2.1 ${\mathsf {AC}}_\omega ( X )$ follows from $\mathsf {DC} ( X )$ together with ${\mathsf {AC}}_\omega ( \mathbb {R} )$

Let us start with the following combinatorial result that might be of independent interest. It is stated for families of sets indexed by an arbitrary set I, but when $I = \omega $ the assumption $\mathsf {AC}_I ( \mathscr {P} ( I ) )$ becomes ${\mathsf {AC}}_\omega ( \mathbb {R} )$ .

Proof Let $F \colon \bigcup _{i \in I} X_i \to \mathscr {P} ( I )$ , $F ( x ) = \mathopen \{{ i \in I }\boldsymbol \mid { x \in X_i } \mathclose \}$ and let $A_i = \mathopen \{{ a \in \operatorname {\mathrm {ran}} ( F ) }\boldsymbol \mid { i \in a } \mathclose \}$ . Observe that for all $x \in \bigcup _{i \in I} X_i$ and all $i \in I$

(1) $$ \begin{align} x \in X_i \Leftrightarrow F ( x ) \in A_i. \end{align} $$

In particular, $\emptyset \neq A_i \subseteq \mathscr {P} ( I )$ for all $i \in I$ . By $\mathsf {AC}_I ( \mathscr {P} ( I ) )$ pick $a_i \in A_i$ , and let $Y_i = F^{-1} ( \mathopen \{ {a_i} \mathclose \} ) \subseteq \bigcup _{i \in I} X_i$ . Then

$$\begin{align*}Y_i = \mathopen \{{ x }\boldsymbol\mid{ F ( x ) = a_i } \mathclose\} = \mathopen \{{ x }\boldsymbol\mid{ \mathopen \{{ j }\boldsymbol\mid{ x \in X_j} \mathclose\} = a_i } \mathclose\}, \end{align*}$$

and since $i \in a_i$ , then $Y_i \subseteq X_i$ . The sets $Y_i$ need not be distinct as the $a_i$ s need not be distinct, but if $a_i \neq a_ j$ , then $Y_i \cap Y_j = \emptyset $ .

By (1) if the $X_i$ s are finite, then so are the $A_i$ s. If, moreover, $\mathscr {P} ( I )$ is linearly orderable (e.g., when I is well-orderable), then the $a_i$ s can be chosen without appealing to any choice axiom. Therefore:

The following result follows from the argument of Theorem 2.8 together with Corollary 2.7.

2.2 Does $\mathsf {DC} ( X )$ imply ${\mathsf {AC}}_\omega ( X )$ ?

By Theorems 2.4 and 2.8,

(2) $$ \begin{align} \forall {X} \, ( \mathsf{DC} ( X ) \Rightarrow {\mathsf{AC}}_\omega ( X ) ) \end{align} $$

follows from either one of the following assumptions:

• • $X \times 2 \precsim X$ for all infinite X,

• • ${\mathsf {AC}}_\omega ( \mathbb {R} )$ .

Sageev in [Reference Sageev10] proved that “ $X \times 2 \precsim X$ for all infinite X” does not imply ${\mathsf {AC}}_\omega ( \mathbb {R} )$ , while Monro in [Reference Monro9] proved that $\mathsf {DC}$ (and hence the weaker ${\mathsf {AC}}_\omega ( \mathbb {R} )$ ) does not imply “ $X \times 2 \precsim X$ for all infinite X”. So neither assumption implies the other.

The obvious question is if (2) is a theorem of $\mathsf {ZF}$ . Suppose that there is a set X such that $\mathsf {DC} ( X ) \wedge \neg {\mathsf {AC}}_\omega ( X )$ . By the proof of Lemma 2.6 the set $A \mathrel {{:}{=}} F [ X ] \subseteq \mathscr {P} ( \omega )$ is such that $\mathsf {DC} ( A )$ holds, as A is the surjective image of X, and $ {\mathsf {AC}}_\omega ( A )$ fails, as otherwise, arguing as in Theorem 2.8, ${\mathsf {AC}}_\omega ( X )$ would hold. Therefore if (2) fails, then the witness of this failure can be taken to be a subset of $\mathbb {R}$ . In Section 4 we construct a model of $\mathsf {ZF}$ in which

(3) $$ \begin{align} \exists { A \subseteq \mathbb{R}} \, \left ( \mathsf{DC} ( A ) \wedge \neg {\mathsf{AC}}_\omega ( A ) \right ) \end{align} $$

showing that (2) is not a theorem of $\mathsf {ZF}$ .

By Lemma 2.5(c), any A as in (3) cannot contain a nonempty perfect set. Moreover, such a set A also needs to be Dedekind-infinite: indeed, A cannot be closed, as otherwise, by the usual Cantor–Bendixson argument, it would either be countable, contradicting $\neg {\mathsf {AC}}_\omega (A)$ , or else it would contain a nonempty perfect set, which we already excluded; therefore A is not closed, and, by Lemma 2.2, A is Dedekind-infinite.

It can be shown that (3) fails both in Cohen’s first model (Proposition 3.4) and in the Feferman–Levy model (Proposition 5.3), and hence in both these models (2) holds.

2.3 An equivalent formulation of $\mathsf {DC}$

A tree on X is a $T \subseteq {}^{< \omega } X$ that is closed under initial segments, that is if $t \in T$ and $s \subseteq t$ then $s \in T$ . A tree T on X is pruned if for every $t \in T$ there is $s \in T$ such that $t \subset s$ . A branch of T is a $b \colon \omega \to X$ such that $\forall {n \in \omega } \, ( b {\mathop {\upharpoonright }} n \in T)$ . A tree T is ill-founded if it has a branch; otherwise it is well-founded. Let

(DC _ω (X)) $$\begin{align} \text{Any nonempty pruned tree on }X\text{ is ill-founded} \end{align}$$

and let $\mathsf {DC}_ \omega $ be $\forall {X} \, \mathsf {DC}_ \omega ( X )$ . As $\mathsf {DC}$ is equivalent to $\mathsf {DC}_ \omega $ (Corollary 2.11), the axiom of dependent choice is often stated as $\mathsf {DC}_\omega $ . The advantage of this formulation is that it can be generalized to ordinals larger than $\omega $ .

Proof ( $\Rightarrow $ ) Suppose R is a binary relation on ${}^{< \omega } X$ such that $\forall {s} \, \exists {t} \, ( s \mathrel {R} t )$ . If $\emptyset \mathrel {R} \emptyset $ , then $\mathopen \langle \emptyset , \emptyset , \dots \mathclose \rangle $ is an R-chain as required, so we may assume otherwise. Let $R' \subseteq R$ be the sub-relation on ${}^{< \omega } X$ defined by

$$\begin{align*}s \mathrel{R}' t \Leftrightarrow s \mathrel{R} t \wedge \forall {t'} \, ( s \mathrel{R} t' \Rightarrow \operatorname{\mathrm{lh}} ( t' ) \ge \operatorname{\mathrm{lh}} ( t ) ). \end{align*}$$

The relation $R'$ is total and any $R'$ -chain is an R-chain. Then

$$\begin{align*}T = \mathopen \{{t \in {}^{< \omega } X}\boldsymbol\mid{ \exists {s_0 , \dots , s_n } \, ( \emptyset \mathrel{R}' s_0 \mathrel{R}' \dots \mathrel{R}' s_n \, \wedge \, t \subseteq s_0 {}^\smallfrown s_1 {}^\smallfrown \dots {}^\smallfrown s_n ) } \mathclose\} \end{align*}$$

is a pruned tree on X, so it has a branch. By the minimality assumption of $R'$ , given a branch b of T one can construct inductively an $R'$ -chain $( s_n )_{ n \in \omega }$ such that $s_0 {}^\smallfrown s_1 {}^\smallfrown \dots {}^\smallfrown s_n \subseteq b$ for all n.

( $\Leftarrow $ ) If T is a pruned tree on X, let $R \subseteq T \times T$ be defined by

$$\begin{align*}s \mathrel{R}t \Leftrightarrow s \subset t \wedge \operatorname{\mathrm{lh}} ( s ) + 1 = \operatorname{\mathrm{lh}} ( t ). \end{align*}$$

As $T \subseteq {}^{< \omega } X$ then $\mathsf {DC} ( T )$ holds, and since R is total, as T is pruned, there is an R-chain. Any such chain yields a branch of T.

In light of Proposition 2.12, our main result, Theorem 4.1, tells us it is consistent with $\mathsf {ZF}$ that there is a set $A\subseteq \mathbb {R}$ for which $\mathsf {DC} (A)$ holds but $ \mathsf {DC}_ \omega ( A )$ fails.

3.1 The first Cohen model

Let $\mathbf { P}_0$ be the forcing that adds countably many Cohen reals, i.e.,

$$\begin{align*}\mathbf{ P}_0 = \mathopen \{{ p }\boldsymbol\mid{ \exists {I \subseteq \omega } \, ( p \colon I \to {}^{<\omega} 2 \text{, and }I \text{ is finite}) } \mathclose\} , \end{align*}$$

with $p \leq q$ if $\operatorname {\mathrm {dom}} ( p ) \supseteq \operatorname {\mathrm {dom}} ( q )$ and $p ( n ) \supseteq q ( n )$ for all $n \in \operatorname {\mathrm {dom}} ( q )$ . Although this is not the standard presentation of such a forcing, this way of defining $\mathbf { P}_0$ will become useful in the Section 4. Let $\dot {a}_n$ be the canonical name for the nth Cohen real, that is,

(4) $$ \begin{align} \dot{a}_n = \mathopen \{{ ( \check{ ( k , i ) } , p ) }\boldsymbol\mid{ p \in \mathbf{ P}_0 \wedge n \in \operatorname{\mathrm{dom}} p \wedge p ( n ) ( k ) = i } \mathclose\}. \end{align} $$

Observe that $\dot {A} \mathrel {{:}{=}} \mathopen \{{ \dot {a}_n }\boldsymbol \mid { n \in \omega } \mathclose \} ^\bullet $ is forced to be a dense subset of ${}^ \omega 2$ .

Every permutation $\pi $ on $\omega $ induces an automorphism of $\mathbf { P}_0$ as follows: given $p \in \mathbf { P}_0$ , we let $\pi p \in \mathbf { P}_0$ be defined by

$$\begin{align*}\forall { n \in \operatorname{\mathrm{dom}} ( p ) } \, \big ( \pi p ( \pi n ) = p ( n ) \big ). \end{align*}$$

We conflate the notation by using the same symbol $\pi $ to denote both the permutation and the automorphism on $\mathbf { P}_0$ it induces. Let $\mathcal {G}_0$ be the group of all such automorphisms. For every finite $E \subset \omega $ , let $\operatorname {\mathrm {Fix}} ( E )$ be the subgroup of $\mathcal {G}_0$ of all those automorphisms induced by permutations that pointwise fix the set E. Let $\mathcal {F}_0$ be the filter on $\mathcal {G}_0$ generated by $\mathopen \{{ \operatorname {\mathrm {Fix}} (E) }\boldsymbol \mid { E \subset \omega \text { finite} } \mathclose \}$ . It is easy to check that $\mathcal {F}_0$ is actually a normal filter on $\mathcal {G}_0$ , and hence $\mathopen \langle \mathbf { P}_0 , \mathcal {G}_0 , \mathcal {F}_0 \mathclose \rangle $ is a symmetric system. Let G be a $\mathord {\mathrm {V}}$ -generic filter for $\mathbf { P}_0$ , and let $\mathcal {N}_0$ be the corresponding symmetric extension, which we call first Cohen model.

Denote by A the realization of the name $\dot {A}$ in $\mathord {\mathrm {V}} [ G ]$ , i.e., the set $\dot {A}_G$ . Note that every $\dot {a}_n$ is in $\mathsf {HS} _{\mathcal {F}_0 }$ and so is $\dot {A}$ .

In $\mathcal {N}_0$ , the set A, being infinite and D-finite, is certainly not separable as a subspace of $\mathbb {R}$ —indeed, every infinite, separable $\mathrm {T}_1$ space is Dedekind-infinite. Moreover, $ \mathsf {DC} ( A )$ also fails, as otherwise A would be Dedekind-infinite (see the penultimate paragraph of Section 2.2).

The simultaneous local failure of both ${\mathsf {AC}}_\omega $ and $\mathsf {DC}$ is not accidental—the next proposition shows that, in the first Cohen model, any X such that $\mathsf {DC} ( X )$ holds must be well-orderable, and hence ${\mathsf {AC}}_\omega ( X )$ holds.

Proof If $\bigcup Y$ has no limit points, then it is discrete, so $\omega \precsim \bigcup Y$ . Suppose otherwise, and let $x \in \mathbb {R}$ be a limit point of $\bigcup Y$ . It is enough to show that $\omega \precsim Y$ and then apply Lemma 3.5 with $X = \mathbb {R}$ . Without loss of generality, we may assume that $\mathopen \{ { x} \mathclose \} , \emptyset \notin Y$ . For all $Z \in Y$ let $d ( x , Z ) = \min \mathopen \{{ | r - x | }\boldsymbol \mid { r \in Z \setminus \mathopen \{ { x } \mathclose \} } \mathclose \}$ be the distance of x from the rest of Z. Let $R \subseteq Y ^2$ be the binary relation defined as follows: for every $Z , W \in Y$ ,

$$\begin{align*}R ( Z , W ) \Leftrightarrow d ( x , W ) < d ( x , Z ). \end{align*}$$

The relation R is acyclic, and, by our hypothesis on x, it is total. It follows from $\mathsf {DC} ( Y )$ that R has an infinite chain, and hence $\omega \precsim Y$ .

4.1 Outline of the proof

We prove the theorem via an iteration of symmetric extensions of length $\omega $ . We start the iteration with the first Cohen model $\mathcal {N}_0$ , with $A \in \mathcal {N}_0$ being the generic D-finite set of reals (see Section 3.1). As noted right after Proposition 3.3, in $\mathcal {N}_0$ the set A is not separable (in particular ${\mathsf {AC}}_\omega ( A )$ fails) and $\mathsf {DC} ( A )$ fails. Next, we define a chain of models $\mathcal {N}_0 \subset \mathcal {N}_1 \subset \dots \subset \mathcal {N}_\omega $ such that, for each n, $\mathcal {N}_{n+1}$ is a symmetric extension of $\mathcal {N}_n$ that contains a generic set of chains for all binary relations in $\mathcal {N}_n$ that are total and acyclic on A. At the final stage, $\mathcal {N}_\omega $ , which is our model, is going to be something resembling “the model of sets definable from finitely many elements of $\bigcup _n \mathcal {N}_n$ ”. If we do the construction properly, we can prove that in $\mathcal {N}_\omega $ we have added enough countable subsets of A (or, equivalently, enough sequences over A) to guarantee $\mathsf {DC} ( A )$ (Theorem 4.10), but A is still not separable, in particular ${\mathsf {AC}}_\omega ( A )$ fails (Corollary 4.8).

Actually, we don’t only show that A is not separable in our model, but we give a topological characterization of its separable subsets: among the subsets of A, the separable ones are precisely those which are scattered with finite scattered height (Definition 4.5 and Theorem 4.7).

4.2 The symmetric system

We define recursively a sequence $\mathopen \langle \mathbf { P}_n , \mathcal {G}_n , \mathcal {F}_n \mathclose \rangle _{ n \in \omega }$ of symmetric systems. Let $\mathopen \langle \mathbf { P}_0 , \mathcal {G}_0 , \mathcal {F}_0 \mathclose \rangle $ be the symmetric system defined in Section 3.1, i.e., the one that induces the first Cohen model. For each n we denote by $\leq _n, \Vdash _n$ the ordering and the forcing relation of $\mathbf { P}_n$ , respectively, and by $\mathsf {HS} _n$ the class $\mathsf {HS} _{ \mathcal {F}_n }$ , i.e., the class of all hereditarily $\mathcal {F}_n$ -symmetric $\mathbf { P} _n$ -names. We also let

(5) $$ \begin{align} \mathcal{R}_n = \mathopen \{{ \dot{R} \in \mathsf{HS} _n }\boldsymbol\mid{ \forall { \dot{x} \in \operatorname{\mathrm{dom}} ( \dot{R} ) } \, \exists { m , k } \, \in \omega \ ( \dot{x} = ( \dot{a}_{ m} , \dot{a}_{k} )^\bullet ) } \mathclose\}, \end{align} $$

where the $\dot {a}_i$ s are as in (4). Then $\mathcal {R}_n$ is the set of all “good” hereditarily $\mathcal {F}_n$ -symmetric $\mathbf { P} _n$ -names for binary relations on $\dot {A}$ .

Recursively on n, we define $\mathbf { P} _{n+1}$ to be the set of all the sequences $p = \mathopen \langle p_k \boldsymbol \mid k \leq n + 1 \mathclose \rangle $ such that:

1. (1) $p {\mathop {\upharpoonright }} n + 1 \in \mathbf { P} _{n}$ .

2. (2) $p_{n+1} \colon \operatorname {\mathrm {dom}} ( p_{n + 1 } ) \to \mathcal {R}_n \times {}^{<\omega } \omega $ with $\operatorname {\mathrm {dom}} ( p_{n+1} )$ a finite subset of $\omega $ .

3. (3) For each $k \in \operatorname {\mathrm {dom}} ( p_{n + 1 } )$ with $p_{n+1} ( k ) = ( \dot {R}, \mathopen \langle m_0 , \dots , m_h \mathclose \rangle )$ we have

   $$\begin{align*}p {\mathop{\upharpoonright}} n + 1 \Vdash _n "\dot{R} \text{ is total, acyclic and }\dot{a}_{ m_0 } \mathrel{\dot{R}} \dot{a}_{ m_1 } \mathrel{\dot{R}} \dots \mathrel{\dot{R}} \dot{a}_{ m_h }", \end{align*}$$

where, at stage $n = 0$ , we identify the conditions $p \in \mathbf { P} _0$ with their singleton sequence  $\mathopen \langle p \mathclose \rangle $ .

For each $p \in \mathbf { P} _{n+1}$ and $k \in \operatorname {\mathrm {dom}} ( p_{ n + 1 } )$ with $p_{ n + 1 } ( k ) = ( \dot {R} , s )$ , we denote $\dot {R}$ and s by $p_{n + 1}^R ( k )$ and $p_{n + 1}^s ( k )$ , respectively. Given $p , q \in \mathbf { P} _{ n + 1 }$ we let $p \leq _{ n + 1 } q$ if and only if:

• • $p {\mathop {\upharpoonright }} n + 1 \leq _n q {\mathop {\upharpoonright }} n + 1$ ,

• • $\operatorname {\mathrm {dom}} ( p_{n + 1} ) \supseteq \operatorname {\mathrm {dom}} ( q_{n + 1 } )$ ,

• • $\forall { k \in \operatorname {\mathrm {dom}} ( q_{n + 1 } ) } \, ( p_{n + 1 }^R ( k ) = q_{ n + 1 }^R ( k )$ and $p_{ n + 1 }^s ( k ) \supseteq q_{n + 1 }^s ( k ))$ .

This defines the forcing $\mathbf { P} _{n + 1}$ . Now we are left to define the subgroup $\mathcal {G}_{ n + 1 }$ of $\operatorname {\mathrm {Aut}} ( \mathbf { P} _{n + 1 } )$ , and the filter $\mathcal {F}_{ n + 1 }$ .

Consider a sequence $\vec {\pi } = \mathopen \langle \pi _0, \dots , \pi _{ n + 1 } \mathclose \rangle $ with each $\pi _i$ being a permutation of $\omega $ . By induction hypothesisFootnote ¹

$$\begin{align*}\vec{\rho} \mathrel{{:}{=}} \mathopen \langle \pi_0, \dots, \pi_n \mathclose \rangle = \vec{\pi} {\mathop{\upharpoonright}} n + 1 \end{align*}$$

induces an automorphism $\vec { \rho } \in \mathcal {G}_n$ . Note that, as in Section 3.1, we conflate the notation by using the same symbol to denote both sequences of permutations and the automorphisms they induce. Now, the sequence $\vec {\pi }$ induces an automorphism on $\mathbf { P} _{ n + 1 }$ as follows: given $p \in \mathbf { P} _{ n + 1 }$ , we let $\vec {\pi } p$ be the condition in $\mathbf { P} _{ n + 1 }$ such that $( \vec {\pi } p ) {\mathop {\upharpoonright }} n + 1 = \vec {\rho } ( p {\mathop {\upharpoonright }} n + 1)$ and, for each $k \in \operatorname {\mathrm {dom}} ( p_{ n + 1 } )$ with $p_{ n + 1 }^s ( k ) = \mathopen \langle m_0 , \dots , m_h \mathclose \rangle $ and $p_{ n + 1 }^R ( k ) = \dot {R}$ ,

$$ \begin{align*} ( \vec{\pi} p )_{ n + 1 }^R ( \pi_{ n + 1 } ( k ) ) & = \vec{\rho} ( \dot{R} ) , \\ ( \vec{\pi} p )_{ n + 1 }^s ( \pi_{ n + 1 } ( k ) ) & = \mathopen \langle \pi_0 ( m_0 ) , \dots, \pi_0 ( m_h ) \mathclose \rangle. \end{align*} $$

Let $\mathcal {G}_{n + 1}$ be the group of all such automorphisms on $\mathbf { P} _{n + 1}$ , i.e., the ones induced by sequences (of length $n + 2$ ) of permutations of $\omega $ . For each sequence ${\vec {H} = \mathopen \langle H_0 , \dots , H_{ n + 1 } \mathclose \rangle }$ of subsets of $\omega $ , we let $\operatorname {\mathrm {Fix}} ( \vec {H} )$ be the subgroup of all those $\vec {\pi } \in \mathcal {G}_{n + 1}$ such that $\pi _k$ pointwise fixes $H_k$ for all $k \leq n + 1$ . We define $\mathcal {F}_{n + 1}$ as the filter on $\mathcal {G}_{n + 1}$ generated by $\mathopen \{{ \operatorname {\mathrm {Fix}} ( \vec {H} ) }\boldsymbol \mid { H_k \text { is finite for all } k \leq n +1 } \mathclose \}$ . From now on, we use the symbol $\vec {H}$ to denote finite sequences of finite subsets of $\omega $ .

This ends the inductive definition of the sequence $\mathopen \langle \mathbf { P} _n , \mathcal {G}_n , \mathcal {F}_n \mathclose \rangle _{ n \in \omega }$ . Note that, for each $n < m$ , there is a natural complete embedding $i_{ n , m } \colon \mathbf { P} _n \to \mathbf { P} _m$ and a natural embedding $j_{ n , m } \colon \mathcal {G}_n \to \mathcal {G}_m$ . Thus we let $\mathbf { P}$ and $\mathcal {G}$ be the direct limits of the forcings $\mathbf { P} _n$ and of the groups $\mathcal {G}_n$ , respectively.

We now define the normal filter $\mathcal {F}$ on $\mathcal {G}$ in the expected way: we let $\mathcal {F}$ be the filter generated by

$$\begin{align*}\mathopen \{{ \operatorname{\mathrm{Fix}} ( \vec{H} ) }\boldsymbol\mid{ H_k \text{ is finite for all } k < \operatorname{\mathrm{lh}} ( \vec{H} ) } \mathclose\} , \end{align*}$$

where, given any $\vec {H}$ , $\operatorname {\mathrm {Fix}} ( \vec {H} )$ is the subgroup of $\mathcal {G}$ made of all those $\vec {\pi }$ such that $\pi _k$ pointwise fixes $H_k$ for all $k < \operatorname {\mathrm {lh}} ( \vec {H} )$ .

A condition p of the direct limit $\mathbf { P}$ is a finite sequence $\mathopen \langle p_0, \dots , p_n \mathclose \rangle $ , and it is identified with $\mathopen \langle p_0, \dots , p_n , \emptyset , \dots , \emptyset \mathclose \rangle $ and with $\mathopen \langle p_0, \dots , p_n, \emptyset , \emptyset \dots \mathclose \rangle $ , that is a sequence obtained by concatenating p with a finite sequence of empty sets or with the infinite sequence of empty sets. We treat analogously the $\vec {H}$ s.

Henceforth $\mathopen \langle \mathbf { P} , \mathcal {G}, \mathcal {F} \mathclose \rangle $ is our symmetric system, with $\mathsf {HS}$ being the class of all $\mathcal {F}$ -symmetric $\mathbf { P}$ -names and $\leq , \Vdash $ being the ordering and the forcing relation of $\mathbf { P}$ , respectively.

Given an $\dot {x} \in \mathsf {HS}$ , we say that $\vec {H}$ is a support of $\dot {x}$ if $\vec {\pi } \dot {x} = \dot {x}$ for all $\vec {\pi } \in \operatorname {\mathrm {Fix}} ( \vec {H} )$ . Also, given $p = \mathopen \langle p_0, \dots , p_{n} \mathclose \rangle \in \mathbf { P}$ and $\vec {H} = \mathopen \langle H_0, \dots , H_n \mathclose \rangle $ , we write $p {\mathop {\upharpoonright }} \vec {H}$ to denote the sequence $\mathopen \langle p_0 {\mathop {\upharpoonright }} H_0 , \dots , p_{n} {\mathop {\upharpoonright }} H_{n} \mathclose \rangle $ . Note that the latter sequence needs not to belong to  $\mathbf { P}$ .

Lemma 4.3 (Restriction Lemma).

Let be a formula in the forcing language, and let $\dot {x}_1,\dots , \dot {x}_n \in \mathsf {HS}$ . For any $p \in \mathbf { P}$ and for any $\vec {H}$ , if $\vec {H}$ is a support for each of the $\dot {x}_i$ ’s and, for all $m> 0$ , for all $k \in H_m \cap \operatorname {\mathrm {dom}} (p_m)$ , $\vec {H} {\mathop {\upharpoonright }} m$ is a support for $ p^R_m ( k )$ and $\operatorname {\mathrm {ran}} ( p^s_m ( k ) ) \subseteq H_0$ , then $p {\mathop {\upharpoonright }} \vec {H} \in \mathbf { P}$ and

Proof We prove the lemma by induction on the length of $\vec {H}$ .

Let’s first assume $\vec {H} = \mathopen \langle H_0 \mathclose \rangle $ for some finite $H_0\subset \omega $ , then $p {\mathop {\upharpoonright }} \vec {H} \in \mathbf { P} _0$ . Assume for a contradiction that , then there is a $q \leq p {\mathop {\upharpoonright }} \vec {H}$ such that . Let $\vec {\pi } \in \mathcal {G}$ such that $\pi _0$ pointwise fixes $H_0$ and such that $\pi _0[ \operatorname {\mathrm {dom}} (q_0)] \cap \operatorname {\mathrm {dom}} (p_0) = H_0 \cap \operatorname {\mathrm {dom}} (p_0)$ and $\pi _m[ \operatorname {\mathrm {dom}} (q_m)] \cap \operatorname {\mathrm {dom}} (p_m) = \emptyset $ for all $m> 0$ . In particular, $\vec {\pi } \in \operatorname {\mathrm {Fix}} ( \vec {H} )$ . By hypothesis, $\vec {H}$ is a support for all the $\dot {x}_i$ ’s. Thus, by the Symmetry Lemma, . However, p and $\vec {\pi } q$ are compatible, contradiction.

Now let’s assume that $\vec {H} = \mathopen \langle H_0, \dots , H_m \mathclose \rangle $ .

Claim 4.3.1. $p {\mathop {\upharpoonright }} \vec {H} \in \mathbf {P}_m$ .

Proof By induction hypothesis, $p {\mathop {\upharpoonright }} ( \vec {H} {\mathop {\upharpoonright }} m ) \in \mathbf {P}_{ m - 1 }$ . Fix a $k \in H_m \cap \operatorname {\mathrm {dom}} (p_m)$ . Let $\dot {R} = p_m^R ( k )$ and $\mathopen \langle n_0 , \dots , n_h \mathclose \rangle = p_m^s ( k )$ . Then, by assumption, $\vec {H}{\mathop {\upharpoonright }} m$ is a support of $\dot {R}$ and $n_i \in H_0$ for every $i \le h$ , or, equivalently, $H_0$ is a support for $\dot {a}_{n_i}$ . By definition of $\mathbf {P}_m$ ,

$$\begin{align*}p {\mathop{\upharpoonright}} m \Vdash "\dot{R} \text{ is total, acyclic and }\dot{a}_{n_0} \mathrel{\dot{R}} \dot{a}_{n_1} \mathrel{\dot{R}} \dots \mathrel{\dot{R}} \dot{a}_{n_h}". \end{align*}$$

By induction hypothesis,

$$\begin{align*}( p {\mathop{\upharpoonright}} \vec{H} ) {\mathop{\upharpoonright}} m = p {\mathop{\upharpoonright}} ( \vec{H} {\mathop{\upharpoonright}} m ) \Vdash "\dot{R} \text{ is total, acyclic and }\dot{a}_{n_0} \mathrel{\dot{R}} \dot{a}_{n_1} \mathrel{\dot{R}} \dots \mathrel{\dot{R}} \dot{a}_{n_h}". \end{align*}$$

Therefore, $p {\mathop {\upharpoonright }} \vec {H} \in \mathbf {P}_m$ .

Assume for a contradiction that , then there is a $q \leq p {\mathop {\upharpoonright }} \vec {H}$ such that . Let $\vec {\pi } \in \mathcal {G}$ such that $\pi _l$ pointwise fixes $H_l$ for each $l \leq m$ and such that $\pi _l [ \operatorname {\mathrm {dom}} ( q_l ) ] \cap \operatorname {\mathrm {dom}} ( p_l ) = H_l \cap \operatorname {\mathrm {dom}} ( p_l )$ for all $l \leq m$ and $\pi _l [ \operatorname {\mathrm {dom}} ( q_l ) ] \cap \operatorname {\mathrm {dom}} ( p_l ) = \emptyset $ for all $l> m$ . In particular, $\vec {\pi } \in \operatorname {\mathrm {Fix}} ( \vec {H} )$ . Thus, by the Symmetry Lemma, .

Claim 4.3.2. p and $\vec {\pi } q$ are compatible.

Proof It suffices to show that $p_l^R ( k ) = ( \vec {\pi } q )_l^R ( k )$ and that the sequence $p_l^s ( k )$ is extended by $( \vec {\pi } q)_l^s ( k )$ , for every $l \le m$ and for every $k \in \operatorname {\mathrm {dom}} ( p_l ) \cap \operatorname {\mathrm {dom}} ( ( \vec { \pi } q )_l )$ . Note that $\operatorname {\mathrm {dom}} ( ( \vec {\pi } q )_l ) = \pi _l [ \operatorname {\mathrm {dom}} ( q_l ) ]$ . Fix an $l \le m$ and a $k \in \operatorname {\mathrm {dom}} ( p_l ) \cap \pi _l [ \operatorname {\mathrm {dom}} ( q_l )]$ . By the way we chose $\vec {\pi }$ , we must have $k \in H_l$ . Also, as we assumed $q \le p$ , we have $q_l^R ( k ) = p_l^R ( k )$ and $q_l^s ( k ) \supseteq p_l^s ( k )$ . Moreover, we assumed $\vec {H} {\mathop {\upharpoonright }} l$ to be a support for $p_l^R ( k )$ , and we have picked $\vec {\pi }$ so that $\vec {\pi } \in \operatorname {\mathrm {Fix}} ( \vec {H} )$ . In particular, $(\vec {\pi } {\mathop {\upharpoonright }} l) \, ( p_l^R ( k ) ) = p_l^R ( k )$ and $\pi _l ( k ) = k$ . Therefore, by the definition of the induced automorphism $\vec {\pi } \in \mathcal {G}$ ,

$$\begin{align*}( \vec{ \pi } q )_l^R ( k ) = (\vec{\pi} {\mathop{\upharpoonright}} l) \, ( q_l^R ( \pi_l^{-1} ( k ) ) ) = (\vec{\pi} {\mathop{\upharpoonright}} l) \,( q_l^R ( k ) ) = (\vec{\pi} {\mathop{\upharpoonright}} l) \, ( p_l^R ( k ) ) = p_l^R ( k ). \end{align*}$$

Moreover, since we assumed $\operatorname {\mathrm {ran}} ( p_l^s ( k ) ) \subseteq H_0$ , and $\pi _0 \in \operatorname {\mathrm {Fix}} ( H_0 )$ , we have, for every $i< \operatorname {\mathrm {lh}} ( p_l^s ( k ) )$ ,

$$\begin{align*}( \vec{\pi} q )_l^s ( k ) ( i ) = \pi_0 ( q_l^s ( \pi_l^{-1} ( k ) ) ( i ) ) = \pi_0 ( q_l^s ( k ) ( i ) ) = \pi_0 ( p_l^s ( k ) ( i ) ) = p_l^s ( k ) ( i ) , \end{align*}$$

and therefore $( \vec {\pi } q )_l^s ( k ) \supseteq p_l^s ( k )$ .

As before, the fact that p and $\vec {\pi } q$ are compatible yields the desired contradiction and concludes the proof.

4.3 The model

For each $n , k \in \omega $ , we let

$$ \begin{align*} \dot{f}_{ n , k } &= \mathopen \{{ ( ( \check{l} , \dot{a}_m )^\bullet , p ) }\boldsymbol\mid{ l , m \in \omega \, \wedge \, p \in \mathbf{ P}_{ n + 1 } \, \wedge \, p^s_{ n + 1 } ( k ) ( l ) = m } \mathclose\} , \\ \dot{F}_n & = \mathopen \{{ \dot{f}_{ n , k } }\boldsymbol\mid{ k \in \omega } \mathclose\}^\bullet. \end{align*} $$

Note that these $\mathbf { P}$ -names are in $\mathsf {HS}$ . These, together with $\dot {A}$ and the $\dot {a}_n$ s, are the (hereditarily symmetric) names for all the generic sets we are interested in. Observe that $\dot {f}_{ n , k }$ is a $\mathbf { P}_{ n + 1 }$ -name for an R-chain belonging to $\mathcal {N}_{ n + 1}$ , where R is the relation with $\mathbf { P}_{n+1}$ -name $\mathopen \{{ ( p^R_{ n + 1 } ( k ) , p ) }\boldsymbol \mid { p \in \mathbf { P}_{ n + 1 } } \mathclose \}$ .

Fix a $\mathord {\mathrm {V}}$ -generic filter G for $\mathbf { P}$ and, for all n, let $\mathcal {N}_n$ be the symmetric extension obtained from $\mathopen \langle \mathbf { P} _n , \mathcal {G}_n , \mathcal {F}_n \mathclose \rangle $ , and $\mathcal {N}$ be the symmetric extension, obtained from $\mathopen \langle \mathbf { P} , \mathcal {G}, \mathcal {F} \mathclose \rangle $ . Clearly we have

$$\begin{align*}\mathord{\mathrm{V}} \subseteq \mathcal{N}_0 \subseteq \mathcal{N}_1 \subseteq \dots \subseteq \mathcal{N} = \mathcal{N}_\omega \subseteq \mathord{\mathrm{V}} [G]. \end{align*}$$

For each $\mathbf { P}$ -name (e.g., $\dot {A}$ ), we let its symbol without the dot (i.e., A) be its evaluation according to G (i.e., $\dot {A}_G$ ).

Proof Let $p \in G$ and $\dot {R} \in \mathsf {HS} _n$ such that

$$\begin{align*}p \Vdash \dot{R}\subseteq \dot{A} \times \dot{A} \text{ total and acyclic,} \end{align*}$$

and without loss of generality we may assume that $p \in \mathbf { P} _n$ . We show that $\dot {R}_G = \dot {S}_G$ for some $\dot {S} \in \mathcal {R}_n$ as in (5). Let

$$\begin{align*}\dot{S} = \mathopen \{{ ( ( \dot{a}_m , \dot{a}_k )^\bullet , q ) }\boldsymbol\mid{ m , k \in \omega, q \in \mathbf{ P} _n \text{, and } q \Vdash \dot{a}_m \mathrel{\dot{R}} \dot{a}_k } \mathclose\}. \end{align*}$$

It readily follows that $\dot {S}$ is in $\mathcal {R}_n$ and $p \Vdash \dot {R} = \dot {S}$ . Fix any $q \in \mathbf { P}_{n + 1}$ with $q \le p$ . Pick an $m \in \omega \setminus \operatorname {\mathrm {dom}} ( q_{ n + 1 } )$ and consider the finite sequence $q'$ such that $q^{\prime }_l = q_l$ for every $l \neq n + 1$ and $q^{\prime }_{ n + 1 } = q_{ n + 1 } \cup \mathopen \{ { ( m , ( \dot {S} , \emptyset ) ) } \mathclose \}$ . Then $q' \in \mathbf { P}_{ n + 1 }$ , $q' \le q$ and

$$\begin{align*}q' \Vdash \dot{f}_{ n , m } \text{ is an } \dot{S} \text{-chain, and } \dot{S} = \dot{R}. \end{align*}$$

By density,

$$\begin{align*}p \Vdash \exists f \in \dot{F}_n \text{ which is an } \dot{R} \text{-chain}. \end{align*}$$

Since $F_n \in \mathcal {N}_{ n + 1 }$ we are done.

Since A is not closed as a set of reals, there exists a total and acyclic binary relation over A in $\mathcal {N}_0$ (see Lemma 2.2). Therefore, by Lemma 4.4, the set A becomes Dedekind-infinite already in $\mathcal {N}_1$ . However, the next proposition tells us that the range of any generic chain introduced by the iteration is far from being dense in A. This result is crucial in showing that A is not separable in $\mathcal {N}$ .

In order to get to the key proposition, we need to recall the well-known notion of scattered space (e.g., see [Reference Semadeni11, Section 8.5]).

Definition 4.5. Given a topological space X, we let by ordinal induction

$$ \begin{align*} X^{ ( 0 ) } &= X, \\ X^{ ( \alpha + 1 ) } &= \mathopen \{{ x \in X^{ ( \alpha ) } }\boldsymbol\mid{ x \text{ is a limit point of } X^{ ( \alpha ) } } \mathclose\}, \\ X^{ ( \lambda ) } &= \bigcap_{\alpha < \lambda} X^{ ( \alpha ) } \qquad \text{for }\lambda\text{ a limit ordinal}. \end{align*} $$

For every space X there is necessarily an ordinal $\alpha $ such that $X^{ ( \alpha ) } = X^{ ( \alpha + 1 ) }$ , and we call the least such ordinal the scattered height of the space. A topological space X is scattered if there is an $\alpha $ such that $X^{ ( \alpha ) } = \emptyset $ .

It is easy to check in $\mathsf {ZF}$ that every second countable scattered space is countable [Reference Semadeni11, Proposition 8.5.5]. If $\emptyset \neq X \subseteq \mathbb {R}$ is dense in itself, then $X^{( 1 )} = X$ and X is not scattered. In particular, A is not scattered.

For each $t \in {}^{<\omega } 2$ , we denote by $\dot {{\boldsymbol N}\!}_t$ the canonical name for the basic open set ${\boldsymbol N}\!_{t}$ , i.e., the set of all infinite binary sequences extending t.

Proof In other words, we want to show that in $\mathcal {N}$ (or, equivalentlyFootnote ² , in $\mathord {\mathrm {V}} [ G ]$ ), for every $n , k$ , the closure with respect to A of the range of $f_{ n , k }$ is scattered of height at most $n + 2$ . For any $H \subseteq \omega $ , let us introduce the $\mathbf { P}$ -name

(6) $$ \begin{align} \dot{A}_H \mathrel{{:}{=}} \mathopen \{{\dot{a}_m }\boldsymbol\mid{ m \in H } \mathclose\}^\bullet. \end{align} $$

Let $( \dagger )$ be the statement $\forall { n \in \omega } \, ( \dagger )_n$ , where $( \dagger )_n$ is the following statement: Let $k \in \omega $ , $p = \mathopen \langle p_0, \dots p_n \mathclose \rangle \in \mathbf { P} _n$ , $\dot {R} \in \mathcal {R}_n$ with support $\vec {H} = \mathopen \langle H_0, \dots , H_n \mathclose \rangle $ such that ${p \Vdash \dot {R}}$ is total and acyclic”. Assume also that, for each $i \leq n$ , $\operatorname {\mathrm {dom}} ( p_i ) = H_i$ , and, for all $0 < i \leq n$ , for all $j \in H_i$ , $\vec {H} {\mathop {\upharpoonright }} i$ is a support for $p_{i}^R ( j )$ . Then

$$ \begin{align*} p {}^\smallfrown \mathopen \langle \mathopen \{ { ( k , ( \dot{R}, \emptyset ) ) } \mathclose \} \mathclose \rangle \Vdash \big ( \operatorname{\mathrm{Cl}}_{\dot{A}} ( \operatorname{\mathrm{ran}} ( \dot{f}_{ n , k } ) ) \big )^{ ( 1 ) } \subseteq \dot{A}_{H_0} \cup \bigcup_{\smash[b]{\substack{i < n \\ j \in H_{ i + 1 } } } } \operatorname{\mathrm{ran}} ( \dot{f}_{ i , j } ). \end{align*} $$

Remarks.

1. (a) The condition $p {}^\smallfrown \mathopen \langle \mathopen \{ { ( k , ( \dot {R} , \emptyset ) ) } \mathclose \} \mathclose \rangle $ in the statement of $( \dagger )_n$ belongs to $\mathbf { P}_{ n + 1 }$ and it is obtained by extending p with the function with domain $\mathopen \{ {k} \mathclose \} \subseteq \omega $ such that $k \mapsto ( \dot {R}, \emptyset ) \in \mathcal {R}_n \times {}^{ < \omega } \omega $ . It forces the limit points of $\operatorname {\mathrm {Cl}}_A ( \operatorname {\mathrm {ran}} ( f_{ n , k }) )$ to belong to the finite union $A_{H_0} \cup \bigcup \mathopen \{{ \operatorname {\mathrm {ran}} ( f_{ i , j } ) }\boldsymbol \mid { i < n \text { and } j \in H_{ i + 1 }} \mathclose \}$ .

2. (b) Note that $p {}^\smallfrown \mathopen \langle \mathopen \{ { ( k , ( \dot {R} , \emptyset ) ) } \mathclose \} \mathclose \rangle $ is the $\le $ -maximum among the conditions $q \le p$ such that $q_{ n + 1 }^R ( k ) = \dot {R}$ . Therefore, for any fixed $n , k$ , the set of conditions $p {}^\smallfrown \mathopen \langle \mathopen \{ { ( k , ( \dot {R}, \emptyset ) ) } \mathclose \} \mathclose \rangle $ we are considering in $( \dagger )_n$ is pre-dense in $\mathbf { P}_{ n + 1 }$ (and also in  $\mathbf { P}$ ).

Claim 4.6.1. Assume $( \dagger )$ . For each $n , k \in \omega $ , $\Vdash \big ( \operatorname {\mathrm {Cl}}_{\dot {A}} ( \operatorname {\mathrm {ran}} ( \dot {f}_{ n , k } ) ) \big )^{( n + 2 ) } = \emptyset $ .

Proof We prove the claim by induction on n. Let $n = 0$ and fix $k \in \omega $ , $p \in \mathbf { P}_0$ , $\dot {R} \in \mathcal {R}_0$ , $\vec {H} = \mathopen \langle H_0 \mathclose \rangle $ satisfying the hypotheses of $( \dagger )_0$ . By $( \dagger )_0$ ,

$$\begin{align*}p {}^\smallfrown \mathopen \langle \mathopen \{ { ( k , ( \dot{R}, \emptyset ) ) } \mathclose \} \mathclose \rangle \Vdash ( \operatorname{\mathrm{Cl}}_{\dot{A}} ( \operatorname{\mathrm{ran}} ( \dot{f}_{ n , k } ) ) )^{ ( 1 ) } \subseteq \dot{A}_{H_0}. \end{align*}$$

Since $H_0$ is finite, we have that $\Vdash "\dot {A}_{H_0}$ is finite”, so our condition forces that $\operatorname {\mathrm {Cl}}_A ( \operatorname {\mathrm {ran}} f_{ n , k } )$ has scattered height $\leq 2$ , that is,

$$\begin{align*}p {}^\smallfrown \mathopen \langle \mathopen \{ { ( k , ( \dot{R}, \emptyset ) ) } \mathclose \} \mathclose \rangle \Vdash \big ( \operatorname{\mathrm{Cl}}_{\dot{A}} ( \operatorname{\mathrm{ran}} ( \dot{f}_{ n , k } ) ) \big ) ^{( 2 ) } = \emptyset. \end{align*}$$

By density, the base case follows.

Now the induction step. Let $n> 0$ and fix $k \in \omega $ , $p \in \mathbf { P}_n$ , $\dot {R} \in \mathcal {R}_n$ , $\vec {H} = \mathopen \langle H_0, \dots , H_n \mathclose \rangle $ satisfying the hypotheses of $( \dagger )_n$ . By $( \dagger )_n$ ,

$$\begin{align*}p {}^\smallfrown \mathopen \langle \mathopen \{ { ( k , ( \dot{R}, \emptyset ) ) } \mathclose \} \mathclose \rangle \Vdash \big ( \operatorname{\mathrm{Cl}}_{\dot{A}} ( \operatorname{\mathrm{ran}} ( \dot{f}_{ n , k } ) ) \big )^{ ( 1) } \subseteq \dot{A}_{H_0} \cup \bigcup_{\smash[b]{\substack{i < n \\ j \in H_{ i + 1 } } }} \operatorname{\mathrm{ran}} ( \dot{f}_{ i , j } ). \end{align*}$$

As the $H_i$ s are all finite,

$$\begin{align*}p {}^\smallfrown \mathopen \langle \mathopen \{ { ( k , ( \dot{R}, \emptyset ) ) } \mathclose \} \mathclose \rangle \Vdash \big ( \operatorname{\mathrm{Cl}}_{\dot{A}} ( \operatorname{\mathrm{ran}} ( \dot{f}_{ n , k } ) ) \big )^{ ( n + 2 ) } \subseteq \bigcup_{\smash[b]{\substack{i < n \\ j \in H_{ i + 1 }}}} \big ( \operatorname{\mathrm{Cl}}_{\dot{A}} ( \operatorname{\mathrm{ran}} ( \dot{f}_{ i , j } ) ) \big )^{ ( n + 1 ) }. \end{align*}$$

By induction hypothesis, for all $i < n$ and all $j \in H_{ i + 1 }$

$$\begin{align*}\Vdash \big ( \operatorname{\mathrm{Cl}}_{\dot{A}} ( \operatorname{\mathrm{ran}} ( \dot{f}_{ i , j } ) ) \big )^{ ( n + 1 ) } = \emptyset , \end{align*}$$

and hence,

$$\begin{align*}p {}^\smallfrown \mathopen \langle \mathopen \{ { ( k , ( \dot{R}, \emptyset ) ) } \mathclose \} \mathclose \rangle \Vdash \big ( \operatorname{\mathrm{Cl}}_{\dot{A}} ( \operatorname{\mathrm{ran}} ( \dot{f}_{ n , k } ) ) \big )^{ ( n + 2 ) } = \emptyset. \end{align*}$$

By density, the induction step follows.

The statement $( \dagger )$ is proved by induction on $n \in \omega $ .

Let $n = 0$ and fix $k \in \omega $ , $p \in \mathbf { P}_0$ , $\dot {R} \in \mathcal {R}_0$ , $\vec {H} = \mathopen \langle H_0 \mathclose \rangle $ satisfying the hypotheses of  $( \dagger )_0$ .

Claim 4.6.2. $p {}^\smallfrown \mathopen \langle \mathopen \{ { ( k , ( \dot {R}, \emptyset ) ) } \mathclose \} \mathclose \rangle \Vdash \operatorname {\mathrm {ran}} ( \dot {f}_{ 0 , k } ) \setminus \dot {A}_{ H_0 } \text { is discrete.}$

Proof Assume for a contradiction that there are $q \leq p {}^\smallfrown \mathopen \langle \mathopen \{ { ( k , ( \dot {R}, \emptyset ) ) } \mathclose \} \mathclose \rangle $ and $l \in \omega $ such that

$$\begin{align*}q \Vdash \dot{f}_{ 0 , k } ( l ) \notin \dot{A}_{H_0} \text{ and } \dot{f}_{ 0 , k } ( l ) \text{ is a limit point of } \operatorname{\mathrm{ran}} ( \dot{f}_{ 0 , k } ) \setminus \dot{A}_{H_0}. \end{align*}$$

Without loss of generality suppose that $\operatorname {\mathrm {lh}} ( q_1^s ( k ) )> l + 1$ and let $m = q_1^s ( k ) ( l )$ , $t = q_0 ( m )$ —in particular, $m \notin H_0$ and $q \Vdash \dot {f}_{ 0 , k } ( l ) = \dot {a}_m \in \dot {{\boldsymbol N}\!}_t$ . From our assumption and from the fact that $H_0$ is a finite set, it follows that there must be a $z \leq q$ and an $h> l$ such that

$$\begin{align*}z \Vdash \dot{f}_{ 0 , k } ( h ) \in \dot{{\boldsymbol N}\!}_t\setminus \dot{A}_{ H_0 }. \end{align*}$$

Assume without loss of generality $\operatorname {\mathrm {lh}} ( z_1^s ( k ) )> h$ and let $m' = z_1^s ( k ) ( h )$ , $t' = z_0 ( m ' )$ —in particular, $m' \notin H_0$ , $t' \supseteq t$ and $z_0 \Vdash \dot {a}_m \mathrel {\dot {R}^+} \dot {a}_{m'}$ . Note that $m' \neq m$ , as otherwise $z_0$ would force $\dot {R}$ to have a cycle, which is a contradiction, as $z_0$ extends p and by hypothesis p forces $\dot {R}$ to be acyclic.

Subclaim 4.6.2.1. $p' = z_0 {\mathop {\upharpoonright }} ( \omega \setminus \mathopen \{ { m } \mathclose \} ) \cup \mathopen \{ { ( m , t' ) } \mathclose \} \Vdash \dot {a}_{m} \mathrel {\dot {R}^+} \dot {a}_{m'}.$

A quick observation: since $z_0 \le q_0$ , $z_0 ( m )$ surely extends $t = q_0 ( m )$ , but a priori $z_0 ( m )$ could be incompatible with $t' = z_0 ( m' ) = p' ( m )$ , making $p'$ incompatible with $ z_0$ . Thus, our subclaim needs some care.

Proof of the Subclaim

Let $m_0 , m_1 , \dots , m_{ h - l } \in \omega $ be such that $m_i = z_1^s ( k ) ( l + i )$ for all $i \le h - l$ . Note that $m_0 = m = q_1^s ( k ) ( l )$ and $m_1 = q_1^s ( k ) ( l + 1 )$ , since $z \le q$ . Moreover, $m_{ h - l } = m'$ , by definition of $m'$ . We can assume that the $m_i$ s are all distinct, as otherwise $z_0$ would force $\dot {R}$ to have a cycle.

Clearly $q_0 \Vdash \dot {a}_{m_0} \mathrel {\dot {R}} \dot {a}_{m_1}$ . By the Restriction Lemma, $q_0 {\mathop {\upharpoonright }} H_0 \cup \mathopen \{ { m_0 , m_1 } \mathclose \}$ forces the same.

On the other hand, $z_0 \Vdash \dot {a}_{m_1} \mathrel {\dot {R}} \dot {a}_{m_{2}} \mathrel {\dot {R}} \dots \mathrel {\dot {R}} \dot {a}_{m_{ h - l }}$ . Again by the Restriction Lemma, $z_0 {\mathop {\upharpoonright }} H_0 \cup \mathopen \{ { m_1 , m_2 , \dots , m_{ h - l } } \mathclose \}$ forces the same.

The condition $p' = z_0 {\mathop {\upharpoonright }} ( \omega \setminus \mathopen \{ { m_0 } \mathclose \} ) \cup \mathopen \{ { ( m_0 , t' ) } \mathclose \}$ extends both $q_0 {\mathop {\upharpoonright }} H_0 \cup \mathopen \{ { m_0 , m_1 } \mathclose \}$ and $z_0 {\mathop {\upharpoonright }} H_0 \cup \mathopen \{ { m_1 , m_2 , \dots , m_{ h - l } } \mathclose \}$ —here use the fact that $m_0 \notin H_0$ and that all the $m_i$ s are distinct. Hence $p'$ forces $\dot {a}_{m_0} \mathrel {\dot {R}} \dot {a}_{m_{1}} \mathrel {\dot {R}} \dot {a}_{m_{2}} \mathrel {\dot {R}} \dots \mathrel {\dot {R}} \dot {a}_{m_{h-l}}$ .

Let $\pi _0 \colon \omega \to \omega $ be the permutation that swaps m and $m'$ fixing everything else—in particular, $\pi _0 \in \operatorname {\mathrm {Fix}} (H_0)$ and $\pi _0 \dot {R} = \dot {R}$ . Then, by the Symmetry Lemma,

$$\begin{align*}\pi_0 p' = p' \Vdash \dot{a}_{m'} \mathrel{\dot{R}^+} \dot{a}_m, \end{align*}$$

but then $p'$ both extends p and forces $\dot {a}_{m} \mathrel {\dot {R}^+} \dot {a}_m$ , which is a contradiction, since we assumed that p forces $\dot {R}$ to be acyclic.

By Claim 4.6.2, condition $p {}^\smallfrown \mathopen \langle \mathopen \{ { ( k , ( \dot {R}, \emptyset ) ) } \mathclose \} \mathclose \rangle $ forces that the limit points of $\operatorname {\mathrm {ran}} ( f_{ 0 , k } )$ belong to the finite set $A_{H_0 }$ . The next claim shows that the same is true for the larger set $\operatorname {\mathrm {Cl}}_{A} ( \operatorname {\mathrm {ran}} ( f_{ 0 , k } ) )$ .

Claim 4.6.3. $p {}^\smallfrown \mathopen \langle \mathopen \{ { ( k , ( \dot {R}, \emptyset ) ) } \mathclose \} \mathclose \rangle \Vdash \big ( \operatorname {\mathrm {Cl}}_{\dot {A}} ( \operatorname {\mathrm {ran}} ( \dot {f}_{ 0 , k } ) ) \big )^{ ( 1 ) } \subseteq \dot {A}_{ H_0 }$ .

Proof Suppose for a contradiction that the claim is false. Then there is a $q \leq p {}^\smallfrown \mathopen \langle \mathopen \{ { ( k , ( \dot {R}, \emptyset ) ) } \mathclose \} \mathclose \rangle $ and an $m \notin H_0$ such that

(7) $$ \begin{align} q \Vdash \dot{a}_m \text{ is a limit point of }\operatorname{\mathrm{ran}} ( \dot{f}_{ 0 , k } ). \end{align} $$

Note that, since $H_0$ is finite, q actually forces $\dot {a}_m$ to be a limit point of $\operatorname {\mathrm {ran}} ( \dot {f}_{ 0 , k } ) \setminus \dot {A}_{ H_0 }$ . Hence, it follows from Claim 4.6.2 that q forces $\dot {a}_m$ not to be in the range of $\dot {f}_{ 0 , k }$ . In particular, $m \notin \operatorname {\mathrm {ran}} ( q_1^s ( k ) )$ .

The condition $q' = \mathopen \langle q_0, q_1 {\mathop {\upharpoonright }} \mathopen \{ { k } \mathclose \} \mathclose \rangle $ extends p and, by the Restriction Lemma, still forces (7). Let t be $q_0 ( m )$ —in particular, $q' \Vdash \dot {a}_m \in \dot {{\boldsymbol N}\!}_t$ .

We now show $q' \Vdash \dot {{\boldsymbol N}\!}_t \subseteq \operatorname {\mathrm {Cl}} ( \operatorname {\mathrm {ran}} ( \dot {f}_{ 0 , k } ) \setminus \dot {A}_{H_0} )$ , which clearly contradicts Claim 4.6.2, as every discrete set of reals is nowhere dense. Pick any $z \leq q'$ and a $t'\supseteq t$ . Fix an $m' \notin H_0 \cup \operatorname {\mathrm {dom}} ( z_0 ) \cup \operatorname {\mathrm {ran}} ( q_1^s ( k ) )$ . Define $z'$ to be the condition such that $z_0' = z_0 \cup \mathopen \{ { ( m' , t' ) } \mathclose \}$ and $z_i' = z_i$ for every $i> 0$ .

Now, $z'$ clearly extends z. Moreover, if we let $\pi _0$ be the permutation of $\omega $ that swaps m and $m'$ , $z'$ also extends $\mathopen \langle \pi _0 \mathclose \rangle q'$ . Indeed, since $t' \supseteq t$ , it’s clear that $z^{\prime }_0$ extends $\pi _0 q_0$ . But since both m and $m'$ do not belong to $H_0 \cup \operatorname {\mathrm {ran}} ( q_1^s ( k ) )$ , we also have $( \mathopen \langle \pi _0 \mathclose \rangle q')_1 = q^{\prime }_1$ , and therefore $\mathopen \langle z^{\prime }_0, z^{\prime }_1 \mathclose \rangle = \mathopen \langle z^{\prime }_0, z_1 \mathclose \rangle $ extends $\mathopen \langle \pi _0 \mathclose \rangle q' = \mathopen \langle \pi _0 q_0 , q^{\prime }_1 \mathclose \rangle $ . Overall, $z'$ extends $\mathopen \langle \pi _0 \mathclose \rangle q'$ .

By (7) and the Symmetry Lemma,

$$\begin{align*}\mathopen \langle \pi_0 \mathclose \rangle q' \Vdash \dot{a}_{m'} \in \operatorname{\mathrm{Cl}} ( \operatorname{\mathrm{ran}} ( \dot{f}_{ 0 , k } ) \setminus \dot{A}_{H_0} ). \end{align*}$$

Since $z'$ extends $\mathopen \langle \pi _0 \mathclose \rangle q'$ and $z' \Vdash \dot {a}_{m'} \in \dot {{\boldsymbol N}\!}_{t'}$ , we have

$$\begin{align*}z' \Vdash \dot{{\boldsymbol N}\!}_{t'} \cap \operatorname{\mathrm{Cl}} ( \operatorname{\mathrm{ran}} ( \dot{f}_{ 0 , k } ) \setminus \dot{A}_{H_0}) \neq \emptyset. \end{align*}$$

By density,

$$\begin{align*}q' \Vdash \dot{{\boldsymbol N}\!}_t \subseteq \operatorname{\mathrm{Cl}} ( \operatorname{\mathrm{ran}} ( \dot{f}_{ 0 , k } ) \setminus \dot{A}_{H_0} ).\\[-36pt] \end{align*}$$

This proves $( \dagger )_0$ .

Here comes the induction step: fix an $n> 0$ and suppose $( \dagger )_i$ holds for every $i < n$ , towards proving $( \dagger )_n$ . Fix $k \in \omega $ , $p \in \mathbf { P}_n$ , $\dot {R} \in \mathcal {R}_n$ , $\vec {H} = \mathopen \langle H_0, \dots , H_n \mathclose \rangle $ satisfying the hypotheses of $( \dagger )_n$ . The next claim is the analogue of Claim 4.6.2.

Claim 4.6.4.

$$\begin{align*}p {}^\smallfrown \mathopen \langle \mathopen \{ { ( k , ( \dot{R}, \emptyset ) ) } \mathclose \} \mathclose \rangle \Vdash \operatorname{\mathrm{ran}} ( \dot{f}_{ n , k }) \setminus \Big ( \dot{A}_{H_0} \cup \bigcup_{\smash[b]{\substack{i < n \\ j \in H_{ i + 1 }}}} \operatorname{\mathrm{ran}} ( \dot{f}_{ i , j }) \Big ) \text{ is discrete}. \end{align*}$$

Proof Suppose for a contradiction that there are $q \leq \mathopen \langle p, \mathopen \{ { ( k , ( \dot {R} , \emptyset ) ) } \mathclose \} \mathclose \rangle $ and $l \in \omega $ such that

$$ \begin{align*} q \Vdash \dot{f}_{ n , k } ( l ) \notin \dot{A}_{H_0} \cup \smash[b]{\bigcup_{\substack{i < n \\ j \in H_{ i + 1 } } }} \operatorname{\mathrm{ran}} ( \dot{f}_{ i , j }) &\text{ and }\dot{f}_{ n , k } ( l ) \text{ is a limit point of } \\ &\quad\operatorname{\mathrm{ran}} ( \dot{f}_{ n , k } ) \setminus \Big ( \dot{A}_{H_0} \cup \bigcup_{\smash[b]{\substack{i < n \\ j \in H_{ i + 1 } }}} \operatorname{\mathrm{ran}} ( \dot{f}_{ i , j } ) \Big ). \end{align*} $$

Suppose without loss of generality that $\operatorname {\mathrm {lh}} ( q_{ n + 1 } ^s ( k ) )> l + 1$ and let $m = q_{ n + 1 }^s ( k ) ( l )$ , and $t = q_0 ( m )$ —in particular, $q \Vdash \dot {f}_{ n , k } ( l ) = \dot {a}_m \in \dot {{\boldsymbol N}\!}_t$ . By assumption there must be a $z \leq q$ and an $h> l$ such that

$$\begin{align*}z \Vdash \dot{f}_{ n , k } ( h ) \in \dot{{\boldsymbol N}\!}_t \setminus \bigg ( \dot{A}_{ H_0 } \cup \bigcup_{\smash[b]{ \substack{i < n \\ j \in H_{i + 1}}}} \operatorname{\mathrm{ran}} ( \dot{f}_{ i , j } ) \bigg). \end{align*}$$

Assume without loss of generality $\operatorname {\mathrm {lh}} ( z_{n + 1 }^s ( k ) )> h$ and let $m' = z_{ n + 1 }^s ( k ) ( h )$ , and $t' = z_0 ( m ' )$ —in particular $t' \supseteq t$ and $z {\mathop {\upharpoonright }} n + 1 \Vdash \dot {a}_m \mathrel {\dot {R}^+ } \dot {a}_{m'}$ . Since

$$\begin{align*}z \Vdash \dot{a}_m, \dot{a}_m' \notin \bigg ( \dot{A}_{H_0} \cup \bigcup_{\smash[b]{ \substack{i < n \\ j \in H_{i + 1}}}} \operatorname{\mathrm{ran}} ( \dot{f}_{ i , j } ) \bigg), \end{align*}$$

then, in particular,

(8) $$ \begin{align} m,m' \notin H_0 \cup \bigcup_{\smash[b]{\substack{i < n \\ j \in H_{i + 1}}} } \operatorname{\mathrm{ran}} \big(z^s_{i+1}(j)\big). \end{align} $$

Now let

$$\begin{align*}p' = \mathopen \langle z_0 {\mathop{\upharpoonright}} ( \omega \setminus \mathopen \{ { m } \mathclose \} ) \cup \mathopen \{ { ( m , t' ) } \mathclose \} , z_1 {\mathop{\upharpoonright}} H_1, \dots, z_n {\mathop{\upharpoonright}} H_n \mathclose \rangle. \end{align*}$$

By an argument analogous to the one used in the proof of Subclaim 4.6.2.1, we can show that

$$\begin{align*}p' \Vdash \dot{a}_m \mathrel{ \dot{R}^+ } \dot{a}_{m'}. \end{align*}$$

If we let $\pi _0 \colon \omega \to \omega $ be the permutation that swaps m and $m'$ , then $\mathopen \langle \pi _0 \mathclose \rangle p' = p'$ . Indeed, it directly follows from the definition of $p'$ that $\pi _0 p^{\prime }_0 = p^{\prime }_0$ . Moreover, by (8), both m and $m'$ do not belong to $H_0$ , hence $\mathopen \langle \pi _0 \mathclose \rangle \in \operatorname {\mathrm {Fix}} ( \vec {H} )$ . As such, $( \mathopen \langle \pi _0 \mathclose \rangle p' )_i^R = ( p' )_i^R$ for every $1 \le i \le n$ . Again by (8), m and $m'$ do not belong to the range of $( p' )_i ^s ( j )$ for any $1 \le i \le n$ and $j \in \operatorname {\mathrm {dom}} ( p^{\prime }_i ) = H_i$ , and therefore $( \mathopen \langle \pi _0 \mathclose \rangle p' )_i ^s = ( p' )_i ^s$ for every $1 \le i \le n$ . Overall, $\mathopen \langle \pi _0 \mathclose \rangle p' = p'$ .

Next note that $\mathopen \langle \pi _0 \mathclose \rangle \dot {R} = \dot {R}$ , as $\mathopen \langle \pi _0 \mathclose \rangle \in \operatorname {\mathrm {Fix}} ( \vec {H} )$ . By the Symmetry Lemma,

$$\begin{align*}\mathopen \langle \pi_0 \mathclose \rangle p' = p' \Vdash \dot{a}_{m'} \mathrel{ \dot{R}^+ } \dot{a}_{m} , \end{align*}$$

but then $p'$ both extends p and forces $\dot {a}_m \mathrel { \dot {R}^+ } \dot {a}_m$ , which is a contradiction, since we assumed that p forces $\dot {R}$ to be acyclic.

Claim 4.6.5.

$$\begin{align*}p\Vdash \dot{A}_{H_0} \cup \bigcup_{\smash[b]{\substack{i < n \\ j \in H_{i+1}}}} \operatorname{\mathrm{ran}} ( \dot{f}_{i,j}) \text{ is closed with respect to } \dot{A}. \end{align*}$$

Proof Fix $q \le p$ and m such that

$$\begin{align*}q \Vdash \dot{a}_m \in \operatorname{\mathrm{Cl}}_{\dot{A}}\Big ( \dot{A}_{H_0} \cup \bigcup_{ \smash[b]{\substack{i < n \\ j \in H_{ i + 1 } } } } \operatorname{\mathrm{ran}} ( \dot{f}_{ i , j } ) \Big ). \end{align*}$$

We would like to prove that there is a condition $z \le q$ such that

$$\begin{align*}z \Vdash \dot{a}_m \in \dot{A}_{H_0} \cup \bigcup_{\smash[b]{ \substack{i < n \\ j \in H_{ i + 1 }}} } \operatorname{\mathrm{ran}} ( \dot{f}_{ i , j } ) , \end{align*}$$

so, to avoid trivialities, we assume

$$\begin{align*}q \Vdash \dot{a}_m \in \Big ( \operatorname{\mathrm{Cl}}_{\dot{A}} \big ( \dot{A}_{H_0} \cup \bigcup_{\smash[b]{ \substack{i < n \\ j \in H_{ i + 1 } } } } \operatorname{\mathrm{ran}} ( \dot{f}_{ i , j } ) \big ) \Big )^{ ( 1 ) }. \end{align*}$$

As the $H_i$ s are finite, there exists a $z \le q$ , an $i < n$ and some $j \in H_{ i + 1 }$ such that $z \Vdash \dot {a}_m \in ( \operatorname {\mathrm {Cl}}_{\dot {A}} ( \operatorname {\mathrm {ran}} ( \dot {f}_{ i , j } ) ) )^{ ( 1 ) }$ . But then, by $( \dagger )_i$ (here we use our induction hypothesis),

$$\begin{align*}z \Vdash \dot{a}_m \in \big ( \operatorname{\mathrm{Cl}}_{\dot{A}} ( \operatorname{\mathrm{ran}} ( \dot{f}_{ i , j } ) ) \big )^{ ( 1 ) } \subseteq \dot{A}_{ H_0 } \cup \bigcup_{\smash[b]{\substack{l < i \\ h \in H_{ l + 1 } } } } \operatorname{\mathrm{ran}} ( \dot{f}_{ l , h } ). \end{align*}$$

By density, the claim follows.

The next claim is the analogue of Claim 4.6.3.

Claim 4.6.6.

$$\begin{align*}p {}^\smallfrown \mathopen \langle \mathopen \{ { ( k , ( \dot{R}, \emptyset ) ) } \mathclose \} \mathclose \rangle \Vdash \big ( \operatorname{\mathrm{Cl}}_{\dot{A}} ( \operatorname{\mathrm{ran}} ( \dot{f}_{ n , k } ) ) \big )^{ ( 1 ) } \subseteq \dot{A}_{H_0} \cup \bigcup_{\smash[b]{ \substack{i < n \\ j \in H_{ i + 1 } } } } \operatorname{\mathrm{ran}} ( \dot{f}_{ i , j } ). \end{align*}$$

Proof Suppose for a contradiction that this is not the case, then there is a $q \leq p {}^\smallfrown \mathopen \langle \mathopen \{ { ( k , ( \dot {R}, \emptyset ) ) } \mathclose \} \mathclose \rangle $ and an m such that

$$\begin{align*}q \Vdash \dot{a}_m \text{ is a limit point of } \operatorname{\mathrm{ran}} ( \dot{f}_{ n , k } ) \text{ and } \dot{a}_m \notin \dot{A}_{ H_0 } \cup \bigcup_{\smash[b]{ \substack{i < n \\ j \in H_{ i + 1 }}}} \operatorname{\mathrm{ran}} ( \dot{f}_{ i , j } ). \end{align*}$$

From Claim 4.6.5 it follows that

(9) $$ \begin{align} q \Vdash \dot{a}_m \text{ is a limit point of } \operatorname{\mathrm{ran}} ( \dot{f}_{ n , k } ) \setminus \Big ( \dot{A}_{ H_0 } \cup \bigcup_{\smash[b]{\substack{i < n \\ j \in H_{ i + 1 } } } } \operatorname{\mathrm{ran}} ( \dot{f}_{ i , j } ) \Big ). \end{align} $$

But then, by Claim 4.6.4, q also forces $\dot {a}_m$ not to be in the range of $\dot {f}_{ n , k }$ . In particular,

(10) $$ \begin{align} m \notin H_0 \cup \operatorname{\mathrm{ran}} ( q_{ n + 1 }^s ( k ) ) \cup \bigcup_{\smash[b]{\substack{i < n \\ j \in H_{ i + 1 } } } } \operatorname{\mathrm{ran}} ( q_{ i + 1 }^s ( j ) ). \end{align} $$

Let

$$\begin{align*}q' = \mathopen \langle q_0, q_1 {\mathop{\upharpoonright}} H_1, \dots, q_n {\mathop{\upharpoonright}} H_n, q_{ n + 1 } {\mathop{\upharpoonright}} \mathopen \{ { k } \mathclose \} \mathclose \rangle. \end{align*}$$

Then $q'$ extends p and, by the Restriction Lemma, still forces (9). Let t be $q_0 ( m )$ —in particular $q' \Vdash \dot {a}_m \in \dot {{\boldsymbol N}\!}_t$ .

We now show that

$$\begin{align*}q' \Vdash \dot{{\boldsymbol N}\!}_t \subseteq \operatorname{\mathrm{Cl}} \Big ( \operatorname{\mathrm{ran}} ( \dot{f}_{ n , k } ) \setminus \big ( \dot{A}_{H_0} \cup \bigcup_{\substack{i < n \\ j \in H_{ i + 1 } } } \operatorname{\mathrm{ran}} ( \dot{f}_{ i , j } ) \big ) \Big ) , \end{align*}$$

which contradicts Claim 4.6.4. Pick any $z \leq q'$ and $t'\supseteq t$ . Fix an $m' \in \omega $ such that

(11) $$ \begin{align} m' \notin H_0 \cup \operatorname{\mathrm{dom}} ( z_0 ) \cup \operatorname{\mathrm{ran}} ( q_{ n + 1 }^s ( k ) ) \cup \bigcup_{\substack{i < n \\ j \in H_{i + 1 } } } \operatorname{\mathrm{ran}} ( q_{ i + 1 }^s ( j ) ). \end{align} $$

Define $z'$ to be the condition such that $z^{\prime }_0 = z_0 \cup \mathopen \{ { ( m' , t' ) } \mathclose \}$ and $z_i' = z_i$ for all $i> 0$ . Now, $z'$ clearly extends z. Moreover, if we let $\pi _0$ be the permutation of $\omega $ that swaps m and $m'$ , $z'$ also extends $\mathopen \langle \pi _0 \mathclose \rangle q'$ . Indeed, since $t' \supseteq t$ , it’s clear that $z^{\prime }_0$ extends $\pi _0 q_0$ . But from (10) and (11), it follows that $( \mathopen \langle \pi _0 \mathclose \rangle q' )_i = q^{\prime }_i$ for every $1 \le i \le n+1$ , and therefore $z'$ extends $\mathopen \langle \pi _0 \mathclose \rangle q'$ .

By (9) and the Symmetry Lemma,

$$\begin{align*}\mathopen \langle \pi_0 \mathclose \rangle q' \Vdash \dot{a}_{m'} \in \operatorname{\mathrm{Cl}} \Big ( \operatorname{\mathrm{ran}} ( \dot{f}_{ n , k } ) \setminus \big ( \dot{A}_{H_0} \cup \bigcup_{\smash[b]{\substack{i < n \\ j \in H_{ i + 1 } } } } \operatorname{\mathrm{ran}} ( \dot{f}_{ i , j } ) \big ) \Big ). \end{align*}$$

Since $z'$ extends $\mathopen \langle \pi _0 \mathclose \rangle q'$ and $z' \Vdash \dot {a}_{m'} \in \dot {{\boldsymbol N}\!}_{t'}$ , we have

$$\begin{align*}z' \Vdash \dot{{\boldsymbol N}\!}_{t'} \cap \operatorname{\mathrm{Cl}} \Big ( \operatorname{\mathrm{ran}} ( \dot{f}_{ n , k } ) \setminus \big ( \dot{A}_{ H_0 } \cup \bigcup_{\smash[b]{\substack{i < n \\ j \in H_{ i + 1 } } } } \operatorname{\mathrm{ran}} ( \dot{f}_{ i , j } ) \big ) \Big ) \neq \emptyset. \end{align*}$$

By density,

$$\begin{align*}q' \Vdash \dot{{\boldsymbol N}\!}_t \subseteq \operatorname{\mathrm{Cl}} \Big ( \operatorname{\mathrm{ran}} ( \dot{f}_{ n , k } ) \setminus \big ( \dot{A}_{ H_0 } \cup \bigcup_{\substack{i < n \\ j \in H_{ i + 1 } } } \operatorname{\mathrm{ran}} ( \dot{f}_{ i , j } ) \big ) \Big ).\\[-50pt] \end{align*}$$

This completes the proof of $( \dagger )_n$ , so by induction $( \dagger )$ holds. By Claim 4.6.1, we are done.

In light of Proposition 4.6, we can prove that in $\mathcal {N}$ every separable subset of A is scattered with finite scattered height.

Proof Let $S \in \mathcal {N}$ be a separable subset of A and fix in $\mathcal {N}$ a function $f \colon \omega \to A$ such that $S \subseteq \operatorname {\mathrm {Cl}} ( \operatorname {\mathrm {ran}} ( f ) )$ . Then there must be a $p \in G$ such that

$$\begin{align*}p \Vdash \dot{f} \colon \check{\omega} \to \dot{A}, \end{align*}$$

where $\dot {f} \in \mathsf {HS}$ is a symmetric name for f, with support $\vec {H} = \mathopen \langle H_0, \dots , H_n \mathclose \rangle $ . We can assume without loss of generality that $\operatorname {\mathrm {dom}} (p_i) = H_i$ for each i, and that for all $i> 0$ , for all $j \in H_i$ , $\vec {H}{\mathop {\upharpoonright }} i$ is a support for $p_{i}^R ( j )$ . We claim that

$$\begin{align*}p \Vdash \operatorname{\mathrm{ran}} ( \dot{f} ) \subseteq \dot{A}_{H_0} \cup \bigcup_{\substack{i < n \\ j \in H_{ i + 1 } } } \operatorname{\mathrm{ran}} ( \dot{f}_{ i , j } ) , \end{align*}$$

where $\dot {A}_{H_0}$ is the $\mathbf { P}$ -name as in (6). If we manage to do so, then Proposition 4.6 ensures that $\operatorname {\mathrm {Cl}}_A ( \operatorname {\mathrm {ran}} ( f ) )$ is scattered of height $\leq n + 2$ , and, a fortiori, that $S^{( n + 2 )} = \emptyset $ , as required.

Suppose that the claim is false, then there exist $q \leq p$ and $l , m \in \omega $ such that

(12) $$ \begin{align} q \Vdash \dot{f} ( l ) = \dot{a}_m \notin \dot{A}_{H_0} \cup\bigcup_{\smash[b]{\substack{i < n \\ j \in H_{i + 1 } } }} \operatorname{\mathrm{ran}} ( \dot{f}_{ i , j } ). \end{align} $$

In particular,

(13) $$ \begin{align} m \notin H_0 \cup \bigcup_{\substack{i < n \\ j \in H_{i + 1 } } } \operatorname{\mathrm{ran}} ( q_{ i + 1 } ^s ( j ) ). \end{align} $$

Let $q' = \mathopen \langle q_0, q_1 {\mathop {\upharpoonright }} H_1, \dots , q_n {\mathop {\upharpoonright }} H_n \mathclose \rangle $ . Then, by the Restriction Lemma, $q'$ still forces (12).

Fix an $m' \in \omega $ such that

(14) $$ \begin{align} m' \notin H_0 \cup \operatorname{\mathrm{dom}} ( q_0 ) \cup \bigcup_{\substack{i < n \\ j \in H_{i + 1 } } } \operatorname{\mathrm{ran}} (q_{ i + 1 } ^s ( j ) ). \end{align} $$

Let $\pi _0$ be the permutation of $\omega $ that swaps m and $m'$ , then $\mathopen \langle \pi _0 \mathclose \rangle q'$ and $q'$ are compatible. Indeed, since $m' \notin \operatorname {\mathrm {dom}} (q_0')$ , then $q^{\prime }_0$ and $\pi _0 q^{\prime }_0$ are clearly compatible. Moreover, it follows from (13) and (14) that $( \mathopen \langle \pi _0 \mathclose \rangle q')_i = q^{\prime }_i$ for every $1 \le i \le n$ , and therefore $\mathopen \langle \pi _0 \mathclose \rangle q'$ and $q'$ are compatible. By the Symmetry Lemma,

$$\begin{align*}\mathopen \langle \pi_0 \mathclose \rangle q' \Vdash \dot{f} ( l ) = \dot{a}_{m'}. \end{align*}$$

So $q'$ and $\mathopen \langle \pi _0 \mathclose \rangle q'$ , while being compatible, force $\dot {f}$ to take different values at l, but they both extend p, which forces $\dot {f}$ to be a function. Contradiction.

Now we are left to prove that $\mathsf {DC} ( A )$ holds in $\mathcal {N}$ . Let $\dot {\mathcal {N}}_n$ be the canonical name for the intermediate model $\mathcal {N}_n$ .

Lemma 4.9. Let $n \in \omega $ and $\dot {x} \in \mathsf {HS}$ with support $\vec {H} = \mathopen \langle H_0, \dots , H_n \mathclose \rangle $ , then

$$\begin{align*}\Vdash \dot{x} \subseteq \dot{\mathcal{N}}_n \Rightarrow \dot{x} \in\dot{\mathcal{N}}_n \,. \end{align*}$$

Proof Fix $( \dot {y} , p ) \in \dot {x}$ . As the set of q such that either $q \Vdash \dot {y} \in \dot { \mathcal {N}}_n$ or else $q \Vdash \dot {y} \notin \dot { \mathcal {N}}_n$ is dense below p, there is a maximal antichain $D_{ ( \dot {y} , p ) }$ below $ p$ and a map $h_{ ( \dot {y} , p ) } \colon D_{ ( \dot {y} , p ) } \to \mathsf {HS} _n$ such that, for each $q \in D_{ ( \dot {y} , p ) }$ , either $q \Vdash \dot {y} = h_{ ( \dot {y} , p ) } ( q )$ or $q \Vdash \dot {y} \notin \dot {\mathcal {N}}_n$ . Let $D^{\prime }_{ ( \dot {y} , p ) } = \mathopen \{{ q \in D_{ ( \dot {y} , p ) } }\boldsymbol \mid { q \Vdash \dot {y} \in \dot {\mathcal {N}}_n } \mathclose \}$ and let

$$\begin{align*}C = \mathopen \{{ \vec{\pi} \big ( h_{ ( \dot{y} , p ) } ( q ) \big ) }\boldsymbol\mid{ ( \dot{y} , p ) \in \dot{x}, \ q \in D^{\prime}_{ ( \dot{y} , p ) } , \ \vec{\pi} \in \operatorname{\mathrm{Fix}} ( \vec{H} ) } \mathclose\}. \end{align*}$$

Consider the following $\mathbf { P} _n$ -name:

$$\begin{align*}\dot{w} = \mathopen \{{ ( \dot{y} , q ) }\boldsymbol\mid{ \dot{y} \in C, \ q \in \mathbf{ P} _n \text{ and } q \Vdash \dot{y} \in \dot{x} } \mathclose\}. \end{align*}$$

Claim 4.9.1. $\dot {w} \in \mathsf {HS} _n$ with support $\vec {H}$ .

Proof Let $\vec {\pi } \in \operatorname {\mathrm {Fix}} ( \vec {H} )$ and $( \dot {y} , q ) \in \dot {w}$ . By definition, $q \Vdash \dot {y} \in \dot {x}$ , hence $\vec {\pi } q \Vdash \vec {\pi } \dot {y} \in \dot {x}$ . Since $\vec {\pi } \dot {y} \in C$ , this means that $( \vec {\pi } \dot {y}, \vec {\pi } q ) \in \dot {w}$ . Hence $\vec {\pi } \dot {w} = \dot {w}$ .

Fix $p \in \mathbf { P}$ such that $p \Vdash \dot {x} \subseteq \dot {\mathcal {N}}_n$ .

Claim 4.9.2. $p \Vdash \dot {w} = \dot {x}$ .

Proof Let $q \leq p$ and $\dot {z} \in \mathsf {HS}$ such that $q \Vdash \dot {z} \in \dot {x}$ . By definition of C and our hypothesis on p, there is an $r \leq q$ and a $\dot {y} \in C$ such that $r \Vdash \dot {z} = \dot {y} \in \dot {x}$ . By the Restriction Lemma, $r {\mathop {\upharpoonright }} n + 1 \Vdash \dot {y} \in \dot {x}$ , hence $( \dot {y} , r {\mathop {\upharpoonright }} n + 1 ) \in \dot {w}$ and, in particular, $r \Vdash \dot {z} = \dot {y} \in \dot {w}$ . By density, $p \Vdash \dot {x} \subseteq \dot {w}$ .

The other inclusion is immediate from the definition of $\dot {w}$ .

Therefore $p \Vdash \dot {x} \in \dot {\mathcal {N}}_n$ . By density, $\Vdash \dot {x} \subseteq \dot {\mathcal {N}}_n \Rightarrow \dot {x} \in \dot { \mathcal {N} }_n$ .

This finishes the proof of Theorem 4.1.

5.1 Dependent choice propagates under finite unions

By Proposition 2.1, the axiom $\mathsf {DC} (X)$ is closed under surjective images and, hence, under subsets. The next result shows that it is also closed under finite unions.

The natural progression from Corollary 5.2 would be to prove that $\mathsf {DC} ( X ) \Rightarrow \mathsf {DC} ( X \times \omega )$ , but this cannot be established in $\mathsf {ZF}$ , since $\mathsf {DC} ( X \times \omega )$ implies ${\mathsf {AC}}_\omega ( X )$ (part (d) of Proposition 2.1) and we know from Theorem 4.1 that $\mathsf {DC} ( X )$ does not necessarily imply ${\mathsf {AC}}_\omega ( X )$ .

If a binary relation R is such that $\operatorname {\mathrm {ran}} ( R ) \subseteq \operatorname {\mathrm {dom}} ( R )$ , then it is total on its domain. The largest $R' \subseteq R$ such that $\operatorname {\mathrm {ran}} ( R' ) \subseteq \operatorname {\mathrm {dom}} ( R' )$ is

$$\begin{align*}\mathcal{D} ( R ) = \bigcup \mathopen \{{S \subseteq R}\boldsymbol\mid{ \operatorname{\mathrm{ran}} ( S ) \subseteq \operatorname{\mathrm{dom}} ( S )} \mathclose\}. \end{align*}$$

By part (a) of Proposition 2.1 it is easy to see that

(15) $$ \begin{align} \mathsf{DC} ( X ) \Leftrightarrow \forall { R \subseteq X^2 } \, ( \mathcal{D} ( R ) \neq \emptyset \Rightarrow \text{there is a }\mathcal{D} ( R )\text{-chain} ). \end{align} $$

Proof of Theorem 5.1

Suppose $\mathsf {DC} ( X )$ and $\mathsf {DC} ( Y )$ , and let $R \subseteq ( X \cup Y )^2$ be total, towards proving that there is an R-chain. Without loss of generality, we may assume that X and Y are nonempty and disjoint. If $\mathcal {D} ( R {\mathop {\upharpoonright }} X ) \neq \emptyset $ , then by $\mathsf {DC} ( X )$ and (15) there is a $\mathcal {D} ( R {\mathop {\upharpoonright }} X )$ -chain, which is, in particular an R-chain. Similarly, if $\mathcal {D} ( R {\mathop {\upharpoonright }} Y ) \neq \emptyset $ , then there is an R-chain. Therefore, without loss of generality, we may assume that R is acyclic, and that

(16) $$ \begin{align} \mathcal{D} ( R {\mathop{\upharpoonright}} X ) = \mathcal{D} ( R {\mathop{\upharpoonright}} Y ) = \emptyset. \end{align} $$

Recall that $R^+$ is the smallest transitive relation containing R. If $x \in X \cup Y$ and $R^+ ( x ) \subseteq X$ , then $R {\mathop {\upharpoonright }} R^+ ( x )$ would witness that $\mathcal {D} ( R {\mathop {\upharpoonright }} X ) \neq \emptyset $ , against (16). Similarly $R^+ ( x )$ cannot be included in Y. Therefore

(17) $$ \begin{align} \forall {x \in X \cup Y} \, ( R^+ ( x ) \nsubseteq X \wedge R^+ ( x ) \nsubseteq Y ). \end{align} $$

Here is the idea of the proof. By (16), any R-chain $( z_n )_{ n \in \omega }$ must visit both X and Y infinitely often, so $( z_n )_{ n \in \omega }$ can be seen as the careful merging of two sequences $( x_n )_{ n \in \omega }$ in X and $( y_n )_{ n \in \omega }$ in Y. The sequence $( x_n )_{ n \in \omega }$ is obtained by applying $\mathsf {DC} ( X )$ to a total relation $R_X$ on X such that $R {\mathop {\upharpoonright }} X \subseteq R_X \subseteq R^+$ . Using $( x_n )_{ n \in \omega }$ , a suitable total relation $R_Y$ on some $Y' \subseteq Y$ is defined, and by $\mathsf {DC} ( Y )$ the required sequence $( y_n )_{ n \in \omega }$ is obtained. Here come the details.

Let $R_X$ be the relation on X given by $R {\mathop {\upharpoonright }} X$ , together with all pairs $( x , x' )$ such that $x \mathrel {R} y_0 \mathrel {R} y_1 \mathrel {R} \dots \mathrel {R} y_n \mathrel {R} x'$ for some finite sequence of elements of Y:

$$ \begin{align*} R_X = ( R {\mathop{\upharpoonright}} X ) \cup &\{ ( x , x' ) \in X^2 \boldsymbol\mid \exists {m \geq 1 } \, \exists {s \in {}^{m } Y} \, \\ &( x \mathrel{R} s ( 0 ) \wedge s ( m - 1 ) \mathrel{R} x' \wedge \forall {i < m - 1 } \, ( s ( i ) \mathrel{R} s ( i + 1 ) ) ) \}. \end{align*} $$

It is immediate that $R_X \subseteq R^+$ .

Claim 5.2.1. $R_X$ is total on X.

Proof We must show that $\operatorname {\mathrm {dom}} ( R_X ) = X$ . Let $x \in X$ . If $R ( x ) \cap X \neq \emptyset $ , then $x \in \operatorname {\mathrm {dom}} ( R {\mathop {\upharpoonright }} X) \subseteq \operatorname {\mathrm {dom}} (R_X)$ .

Now suppose otherwise. By (3) $R^+ ( x ) \nsubseteq Y$ , so there are $y_0 , \dots , y_n \in Y$ and $x' \in X$ such that $x \mathrel {R} y_0 \mathrel {R} \dots \mathrel {R} y_n \mathrel {R} x'$ . Thus $( x , x' ) \in R_X$ , so $x \in \operatorname {\mathrm {dom}} ( R_X )$ .

By $\mathsf {DC} ( X )$ there is an $R_X$ -chain $( x_n )_{ n \in \omega }$ .

Claim 5.2.2. $\forall {n} \, \exists { m> n } \, \neg ( x_{m} \mathrel {R} x_{m+1} )$ .

Proof Towards a contradiction, suppose that there is $\bar {n}\in \omega $ such that $x_{m} \mathrel {R} x_{ m + 1 }$ for every $m \geq \bar {n}$ . Then $R {\mathop {\upharpoonright }} \mathopen \{{ x_m}\boldsymbol \mid { m \geq \bar {n}} \mathclose \}$ is total on $\mathopen \{{ x_m}\boldsymbol \mid { m \geq \bar {n}} \mathclose \}$ and contained in $R {\mathop {\upharpoonright }} X$ , against (16).

Let $( n_k )_{ k \in \omega }$ be the sequence enumerating the set of ms such that $\neg ( x_{m} \mathrel {R} x_{ m + 1 } )$ . By the definition of $R_X$ , each $x_{n_k}$ is linked to $x_{ n_k + 1}$ via R through some finite path in Y, and let $Y_k$ be the collection of all places visited by these paths:

$$ \begin{align*} Y_k \mathrel{{:}{=}} \bigcup \big \{ \operatorname{\mathrm{ran}} ( s ) \boldsymbol\mid \exists {m } \, \big ( s \in {}^{m+ 1} Y \wedge x_{n_k} \mathrel{R} s ( 0 ) \wedge &s ( m ) \mathrel{R} x_{ n_k + 1 } \wedge{} \\ &\forall {i < m } \, ( s ( i ) \mathrel{R} s ( i + 1 ) ) \big ) \big \}. \end{align*} $$

Claim 5.2.3. The $Y_k$ s are nonempty, pairwise disjoint subsets of Y.

Proof For each k we have $( x_{ n_k } , x_{ n_k + 1 } ) \in R_X \setminus R$ . This means that there is some $\mathopen \langle y_0 , \dots , y_m \mathclose \rangle \in {}^{<\omega } Y$ such that $x_{n_k} \mathrel {R} y_0 \mathrel {R} \dots \mathrel {R} y_m \mathrel {R} x_{n_k + 1 }$ . In particular, $Y_k \neq \emptyset $ .

Towards a contradiction, suppose there are indices $k < j$ such that $Y_k \cap Y_j \neq \emptyset $ . Pick $y \in Y_k \cap Y_j$ . Then $y \mathrel {R^+} x_{ n_k + 1 } \mathrel {R^+} x_{n_j} \mathrel {R^+} y$ , if $x_{ n_k + 1 } \neq x_{n_j}$ , or $y \mathrel {R^+} x_{ n_k + 1 } = x_{n_j} \mathrel {R^+} y$ otherwise. Either way, this contradicts our assumption that R is acyclic.

Now we let $R_Y$ be the relation on $\bigcup _{k \in \omega } Y_k$

$$\begin{align*}\bigcup_{ k \in \omega } ( R {\mathop{\upharpoonright}} Y_k ) \cup \bigcup_{ k \in \omega } \mathopen \{{ ( y , y' ) \in Y_k \times Y_{ k + 1 }}\boldsymbol\mid{ } \mathclose{ y \mathrel{R} x_{ n_k + 1 } \text{ and } x_{ n_{ k + 1 } } \mathrel{R} y'}\}. \end{align*}$$

It readily follows from the definition that $R_Y \subseteq R^+$ .

Claim 5.2.4. $R_Y$ is total on $\bigcup _{k \in \omega } Y_k$ .

Proof Pick $k \in \omega $ and $y \in Y_k$ , towards proving that $y \in \operatorname {\mathrm {dom}} ( R_Y )$ . Then there is a finite sequence $\mathopen \langle y_0 , \dots , y_m \mathclose \rangle $ of elements of $Y_k$ such that $x_{n_k} \mathrel {R} y_0 \mathrel {R} \dots \mathrel {R} y_m \mathrel {R} x_{n_k + 1}$ , and $y = y_i$ for some $0 \leq i \leq m$ . If $i < m$ , then $y \mathrel {R} y_{i + 1}$ . If $i = m$ then $y \mathrel {R_Y} y'$ for any $y' \in Y_{ k + 1}$ such that $x_{n_{k+1}} \mathrel {R} y'$ . In either case $y \in \operatorname {\mathrm {dom}} ( R_Y )$ .

By $\mathsf {DC} ( Y )$ , there is an $R_Y$ -chain $( y_n )_{ n \in \omega }$ . By part (b) of Proposition 2.1 we can suppose that $y_0 \in Y_0$ and that $x_{n_0} \mathrel {R} y_0$ . As the $Y_k$ s are disjoint, for every n there is a unique k such that $y_n \in Y_k$ , and let $i ( n )$ be this k.

Claim 5.2.5. The set $I_k = \mathopen \{{ n \in \omega }\boldsymbol \mid { i ( n ) = k } \mathclose \}$ is a finite interval of natural numbers.

Proof By definition of $R_Y$ it follows that either $i ( n + 1 ) = i ( n )$ or else $i ( n + 1 ) = i ( n ) + 1$ , so it is enough to show that $I_k$ is finite. Towards a contradiction, suppose $I_{\bar {k}}$ is infinite, for some $\bar {k} \in \omega $ . This means that there is $\bar {n}$ such that $i ( n ) = i ( \bar {n} )$ for all $n \geq \bar {n}$ , that is $\mathopen \{{ y_n }\boldsymbol \mid { n \geq \bar {n} } \mathclose \} \subseteq Y_{\bar {k}}$ . But then $R {\mathop {\upharpoonright }} \mathopen \{{ y_n }\boldsymbol \mid { n \geq \bar {n} } \mathclose \}$ would be a total on $\mathopen \{{ y_n }\boldsymbol \mid { n \geq \bar {n} } \mathclose \}$ and contained in $R {\mathop {\upharpoonright }} Y$ , against (16).

Let $m_k = \max ( I_k )$ so that $I_0 = [ 0; m_0 ]$ and $I_{k + 1} = [ m_k + 1; m_{ k+1} ]$ . Then

$$ \begin{align*} \mathopen \langle x_0 , \dots , x_{ n_0 } \mathclose \rangle &{}^\smallfrown \mathopen \langle y_0 , \dots , y_{ m_0 } \mathclose \rangle {}^\smallfrown \mathopen \langle x_{ n_0 + 1 } , \dots , x_{ n_1 } \mathclose \rangle{}^\smallfrown \mathopen \langle y_{ m_0 + 1 } , \dots , y_{ m_1 } \mathclose \rangle {}^\smallfrown \dots \\ &\dots {}^\smallfrown \mathopen \langle x_{ n_k + 1 } , \dots , x_{ n_{ k + 1 } } \mathclose \rangle{}^\smallfrown \mathopen \langle y_{ m_k + 1 } , \dots , y_{ m_{ k + 1 } } \mathclose \rangle {}^\smallfrown \dots \end{align*} $$

is the required R-chain.

5.2 The Feferman–Levy model

Feferman and Levy showed that the following is consistent relative to $\mathsf {ZF}$ :

(FL ) $$\begin{align} \mathbb{R} \text{ is the countable union of countable sets.} \end{align}$$

(See [Reference Jech2, p. 142] for an exposition of the Feferman–Levy model.)

The next result shows that in the Feferman–Levy model, the statement of Theorem 4.1 fails, that is, there is no set $A \subseteq \mathbb {R}$ such that $\mathsf {DC} ( A )$ and $\neg {\mathsf {AC}}_\omega ( A )$ .

We need a preliminary result.

Proof Fix a bijection $\pi \colon \mathbb {R} \to \mathbb {R}^\omega $ , and for each $m \in \omega $ let $\pi _m \colon \mathbb {R} \to \mathbb {R}$ be defined as $\pi _m ( x ) = \pi ( x )_m$ . If $Y \subseteq \mathbb {R}$ and $f \colon \omega \to Y$ is surjective, then $\tilde {Y}$ , the closure of Y under the $\pi _m$ s, is also countable, as

$$\begin{align*}\tilde{f} \colon {}^{< \omega } \omega \times \omega \to \tilde{Y} \qquad ( \mathopen \langle n_0 , \dots , n_k \mathclose \rangle , m ) \mapsto \pi_{n_k} \circ \cdots \circ \pi_{n_0}\circ f ( m ) \end{align*}$$

is surjective. By $\mathsf {FL}$ let $( Y_n )_{ n \in \omega }$ be a sequence of countable sets such that $\mathbb {R} = \bigcup _n Y_n$ , and without loss of generality we may assume that each $Y_n$ is closed under every $\pi _m$ . Then let $X_n = Y_n \setminus \bigcup _{m < n} Y_m$ for each $n \in \omega $ . If necessary, we can pass to a subsequence to get them to be nonempty.

We claim that no infinite subsequence of $( X_n )_{ n \in \omega }$ has a choice function. Otherwise there would be an infinite sequence $( x_n )_{ n \in \omega } \in \mathbb {R}^\omega $ whose range intersects infinitely many $X_n$ s. Let $x \in \mathbb {R}$ be such that $\pi ( x) = ( x_n )_{n \in \omega }$ . Then $x \in X_k \subseteq Y_k$ for some $k \in \omega $ , and hence

$$\begin{align*}\forall {n \in \omega } \, \left ( x_n = \pi_n ( x ) \in Y_k \subseteq X_0 \cup \dots \cup X_k \right ) \end{align*}$$

as $Y_k$ is closed under the $\pi _n$ s. But this contradicts the assumption that $\mathopen \{{ x_n }\boldsymbol \mid { n \in \omega } \mathclose \}$ intersects infinitely many $X_n$ s.

5.3 Definability of the counterexample

Theorem 4.1 shows that the statement (3) is consistent with $\mathsf {ZF}$ , that is to say, it is consistent that there is a set $A \subseteq \mathbb {R}$ such that $\mathsf {DC} ( A )$ and $\neg {\mathsf {AC}}_\omega ( A )$ . The set A constructed in the proof of Theorem 4.1 is a set of Cohen reals, so it is not ordinal definable. But what is the possible descriptive complexity of a set A as above?

By part (c) of Proposition 2.5, the set A cannot contain a perfect set. Recall that a set has the perfect set property if it is either countable or else it contains a perfect subset. Assuming ${\mathsf {AC}}_\omega ( \mathbb {R} )$ , every Borel set has the perfect-set property. In a choice-less context the situation becomes murky. Assuming $\mathsf {FL}$ , every set of reals is $\mathbf {F}_{\sigma \sigma }$ (i.e., countable union of $\mathbf {F}_\sigma $ sets), and by taking complements it is also $\mathbf {G}_{\delta \delta }$ (i.e., countable intersection of $\mathbf {G}_\delta $ sets), so every set is $\boldsymbol {\Delta }^{0}_{4}$ , as $\mathbf {F}_\sigma = \boldsymbol {\Sigma }^{0}_{2} \subset \boldsymbol {\Pi }^{0}_{3}$ , and hence $\mathbf {F}_{\sigma \sigma } \subseteq \boldsymbol {\Sigma }^{0}_{4}$ . Therefore $\mathsf {FL}$ collapses the Borel hierarchy at level $4$ . Moreover $\mathsf {FL}$ implies that there is an uncountable set in $\boldsymbol {\Pi }^{0}_{3}$ without a perfect subset [Reference Miller8, Theorem 1.3].

On the other hand A. Miller has shown in $\mathsf {ZF}$ that $\boldsymbol {\Sigma }^{0}_{3} \neq \boldsymbol {\Pi }^{0}_{3}$ [Reference Miller7, Theorem 2.1], and that every set in $ \boldsymbol {\Sigma }^{0}_{3}$ has the perfect-set property [Reference Miller8, Theorem 1.2].

Recall that a subset of $\mathbb {R}$ is $\boldsymbol {\Pi }^{1}_{n}$ if it is the complement of a $\boldsymbol {\Sigma }^{1}_{n}$ , and it is $\boldsymbol {\Sigma }^{1}_{n}$ if it is the projection of a $\boldsymbol {\Pi }^{1}_{n - 1}$ set $C \subseteq \mathbb {R} \times \mathbb {R}$ , where $\boldsymbol {\Pi }^{1}_{0}$ is the collection of closed sets. The lightface hierarchy $\varSigma ^1_n , \varPi ^1_n$ is obtained by replacing $\boldsymbol {\Pi }^{1}_{0}$ with $\varPi ^1_0$ , the collection of recursively-closed sets (see [Reference Kanamori4, Chapter 3, Section 12]). Working in $\mathsf {ZF}$ , every $\boldsymbol {\Sigma }^{1}_{1}$ set has the perfect set property, and by a theorem of Mansfield and Solovay (see [Reference Kanamori4, Chapter 3 and Corollary 14.9]) every $ \boldsymbol {\Sigma }^{1}_{2}$ set is either well-orderable, being included in $\mathord {\mathrm {L}} [ a ]$ for some real a, or else it contains a perfect set.