not. Then there are some y and a such that $y \in v$ , $a \in ker(y)$ and $a \notin ker(w) \cup ker(u)$ . Let $A = ker(w) \cup ker(u) \cup ker(y)$ , which is in $\mathscr {I}$ . By Lemma 2.15, there is an automorphism $\pi $ such that (i) $\pi \mathscr {I} = \mathscr {I}$ , (ii) $\pi a \neq a$ and (iii) $\pi $ point-wise fixes $A - \{a\}$ . So $\pi $ point-wise fixes w and u. Since $y \in v$ , there is some $x \in w$ with $\varphi ^{U^{\mathscr {I}}}(x, y, u)$ . It follows that $\varphi ^{U^{\mathscr {I}}}(x, \pi y, u)$ , but $\pi y \neq y$ because $\pi a \in ker(\pi y)$ but $\pi a \notin ker(y)$ , which contradicts the uniqueness of y.

Proof. In each case, U is a model of $\text {ZFCU}_{\text {R}}$ . These models exist if we assume the consistency of ZF (see [Reference Yao18, Theorem 10]), which follows from the consistency of $\text {ZFCU}_{\text {R}}$ .

(1) Assume that in U, $\mathscr {A} \sim \omega _1$ . Let $\mathscr {I}_1$ be the ideal of all countable subsets of $\mathscr {A}$ . It is clear that in $U^{\mathscr {I}_1}$ , Closure but Duplication hold. Plenitude fails in $U^{\mathscr {I}_1}$ because $\omega _1$ is not realized; since $\mathscr {A}$ is a proper class in $U^{\mathscr {I}_1}$ , the DC $_{\omega _1}$ -scheme fails by Lemma 2.10.

(2) Assume that in U, $\mathscr {A} \sim \omega _1$ . Fix an $A \subseteq \mathscr {A}$ such that $A \sim \omega _1$ and $\mathscr {A} - A \sim \omega _1$ . Let $\mathscr {I}_2 = \{ B \subseteq \mathscr {A}: B - A \text { is countable} \}$ . For every $B \in U^{\mathscr {I}_2}$ , there will be a countable C that is disjoint from $A\cup B$ . Then $C \cup (A- B)$ is a tail of B, so Collection holds in $U^{\mathscr {I}_2}$ by Theorem 2.9. Duplication fails because A has no duplicates in $U^{\mathscr {I}_2}$ .

(3) Assume that in U, $\mathscr {A} \sim \omega $ . Let $\mathscr {I}_3$ be the ideal of finite subsets on $\mathscr {A}$ . It is cleat that in $U^{\mathscr {I}_3}$ Duplication holds and Closure fails. The DC $_{\omega }$ -scheme fails in $U^{\mathscr {I}_3}$ by Lemma 2.10.

(4) Assume that in U, $\mathscr {A} \sim \omega $ and fix an infinite and co-infinite $A \subseteq \mathscr {A}$ . Let $\mathscr {I}_4 = \{ B \subseteq \mathscr {A}: B - A \text { is finite}\}$ . Closure holds in $U^{\mathscr {I}_4}$ because $\omega $ is the greatest realized cardinal. The DC $_{\omega }$ -scheme fails in $U^{\mathscr {I}_4}$ since every set of urelements can be properly extended by another set of urelements disjoint from A, yet there cannot be a corresponding infinite sequence.

(5) Let $\kappa $ be an infinite cardinal. Assume that in U, $\mathscr {A} \sim \omega _{\kappa ^+}$ . Let $\mathscr {I}_5$ be the set of sets of urelements of size less than $\omega _{\kappa ^+}$ . Closure fails in $U^{\mathscr {I}_5}$ because $ \omega _{\kappa ^+}$ is not realized while every cardinal below $\omega _{\kappa ^+}$ is. To show that the DC $_\kappa $ -scheme holds, suppose that for every $x \in U^{\mathscr {I}_5}$ , there is some $y \in U^{\mathscr {I}_5}$ such that $\varphi ^{U^{\mathscr {I}_5}}(x, y, u)$ . Since the DC $_\kappa $ -scheme holds in U by Theorem 2.12(1) and $U^{\mathscr {I}_5}$ is closed under $\kappa $ -sequences, in U there is a function $f: \kappa \rightarrow U^{\mathscr {I}_5}$ such that $\varphi ^{U^{\mathscr {I}_5}}(f\mathord {\upharpoonright } \alpha , f(\alpha ), u)$ for every $\alpha < \kappa $ , which also exists in $U^{\mathscr {I}_5}$ .

(6) It suffices to show that for any infinite cardinal $\kappa $ , $\text {ZFCU}_{\text {R}}$ + Collection + DC $_\kappa $ -scheme does not prove the DC $_{\kappa ^+}$ -scheme. Assume that in U, $\mathscr {A} \sim \kappa ^+$ and let $\mathscr {I}_6$ be the ideal of all sets of urelements of size less than $\kappa ^+$ . $\kappa ^+$ is not realized so by Lemma 2.10, the DC $_{\kappa ^+}$ -scheme fails in $U^{\mathscr {I}_6}$ . Every set of urelements in $U^{\mathscr {I}_6}$ has tail cardinal $\kappa $ , so Collection holds by Theorem 2.9 and the DC $_\kappa $ -scheme holds by Lemma 2.11.

Proof. The proof of (1) $\rightarrow $ (2) is standard, and the point here is that the use of Collection is essential.

(2) $\rightarrow $ (1). Suppose that Collection fails in U. By Figure 1, it follows that both Plenitude and Tail fail in U. In U, fix some $A \subseteq \mathscr {A}$ that does not have a tail and let $\kappa $ be the least cardinal that is not realized by any $B\subseteq \mathscr {A}$ disjoint from A, which is an infinite limit cardinal. Let $F \in U$ be an ultrafilter on $\kappa $ containing all the unbounded subsets of $\kappa $ . Suppose for reductio that the Łoś theorem holds for $U/F$ . Let $id$ be the identity function on $\kappa $ and $c_A$ be the constant function sending every $\alpha < \kappa $ to A. Since $\{ \alpha < \kappa : \exists B \subseteq \mathscr {A} \ (B \sim \alpha \land B \cap A =\emptyset )\} \in F$ , by the Łoś theorem $U/F \models \exists B \subseteq \mathscr {A} (B \sim [id] \land B \cap [C_A] =\emptyset )$ . Thus, there is some $g \in U$ such that

$$ \begin{align*}U/F \models [g] \subseteq \mathscr{A} \land [g] \sim [id] \land ([g] \cap [C_A] =\emptyset).\end{align*} $$

Let $x= \{\alpha < \kappa : g(\alpha ) \subseteq \mathscr {A} \land g(\alpha ) \sim \alpha \land (g(\alpha ) \cap A = \emptyset ) \}$ , which is in F by the Łoś theorem again. Then $D = \bigcup _{\alpha \in x} g(\alpha )$ has size $\kappa $ and is disjoint from A—contradiction.

Proof. Suppose that ${\mathbb P} \in M$ has a maximal antichain ${\left \langle p_i : i \in I\right \rangle }$ indexed by some I ( $|I|> 1$ ). Let ${\left \langle a_i : i \in I\right \rangle }$ be some urelements such that at least two of them are distinct. Consider the ${\mathbb P}$ -name $\dot {x} = \{{\left \langle a_i, p_i\right \rangle } : i \in I \}$ . It follows that . But if for some , then $\dot {y}$ must be some $a_i$ , which is impossible since one can take an M-generic filter containing $p_j$ for some $j \neq i$ .

Proof. (1)–(5) are all proved by standard text-book arguments as in [Reference Kunen11, Chapter VII].

(6) This is clear by the construction of $M[G]$ because every urelement in $M[G]$ must come from $ker(\dot {x})$ for some $\dot {x} \in M^{\mathbb P}$ .

(7) This is proved by induction on the rank of $\dot {x}$ . We may assume that $\dot {x}_G$ is a set. Since $ker(\dot {x}_{G}) \subseteq \bigcup \{ ker(\dot {y}_G) : \dot {y} \in dom(\dot {x})\}$ and by the induction hypothesis we have $ker(\dot {y}_G) \subseteq ker(\dot {y}) \subseteq ker(\dot {x})$ for every $\dot {y} \in dom(\dot {x})$ , it follows that $ker(\dot {x}_G) \subseteq ker(\dot {x})$ .

(8) For every $A \in M[G]$ , $A = \dot {x}_G$ for some $\dot {x} \in M^{\mathbb P}$ so by (7) we have $A = ker(\dot {x}_G) \subseteq ker(\dot {x})$ .

(9) The left-to-right direction follows from (4) and (6). For the other direction, let A be the set of all urelements in $M[G]$ . By (8), $A \subseteq A'$ for some $A' \in M$ . So if a is an urelement in M, it follows from (6) that $a \in A'$ .

Proof. Since the Boolean and quantifier cases can be proved in the same way as in [Reference Kunen11, Chapter VII, Theorem 3.5], we omit their proofs. It remains to show that the lemma holds for all atomic formulas, which we shall prove by induction on the rank of the ${\mathbb P}$ -names.

Case 1. $\varphi $ is $\dot {x}_1 \in \dot {x}_2$ . The argument is the same as in [Reference Kunen11, Chapter VII, Theorem 3.5].

Case 2. $\varphi $ is $\mathscr {A}(\dot {x})$ . For (2), suppose that $\dot {x}_G$ is some urelement b. Then ${\left \langle b, p\right \rangle } \in \dot {x}$ for some $p \in G$ , so $\{q \in {\mathbb P} : \exists {\left \langle a, r\right \rangle } \in \dot {x} \ (\mathscr {A}(a) \land q \leq r)\}$ is dense below p and hence $p \Vdash ^* \mathscr {A}(\dot {x})$ . For (1), suppose that $p \Vdash ^* \mathscr {A}(\dot {x})$ for some $p \in G$ . Then there is some $q \in G$ such that ${\left \langle b, r\right \rangle } \in \dot {x}$ for some $r \geq q$ and urelement b. Thus, $\dot {x}_G = b$ .

Case 3. $\varphi $ is $\dot {x}_1 = \dot {x}_2$ . For (2), suppose that $\dot {x}_{1_G} = \dot {x}_{2_G}$ .

Subcase 3.1. $\dot {x}_{1_G} = \dot {x}_{2_G} = b$ for some urelement b. Then ${\left \langle b, s_1\right \rangle } \in \dot {x}_1$ and ${\left \langle b, s_2\right \rangle } \in \dot {x}_2$ for some $s_1, s_2 \in G$ . Fix some $p \in G$ such that $p \leq s_1, s_2$ . Observe first that $p \Vdash ^* \dot {x}_1 \subseteq \dot {x}_2$ and $p \Vdash ^* \dot {x}_2 \subseteq \dot {x}_1$ trivially hold: for any ${\mathbb P}$ -name $\dot {y}$ and $r \in {\mathbb P}$ such that ${\left \langle \dot {y}, r\right \rangle } \in \dot {x}_1 (\text {or } \dot {x}_2)$ , p must be incompatible with r because r is incompatible with $s_1(\text {or }s_2)$ . Moreover, $p \Vdash ^* \dot {x}_1 \overset {\mathscr {A}}{=} \dot {x}_2$ because $\{q \in {\mathbb P} : \exists a, r_1, r_2 (\mathscr {A}(a) \land {\left \langle a, r_1\right \rangle } \in \dot {x}_1 \land {\left \langle a, r_2\right \rangle } \in \dot {x}_2 \land q \leq r_1, r_2 )\}$ is clearly dense below p. Hence, $p \Vdash ^* \dot {x}_1 = \dot {x}_2$ .

Subcase 3.2. $\dot {x}_{1_G}$ is a set. We first use a standard text-book argument to show that $p \Vdash ^* \dot {x}_1 \subseteq \dot {x}_2$ and $p \Vdash ^* \dot {x}_2 \subseteq \dot {x}_1$ for some $p \in G$ . Define:

• $D_1= \{p \in {\mathbb P} : p \Vdash ^* \dot {x}_1 \subseteq \dot {x}_2 \land p \Vdash ^* \dot {x}_2 \subseteq \dot {x}_1 \}$ ;

• $D_2 = \{p \in {\mathbb P} : \exists {\left \langle \dot {y}_1, q_1\right \rangle } \in \dot {x}_1 \ (p \leq q_1 \land \forall {\left \langle \dot {y}_2, q_2\right \rangle } \in \dot {x}_2 \ \forall r \leq q_2 \ (r \Vdash ^* \dot {y}_1 = \dot {y}_2 \rightarrow p \bot r )) \}$ ;

• $D_3 = \{p \in {\mathbb P} : \exists {\left \langle \dot {y}_2, q_2\right \rangle }\in \dot {x}_2 \ (p \leq q_2 \land \forall {\left \langle \dot {y}_1, q_1\right \rangle } \in \dot {x}_1 \ \forall r \leq q_1 \ (r \Vdash ^* \dot {y}_2 = \dot {y}_1 \rightarrow p \bot r )) \}$ .

If $p \nVdash ^* \dot {x}_1 \subseteq \dot {x}_2$ , then there are ${\left \langle \dot {y}_1, q_1\right \rangle } \in \dot {x}_1$ and $r \leq p, q_1$ such that $r \nVdash ^* \dot {y}_1 \in \dot {x}_2$ ; so there is an $s \leq r$ such that for every ${\left \langle \dot {y}_2, q_2\right \rangle } \in \dot {x}_2$ and $s' \leq q_2$ . If $s' \Vdash ^* \dot {y}_1 = \dot {y}_2$ , then $s \bot s'$ . Hence, $s \leq p$ and $s \in D_2$ . Similarly, if $p \nVdash ^* \dot {x}_2 \subseteq \dot {x}_1$ , then p will have an extension in $D_3$ . This shows that $D_1 \cup D_2 \cup D_3$ is dense. However, $G \cup (D_2 \cup D_3)$ must be empty. Suppose for reductio that $p \in G\cap D_2$ . Fix some ${\left \langle \dot {y}_1, q_1\right \rangle }\in \dot {x}_1$ with $p \leq q_1$ that witnesses $p \in D_2$ . It follows that $\dot {y}_1{_G} = \dot {y}_2{_G}$ for some ${\left \langle \dot {y}_2, q_2\right \rangle } \in \dot {x}_2$ with $q_2 \in G$ . By the induction hypothesis, there is some $r \in G$ such that $r \leq q_2$ and $r \Vdash ^* \dot {y}_1 = \dot {y}_2$ . But p must be incompatible with such r, which is a contradiction. The same argument shows that $G \cap D_3$ is empty. Therefore, there is some $p \in G$ such that $p \Vdash ^* \dot {x}_1 \subseteq \dot {x}_2$ and $p \Vdash ^* \dot {x}_2 \subseteq \dot {x}_1$ .

Now we wish to find some $q \in G$ such that $q \Vdash ^* \dot {x}_1 \overset {\mathscr {A}}{=} \dot {x}_2$ . Define:

• $E_1 {\kern-1pt}={\kern-1pt} \{ q \in {\mathbb P} : \forall r {\kern-1pt}\leq{\kern-1pt} q \ [\forall {\left \langle a_1, s_1\right \rangle } {\kern-1pt}\in{\kern-1pt} \dot {x}_1 \ (\mathscr {A}(a) \rightarrow r \bot s_1) \land \forall {\left \langle a_2, s_2\right \rangle } \in \dot {x}_2 \ (\mathscr {A} (a_2) \rightarrow r \bot s_2)]\}$ ;

• $E_2 = \{ q \in {\mathbb P} : \exists {\left \langle a, r\right \rangle } \in \dot {x}_1 \ (\mathscr {A}(a) \land q \leq r) ) \}$ ;

• $E_3 = \{ q \in {\mathbb P} : \exists {\left \langle a, r\right \rangle } \in \dot {x}_2 \ (\mathscr {A}(a) \land q \leq r)\}$ .

$E_1 \cup E_2 \cup E_3$ is dense. But if there is some $q \in G \cap (E_2 \cup E_3)$ , either $\dot {x}_{1_G}$ or $\dot {x}_{2_G}$ would be a urelement. Thus there is some $q \in G \cap E_1$ such that the set

• $\{r \in {\mathbb P} : \forall {\left \langle a_1, s_1\right \rangle } \in \dot {x}_1 \ (\mathscr {A}(a_1) \rightarrow r \bot s_1) \land \forall {\left \langle a_2, s_2\right \rangle } \in \dot {x}_2 \ (\mathscr {A} (a_2) \rightarrow r \bot s_2)\}$

is dense below q. Therefore, $q \Vdash ^* \dot {x}_1 \overset {\mathscr {A}}{=} \dot {x}_2$ . A common extension of p and q in G will then force $\dot {x}_1 = \dot {x}_2$ .

To show that (1) holds for Case 3, suppose that for some $p \in G$ , $p \Vdash ^* \dot {x}_1 = \dot {x}_2$ .

Subcase 3.3. $\dot {x}_{1_G} = b$ for some urelement b. Then ${\left \langle b, r\right \rangle } \in \dot {x}_1$ for some $r \in G$ . Define:

• $F_1 = \{ q \in {\mathbb P} : \exists a, s_1, s_2 (\mathscr {A} (a) \land {\left \langle a, s_1\right \rangle } \in \dot {x}_1 \land {\left \langle a, s_2\right \rangle } \in \dot {x}_2 \land q \leq s_1, s_2) \}$ .

• $F_2 = \{q \in {\mathbb P} : \forall {\left \langle a, s_1\right \rangle } \in \dot {x}_1 \ (\mathscr {A}(a) \rightarrow q \bot s_1) \land \forall {\left \langle a, s_2\right \rangle } \in \dot {x}_2 \ (\mathscr {A}(a) \rightarrow q \bot s_2 )\}$ .

Since $p \Vdash ^* \dot {x}_1 \overset {\mathscr {A}}{=} \dot {x}_2$ , $F_1 \cup F_2$ is dense below p. But clearly $F_2 \cap G$ is empty as ${\left \langle b, r\right \rangle } \in \dot {x}_1$ , so there is some $q \in F_1 \cap G$ . It follows that ${\left \langle b, s_1\right \rangle } \in \dot {x}_1$ and ${\left \langle b, s_2\right \rangle } \in \dot {x}_2$ for some $s_1, s_2 \in G$ . Therefore, $\dot {x}_{2_G} = b = \dot {x}_{2_G}$ .

Subcase 3.4. $\dot {x}_{1_G}$ is a set. Suppose for reductio that $\dot {x}_{2_G}$ is some urelement b and so ${\left \langle b, r\right \rangle } \in \dot {x}_2$ for some $r \in G$ . Since $p \Vdash ^* \dot {x}_1 \overset {\mathscr {A}}{=} \dot {x}_2$ , it follows that there are some urelement a and $s \in G$ such that ${\left \langle a, s\right \rangle } \in \dot {x}_1$ . This implies that $\dot {x}_{1_G} = a$ , which is a contradiction. Hence, $\dot {x}_{2_G}$ is a set, so it remains to show that $\dot {x}_{1_G}$ and $\dot {x}_{2_G}$ have the same members. If $\dot {y}_G \in \dot {x}_{1_G}$ , then ${\left \langle \dot {y}, r\right \rangle } \in \dot {x}_1$ for some $r \in G$ . So there is some $q \in G$ with $q \leq p, r$ , and $q \Vdash ^* \dot {y} \in \dot {x}_2$ because $p \Vdash ^* \dot {x}_1 \subseteq \dot {x}_2$ . By the induction hypothesis, $\dot {y}_G \in \dot {x}_{2_G}$ . The same argument will show that $\dot {x}_{2_G} \subseteq \dot {x}_{1_G}$ .

Proof. By a standard argument as in [Reference Kunen11, Chapter VII, Theorem 3.6] using Lemma 3.9.

Proof. In M, we define $\dot {v}$ as follows:

$$ \begin{align*} \dot{v}= \bigcup_{p \in dom(f)}\{{\left\langle y, r\right\rangle} \in dom(f(p)) \times {\mathbb P} : \exists q \ ( {\left\langle y, q\right\rangle} \in f(p) \land r \leq p, q )\}. \end{align*} $$

We first check that $\dot {v}$ satisfies the incompatibility condition (ii) in Definition 3.4. Consider any ${\left \langle a, r_1\right \rangle }\in \dot {v}$ for some urelement a. Then there are $p_1, q_1$ such that $p_1 \in dom(f)$ and ${\left \langle a, q_1\right \rangle } \in f(p_1)$ and $r_1 \leq p_1, q_1$ . For any ${\left \langle x, r_2\right \rangle } \in \dot {v}$ with $x \neq a$ , there are $p_2, q_2$ such that $p_2 \in dom(f)$ and ${\left \langle x, q_2\right \rangle } \in f(p_2)$ and $r_2 \leq p_2, q_2$ . If $p_1 = p_2$ , then $r_1$ is incompatible with $r_2$ because $f(p_1)$ is a ${\mathbb P}$ -name. If not, $r_1$ is incompatible with $r_2$ because $dom(f)$ is an antichain.

Fix a $p \in dom(f)$ . We show that $p \Vdash \dot {v} = f(p)$ . Let G be an M-generic filter over ${\mathbb P}$ that contains p.

Case 1. $\dot {v}_{G}$ is some urelement a. Then ${\left \langle a, r\right \rangle } \in \dot {v}$ for some $r \in G$ . So for some $p' \in dom(f)$ and q, ${\left \langle a, q\right \rangle } \in f(p')$ and $r \leq p', q$ . So $p', q \in G$ and $p' = p$ . Therefore, $\dot {v}_{G} = f(p)_{G}$ .

Case 2. $\dot {v}_{G}$ is a set. Then $f(p)_{G}$ must be a set. Otherwise, $f(p)_{G}$ is some urelement a and there will be some $q \in G$ such that ${\left \langle a, q\right \rangle } \in f(p)$ ; then there is some $s \in G$ such that $s \leq q, p$ so ${\left \langle a, s\right \rangle } \in \dot {v}$ , which means $\dot {v}_{G}$ is a urelement—contradiction. For every $ \dot {x}_{G} \in \dot {v}_{G}$ with ${\left \langle \dot {x}, r\right \rangle } \in \dot {v}$ and $r \in G$ , ${\left \langle \dot {x}, q\right \rangle } \in f(p')$ and $r \leq p', q$ for some $p' \in dom(f)$ and q; so $p'= p$ and $ \dot {x}_{G} \in f(p)_{G}$ . This shows that $\dot {v}_{G} \subseteq f(p)_{G}$ . Consider any $\dot {x}_{G} \in f(p)_{G}$ such that ${\left \langle \dot {x}, q\right \rangle } \in f(p)$ for some $q \in G$ . Let $r \in G$ be a common extension of p and q. It follows that ${\left \langle \dot {x}, r\right \rangle } \in \dot {v}$ and so $\dot {x}_{G} \in \dot {v}_{G}$ . This shows that $f(p)_{G} \subseteq \dot {v}_{G}$ and the proof is completed.

Proof. (1) $\rightarrow $ (2). This is proved by a standard argument, and the point here is that we can now mix ${\mathbb P}$ -names using the new definition. Assume that $M \models $ Collection and we now work in M. Fix some ${\mathbb P}$ and suppose that $p \Vdash \exists y \varphi (y)$ for some $p \in {\mathbb P}$ . By AC there exists a maximal antichain X in the subposet $\mathbb {Q} = \{ q \in {\mathbb P} : q \leq p \land \exists \dot {y} \in M^{\mathbb P} q \Vdash \varphi (\dot {y})\}$ . It follows from Collection that there is some v such that for every $q \in X$ , there is some name $\dot {w} \in v$ with $q \Vdash \varphi (\dot {w})$ . By well-ordering v we can pick a $\dot {w}_q \in M^{\mathbb P}$ such that $q \Vdash \varphi (\dot {w}_q)$ for every $q \in X$ . Then by Lemma 3.11, there is a $\dot {v} \in M^{\mathbb P}$ such that $q \Vdash \dot {w}_q = \dot {v}$ for every $q \in X$ . Suppose that $p \nVdash \varphi (\dot {v})$ for reductio. Then there will be some $r \in \mathbb {Q}$ such that $r \Vdash \neg \varphi (\dot {v})$ , which means r is incompatible with every $q \in X$ , but this contradicts the maximality of X.

(2) $\rightarrow $ (1). Suppose that $M \models \forall x \in w \exists y \varphi (x, y, u)$ . Let ${\mathbb P}$ be the forcing poset $w \cup \{w\}$ such that for every $p, q \in {\mathbb P}$ , $p \leq q$ if and only if $p = q$ or $q = w$ . That is, w is $1_{\mathbb P}$ , while the members of w constitute the only maximal antichain. Thus, $M[G] = M$ for every generic filter G over ${\mathbb P}$ . Define $\dot {x} \in M^{\mathbb P}$ to be $\{ {\left \langle \check {z}, x\right \rangle } : z \in x \land x \in w\}$ . For every $x \in w$ and generic filter G containing x, since $\dot {x}_{G} = x$ it follows that $M[G] \models \exists y \varphi (\dot {x}_{G}, y, u)$ . Thus, $1_{\mathbb P} \Vdash \exists y \varphi (\dot {x}, y, \check {u})$ and by (2), $1_{\mathbb P} \Vdash \varphi (\dot {x}, \dot {y}, \check {u})$ for some $\dot {y} \in M^{\mathbb P}$ . For every $x \in w$ , let G be the generic filter containing x. Then $M[G] \models \varphi (x, \dot {y}_{G}, u)$ ; so $M \models \varphi (x, \dot {y}_{G}, u)$ and $ker(\dot {y}_{G}) \subseteq ker(\dot {y})$ by Lemma 3.6(7). It follows that $M \models \forall x \in w \ \exists y \in V(ker(\dot {y})) \ \varphi (x, y, u)$ , which suffices for Collection by Proposition 1.2.

Proof. The proofs of (1) and (2) are standard text-book arguments as in Kunen [Reference Kunen11, Chapter VII]. For (3), suppose that $M[G] \models \forall v \in \dot {w}_G\ \exists y \varphi (v, y, \dot {u}_G)$ for some $\dot {w}_G$ and $\dot {u}_G$ . In M, define

$$ \begin{align*} x = \{\langle \dot{x}, p \rangle \in (dom(\dot{w}) \cap M^{\mathbb P}) \times {\mathbb P} : \exists \dot{y} \in M^{\mathbb P} p \Vdash \varphi(\dot{x}, \dot{y}, \dot{u}) \}. \end{align*} $$

By Collection in M, there is a set of ${\mathbb P}$ -names v such that for every $\langle \dot {x}, p \rangle \in x$ , there is a $\dot {y} \in v$ with $p \Vdash \varphi (\dot {x}, \dot {y}, \dot {u})$ . Define $\dot {v}$ to be $v \times \{1_{\mathbb P}\}$ . It’s routine to check that $M[G] \models \forall x \in \dot {w}_G \ \exists y \in \dot {v}_G \ \varphi (x, y, \dot {u}_G)$ .

Proof. By induction on the rank of $\dot {x}$ . To show that $\overset {A}{\dot {x}}$ is always a ${\mathbb P}$ -name, we only need to check the incompatibility condition in Definition 3.4 holds. Suppose that ${\left \langle a, p\right \rangle }, {\left \langle y, q\right \rangle } \in \overset {A}{\dot {x}}$ , where a is a urelement and $y \neq a$ . If y is another urelement in $dom(\dot {x})$ , then p and q are incompatible; otherwise y is some $\overset {A}{\dot {z}}$ , where ${\left \langle \dot {z}, q\right \rangle } \in \dot {x}$ and $\dot {z}$ is a ${\mathbb P}$ -name, then p and q are incompatible because no urelement is a ${\mathbb P}$ -name. $ker(\overset {A}{\dot {x}}) \subseteq A$ because $ker(\overset {A}{\dot {x}})$ is contained in $\bigcup _{y \in dom(\dot {x})}ker(\overset {A}{y}) \cup ker({\mathbb P})$ , which is a subset of A by the induction hypothesis.

Proof. Suppose that $M[G] \models \forall v \in \dot {w}_G \exists ! y \varphi (v, y, \dot {u}_G)$ . Let $A = ker(\dot {w}) \cup ker({\mathbb P}) \cup ker(\dot {u})$ . By Lemma 3.13, we may assume M does not satisfy Collection so it has a proper class of urelements by Proposition 1.2.

Proof. Fix a $\dot {v}_G \in \dot {w}_G$ for some $\dot {v} \in dom (\dot {w}) \cap M^{\mathbb P}$ . There is a ${\mathbb P}$ -name $\mu $ and a $p \in G$ such that $ p \Vdash \varphi (\dot {v}, \mu , \dot {u}) \land \forall z (\varphi (\dot {v}, z, \dot {u}) \rightarrow \mu = z)$ .

Claim 3.17.1. For every M-generic filter H over ${\mathbb P}$ such that $p \in H$ , $ker(\mu _H) \subseteq A$ .

Proof of the Claim. Suppose not. Then there is some $b \in ker(\mu _H) - A$ . Since M has a proper class of urelements, there is some urelement $c \in M$ such that $c \notin A \cup ker(\mu )$ . In M, let $\pi $ be an automorphism that only swaps b and c. Since $\pi $ point-wise fixes A, it follows that

$$ \begin{align*} p \Vdash \varphi (\dot{v}, \pi \mu, \dot{u}) \land \forall z (\varphi (\dot{v}, z, \dot{u}) \rightarrow \pi \mu = z). \end{align*} $$

Thus, $M[H] \models \mu _H = (\pi \mu )_H$ . Since $b \in ker(\mu _H)$ , $\pi b \in ker(\pi \mu _H)$ ; but $\pi b = c \notin ker(\mu )$ and $ker(\mu _H) \subseteq ker(\mu )$ , so $\pi b \notin ker(\mu _H)$ , which is a contradiction.

Note that we cannot hope to show that $ker(\mu ) \subseteq A$ in general. For if $\mu ^*$ is some ${\mathbb P}$ -name such that $\mu ^* = \mu \cup \{\langle \{{\left \langle b, 1_{\mathbb P}\right \rangle }\}, q \rangle \}$ , where b is a urelement not in A and q is incompatible with p, we would still have $p \Vdash \mu = \mu ^*$ .

Claim 3.17.2. Let H be an M-generic filter over ${\mathbb P}$ such that $p \in H$ . For every $\dot {x}, \dot {y} \in M^{{\mathbb P}}$ , if $\dot {x}_H, \dot {y}_H \in TC(\{\mu _H\})$ , then $\dot {x}_H = \dot {y}_H$ if and only if $(\overset {A}{\dot {x})}_H = (\overset {A}{\dot {y}})_H$ .

Proof of the Claim. If $\dot {x}_H = \dot {y}_H = a$ for some urelement a, then by Claim 3.17.1 $a \in A$ . It is easy to check that $(\overset {A}{\dot {y})}_H = (\overset {A}{\dot {x}})_H = a$ . If $(\overset {A}{\dot {y})}_H = (\overset {A}{\dot {x}})_H = b$ for some urelement b, then $b \in A$ and it follows that $\dot {x}_H = \dot {y}_H = b$ .

So suppose $\dot {x}_H = \dot {y}_H$ are sets in $TC(\{\mu _H\})$ and the claim holds for every $\dot {z} \in dom(\dot {x}) \cup dom(\dot {y})$ . Clearly, $(\overset {A}{\dot {x})}_H$ and $(\overset {A}{\dot {y}})_H$ must also be sets. If $\overset {A}{\dot {z}}_H \in \overset {A}{\dot {x}}_H$ for some $\dot {z} \in M^{\mathbb P} \cap dom(\dot {x})$ , we have $\dot {z}_H \in \dot {y}_H = \dot {x}_H$ . So there is some $\dot {w} \in M^{\mathbb P} \cap dom(\dot {y})$ such that $\dot {w}_H = \dot {z}_H$ . $\dot {z}_H \in TC(\{\mu _H\})$ so by the induction hypothesis $\overset {A}{\dot {z}}_H = \overset {A}{\dot {w}}_H \in (\overset {A}{\dot {y}})_H$ . This shows that $\overset {A}{\dot {x}}_H \subseteq \overset {A}{\dot {y}}_H$ , and we will have $\overset {A}{\dot {x}}_H = \overset {A}{\dot {y}}_H$ by the same argument.

Now suppose that $\dot {x}_H, \dot {y}_H \in TC(\{\mu _H\})$ and $\overset {A}{\dot {x}}_H = \overset {A}{\dot {y}}_H$ are sets. Then $\dot {x}_H$ and $\dot {y}_H$ must be sets. For if, say, $\dot {x}_H = a$ for some urelement a, then $a \in A$ by Claim 3.17.1, which would yield $\overset {A}{\dot {x}}_H = a$ . Let $\dot {z}_H \in \dot {x}_H$ for some $\dot {z} \in M^{\mathbb P} \cap dom(\dot {x})$ . Then $\overset {A}{\dot {z}}_H \in \overset {A}{\dot {y}}_H$ and so $\overset {A}{\dot {z}}_H = \overset {A}{\dot {w}}_H$ for some $\dot {w}_H \in \dot {y}_H$ . By the induction hypothesis, it follows that $\dot {z}_H = \dot {w}_H$ . This shows that $\dot {x}_H \subseteq \dot {y}_H$ and consequently, $\dot {x}_H = \dot {y}_H$ .

Claim 3.17.3. $p \Vdash \overset {A}{\mu } = \mu $ .

Proof of the Claim. Let H be an M-generic filter on ${\mathbb P}$ that contains p. We show that $\overset {A}{\mu }_H = \mu _H$ . Let f be the function on $TC(\{\mu _H\})$ that sends every $\dot {y}_H$ to $\overset {A}{\dot {y}}_H$ , which is well-defined by Claim 3.17.2. Note that every $\in $ - isomorphism of transitive sets that fixes the urelements can only be the identity map. So it suffices to show that f maps $TC(\{\mu _H\})$ onto $TC(\{\overset {A}{\mu }_H\})$ , preserves $\in $ and fixes all the urelements.

f preserves $\in $ . Consider any $\dot {y}{_H}, \dot {x}{_H} \in TC(\{\mu _H\})$ . Suppose that $\dot {y}{_H} \in \dot {x}{_H}$ . Then $\dot {y}{_H} = \dot {z}_H$ for some $\dot {z} \in M^{\mathbb P} \cap dom(\dot {x})$ so $ \overset {A}{\dot {z}}_H \in \overset {A}{\dot {x}}_H$ ; by Claim 3.17.2, it follows that $\overset {A}{\dot {y}}_H = \overset {A}{\dot {z}}_H \in \overset {A}{\dot {x}}_H$ . Suppose that $\overset {A}{\dot {y}}_H \in \overset {A}{\dot {x}}_H$ . Then $\overset {A}{\dot {y}}_H = \overset {A}{\dot {z}}_H$ for some $\dot {z}_H \in \dot {x}_{H}$ so $\dot {y}{_H} = \dot {z}_H \in \dot {x}_{H}$ by Claim 3.17.2 again.

f maps $TC(\{\mu _H\})$ onto $TC(\{\overset {A}{\mu }_H\})$ . If $\dot {y}_H \in TC(\{\mu _H\})$ , then $\dot {y}_H \in \dot {y}_{1_H} \in \dots \in \dot {y}_{n_H} \in \mu _H$ for some n. Since f is $\in $ -preserving, it follows that $\overset {A}{\dot {y}}_H \in \overset {A}{\dot {y}_1}_H \in \dots \in \overset {A}{\dot {y}_n}_H \in \overset {A}{\mu }_H$ and hence $\overset {A}{\dot {y}}_H \in TC(\{{\overset {A}{\mu } }_H\})$ . To see it is onto, let $x \in x_1 \in \dots \in x_n \in \overset {A}{\mu }_H$ . Then $x = \overset {A}{\dot {y}}_H \in \overset {A}{\dot {y}_1}_H \in \dots \in \overset {A}{\dot {y}_n}_H \in \overset {A}{\mu }_H$ , but then $\dot {y}_H \in \dot {y}_{1_H} \in \dots \in \dot {y}_{n_H} \in \mu _H$ and hence $\dot {y}_H \in TC(\{\mu _H\})$ .

f fixes all the urelements in $TC(\{\mu _H\})$ . Suppose $\dot {x}_H = a \in TC(\{\mu _H\})$ for some urelement a. Then by Claim 3.17.1, $a \in A$ and hence $\overset {A}{\dot {x}}_H = a$ .

Lemma 3.17 is now proved by letting $\mu '$ be $\overset {A}{\mu }$ .

Now in M, we define

$$ \begin{align*} \bar{x} = \{\langle \dot{v}, p\rangle \in (dom(\dot{w}) \cap M^{\mathbb P}) \times {\mathbb P} : \exists \mu \in M^{{\mathbb P}} (ker(\mu) \subseteq A \land p \Vdash \varphi(\dot{v}, \mu, \dot{u})) \}. \end{align*} $$

For every $\langle \dot {v}, p \rangle \in \bar {x}$ , let $\alpha _{ \dot {v}, p }$ be the least $\alpha $ such that there is some $\mu \in V_\alpha (A) \cap M^{{\mathbb P}}$ such that $p \Vdash \varphi (\dot {v}, \mu , \dot {u})$ . Let $\beta = Sup_{\langle \dot {v}, p \rangle \in \bar {x}} \alpha _{\dot {v}, p}$ and set $\rho = (V_\beta (A) \cap M^{{\mathbb P}}) \times \{1_{\mathbb P}\}$ . It remains to show that $M[G] \models \forall x \in \dot {w}_G \ \exists y \in \rho _G \ \varphi (x, y, \dot {u}_G)$ . Let $\dot {v}_G \in \dot {w}_G$ . By Lemma 3.17, there is some $p \in G$ such that $\langle \dot {v}, p \rangle \in \bar {x}$ . So there is some ${\mathbb P}$ -name $\mu \in dom(\rho )$ such that $p \Vdash \varphi (\dot {v}, \mu , \dot {u})$ . Thus, $M[G] \models \varphi (\dot {v}_G, \mu _G, \dot {u}_G)$ and $\mu _G \in \rho _G$ .