Proof By a theorem of Kuratowski (see [Reference Kechris14, Theorem 15.6]), any two uncountable Polish spaces are Borel isomorphic: in other words, there is a bijection $f: \mathbb {R} \to X$ such that $A \subseteq \mathbb {R}$ is Borel if and only if $f[A]$ is Borel. Thus if $\mathcal {P}$ is any partition of $\mathbb {R}$ into Borel sets, then $\left \lbrace f[B] \colon B \in \mathcal {P} \right \rbrace $ is a partition of X into Borel sets, and if $\mathcal Q$ is any partition of X into Borel sets, then $\left \lbrace f^{-1}[B] \colon B \in \mathcal Q \right \rbrace $ is a partition of $\mathbb {R}$ into Borel sets.

Proof Throughout the proof, if $X \subseteq Y$ and $\mathcal {P}$ is a partition of Y, then $\mathcal {P} \!\restriction \! X = \left \lbrace K \cap X \colon K \in \mathcal {P} \right \rbrace $ denotes the restriction of $\mathcal {P}$ to X. We prove that $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (4) \Rightarrow (1)$ .

$(1) \Rightarrow (2)$ : Suppose $\mathcal {P}$ is a partition of $2^{\omega }$ into $\kappa $ closed sets. We claim first that there is a closed $X \subseteq 2^{\omega }$ such that $\mathcal {P} \!\restriction \! X$ is a partition of X into $\kappa $ nowhere dense closed sets (nowhere dense in X, that is).

To see this, we define a descending transfinite sequence of closed subsets of $2^{\omega }$ . Let $X_0 = 2^{\omega }$ , and if $\alpha $ is a limit ordinal, take $X_{\alpha } = \bigcap _{\xi < \alpha }X_{\xi }$ . At stage $\alpha $ , given $X_{\alpha }$ , form $X_{\alpha +1}$ by removing any open subset of $X_{\alpha }$ contained in a single member of $\mathcal {P}$ : $X_{\alpha +1} = X_{\alpha } \setminus \bigcup \left \lbrace U \colon U \text { is open in }X_{\alpha } \text { and } \left \lvert \vphantom {f^f}\mathcal {P} \!\restriction \! U \right \rvert = 1 \right \rbrace $ . Because $2^{\omega }$ is second countable, there is some $\beta < \omega _1$ such that $X_{\beta } = X_{\gamma }$ for all $\gamma \geq \beta $ . Let $X = X_{\beta }$ . Clearly, $\mathcal {P} \!\restriction \! X$ is a partition of X into compact nowhere dense sets. Furthermore, at any stage $\alpha < \beta $ of our recursion, $\left \lbrace K \in \mathcal {P} \colon K \cap X_{\alpha } \neq K \cap X_{\alpha +1} \right \rbrace $ is countable. Thus $\left \lbrace K \in \mathcal {P} \colon K \cap X \neq K \right \rbrace $ is countable. It follows that $\left \lvert \vphantom {f^f}\mathcal {P} \!\restriction \! X \right \rvert = \left \lvert \mathcal {P} \right \rvert = \kappa $ .

Now, we claim there is a subspace Y of X such that $Y \approx \omega ^{\omega }$ and $\mathcal {P} \!\restriction \! Y$ is a partition of Y into $\kappa $ compact sets. Fix a countable basis $\mathcal {B}$ for X, and for every $U \in \mathcal {B}$ fix some $K_U \in \mathcal {P}$ such that $K_U \cap U \neq \emptyset $ . Let $Y = X \setminus \bigcup \left \lbrace K_U \colon U \in \mathcal {B} \right \rbrace $ . Clearly $\mathcal {P} \!\restriction \! Y = \mathcal {P} \setminus \left \lbrace K_U \colon U \in \mathcal {B} \right \rbrace $ , so $\mathcal {P}$ partitions Y into $\kappa $ compact sets. The set $\left \lbrace K_U \colon U \in \mathcal {B} \right \rbrace $ is both dense and meager in X. It follows that no relatively (cl)open subset of Y is closed in X, and therefore no (cl)open subset of Y is compact. Furthermore Y is $G_{\delta }$ in X, and therefore Polish. That $Y \approx \omega ^{\omega }$ now follows from the Alexander–Urysohn characterization of $\omega ^{\omega }$ as the unique nowhere compact, zero-dimensional Polish space (see [Reference Kechris14, Theorem 7.7]).

$(2) \Rightarrow (3)$ : Suppose $\mathcal {P}$ is a partition of $\omega ^{\omega }$ into $\kappa $ compact sets, and let X be any uncountable Polish space. Decompose X into its scattered part and perfect part: i.e., let $X = Y \cup Z$ , where Y is countable and Z is closed in X (hence still Polish) and Z has no isolated points. By [Reference Kechris14, Exercise 7.15], there is a continuous bijection $f: \omega ^{\omega } \to Z$ . But then $\left \lbrace f[K] \colon K \in \mathcal {P} \right \rbrace \cup \left \lbrace \{y\} \colon y \in Y \right \rbrace $ is a partition of X into $\kappa $ compact sets.

$(3) \Rightarrow (4)$ : This implication is obvious, since “every uncountable Polish space” includes some compact spaces, and compact sets are $F_{\sigma }$ .

$(4) \Rightarrow (1)$ : Suppose X is an uncountable compact Polish space. By [Reference Kechris14, Theorem 7.4], there is a continuous surjection $f: 2^{\omega } \to X$ . If $\mathcal {P}$ is any partition of X into $\kappa F_{\sigma }$ sets, then $\mathcal {Q} = \left \lbrace f^{-1}[K] \colon K \in \mathcal {P} \right \rbrace $ is a partition of $2^{\omega }$ into $\kappa F_{\sigma }$ sets. But every $F_{\sigma }$ subset of $2^{\omega }$ can be partitioned into countably many closed sets. (This observation is attributed to Luzin in [Reference Miller, Barwise, Keisler and Kunen17, Theorem 2].) Thus, by breaking up any non-closed members of $\mathcal Q$ into countably many closed pieces, we can refine $\mathcal Q$ to obtain a partition of $2^{\omega }$ into $\kappa $ closed sets.

Proof Suppose $\kappa $ is a singular cardinal and $S \cap \kappa $ is unbounded in $\kappa $ . Let $\nu = \mathrm {cf}(\kappa ) < \kappa $ , and let $\left \langle \alpha _{\xi } \colon \xi < \nu \right \rangle $ be a sequence of cardinals in S increasing up to $\kappa $ . Fix some $\lambda \in S$ with $\nu \leq \lambda < \kappa $ .

Let $\left \lbrace K_{\xi } \colon \xi < \lambda \right \rbrace $ be a partition of $2^{\omega }$ into closed sets. Observe that $\left \lbrace 2^{\omega } \times K_{\xi } \colon \xi < \lambda \right \rbrace $ is a partition of $2^{\omega } \times 2^{\omega } \approx 2^{\omega }$ into $\lambda $ copies of $2^{\omega }$ . Thus we may (and do) assume that $K_{\xi } \approx 2^{\omega }$ for each $\xi < \lambda $ . For each $\xi < \nu $ , let $\mathcal {P}_{\xi }$ be a partition of $K_{\xi }$ into $\alpha _{\xi }$ closed sets (using the fact that $\alpha _{\xi } \in S$ ). Then $\bigcup _{\xi < \nu }\mathcal {P}_{\xi } \cup \left \lbrace K_{\xi } \colon \nu \leq \xi < \lambda \right \rbrace $ is a partition of $2^{\omega }$ into $\kappa $ closed sets.

Proof It is clear that (every-compact) $\,\Rightarrow \,$ (every-closed) $\,\Rightarrow \,$ (every- $F_{\sigma }$ ), and that (some-compact) $\,\Rightarrow \,$ (some-closed) $\,\Rightarrow \,$ (some- $F_{\sigma }$ ). Also, Lemma 2.2 implies that (every- $F_{\sigma }$ ) $\,\Rightarrow \,$ (some-compact): because if every uncountable Polish space can be partitioned into $\kappa F_{\sigma }$ sets, then in particular some uncountable compact Polish space can be, and by Lemma 2.2 this implies $\omega ^{\omega }$ can be partitioned into $\kappa $ compact sets. Thus, to prove the theorem, we need to show (some- $F_{\sigma }$ ) $\,\Rightarrow \,$ (every-compact).

So let X be some uncountable Polish space, and suppose $\mathcal {P}$ is a partition of X into $\kappa F_{\sigma }$ sets. If $\kappa = \acute {\mathfrak {n}}$ , then by definition, there is a partition of $\omega ^{\omega }$ into $\acute {\mathfrak {n}}$ compact sets, and by Lemma 2.2 every uncountable Polish space can be partitioned into $\kappa $ compact sets, and we are done. So let us suppose $\kappa> \acute {\mathfrak {n}}$ . We consider two cases.

For the first case, suppose $\kappa $ is regular. By the definition of $\acute {\mathfrak {n}}$ and Lemma 2.2, there is a partition $\mathcal Q$ of X into $\acute {\mathfrak {n}}$ compact sets. Because $|\mathcal {P}| = \kappa> \acute {\mathfrak {n}}$ and $\kappa $ is regular, there is some $K \in \mathcal Q$ such that $\left \lvert \vphantom {f^f}\mathcal {P} \!\restriction \! K \right \rvert = \kappa $ . Thus $\mathcal {P} \!\restriction \! K$ is a partition of K into $\kappa $ compact subsets. But K is a compact Polish space, and uncountable because $|K| \geq \left \lvert \vphantom {f^f}\mathcal {P} \!\restriction \! K \right \rvert = \kappa $ . Thus some compact uncountable Polish space can be partitioned into $\kappa F_{\sigma }$ sets. Invoking Lemma 2.2 again, this implies every uncountable Polish space can be partitioned into $\kappa $ compact sets.

For the second case, suppose $\kappa $ is singular. As in the first case, there is a partition $\mathcal Q$ of X into $\acute {\mathfrak {n}}$ compact sets. If $\left \lvert \vphantom {f^f}\mathcal {P} \!\restriction \! K \right \rvert = \kappa $ for some $K \in \mathcal Q$ , then we may argue as in the first case and conclude that every uncountable Polish space can be partitioned into $\kappa $ compact sets, and we are done. So let us suppose instead that $\left \lvert \vphantom {f^f}\mathcal {P} \!\restriction \! K \right \rvert < \kappa $ for every $K \in \mathcal Q$ . Let $\lambda $ be any cardinal with $\acute {\mathfrak {n}} < \lambda < \kappa $ . Because $|\mathcal {P}| = \kappa> \lambda > \acute {\mathfrak {n}} = |\mathcal Q|$ , there is some $K \in \mathcal Q$ such that $\left \lvert \vphantom {f^f}\mathcal {P} \!\restriction \! K \right \rvert \geq \lambda $ . Let $\mu = \left \lvert \vphantom {f^f}\mathcal {P} \!\restriction \! K \right \rvert $ , and note that $\mu < \kappa $ (by the third sentence of this paragraph). Now, as in the previous paragraph, K is a compact uncountable Polish space that can be partitioned into $\mu F_{\sigma }$ sets. By Lemma 2.2, this implies $2^{\omega }$ can be partitioned into $\mu $ closed sets. Because $\lambda $ was an arbitrary cardinal below $\kappa $ and $\lambda \leq \mu < \kappa $ , this shows that $S = \left \lbrace \mu \colon \text {there is a partition of }2^{\omega } \text { into }\mu \text { closed sets} \right \rbrace $ is unbounded below $\kappa $ . By Lemma 2.3, $\kappa \in S$ . By Lemma 2.2, this implies every uncountable Polish space can be partitioned into $\kappa $ compact sets.

Proof This follows from Corollary 2.5 and Lemma 2.3.

Proof Suppose $\kappa $ is a singular cardinal and $\mathfrak {sp}({\small\text{Borel}}) \cap \kappa $ is unbounded in $\kappa $ . Let $\nu = \mathrm {cf}(\kappa ) < \kappa $ , and let $\left \langle \alpha _{\xi } \colon \xi < \nu \right \rangle $ be a sequence of cardinals in $\mathfrak {sp}({\small\text{Borel}})$ increasing up to $\kappa $ . Fix some $\lambda \in \mathfrak {sp}({\small\text{Borel}})$ with $\nu \leq \lambda < \kappa $ .

Let $\left \lbrace B_{\xi } \colon \xi < \lambda \right \rbrace $ be a partition of $2^{\omega }$ into $\lambda $ Borel sets. Observe that $\left \lbrace 2^{\omega } \times B_{\xi } \colon \xi < \lambda \right \rbrace $ is a partition of $2^{\omega } \times 2^{\omega } \approx 2^{\omega }$ into $\lambda $ uncountable Borel sets. Thus we may (and do) assume that $B_{\xi }$ is uncountable for each $\xi < \lambda $ . Every uncountable Borel set contains a closed subspace homeomorphic to $2^{\omega }$ . For each $\xi < \nu $ , fix $C_{\xi } \subseteq B_{\xi }$ with $C_{\xi } \approx 2^{\omega }$ , and let $\mathcal {P}_{\xi }$ be a partition of $C_{\xi }$ into $\alpha _{\xi }$ Borel sets. Then $\bigcup _{\xi < \nu }\mathcal {P}_{\xi } \cup \left \lbrace B_{\xi } \setminus C_{\xi } \colon \xi < \nu \right \rbrace \cup \left \lbrace B_{\xi } \colon \nu \leq \xi < \lambda \right \rbrace $ is a partition of $2^{\omega }$ into $\kappa $ Borel sets.

Question 2.9. Is it consistent that $\acute {\mathfrak {n}}$ has countable cofinality?

Question 2.10. Is $\mathfrak {a} \leq \acute {\mathfrak {n}}$ ?

Proof sketch of Theorem 3.1

Given some $X \subseteq 2^{\omega }$ with empty interior, let $\mathbb {T}_{\!X}$ be the poset whose conditions are pairs $(t,B)$ , where

1. ∘ There is some $k \in \omega $ such that t is a subtree of $2^{<k}$ , and t is pruned: i.e., if $\sigma \in t$ with $|\sigma | < k$ , then $\sigma $ has a proper extension in t.

2. ∘ B is a finite subset of $2^{\omega } \setminus X$ , and $b \!\restriction \! k$ is a branch of t for each $b \in B$ .

The ordering on $\mathbb {T}_{\!X}$ is defined by: $(t',B')$ extends $(t,B)$ if and only if $B' \supseteq B$ and $t'$ is an end extension of t (meaning that $t = t' \cap 2^{<k}$ for some k).

Roughly, a condition $(t,B)$ can be thought of as a finite approximation t to a subtree T of $2^{<\omega }$ that we are trying to build generically, and a promise that some finite set B of reals will be branches of T.

For any $X \subseteq 2^{\omega }$ , $\mathbb {T}_{\!X}$ is $\sigma $ -centered. The poset $\mathbb {T}_{\!X}$ generically adds an infinite pruned subtree T of $2^{<\omega }$ , defined from a generic filter G on $\mathbb {T}_{\!X}$ as $T = \bigcup \left \lbrace t \colon (t,B) \in G \text { for some } B \right \rbrace $ . (Equivalently, T is the evaluation in $V[G]$ of the name $\dot T = \left \lbrace \langle \sigma ,q \rangle \colon q = (t,B) \text { for some } B \text {, and } \sigma \in t \right \rbrace .$ ) In the extension, $[\![T]\!]$ is a closed subset of $2^{\omega }$ disjoint from X.

Let $\mathbb {T}_{\!X}^{\,\omega }$ denote the finite support product of countably many copies of $\mathbb {T}_{\!X}$ . For any $X \subseteq 2^{\omega }$ , $\mathbb {T}_{\!X}^{\,\omega }$ is $\sigma $ -centered. The poset $\mathbb {T}_{\!X}^{\,\omega }$ generically adds countably many infinite pruned subtrees $T_0,T_1,T_2,\dots $ of $2^{<\omega }$ , and in the extension,

$$ \begin{align*}\textstyle (2^{\omega})^V \cap \bigcup_{n < \omega}[\![T_n]\!] \,=\, (2^{\omega})^V \setminus X.\end{align*} $$

In other words, $\mathbb {T}_{\!X}^{\,\omega }$ generically adds an $F_{\sigma }$ subset of $2^{\omega }$ that intersects the ground model reals in precisely the complement of X. (Note: this poset may look familiar: it is the one usually used for showing that $\mathsf {M}$ artin’s $\mathsf {A}$ xiom implies every $<\!\mathfrak {c}$ -sized subset of $2^{\omega }$ is a Q-set.)

Let C denote the set of cardinals in $[\aleph _1,\kappa ]$ . We now define a finite support iteration of length $\omega _1$ as follows. At stage $0$ , force with the poset $\mathbb {Q}_0 = \mathbb {C}_{\kappa }$ of finite partial functions $\kappa \to 2$ , in order to add a set of $\kappa $ mutually generic Cohen reals. Let $\left \lbrace c_{\xi } \colon \xi < \kappa \right \rbrace $ be an enumeration of these Cohen reals in $V^{\mathbb {Q}_0}$ , and for each $\mu \in C$ , let $\mathcal {P}_0^{\mu } = \left \lbrace \{c_{\xi }\} \colon \xi < \mu \right \rbrace $ . Note that $\mathcal {P}^{\mu }_0$ is a $\mu $ -sized collection of disjoint subsets of $\mathbb {R}$ : we think of $\mathcal {P}^{\mu }_0$ as a first approximation to a $\mu $ -sized partition we are trying to build. At a later stage $\alpha $ of the iteration, suppose we have already obtained, for each $\mu \in C$ , a $\mu $ -sized collection $\mathcal {P}^{\mu }_{\alpha }$ of disjoint subsets of $\mathbb {R}$ . In $V^{\mathbb {Q}_{\alpha }}$ , define $X^{\mu }_{\alpha } = \bigcup \mathcal {P}^{\mu }_{\alpha }$ for each $\mu \in C$ , and then obtain $V^{\mathbb {Q}_{\alpha +1}}$ from $V^{\mathbb {Q}_{\alpha }}$ by forcing with $\prod _{\mu \in C}\mathbb {T}^{\,\omega }_{\!X_{\alpha }^{\mu }}$ . This adds countably many generic trees $T^{\mu }_{\alpha ,0},T^{\mu }_{\alpha ,1},T^{\mu }_{\alpha ,2},\dots $ for each $\mu \in C$ , and in $V^{\mathbb {Q}_{\alpha +1}}$ we define $\mathcal {P}^{\mu }_{\alpha +1} = \mathcal {P}^{\mu }_{\alpha } \cup \{ \bigcup _{n \in \omega } [\![T^{\mu }_{\alpha ,n}]\!] \}$ .

At the end of the iteration, in $V^{\mathbb {Q}_{\omega _1}}$ , let $\mathcal {P}^{\mu } = \bigcup _{\alpha < \omega _1}\mathcal {P}_{\alpha }^{\mu }$ . This is a $\mu $ -sized collection of disjoint $F_{\sigma }$ subsets of $\mathbb {R}$ . Furthermore, if x is a real in $V^{\mathbb {Q}_{\omega _1}}$ , then there is some $\alpha < \omega _1$ with $x \in V^{\mathbb {Q}_{\alpha }}$ . At that stage of the iteration, either $x \in \bigcup \mathcal {P}^{\mu }_{\alpha }$ , or if not, then $x \in \bigcup _{n \in \omega } [\![T^{\alpha }_{\mu ,n}]\!]$ because $\textstyle (2^{\omega })^{V^{\mathbb {P}_{\alpha }}} \cap \bigcup _{n < \omega }[\![T^{\alpha }_{\mu ,n}]\!] \,=\, (2^{\omega })^{V^{\mathbb {P}_{\alpha }}} \setminus X_{\alpha }^{\mu }$ . Either way, $x \in \bigcup \mathcal {P}^{\mu }_{\alpha +1} \subseteq \bigcup \mathcal {P}^{\mu }$ . Thus $\mathcal {P}^{\mu }$ is a partition of $2^{\omega }$ into $F_{\sigma }$ sets in $V^{\mathbb {Q}_{\omega _1}}$ , and this means $\mu \in \mathfrak {sp}({\small\text{closed}})$ by Corollary 2.5.

Proof This follows immediately from the previous theorem.

Proof Fix some $A \subseteq \omega \setminus \{0\}$ . First pass to a forcing extension in which $\mathsf {GCH}$ holds up to $\aleph _{\omega +1}$ . Then, for the result about $\mathfrak {sp}({\small\text{closed}})$ , apply Theorem 3.2 with $C = \left \lbrace \aleph _n \colon n \in A \right \rbrace \cup \{\aleph _{\omega },\aleph _{\omega +1}\}$ . Or, for the result about $\mathfrak {sp}({\small\text{Borel}})$ , apply Corollary 3.3 with $C = \left \lbrace \aleph _n \colon n \in A \right \rbrace \cup \{\aleph _1,\aleph _{\omega },\aleph _{\omega +1}\}$ .

Proof This is proved in exactly the same way as the previous corollary, but without $\aleph _{\omega }$ and $\aleph _{\omega +1}$ put into C.

Proof of Theorem 3.2

Let C be a set of uncountable cardinals satisfying the hypotheses listed in the statement of the theorem, and assume $\mathsf {GCH}$ holds up to $\max (C)$ . Let $\kappa = \min (C)$ and let $\theta = \max (C)$ .

For each $\mu \in C$ , let $I_{\mu } = \mu \times \{\mu ,\theta ^{+}\}$ . (These are merely indexing sets, and for all practical purposes one may think of each $I_{\mu }$ as a set of atoms, or urelements, in the set-theoretic universe. The relevant properties of these $I_{\mu }$ ’s are that they are pairwise disjoint, each $I_{\mu }$ has size $\mu $ , and the $I_{\mu }$ do not “interact” in any accidental way with any other sets in the proof.)

Let $I = \bigcup _{\mu \in C}I_{\mu }$ , and let $\mathbb {P}_0$ denote the poset of finite partial functions from $I \times \omega $ to $2$ . Note that this is equivalent to the standard poset $\mathbb {C}_{\theta }$ for adding $\theta $ mutually generic Cohen reals. For each $i \in I$ , let $c_i$ denote in $V^{\mathbb {P}_0}$ the Cohen real added by $\mathbb {P}_0$ in coordinate i. More formally, $c_i$ denotes in $V^{\mathbb {P}_0}$ the evaluation of the name $\dot c_i = \left \lbrace \langle (n,j),p\rangle \colon p(i,n) = j \right \rbrace .$

Next we define, recursively, a poset $\langle \mathbb {P}_{\alpha },\leq _{\mathbb {P}_{\alpha }} \rangle $ for each $\alpha \leq \kappa $ with $\alpha> 0$ . At every stage of the recursion, $\mathbb {P}_{\alpha }$ is defined so that $\mathbb {P}_{\beta }$ is a sub-poset of $\mathbb {P}_{\alpha }$ for each $\beta < \alpha $ . Conditions in $\mathbb {P}_{\alpha }$ are finite partial functions on $(I \times \omega ) \cup (\alpha \times C \times \omega )$ , and the $\mathbb {P}_{\alpha }$ are defined so that for any $\beta < \alpha $ , the restriction of a condition in $\mathbb {P}_{\alpha }$ to $(I \times \omega ) \cup (\beta \times C \times \omega )$ is a condition in $\mathbb {P}_{\beta }$ .

At limit stages, we take $\mathbb {P}_{\alpha } = \bigcup _{\xi < \alpha }\mathbb {P}_{\xi }$ and $\leq _{\mathbb {P}_{\alpha }} = \bigcup _{\xi < \alpha }\leq _{\mathbb {P}_{\xi }}$ . In other words, $\langle \mathbb {P}_{\alpha },\leq _{\mathbb {P}_{\alpha }} \rangle $ is the direct limit of $\left \langle \langle \mathbb {P}_{\xi },\leq _{\mathbb {P}_{\xi }} \rangle \colon \xi < \alpha \right \rangle $ for limit $\alpha $ .

At successor stages, suppose $\mathbb {P}_{\xi }$ is given for every $\xi \leq \beta $ , and $\mathbb {P}_{\xi } \supseteq \mathbb {P}_{\zeta }$ whenever $\zeta \leq \xi \leq \beta $ . Let $\alpha = \beta +1$ . Conditions in $\mathbb {P}_{\alpha }$ are finite partial functions p on $(I \times \omega ) \cup (\alpha \times C \times \omega )$ such that:

1. ∘ If $(i,n) \in (I \times \omega ) \cap \mathrm {dom}(p)$ , then $p(i,n) \in \{0,1\}$ .

2. ∘ If $(\gamma ,\mu ,n) \in \mathrm {dom}(p)$ for some $\gamma \leq \beta $ , $\mu \in C$ , and $n \in \omega $ , then $p(\gamma ,\mu ,n) = (t,B)$ , where:

   1. ∘ t is a pruned subtree of $2^{<k}$ for some $k \in \omega $ (in this context, “pruned” means that if $\sigma \in t$ and $|\sigma | < k$ , then $\sigma $ has an extension in t).

   2. ∘ B is a finite set of nice $\mathbb {P}_{\gamma }$ -names for members of $2^{\omega }$ such that for every $\dot b \in B$ ,

      $$ \begin{align*} \quad \quad \quad \quad \quad p \!\restriction\! ((I \times \omega) \cup (\gamma \times C \times \omega)) \ \Vdash_{\mathbb{P}_{\gamma}} \quad & \dot b \!\restriction\! j \in t \text{ for every } j < k, \\ & \dot b \neq \dot c_i \text{ for every } i \in I_{\mu}, \text{ and} \\ & \dot b \text{ is not a branch of } \dot T^{\xi}_{\mu,m} \\ & \quad \text{ for any } \xi < \gamma \text{ and } m \in \omega, \end{align*} $$

      where $\dot c_i$ is the $\mathbb {P}_0$ -name described above, but interpreted as a $\mathbb {P}_{\gamma }$ -name, and where for every $\xi < \gamma $ and $m \in \omega $ , $\dot T^{\xi }_{\mu ,m}$ is the $\mathbb {P}_{\gamma }$ -name for a subtree of $2^{<\omega }$ defined as follows:

      $$ \begin{align*}\quad \quad \quad \quad \ \ \dot T^{\xi}_{\mu,m} = \left\lbrace \langle \sigma,q \rangle \colon q(\xi,\mu,m) = (t',B') \text{ for some } B' \text{, and } \sigma \in t' \right\rbrace.\end{align*} $$

The extension relation $\leq _{\mathbb {P}_{\alpha }}$ on $\mathbb {P}_{\alpha }$ is defined as follows: given $p,q \in \mathbb {P}_{\alpha }$ , we write $q \leq _{\mathbb {P}_{\alpha }} p$ (meaning that q extends p) if and only if:

1. ∘ $\mathrm {dom}(q) \supseteq \mathrm {dom}(p)$ ,

2. ∘ $q \!\restriction \! (I \times \omega ) \supseteq p \!\restriction \! (I \times \omega )$ , and

3. ∘ if $(\gamma ,\mu ,n) \in \mathrm {dom}(p)$ for some $\gamma \leq \beta $ , $\mu \in C$ , and $n \in \omega $ , and if $p(\gamma ,\mu ,n) = (t,B)$ and $q(\gamma ,\mu ,n) = (t',B')$ , then $B' \supseteq B$ and $t'$ is an end extension of t (meaning that $t = t' \cap 2^{<k}$ for some k).

Naturally, we abbreviate “ $\leq _{\mathbb {P}_{\alpha }}$ ” with “ $\leq $ ” in situations where this creates no ambiguity. We also write $p \!\restriction \! \alpha $ as an abbreviation for $p \!\restriction \! ((I \times \omega ) \cup (\alpha \times C \times \omega ))$ where doing so creates no ambiguity.

This completes the recursive definition of the $\mathbb {P}_{\alpha }$ and $\leq _{\mathbb {P}_{\alpha }}$ . It is easy to see that $\leq _{\mathbb {P}_{\alpha }}$ is a partial order on $\mathbb {P}_{\alpha }$ for all $\alpha \leq \kappa $ , and that $\mathbb {P}_{\beta }$ is a sub-poset of $\mathbb {P}_{\alpha }$ whenever $\beta \leq \alpha \leq \kappa $ .

Proof Let $\mathcal {A}_0$ be an uncountable subset of $\mathbb {P}_{\kappa }$ . By the $\Delta $ -system lemma, there is an uncountable $\mathcal {A}_1 \subseteq \mathcal {A}_0$ and a finite $R \subseteq (I \times \omega ) \cup (\kappa \times C \times \omega )$ such that for any $p,q \in \mathcal {A}_1$ , $\mathrm {dom}(p) \cap \mathrm {dom}(p) = R$ . If $i \in R \cap (I \times \omega )$ , then $p(i) \in \{0,1\}$ for every $p \in \mathcal {A}_1$ . As there are only finitely many functions $R \cap (I \times \omega ) \to 2$ , this implies there is some uncountable $\mathcal {A}_2 \subseteq \mathcal {A}_1$ such that $p \!\restriction \! (R \cap (I \times \omega )) = q \!\restriction \! (R \cap (I \times \omega ))$ for any $p,q \in \mathcal {A}_2$ . For every $p \in \mathcal {A}_2$ , if $j \in R \setminus (I \times \omega )$ , then $p(j) = (t,B)$ for some finite subtree t of $2^{<\omega }$ . As there are only countably many finite subtrees of $2^{<\omega }$ , there is some uncountable $\mathcal {A}_3 \subseteq \mathcal {A}_2$ such that for each $j \in R \setminus (I \times \omega )$ , there is some fixed $t_j$ such that for any $p \in \mathcal {A}_3$ , $p(j) = (t_j,B^p_j)$ for some $B^p_j$ . But any two members of $\mathcal {A}_3$ are compatible: if $p,q \in \mathcal {A}_3$ , then

$$ \begin{align*}r(j)= \begin{cases} p(j), & \text{ if } j \in \mathrm{dom}(p) \setminus R, \\ q(j), & \text{ if } j \in \mathrm{dom}(q) \setminus R, \\ (t_j,B^p_j \cup B^q_j), & \text{ if } j \in \mathrm{dom}(p) \cap \mathrm{dom}(q) = R \end{cases} \end{align*} $$

is a common extension of p and q. Thus $\mathbb {P}_{\kappa }$ has no uncountable antichains, and is ccc. (In fact we have shown a bit more: $\mathbb {P}_{\kappa }$ has property K.)

If $\beta \leq \alpha $ , then because $\mathbb {P}_{\beta } \subseteq \mathbb {P}_{\alpha }$ , we may (and do) consider every $\mathbb {P}_{\beta }$ -name to be a $\mathbb {P}_{\alpha }$ -name as well. Let us also set the convention that in $V^{\mathbb {P}_{\alpha }}$ , the evaluation of a $\mathbb {P}_{\alpha }$ -name is indicated by removing its dot. So, for example, $c_i$ denotes in $V^{\mathbb {P}_{\alpha }}$ (for any $\alpha \leq \kappa $ ) the Cohen real added by $\mathbb {P}_0$ in coordinate i, and $T^{\xi }_{\mu ,n}$ denotes in $V^{\mathbb {P}_{\alpha }}$ , for any $\alpha \geq \xi $ , the evaluation of the name $\dot T^{\xi }_{\mu ,n}$ .

Proof Let $\mu \in C$ . We claim that in $V^{\mathbb {P}_{\kappa }}$ ,

$$ \begin{align*}\mathcal{P} = \left\lbrace \{c_i\} \colon \vphantom{2^i}i \in I_{\mu} \right\rbrace \cup \textstyle \left\lbrace \bigcup_{n \in \omega}[\![T^{\alpha}_{\mu,n}]\!] \colon \alpha < \kappa \right\rbrace\end{align*} $$

is a partition of $2^{\omega }$ . Observe that every member of $\mathcal {P}$ is an $F_{\sigma }$ subset of $2^{\omega }$ , and $|\mathcal {P}| = \mu $ . (It is clear that each $T^{\alpha }_{\mu ,n}$ is a subtree of $2^{<\omega }$ , which means the sets of the form $\bigcup _{n \in \omega } [\![T^{\alpha }_{\mu ,n}]\!]$ are all $F_{\sigma }$ .) Thus if $\mathcal {P}$ is a partition, then $\mu \in \mathfrak {sp}({\small\text{closed}})$ by Theorem 2.4.

The members of $\mathcal {P}$ come in two types, so in order to show they are pairwise disjoint, we have three things to prove.

First, if $i,i' \in I_{\mu }$ and $i \neq i'$ , then clearly $c_i \neq c_{i'}$ , i.e., $\{c_i\} \cap \{c_{i'}\} = \emptyset $ .

Second, fix $i \in I_{\mu }$ , $\alpha < \kappa $ , and $n \in \omega $ . We claim $\{c_i\} \cap [\![T^{\alpha }_{\mu ,n}]\!] = \emptyset $ , or equivalently, $c_i \notin [\![T^{\alpha }_{\mu ,n}]\!]$ . To see this, fix some $p \in \mathbb {P}_{\kappa }$ : we will find an extension r of p forcing $c_i \notin [\![T^{\alpha }_{\mu ,n}]\!]$ . Extending p to $p \cup \{\langle (\alpha ,\mu ,n),(\emptyset ,\emptyset ) \rangle \}$ , if necessary, we may (and do) assume $(\alpha ,\mu ,n) \in \mathrm {dom}(p)$ . Let $(t,B) = p(\alpha ,\mu ,n)$ , with t a pruned subtree of $2^{<k}$ . From the definition of $\mathbb {P}_{\kappa }$ , we know $p \!\restriction \! \alpha \Vdash _{\mathbb {P}_{\alpha }} \dot b \neq \dot c_i$ for every $\dot b \in B$ . Using this, and the fact that B is finite, there is an extension q of p in $\mathbb {P}_{\alpha }$ , and some finite sequence $\sigma \in 2^{\ell }$ for some $\ell> k$ , such that $q \Vdash _{\mathbb {P}_{\alpha }}$ “ $\dot c_i \!\restriction \! \ell = \sigma $ but $\dot b \!\restriction \! \ell \neq \sigma $ for all $\dot b \in B$ .” Let $t'$ be the largest subtree of $2^{<\ell +1}$ that is an end extension of t and that does not contain $\sigma $ . (In other words, $\tau \in t'$ if and only if $\tau \in 2^{<\ell +1} \setminus \{\sigma \}$ and $\tau \!\restriction \! k \in t$ . This is a subtree of $2^{<\ell +1}$ , and it is an end extension of t because $\ell> k$ and $\sigma \in 2^{\ell }$ .) Define a function r on $(I \times \omega ) \cup (\kappa \times \mathcal {C} \times \omega )$ by setting

$$ \begin{align*}r(j)= \begin{cases} q(j), & \text{ if } j \in (I \times \omega) \cup (\alpha \times \mathcal{C} \times \omega), \\ p(j), & \text{ if } j \in (I \times \omega) \cup ((\kappa \setminus \alpha) \times \mathcal{C} \times \omega) \text{ but } j \neq (\alpha,\mu,n)), \\ (t',B), & \text{ if } j = (\alpha,\mu,n). \end{cases} \end{align*} $$

Then $r \in \mathbb {P}_{\kappa }$ , and $r \Vdash \sigma \notin \dot T^{\alpha }_{\mu ,n}$ and $\dot c_i \!\restriction \! \ell = \sigma $ , which means $r \Vdash \dot c_i \notin [\![\dot T^{\alpha }_{\mu ,n}]\!]$ . As p was arbitrary, this shows that $c_i \notin [\![T^{\alpha }_{\mu ,n}]\!]$ in $V^{\mathbb {P}_{\kappa }}$ .

Third, fix $\beta < \alpha < \kappa $ , and $m,n \in \omega $ . We claim that $[\![T^{\beta }_{\mu ,m}]\!] \cap [\![T^{\alpha }_{\mu ,n}]\!] = \emptyset $ . To see this, fix $p \in \mathbb {P}_{\kappa }$ : we will find an extension of p forcing $[\![T^{\beta }_{\mu ,m}]\!] \cap [\![T^{\alpha }_{\mu ,n}]\!] = \emptyset $ . Extending p to $p \cup \{\langle (\beta ,\mu ,m),(\emptyset ,\emptyset ) \rangle ,\langle (\alpha ,\mu ,n),(\emptyset ,\emptyset ) \rangle \}$ , if necessary, we may (and do) assume $(\beta ,\mu ,m),(\alpha ,\mu ,n) \in \mathrm {dom}(p)$ . Let $(t_{\beta },B_{\beta }) = p(\beta ,\mu ,m)$ , where $t_{\beta }$ is a pruned subtree of $2^{<k_{\beta }}$ for some $k_{\beta } \in \omega $ , and let $(t_{\alpha },B_{\alpha }) = p(\alpha ,\mu ,n)$ , where $t_{\alpha }$ is a pruned subtree of $2^{<k_{\alpha }}$ for some $k_{\alpha } \in \omega $ . From the definition of $\mathbb {P}_{\kappa }$ , we know that $p \!\restriction \! ((I \times \omega ) \cup (\alpha \times C \times \omega )) \Vdash _{\mathbb {P}_{\alpha }} \dot b \neq \dot a$ for every $\dot b \in B_{\beta }$ and $\dot a \in B_{\alpha }$ . Using this, and the fact that $B_{\beta }$ and $B_{\alpha }$ are both finite, there is an extension q of p in $\mathbb {P}_{\alpha }$ , and some $\ell> k_{\beta },k_{\alpha }$ , such that q “decides” all the $\dot b$ and $\dot a$ up to $\ell $ , and in such a way that witnesses $\dot b \neq \dot a$ for all $\dot b \in B_{\beta }$ and $\dot a \in B_{\alpha }$ . More precisely: for each $\dot b \in B_{\beta }$ , there is a particular branch $c(\dot b)$ of $2^{<\ell +1}$ such that $q \Vdash _{\mathbb {P}_{\alpha }}$ “ $\dot b \!\restriction \! \ell = c(\dot b)$ ”; and similarly, for each $\dot a \in B_{\alpha }$ , there is a particular branch $c(\dot a)$ of $2^{<\ell +1}$ such that $q \Vdash _{\mathbb {P}_{\alpha }}$ “ $\dot a \!\restriction \! \ell = c(\dot a)$ ”; and furthermore, $c(\dot b) \!\restriction \! \ell \neq c(\dot a) \!\restriction \! \ell $ for all $\dot b \in B_{\beta }$ and $\dot a \in B_{\alpha }$ . Let $t^{\prime }_{\beta }$ be the largest end extension of $t_{\beta }$ that is a pruned subtree of $2^{<\ell }$ , and let $t^{\prime \prime }_{\beta }$ be the end extension of $t^{\prime }_{\beta }$ to a pruned subtree of $2^{<\ell +1}$ containing on level $\ell $ the nodes:

$$ \begin{align*} \sigma \!\,^{\frown} 0 \quad & \text{ if } \sigma \in t^{\prime}_{\beta}, \sigma = c(\dot a) \text{ for some } \dot a \in B_{\alpha}, \text{ and } c(\dot a)(\ell) = 1, \\ \sigma \!\,^{\frown} 1 \quad & \text{ if } \sigma \in t^{\prime}_{\beta}, \sigma = c(\dot a) \text{ for some } \dot a \in B_{\alpha}, \text{ and } c(\dot a)(\ell) = 0, \\ \sigma \!\,^{\frown} 0 \quad & \text{ if } \sigma \in t^{\prime}_{\beta} \text{ and } \sigma \neq c(\dot a) \text{ for any } \dot a \in B_{\alpha}. \end{align*} $$

Similarly, let $t^{\prime }_{\alpha }$ be the largest end extension of $t_{\alpha }$ that is a pruned subtree of $2^{<\ell }$ , and let $t^{\prime \prime }_{\alpha }$ be the end extension of $t^{\prime }_{\alpha }$ to a pruned subtree of $2^{<\ell +1}$ containing on level $\ell $ the nodes:

$$ \begin{align*} \sigma \!\,^{\frown} 0 \quad & \text{ if } \sigma \in t^{\prime}_{\alpha}, \sigma = c(\dot b) \text{ for some } \dot b \in B_{\beta}, \text{ and } c(\dot b)(\ell) = 1, \\ \sigma \!\,^{\frown} 1 \quad & \text{ if } \sigma \in t^{\prime}_{\alpha}, \sigma = c(\dot b) \text{ for some } \dot b \in B_{\beta}, \text{ and } c(\dot b)(\ell) = 0, \\ \sigma \!\,^{\frown} 1 \quad & \text{ if } \sigma \in t^{\prime}_{\alpha} \text{ and } \sigma \neq c(\dot b) \text{ for any } \dot b \in B_{\beta}. \end{align*} $$

Observe that $t^{\prime \prime }_{\beta }$ and $t^{\prime \prime }_{\alpha }$ have no common nodes on level $\ell $ . Define a function r on $(I \times \omega ) \cup (\kappa \times \mathcal {C} \times \omega )$ by setting

$$ \begin{align*}r(j)= \begin{cases} q(j), & \text{ if } j \in (I \times \omega) \cup (\alpha \times \mathcal{C} \times \omega) \text{ but } j \neq (\beta,\mu,m), \\ p(j), & \text{ if } j \in (I \times \omega) \cup ((\kappa \setminus \alpha) \times \mathcal{C} \times \omega)) \text{ but } j \neq (\alpha,\mu,n), \\ (t^{\prime\prime}_{\beta},B_{\beta}), & \text{ if } j = (\beta,\mu,m), \\ (t^{\prime\prime}_{\alpha},B_{\alpha}), & \text{ if } j = (\alpha,\mu,n). \end{cases} \end{align*} $$

Then $r \in \mathbb {P}_{\kappa }$ , and $r \Vdash t^{\prime \prime }_{\beta }$ is a subtree of $\dot T^{\beta }_{\mu ,m}$ and $r \Vdash t^{\prime \prime }_{\alpha }$ is a subtree of $\dot T^{\alpha }_{\mu ,n}$ . Because $t^{\prime \prime }_{\beta }$ and $t^{\prime \prime }_{\alpha }$ share no nodes on level $\ell $ , $r \Vdash [\![\dot T^{\beta }_{\mu ,m}]\!] \cap [\![\dot T^{\alpha }_{\mu ,n}]\!] = \emptyset $ . As p was arbitrary, this shows that $[\![T^{\beta }_{\mu ,m}]\!] \cap [\![T^{\alpha }_{\mu ,n}]\!] = \emptyset $ in $V^{\mathbb {P}_{\kappa }}$ .

Thus $\mathcal {P}$ is a pairwise disjoint collection of $F_{\sigma }$ subsets of $2^{\omega }$ in $V^{\mathbb {P}_{\kappa }}$ . To finish the proof, we must show also that $\bigcup \mathcal {P} = 2^{\omega }$ .

Fix $x \in 2^{\omega }$ in $V^{\mathbb {P}_{\kappa }}$ . Because $\mathbb {P}_{\kappa }$ has the ccc, and each $\mathbb {P}_{\alpha }$ is a complete sub-poset of $\mathbb {P}_{\kappa }$ , there is some $\alpha < \kappa $ such that $x \in V^{\mathbb {P}_{\alpha }}$ . If $x = c_i$ for some $i \in I_{\mu }$ , or if $x \in [\![T^{\xi }_{\mu ,n}]\!]$ for some $\xi < \alpha $ , then we’re done. If not, let $\dot x$ be a nice $\mathbb {P}_{\alpha }$ -name for x, and fix some $p \in \mathbb {P}_{\kappa }$ such that $p \Vdash $ “ $\dot x \neq \dot c_i$ for all $i \in I_{\mu }$ , and $\dot x \notin [\![\dot T^{\xi }_{\mu ,n}]\!]$ for all $\xi < \alpha $ and $n \in \omega $ .” Because p has finite support, there is some $N \in \omega $ such that $(\alpha ,\mu ,N) \notin \mathrm {dom}(p)$ . But then $q = p \cup \{ \langle (\alpha ,\mu ,N),(\emptyset ,\{\dot x\}) \rangle \}$ is in $\mathbb {P}_{\alpha }$ , and $q \Vdash \dot x \in [\![\dot T^{\alpha }_{\mu ,N}]\!]$ . Thus for every condition p forcing $x \notin \bigcup _{i \in I_{\mu }}\{c_i\} \cup \bigcup _{\xi < \alpha }\bigcup _{n \in \omega }[\![T^{\xi }_{\mu ,n}]\!]$ , there is an extension q of p forcing $x \in \bigcup _{n \in \omega }[\![T^{\alpha }_{\mu ,n}]\!]$ . Hence $x \in \bigcup _{i \in I_{\mu }}\{c_i\} \cup \bigcup _{\xi \leq \alpha }\bigcup _{n \in \omega }[\![T^{\xi }_{\mu ,n}]\!] \subseteq \bigcup \mathcal {P}$ . As x was arbitrary, $\bigcup \mathcal {P} = 2^{\omega }$ as claimed.

Proof By Lemma 3.8 and Corollary 2.6, $\kappa \geq \acute {\mathfrak {n}} \geq {\mathfrak d} \geq \mathrm {cov}(\mathcal M)$ in $V^{\mathbb {P}_{\kappa }}$ . So to prove the lemma, it suffices to show that $2^{\omega }$ cannot be covered with $<\!\kappa $ meager sets in $V^{\mathbb {P}_{\kappa }}$ .

Suppose $\lambda < \kappa $ and $\left \lbrace C_{\xi } \colon \xi < \lambda \right \rbrace $ is a collection of closed nowhere dense subsets of $2^{\omega }$ in $V^{\mathbb {P}_{\kappa }}$ . For each $C_{\xi }$ , let $A_{\xi }$ be a Borel code that evaluates to $C_{\xi }$ . Because $\mathbb {P}_{\kappa }$ has the ccc and each $\mathbb {P}_{\alpha }$ is a complete sub-poset of $\mathbb {P}_{\kappa }$ , there is for each $\xi $ some $\alpha < \kappa $ such that $A_{\xi } \in V^{\mathbb {P}_{\alpha }}$ . Because $\kappa $ is regular and $\lambda < \kappa $ , there is some particular $\alpha < \kappa $ such that $A_{\xi } \in V^{\mathbb {P}_{\alpha }}$ for all $\xi < \lambda $ . So to prove the lemma, it suffices to show that for each $\alpha < \kappa $ there is some $c \in V^{\mathbb {P}_{\kappa }}$ not contained in any closed nowhere dense subset of $2^{\omega }$ coded in $V^{\mathbb {P}_{\alpha }}$ (i.e., we want c to be Cohen-generic over $V^{\mathbb {P}_{\alpha }}$ ).

Fix $\alpha < \kappa $ , and define $c: \omega \to 2$ by setting

$$ \begin{align*}c(n) = 0 \quad \text{if and only if} \quad \langle 0 \rangle \in T^{\alpha}_{\kappa,n}.\end{align*} $$

We claim that c is Cohen-generic over $V^{\mathbb {P}_{\alpha }}$ . To see this, let $U \subseteq 2^{\omega }$ be a dense open set whose Borel code is in $V^{\mathbb {P}_{\alpha }}$ . In particular, $\tilde U = \left \lbrace \tau \in 2^{<\omega } \colon [\![\tau ]\!] \subseteq U \right \rbrace \in V^{\mathbb {P}_{\alpha }}$ (where, abusing notation slightly, $[\![\tau ]\!]$ denotes all those x in $2^{\omega }$ with $x \!\restriction \! \mathrm {dom}(\tau ) = \tau $ ). Now fix $p \in \mathbb {P}_{\kappa }$ . We will find an extension r of p forcing that $c \in U$ . Extending p if necessary, we may (and do) assume that there is some $N \in \mathbb {N}$ such that $(\alpha ,\mu ,n) \in \mathrm {dom}(p)$ for all $n < N$ , but $(\alpha ,\mu ,n) \notin \mathrm {dom}(p)$ for all $n \geq N$ . For all $n < N$ , let $p(\alpha ,\mu ,n) = (t_n,B_n)$ . Define $\sigma \in 2^N$ by setting

$$ \begin{align*}\sigma(n) = 0 \quad \text{if and only if} \quad \langle 0 \rangle \in t_n\end{align*} $$

for all $n < N$ . Because U is dense in $2^{\omega }$ , there is some $\tau \in \tilde U$ such that $\tau \!\restriction \! N = \sigma $ . Because $\tilde U \in V^{\mathbb {P}_{\alpha }}$ , there is some $q \in \mathbb {P}_{\alpha }$ with $q \leq _{\mathbb {P}_{\alpha }} p \!\restriction \! \alpha $ forcing that $\tau \in \tilde U$ . Define $r \in \mathbb {P}_{\kappa }$ by

$$ \begin{align*}r(j) = \begin{cases} r(j), & \text{if } j \in (I \times \omega) \cup (\alpha \times C \times \omega), \\ (\langle\tau(n)\rangle,\emptyset), & \text{if } j = (\alpha,\mu,n) \text{ for some } n \in \mathrm{dom}(\tau) \setminus \mathrm{dom}(\sigma), \\ p(j), & \text{otherwise.} \end{cases}\end{align*} $$

Then $r \in \mathbb {P}_{\kappa }$ , and r forces $c \in U$ . As U was an arbitrary open dense subset of $2^{\omega }$ coded in $V^{\mathbb {P}_{\alpha }}$ , this shows c is Cohen-generic over $V^{\mathbb {P}_{\alpha }}$ .

From the last two lemmas, it follows that $\mathfrak {sp}({\small\text{closed}}) \subseteq [\kappa ,\theta ]$ in $V^{\mathbb {P}_{\kappa }}$ , and if $\lambda> \theta $ then $\lambda \notin \mathfrak {sp}({\small\text{Borel}})$ . To finish proving the theorem, it remains only to show that if $\lambda \in [\kappa ,\theta ]$ and $\lambda \notin C$ , then $\lambda \notin \mathfrak {sp}({\small\text{Borel}})$ . This is where the isomorphism-of-names argument comes in. This argument requires a rich supply of automorphisms of $\mathbb {P}_{\kappa }$ , which we describe next.

If $\phi : I \to I$ is a permutation, then for every set x we define $\bar \phi (x)$ to be the set obtained from x by replacing every $i \in I$ in the transitive closure of x with $\phi (i)$ . This is well defined, because if $i,j \in I$ then i is not in the transitive closure of j. (One may think of I as a set of urelements, $\phi $ as a permutation of them, and $\bar \phi $ as the automorphism of the universe induced by $\phi $ .) Alternatively, $\bar \phi $ is described via well-founded recursion by the relation

$$ \begin{align*}\bar \phi(x) = \left\lbrace \bar \phi(y) \colon y \in x \right\rbrace\end{align*} $$

if $x \notin I$ , and $\bar \phi \!\restriction \! I = \phi $ . In particular, if $p \in \mathbb {P}_{\kappa }$ , $\mathrm {dom}(\bar \phi (p)) = \bar \phi (\mathrm {dom}(p))$ and

$$ \begin{align*}\bar \phi(p)(j) = \begin{cases} p(i), & \text{if } i \in (I \times \omega) \cap \mathrm{dom}(p) \text{ and } \bar \phi(i) = j, \\ \big(t,\{ \bar \phi(\dot b) :\, \dot b \in B \}\big), & \text{if } j \in \mathrm{dom}(p) \setminus (I \times \omega) \text{ and } p(j) = (t,B), \end{cases}\end{align*} $$

and this expression can be taken as a recursive definition of $\bar \phi $ on $\mathbb {P}_{\kappa }$ , and the natural extension of $\bar \phi $ to the class of $\mathbb {P}_{\kappa }$ -names.

Proof First, note that for all $p,q \in \mathbb {P}_{\alpha }$ , we have

$$ \begin{align*}\langle p,q\rangle \in \ \leq_{\mathbb{P}_{\alpha}} \quad \text{if and only if } \quad \bar \phi(\langle p,q\rangle) = \langle\bar \phi(p),\bar \phi(q)\rangle \in \bar \phi(\leq_{\mathbb{P}_{\alpha}}).\end{align*} $$

Together with the fact that $\bar \phi $ is invertible (with inverse $\bar \phi ^{-1} = \overline {\phi ^{-1}}$ ), this shows that the image of $\mathbb {P}_{\alpha }$ under $\bar \phi $ is a poset, and is naturally isomorphic to $\mathbb {P}_{\alpha }$ as witnessed by $\bar \phi $ . But it is not immediately clear that the image of $\mathbb {P}_{\alpha }$ under $\bar \phi $ is equal to $\mathbb {P}_{\alpha }$ , which is of course required for our claim that $\bar \phi \!\restriction \! \mathbb {P}_{\alpha }$ is an automorphism of $\mathbb {P}_{\alpha }$ . (In fact, one may show that this would not be true for $\alpha \geq 1$ if $\phi $ were not required to fix all the $I_{\mu }$ .) This is proved for $\bar \phi $ and its inverse, simultaneously, by induction on  $\alpha $ .

The base case $\alpha =0$ is clear. And for limit $\alpha $ , if $\bar \phi \!\restriction \! \mathbb {P}_{\xi }$ is an automorphism of $\mathbb {P}_{\xi }$ , for all $\xi < \alpha $ , then $\bar \phi \!\restriction \! \mathbb {P}_{\alpha }$ is an automorphism of $\mathbb {P}_{\alpha }$ because $\langle \mathbb {P}_{\alpha },\leq _{\mathbb {P}_{\alpha }} \rangle $ is the direct limit of $\left \langle \langle \mathbb {P}_{\xi },\leq _{\mathbb {P}_{\xi }} \rangle \colon \xi < \alpha \right \rangle $ . The same applies to $\bar \phi ^{-1}$ .

For the successor case, fix some $\beta < \kappa $ and suppose $\bar \phi \!\restriction \! \mathbb {P}_{\xi }$ is an automorphism of $\mathbb {P}_{\xi }$ for all $\xi \leq \beta $ (the inductive hypothesis). Let $\alpha = \beta +1$ , and fix $p \in \mathbb {P}_{\alpha }$ . We claim that $\bar \phi (p) \in \mathbb {P}_{\alpha }$ . Recall that

$$ \begin{align*}\bar \phi(p)(j) = \begin{cases} p(i), & \text{if } i \in (I \times \omega) \cap \mathrm{dom}(p) \text{ and } \bar \phi(i) = j, \\ \big(t,\{ \bar \phi(\dot b) :\, \dot b \in B \}\big), & \text{if } j \in \mathrm{dom}(p) \setminus (I \times \omega) \text{ and } p(j) = (t,B). \end{cases}\end{align*} $$

It is clear that $\bar \phi (p)$ is a finite partial function on $(I \times \omega ) \cup (\alpha \times C \times \omega )$ , and that if $i \in (I \times \omega ) \cap \mathrm {dom}(\bar \phi (p))$ , then $\bar \phi (p)(i) = p(\phi ^{-1}(i)) \in \{0,1\}$ . It remains to check that the last bullet point in the definition of the conditions in $\mathbb {P}_{\alpha }$ is satisfied by $\bar \phi (p)$ .

Suppose $(\gamma ,\mu ,n) \in \mathrm {dom}(\bar \phi (p))$ for some $\gamma \leq \beta $ , $\mu \in C$ , and $n \in \omega $ . This means $(\gamma ,\mu ,n) \in \mathrm {dom}(p)$ as well: let us denote $p(\gamma ,\mu ,n) = (t,B)$ . Then

$$ \begin{align*}\bar \phi(p)(\gamma,\mu,n) = \big( t, \{ \bar \phi(\dot b) :\, \dot b \in B \} \big).\end{align*} $$

Because $p \in \mathbb {P}_{\alpha }$ , t is a pruned subtree of $2^{<k}$ for some $k \in \omega $ , and (by the inductive hypothesis, that $\bar \phi \!\restriction \! \mathbb {P}_{\gamma }$ is an automorphism of $\mathbb {P}_{\gamma }$ ), $\{ \bar \phi (\dot b) :\, \dot b \in B \}$ is a finite set of nice $\mathbb {P}_{\gamma }$ -names for members of $2^{\omega }$ . Furthermore, by applying the automorphism $\bar \phi $ of $\mathbb {P}_{\gamma }$ to the displayed statement in the last bullet point in the definition of the conditions in $\mathbb {P}_{\alpha }$ , we get that for every $\dot b \in B$ ,

$$ \begin{align*} \bar \phi(p) \restriction ((I \times \omega) \cup (\gamma \times C \times \omega)) \ \Vdash_{\mathbb{P}_{\gamma}} \quad & \bar \phi(\dot b) \!\restriction\! j \in t \text{ for every } j < k, \\ & \bar \phi(\dot b) \neq \bar \phi(\dot c_i) \text{ for every } i \in I_{\mu}, \text{ and} \\ & \bar \phi(\dot b) \text{ is not a branch of } \bar \phi(\dot T^{\xi}_{\mu,m}) \\ & \quad \text{ for any } \xi < \gamma \text{ and } m \in \omega. \end{align*} $$

Considering the second of these three statements, note that $\bar \phi (\dot c_i) = \dot c_{\phi (i)}$ for every $i \in I$ . Because $\phi \!\restriction \! I_{\mu }$ is a permutation of $I_{\mu }$ , this means that the assertion “ $\bar \phi (\dot b) \neq \bar \phi (\dot c_i) \text { for every } i \in I_{\mu }$ ” is equivalent to the assertion “ $\bar \phi (\dot b) \neq \dot c_i \text { for every } i \in I_{\mu }$ .” Thus

$$ \begin{align*} \bar \phi(p) \restriction ((I \times \omega) \cup (\gamma \times C \times \omega)) \ \Vdash_{\mathbb{P}_{\gamma}} \quad & \bar \phi(\dot b) \neq \dot c_i \text{ for every } i \in I_{\mu}. \end{align*} $$

Considering the third of these three statements, observe that

$$ \begin{align*} \bar \phi(\dot T^{\xi}_{\mu,n}) &= \left\lbrace \bar \phi( \langle \sigma, p \rangle ) \colon p(\xi,\mu,m) = (t,B) \text{ for some } B \text{, and } \sigma \in t \right\rbrace \\ & = \left\lbrace \langle \sigma, \bar \phi(p) \rangle \colon p(\xi,\mu,m) = (t,B) \text{ for some } B \text{, and } \sigma \in t \right\rbrace \\ & = \left\lbrace \langle \sigma, \bar \phi(p) \rangle \colon \bar \phi(p)(\xi,\mu,m) = (t,\bar \phi(B)) \text{ for some } B \text{, and } \sigma \in t \right\rbrace \\ & = \left\lbrace \langle \sigma, \bar \phi(p) \rangle \colon \bar \phi(p)(\xi,\mu,m) = (t,B') \text{ for some } B' \text{, and } \sigma \in t \right\rbrace \\ & = \left\lbrace \langle \sigma, q \rangle \colon q(\xi,\mu,m) = (t,B') \text{ for some } B' \text{, and } \sigma \in t \right\rbrace \\ & = \dot T^{\xi}_{\mu,n}. \end{align*} $$

The third equality is true because for any condition p, $p(\xi ,\mu ,m) = (t,B)$ if and only if $\bar \phi (p)(\xi ,\mu ,m) = (t,\bar \phi (B))$ . The fourth equality uses the inductive hypothesis, that $\bar \phi \!\restriction \! \mathbb {P}_{\gamma }$ is an automorphism: $p(\xi ,\mu ,m) = (t,\bar \phi (B))$ for some B if and only if $p(\xi ,\mu ,m) = (t,B')$ for some $B'$ , specifically for $B' = \bar \phi ^{-1}(B)$ . The fifth equality also uses the fact that $\bar \phi \!\restriction \! \mathbb {P}_{\gamma }$ is an automorphism. Thus

$$ \begin{align*} \bar \phi(p) \restriction ((I \times \omega) \cup (\gamma \times C \times \omega)) \ \Vdash_{\mathbb{P}_{\gamma}} \quad & \bar \phi(\dot b) \text{ is not a branch of } \dot T^{\xi}_{\mu,m} \\ & \quad \text{ for any } \xi < \gamma \text{ and } m \in \omega. \end{align*} $$

Putting these together, we obtain

$$ \begin{align*} \bar \phi(p) \restriction ((I \times \omega) \cup (\gamma \times C \times \omega)) \ \Vdash_{\mathbb{P}_{\gamma}} \quad & \bar \phi(\dot b) \!\restriction\! j \in t \text{ for every } j < k, \\ & \bar \phi(\dot b) \neq c_i \text{ for every } i \in I_{\mu}, \text{ and} \\ & \bar \phi(\dot b) \text{ is not a branch of } \dot T^{\xi}_{\mu,m} \\ & \quad \text{ for any } \xi < \beta \text{ and } m \in \omega. \end{align*} $$

In other words, $\bar \phi (p)$ satisfies the final bullet point in the definition of the $\mathbb {P}_{\alpha }$ conditions. Hence $\bar \phi (p) \in \mathbb {P}_{\alpha }$ , as claimed.

This shows that (the restriction of) $\bar \phi $ is an injective morphism from $\mathbb {P}_{\alpha }$ to $\mathbb {P}_{\alpha }$ . To get surjectivity, simply note that the same argument applies to $\bar \phi ^{-1} = \overline {\phi ^{-1}}$ as well.

From now on, we work in the ground model. Let $\lambda $ be a cardinal such that $\lambda \in [\kappa ,\theta ]$ and $\lambda \notin C$ . Suppose $\big \{ \dot B_{\alpha } :\, \alpha < \lambda \big \}$ is a set of nice $\mathbb {P}_{\kappa }$ -names for Borel codes for subsets of $\mathbb {R}$ . (Any standard method of constructing Borel codes can be used for the proof, provided only that the codes are hereditarily countable sets. But for concreteness, let us take a Borel code to be a subset of $\omega $ .) We let $B_{\alpha } \subseteq \omega $ denote the evaluation of the name $\dot B_{\alpha }$ in $V^{\mathbb {P}_{\kappa }}$ , and we let $\tilde B_{\alpha } \subseteq \mathbb {R}$ denote the interpretation of $B_{\alpha }$ .

We aim to show that $\big \{ \tilde B_{\alpha } :\, \alpha < \lambda \big \}$ is not a partition of $\mathbb {R}$ in $V^{\mathbb {P}_{\kappa }}$ . To this end, suppose $q \in \mathbb {P}_{\kappa }$ and q forces each of the $\tilde B_{\alpha }$ is nonempty, and $\tilde B_{\alpha } \cap \tilde B_{\beta } = \emptyset $ whenever $\alpha \neq \beta $ . We will show that q also forces $\bigcup _{\alpha < \lambda } \tilde B_{\alpha } \neq \mathbb {R}$ .

Fix a cardinal $\nu \leq \lambda $ such that $\alpha ^{\aleph _0} < \nu $ for all cardinals $\alpha < \nu $ , and such that C contains no cardinals in the interval $[\nu ,\lambda ]$ . If $\lambda $ is neither singular nor the successor of a singular cardinal, then we may simply take $\lambda = \nu $ . Otherwise, using the last two bullet points in our description of C, there is an infinite interval of cardinals below $\lambda $ and disjoint from C, and we may take $\nu $ to be any successor-of-a-successor cardinal in this interval. In either case, $\alpha ^{\aleph _0} < \nu $ for all cardinals $\alpha < \nu $ by the $\mathsf {GCH}$ .

Given $\alpha \leq \kappa $ and a condition $p \in \mathbb {P}_{\alpha }$ , for each $\mu \in C$ define

$$ \begin{align*}\mathrm{hdom}_{\mu}(p) = \textstyle \text{TC}(p) \cap I_{\mu},\end{align*} $$

where $\text {TC}(p)$ denotes the transitive closure of p. We think of $\mathrm {hdom}_{\mu }(p)$ as the “hereditary domain” of p on $I_{\mu }$ : all those $i \in I_{\mu }$ that are used at any stage in building the condition p. A straightforward transfinite induction on $\alpha $ shows that $\left \lvert \mathrm {hdom}_{\mu }(p) \right \rvert \leq \aleph _0$ for every $p \in \mathbb {P}_{\alpha }$ and every $\mu \in C$ . (The base case and the limit case are clear. For the successor case, use the fact that $\mathbb {P}_{\alpha }$ has the ccc.) Similarly, for each $\mathbb {P}_{\kappa }$ -name $\dot x$ and each $\mu \in C$ , let $\mathrm {hdom}_{\mu }(\dot x) = \textstyle \text {TC}(\dot x) \cap I_{\mu }$ . As before, it is not difficult to see that if $\dot x$ is a nice $\mathbb {P}_{\kappa }$ -name for a Borel code (or for any hereditarily countable set), then $\mathrm {hdom}_{\mu }(\dot x)$ is countable for every $\mu \in C$ .

For each $\alpha < \lambda $ and each $\mu \in C$ , let $D^{\alpha }_{\mu } = \mathrm {hdom}_{\mu }(\dot B_{\alpha }).$ Note that $D^{\alpha }_{\mu }$ is countable for each $\alpha < \lambda $ and $\mu \in C$ . Expanding some of the $D^{\alpha }_{\mu }$ if necessary, we may (and do) assume each $D^{\alpha }_{\mu }$ is countably infinite and includes $\mathrm {hdom}_{\mu }(q)$ . For each $\alpha < \lambda $ , let $D^{\alpha } = \bigcup _{\mu \in C}D^{\alpha }_{\mu }$ . We note that $|D^{\alpha }| = \sum _{\mu \in C}|D^{\alpha }_{\mu }| = |C| \cdot \aleph _0$ for all $\alpha < \lambda $ , and $|C| \cdot \aleph _0 < \min (C) = \kappa $ . (Note: the inequality $|C| \cdot \aleph _0 < \min (C)$ uses the second bullet point in our description of C.)

By our choice of $\nu $ , together with the aforementioned fact that $|D^{\alpha }| < \kappa $ for all $\alpha < \lambda $ , $\left \lbrace D^{\alpha } \colon \alpha < \nu \right \rbrace $ meets the conditions of the generalized $\Delta $ -system lemma [Reference Kunen15, Lemma III.6.15]. Thus there is some $\mathcal {A}_0 \subseteq \nu $ with $\left \lvert \mathcal {A}_0 \right \rvert = \nu $ such that $\left \lbrace D^{\alpha } \colon \alpha \in \mathcal {A}_0 \right \rbrace $ is a $\Delta $ -system with root R.

Let $\mathcal {A}_1 = \left \lbrace \alpha \in \mathcal {A}_0 \colon D^{\alpha }_{\mu } \setminus R = \emptyset \text { for all } \mu \in C \text { with } \mu < \lambda \right \rbrace $ . For all $\mu \in C$ , $\left \lbrace D^{\alpha }_{\mu } \setminus R \colon \alpha \in \mathcal {A}_0 \right \rbrace $ is a $\nu $ -size collection of pairwise disjoint subsets of $I_{\mu }$ . If also $\mu < \lambda $ , then $\mu < \nu $ and it follows that $D^{\alpha }_{\mu } \setminus R = \emptyset $ for all but (at most) $\mu $ members of $\mathcal {A}_0$ . Furthermore, $|\left \lbrace \mu \in C \colon \mu < \lambda \right \rbrace | \leq |C| < \kappa < \nu $ . It follows that $\left \lvert \mathcal {A}_1 \right \rvert = \nu $ .

For each $\alpha < \lambda $ and $\mu \in C$ , fix a bijection $\phi ^{\alpha }_{\mu }: D^{\alpha }_{\mu } \to \omega $ , in such a way that $\phi ^{\alpha }_{\mu } \!\restriction \! R = \phi ^{\beta }_{\mu } \!\restriction \! R$ for all $\alpha ,\beta < \lambda $ . For each $\alpha ,\beta < \lambda $ and $\mu \in C$ , let $\phi ^{\alpha ,\beta }_{\mu }$ be the involution of $I_{\mu }$ given by

$$ \begin{align*}\phi^{\alpha,\beta}_{\mu}(i) = \begin{cases} (\phi^{\beta}_{\mu})^{-1} \circ \phi^{\alpha}_{\mu}(i), \quad & \text{if } i \in D^{\alpha}_{\mu}, \\ (\phi^{\alpha}_{\mu})^{-1} \circ \phi^{\beta}_{\mu}(i), & \text{if } i \in D^{\beta}_{\mu}, \\ i, &\text{otherwise.} \end{cases}\end{align*} $$

(Recall that an “involution” of a set S means a bijection $f: S \to S$ such that $\psi \circ \psi = \mathrm {id}_S$ .) This function is well defined, because $D^{\alpha }_{\mu } \cap D^{\beta }_{\mu } = R$ , and $(\phi ^{\beta }_{\mu })^{-1} \circ \phi ^{\alpha }_{\mu }(i) = i = (\phi ^{\alpha }_{\mu })^{-1} \circ \phi ^{\beta }_{\mu }(i)$ whenever $i \in R$ . This gives us a collection $\left \lbrace \phi ^{\alpha ,\beta }_{\mu } \colon \alpha ,\beta < \lambda \right \rbrace $ of involutions of $I_{\mu }$ such that for any $\alpha ,\beta < \lambda $ ,

1. ∘ $\phi ^{\alpha ,\beta }_{\mu }$ maps $D^{\alpha }_{\mu }$ onto $D^{\beta }_{\mu }$ and $D^{\beta }_{\mu }$ onto $D^{\alpha }_{\mu }$ , but acts as the identity on the rest of $I_{\mu }$ ,

2. ∘ $\phi ^{\alpha ,\beta }_{\mu }$ acts as the identity on $R \cap I_{\mu }$ , and

3. ∘ $\phi ^{\alpha ,\beta }_{\mu } = \phi ^{\gamma ,\beta }_{\mu } \circ \phi ^{\alpha ,\gamma }_{\mu }$ for any $\gamma < \lambda $ .

For each $\alpha ,\beta < \lambda $ , let $\phi ^{\alpha ,\beta }$ denote the product map $\bigotimes _{\mu \in C}\phi ^{\alpha ,\beta }_{\mu }$ (that is, the map defined by setting $\phi ^{\alpha ,\beta }(i) = \phi ^{\alpha ,\beta }_{\mu }(i)$ whenever $i \in I_{\mu }$ ). This is an involution of I, and restricts to $\phi ^{\alpha ,\beta }_{\mu }$ on each $I_{\mu }$ . In particular, Lemma 3.11 applies, and $\bar \phi ^{\alpha ,\beta } \!\restriction \! \mathbb {P}_{\kappa }$ is an automorphism of $\mathbb {P}_{\kappa }$ for each $\alpha ,\beta < \lambda $ .

In particular, there are $\kappa $ nice $\mathbb {P}_{\kappa }$ -names $\dot x$ for subsets of $\omega $ having the property that $\mathrm {hdom}(\dot x) \subseteq D^0$ . But for each $\alpha \in \mathcal {A}_1$ , $\bar \phi ^{\alpha ,0}(\dot B_{\alpha })$ is just such a name. By the pigeonhole principle, and the fact that $\kappa < \nu $ , this implies that there is some $\mathcal {A}_2 \subseteq \mathcal {A}_1$ with $\left \lvert \mathcal {A}_2 \right \rvert = \nu $ such that $\bar \phi ^{\alpha ,0}(\dot B_{\alpha }) = \bar \phi ^{\beta ,0}(\dot B_{\beta })$ whenever $\alpha ,\beta \in \mathcal {A}_2$ . Reindexing the $\dot B_{\alpha }$ ’s if necessary, we may (and do) assume $0 \in \mathcal {A}_2$ .

We now proceed to define a new $\mathbb {P}_{\kappa }$ -name $\dot B_{\lambda }$ for a subset of $\omega $ .

First, for all $\mu \in C$ with $\mu < \lambda $ , let $D^{\lambda }_{\mu } = R \cap I_{\mu }$ . (Recall that $D^{\alpha }_{\mu } \cap I_{\mu } \subseteq R$ whenever $\mu \in C$ and $\mu < \lambda $ and $\alpha \in \mathcal {A}_1$ . Thus, in this case, $D^{\lambda }_{\mu } \cap D^{\alpha }_{\mu } = R \cap I_{\mu }$ for all $\alpha \in \mathcal {A}_1 \supseteq \mathcal {A}_2$ .) Next, for each $\mu \in C$ with $\mu> \lambda $ , let $D^{\lambda }_{\mu }$ be a countable subset of $I_{\mu }$ with $D^{\lambda }_{\mu } \cap \bigcup \left \lbrace D^{\alpha }_{\mu } \colon \alpha < \lambda \right \rbrace = R \cap I_{\mu }$ , and with $|D^{\lambda }_{\mu } \setminus R| = |D^0_{\mu } \setminus R|$ . Some such set exists because $\left \lvert \bigcup \left \lbrace D^{\alpha }_{\mu } \colon \alpha < \lambda \right \rbrace \right \rvert = \lambda < \mu = |I_{\mu }|$ , so we may take $D^{\lambda }_{\mu }$ to be any subset (of the appropriate size) of $I_{\mu } \setminus \bigcup \left \lbrace D^{\alpha }_{\mu } \colon \alpha < \lambda \right \rbrace $ , together with $R \cap I_{\mu }$ .

For each $\mu \in C$ , fix a bijection $\phi ^{\lambda }_{\mu }: D^{\lambda }_{\mu } \to \omega $ such that $\phi ^{\lambda }_{\mu } \!\restriction \! R = \phi ^0_{\mu } \!\restriction \! R$ . For each $\alpha < \lambda $ and $\mu \in C$ , let $\phi ^{\alpha ,\lambda }_{\mu }$ and $\phi ^{\lambda ,\alpha }_{\mu }$ be the involutions of $I_{\mu }$ defined just like the $\phi ^{\alpha ,\beta }_{\mu }$ above, but using $\phi ^{\lambda }_{\mu }$ in place of $\phi ^{\beta }_{\mu }$ . This naturally extends the system of involutions described above: for any $\alpha ,\beta \leq \lambda $ ,

1. ∘ $\phi ^{\alpha ,\beta }_{\mu }$ maps $D^{\alpha }_{\mu }$ onto $D^{\beta }_{\mu }$ and $D^{\beta }_{\mu }$ onto $D^{\alpha }_{\mu }$ , but acts as the identity on the rest of $I_{\mu }$ ,

2. ∘ each $\phi ^{\alpha ,\beta }_{\mu }$ acts as the identity on $R \cap I_{\mu }$ , and

3. ∘ $\phi ^{\alpha ,\beta }_{\mu } = \phi ^{\gamma ,\beta }_{\mu } \circ \phi ^{\alpha ,\gamma }_{\mu }$ for any $\gamma \leq \lambda $ .

For each $\alpha ,\beta \leq \lambda $ , let $\phi ^{\alpha ,\beta } = \bigotimes _{\mu \in C}\phi ^{\alpha ,\beta }_{\mu }$ . By Lemma 3.11, $\bar \phi ^{\alpha ,\beta } \!\restriction \! \mathbb {P}_{\kappa }$ is an automorphism of $\mathbb {P}_{\kappa }$ for each $\alpha ,\beta \leq \lambda $ .

Because $\mathrm {hdom}_{\mu }(q) \subseteq R$ for all $\mu $ and $\phi ^{0,\lambda }$ fixes R, $\bar \phi ^{0,\lambda }(q) = q$ . Recall

$$ \begin{align*} q \Vdash \text{ the evaluation } B_0 \text{ of } \dot B_0 \text{ codes a nonempty Borel subset of } \mathbb{R}. \end{align*} $$

Applying standard facts about automorphisms of forcing posets,

$$ \begin{align*}\bar \phi^{0,\lambda}(q) \Vdash \text{ the evaluation of } \bar \phi^{0,\lambda}(\dot B_{\lambda}) \text{ codes a nonempty Borel subset of } \mathbb{R}. \end{align*} $$

Let $B_{\lambda }$ denote the evaluation of $\dot B_{\lambda }$ in $V^{\mathbb {P}_{\kappa }}$ , and let $\tilde B_{\lambda }$ be the Borel set that it codes. So, in particular, the displayed statement above implies that $q = \bar \phi ^{0,\lambda }(q) \Vdash \tilde B_{\lambda } \neq \emptyset $ . To show $q \Vdash \bigcup _{\alpha < \lambda }\tilde B_{\alpha } \neq \mathbb {R}$ , it now suffices to show $q \Vdash \tilde B_{\lambda } \cap \tilde B_{\beta } = \emptyset $ for all $\beta < \lambda $ .

Fix $\beta < \lambda $ . By our choice of the sets $D^{\lambda }_{\mu }$ , we have $D^{\lambda }_{\mu } \cap D^{\beta }_{\mu } \subseteq R$ for all $\mu \in C$ , and therefore $D^{\lambda } \cap D^{\beta } \subseteq R$ . Because $\left \lbrace D^{\alpha } \colon \alpha \in \mathcal {A}_2 \right \rbrace $ is a $\Delta $ -system of size $\nu $ , and because $|D^{\beta }| \leq |C| \cdot \aleph _0 < \kappa < \nu $ , there is some $\alpha \in \mathcal {A}_2$ with $D^{\alpha } \cap D^{\beta } \subseteq R$ (regardless of whether $\beta \in \mathcal {A}_2$ ). Fix some such $\alpha $ .

Note that $\phi ^{\alpha ,\lambda }$ sends $D^{\alpha }$ to $D^{\lambda }$ , while acting as the identity on $D^{\beta }$ . And because $0,\alpha \in \mathcal {A}_2$ , our choice of $\mathcal {A}_2$ implies $\bar \phi ^{0,\alpha }(\dot B_0) = \dot B_{\alpha }$ . It follows that

$$ \begin{align*}\bar{\phi}^{\alpha,\lambda}(\dot B_{\alpha}) = \bar \phi^{\alpha,\lambda} \circ \bar \phi^{0,\alpha}(\dot B_0) = \bar \phi^{0,\lambda}(\dot B_0) = \dot B_{\lambda},\end{align*} $$

while $\bar \phi ^{\alpha ,\lambda }(\dot B_{\beta }) = \dot B_{\beta }$ . But

$$ \begin{align*}q \Vdash \ B_{\alpha} \text{ and } B_{\beta} \text{ code disjoint Borel sets}.\end{align*} $$

Applying the automorphism $\bar \phi ^{\alpha ,\lambda }$ of $\mathbb {P}_{\kappa }$ ,

$$ \begin{align*}\bar \phi^{\alpha,\lambda}(q) = q \Vdash \ B_{\lambda} \text{ and } B_{\beta} \text{ code disjoint Borel sets}.\end{align*} $$

Because $\beta $ was arbitrary, this shows that $q \Vdash \tilde B_{\lambda } \cap \tilde B_{\beta } = \emptyset $ for all $\beta < \lambda $ . In other words, $q \Vdash \bigcup _{\alpha < \lambda }\tilde B_{\alpha } \neq \mathbb {R}$ .

Question 3.16. Given $\alpha < \omega _1$ , is it consistent that $\mathfrak {sp}({\small\text{\(\mathbf {\Pi }^{0}_{\alpha}\)}}) \neq \mathfrak {sp}({\small\text{\(\mathbf {\Pi}^{0}_{\alpha +1}\)}})$ ?