Different covering numbers of compact tree ideals

Abstract

We investigate the covering numbers of some ideals on ${^{\omega }}{2}{}$ associated with tree forcings. We prove that the covering of the Sacks ideal remains small in the Silver and uniform Sacks model, respectively, and that the coverings of the uniform Sacks ideal and the Mycielski ideal, ${\mathfrak {C}_{2}}$, remain small in the Sacks model.

1 Introduction

Let $\mathbb {T}$ be a compact tree-forcing, i.e. a forcing whose conditions are subtrees of ${^{<\omega }}{2}{}$. Associate with $\mathbb {T}$ the $\sigma$-ideal

$$\begin{aligned} t^0:= \left\{ {A\subseteq {^{\omega }}{2}{}:\forall p\in \mathbb {T}\exists q\in \mathbb {T}\quad (q\le p\;\wedge \; [q]\cap A = \emptyset )}\right\} . \end{aligned}$$

(1)

In several cases it has been shown that in the forcing extension obtained by a countable support iteration of such a $\mathbb {T}$ of length $\omega _2$, the covering number of $t^0$ is equal to the continuum. We will continue some of Brendle’s work. In his unpublished habilitation thesis, he proves the following theorem:

Theorem 1.1

([1, Sec. 7.1 Theorem 12.]) In the model obtained by adding iteratively $\omega _2$ Miller reals with countable support to a model of $\textrm{CH}$, we have

• (1) $\textrm{cov}({m^{0}}) = \omega _2$ but

• (2) $\textrm{cov}(r^0) = \textrm{cov}({l^{0}}) = \textrm{cov}({s^{0}}) = \textrm{cov}({v^{0}}) = \omega _1$.

Here, ${m^{0}}$, $r^0$, $l^0$, $s^0$, and $v^0$ are the respective ideals associated via (1) with Miller, Mathias, Laver, Sacks, and Silver forcing. Without a proof he mentions the following result concerning iterations of Laver forcing:

Theorem 1.2

([1]) In the Laver model, $\textrm{cov}({m^{0}}) < \textrm{cov}({l^{0}})$.

In contrast to Brendle, who worked on iterations of non-compact tree forcings, we will focus on the case of compact tree ideals. We will extend the scope of our investigation to the covering numbers of the Mycielski ideals that are closely related with Silver forcing and uniform Sacks forcing.

Definition 1.3

We write $\mathbb {S}$ for Sacks forcing, i.e. the forcing whose conditions are perfect subtrees of ${^{<\omega }}{2}{}$ ordered by inclusion.

A uniform tree is a perfect tree u such that

$$\begin{aligned} \exists a\in [\omega ]^\omega \forall \sigma \in u(\sigma \in \textrm{split}(u)\iff \left| {\sigma }\right| \in a).\end{aligned}$$

(*)

Let $\mathbb {U}$ be the set of all uniform trees. We call this notion of forcing uniform Sacks forcing. We call the unique set a from $(*)$ the split levels of u and write $a^{u}$ for this set.

A uniform tree v is called a Silver tree, if there exists a function $f:\omega \setminus a^{v}\rightarrow 2$ such that ${\sigma }{\upharpoonright }{(\omega \setminus a^{v})}\subseteq f$ for every $\sigma \in v$. We write $\mathbb {V}$ for the set of all Silver trees and call it Silver forcing.

We write ${v^{0}}$, ${u^{0}}$, ${s^{0}}$ for the ideals defined in (1) for $\mathbb {V}$, $\mathbb {U}$, and $\mathbb {S}$, respectively.

Definition 1.4

For an ideal $\mathcal {I}$ on ${^{\omega }}{2}{}$ we call

$$\begin{aligned}\textrm{cov}(\mathcal {I}):= \min \left\{ {\left| {\mathcal {F}}\right| : \mathcal {F}\subseteq \mathcal {I}\;\wedge \; \bigcup \mathcal {F}= {^{\omega }}{2}{}}\right\} \end{aligned}$$

the covering number of $\mathcal {I}$.

We will show that consistently, $\textrm{cov}(j^0) < \textrm{cov}(i^0)$ for different $i^0,j^0\in \lbrace {{v^{0}},{u^{0}},{s^{0}}}\rbrace$. In most cases we are able to separate the covering numbers of another type of ideals, which are also closely related to tree forcings:

Definition 1.5

We call the ideals

$$\begin{aligned}{\mathfrak {P}_{2}}:= \left\{ {A\subseteq {^{\omega }}{2}{}:\forall b\in [\omega ]^\omega ({A}{\upharpoonright }{b}\ne {^{b}}{2}{})}\right\} \end{aligned}$$

and

$$\begin{aligned}{\mathfrak {C}_{2}}:= \left\{ {A\subseteq {^{\omega }}{2}{}: \forall b\in [\omega ]^\omega \exists u\in \mathbb {U}(a^{u} = \omega \setminus b\;\wedge \; [u]\cap A = \emptyset )}\right\} \end{aligned}$$

Mycielski ideals. Note that

$$\begin{aligned}{\mathfrak {P}_{2}} = \left\{ {A\subseteq {^{\omega }}{2}{}: \forall b\in [\omega ]^\omega \exists v\in \mathbb {V}(a^{v} = \omega \setminus b\;\wedge \; [v]\cap A = \emptyset )}\right\} .\end{aligned}$$

Note further that it is not necessary to quantify over all $b\in [\omega ]^\omega$. It suffices to consider a dense set $D\subseteq [\omega ]^\omega$.

The inclusions ${\mathfrak {C}_{2}}\subseteq {u^{0}}$ and ${\mathfrak {P}_{2}}\subseteq {v^{0}}$ can be proved in $\textrm{ZFC}$ (see [2, Lemma 4.1]). Trivially, ${\mathfrak {P}_{2}}\subseteq {\mathfrak {C}_{2}}$. Thus, one obtains the $\textrm{ZFC}$-inequalities

$$\begin{aligned}\textrm{cov}({u^{0}})\le \textrm{cov}({\mathfrak {C}_{2}}),\quad \textrm{cov}({v^{0}})\le \textrm{cov}({\mathfrak {P}_{2}}),\quad \text {and}\quad \textrm{cov}({\mathfrak {C}_{2}})\le \textrm{cov}({\mathfrak {P}_{2}}).\end{aligned}$$

It is common knowledge that $\textrm{cov}({v^{0}})\le \mathfrak {r}$, where $\mathfrak {r}$ is the smallest cardinality of a reaping family (cf. [3, Lemma 3 in 3.2]). Since both $\mathbb {S}$ and $\mathbb {U}$ are P-point preserving (see [3, Lemma 1 in 3.2] for the proof that $\mathbb {U}$ preserves P-points), $\mathfrak {r}$, and hence $\textrm{cov}({v^{0}})$, remains small in the Sacks model and the uniform Sacks model.

A standard Löwenheim–Skolem argument shows that $\textrm{cov}(t^0) = \aleph _2$ in the model obtained by a countable support iteration of $\mathbb {T}\in \left\{ {\mathbb {V},\mathbb {U},\mathbb {S}}\right\}$ of length $\omega _2$ over a model for $\textrm{ZFC}+ \textrm{CH}$ (see e.g. [4, Theorem 1.2] for the proof in the case of $\mathbb {T}= \mathbb {S}$). Together with Brendle’s inequality, this yields the (respective) relative consistency of $\textrm{cov}({v^{0}}) < \textrm{cov}({s^{0}})$ and $\textrm{cov}({v^{0}}) < \textrm{cov}({u^{0}})$. Since $\textrm{cov}({u^{0}})\le \textrm{cov}({\mathfrak {C}_{2}})\le \textrm{cov}({\mathfrak {P}_{2}})$ is provable in $\textrm{ZFC}$, $\textrm{cov}({u^{0}}) = \textrm{cov}({\mathfrak {C}_{2}}) = \textrm{cov}({\mathfrak {P}_{2}}) = \aleph _2$ in the uniform Sacks model. Thus, we also obtain the relative consistency of $\textrm{cov}({v^{0}}) < \textrm{cov}({\mathfrak {P}_{2}}) = \textrm{cov}({\mathfrak {C}_{2}})$.

Finally, the second author has recently shown in [5] that in the Silver model

$$\begin{aligned}\textrm{cov}({u^{0}}) \overset{(\textrm{ZFC})}{\le }\ \textrm{cov}({\mathfrak {C}_{2}}) < \textrm{cov}({v^{0}}) \overset{(\textrm{ZFC})}{\le }\ \textrm{cov}({\mathfrak {P}_{2}}).\end{aligned}$$

We obtain the following table, summarizing all known results and the results proved in this paper:

<	$\textrm{cov}({v^{0}})$	$\textrm{cov}({u^{0}})$	$\textrm{cov}({s^{0}})$	$\textrm{cov}({\mathfrak {P}_{2}})$	$\textrm{cov}({\mathfrak {C}_{2}})$
$\textrm{cov}({v^{0}})$	$=$	[3]	[3]	[3]	[3]
$\textrm{cov}({u^{0}})$	[5] + [2]	$=$	4.13	[5]	???
$\textrm{cov}({s^{0}})$	3.10	3.10	$=$	[5]	3.10
$\textrm{cov}({\mathfrak {P}_{2}})$	$\textrm{ZFC}\vdash \ge$,	$\textrm{ZFC}\vdash \ge$,	???	$=$	$\textrm{ZFC}\vdash \ge$,
	[2, 4.1]	[2]			[2]
$\textrm{cov}({\mathfrak {C}_{2}})$	[5]	$\textrm{ZFC}\vdash \ge$, [2, 4.1]	4.12	[5]	$=$

Except for two pairs of ideals, all coverings have been separated. The value of $\textrm{cov}({\mathfrak {P}_{2}})$ in the Sacks model remains an open question. The value of $\textrm{cov}({\mathfrak {C}_{2}})$ can be calculated in all our models; however, it is always the same as $\textrm{cov}({u^{0}})$.

In each case, we will construct a family of ideal sets along the iteration. Some care must be made to assure that only $\aleph _1$ of these sets are required. This is ensured by some form of continuous reading of names that allows us to capture the reals in the final model with some trees in the ground model in such a way that names of the same type are captured by the same tree and there are only $\aleph _1$ many types (see Definitions 3.5 and 4.9 to understand what we mean by types).

Our method is somewhat similar to the one used by Brendle. He, too, constructed the covering sets along the iteration. However, due to the fact that he forced with non-compact tree forcings, he had to use the Laver property of the respective forcing to capture the reals. Our construction does not rely on the Laver property and we obtain much more control over the reals that are added.

2 Some notation and preliminary results

In this section we will introduce some concepts and notation that will be useful throughout the paper.

Definition 2.1

Let $T\subseteq {^{<\omega }}{2}{}$ be a perfect tree. We write $\textrm{split}(T)$ for the set of split nodes of T. For $n<\omega$, we write $\textrm{split}_n(T)$ for the set of split nodes that properly contain exactly n split nodes. We call the unique element of $\textrm{split}_0(T)$ the stem of T and write $\textrm{stem}(T)$ for this element. The canonical enumeration of $\textrm{split}(T)$ is the enumeration $\left\langle {{\sigma _s} : {s\in {^{<\omega }}{2}{}}}\right\rangle$ that satisfies $\sigma _{\emptyset }:= \textrm{stem}(T)$ and, if $\sigma _s\in \textrm{split}_n(T)$ is defined for some $s\in {^{n}}{2}{}$ and $j<2$, then $\sigma _{s^\frown j}$ is the unique element of $\textrm{split}_{n+1}(T)$ that extends $\sigma _s^\frown j$. For $\sigma \in \textrm{split}(T)$, we put

$$\begin{aligned}T_\sigma := \left\{ {\tau \in T: \sigma \subseteq \tau \;\vee \; \tau \subseteq \sigma }\right\} .\end{aligned}$$

For $s\in {^{<\omega }}{2}{}$, put

$$\begin{aligned}T*s:= T_{\sigma _s} = \left\{ {\tau \in T: \sigma _s\subseteq \tau \;\vee \; \tau \subseteq \sigma _s}\right\} .\end{aligned}$$

Let $p,q\in \mathbb {S}$ and $n<\omega$, put

$$\begin{aligned}p\le _n q\iff (p\le q\;\wedge \; \textrm{split}_n(p) = \textrm{split}_n(q)).\end{aligned}$$

A fusion sequence is a sequence $\left\langle {{p_n} : {n<\omega }}\right\rangle$ of perfect trees such that $p_{n+1}\le _n p_n$ for every $n<\omega$. It is easy to see that $p_{\omega }:= \bigcap _{n<\omega }p_n$ is a perfect tree, that satisfies $p_\omega \le _n p_n$ for every $n<\omega$. Furthermore, if the $p_n$ are all uniform (or Silver), then $p_\omega$ is uniform (or Silver). We call $p_\omega$ the fusion limit of $\left\langle {{p_n} : {n<\omega }}\right\rangle$.

For a forcing notion $\mathbb {P}$ and $\alpha \in \textbf{Ord}$ we will write $\mathbb {P}_\alpha$ for the countable support iteration of $\mathbb {P}$ of length $\alpha$ over some ground model V for $\textrm{ZFC}+ \textrm{CH}$.

Definition 2.2

Let $\mathbb {P}\in \left\{ {\mathbb {V},\mathbb {U},\mathbb {S}}\right\}$ and $\alpha \in \textbf{Ord}$. A triple $(p,F,\eta )$ such that $p\in \mathbb {P}_\alpha$, $F\in [\textrm{supp}(p)]^{<\aleph _0}$, and $\eta : F\rightarrow \omega$ is called a fusion condition. Given $p'\in \mathbb {P}_\alpha$ we define

$$\begin{aligned}p'\le _{F,\eta }p\iff (p'\le p\;\wedge \; \forall \beta \in F ({p'}{\upharpoonright }{\beta }\Vdash p'(\beta )\le _{\eta (\beta )}p(\beta ))).\end{aligned}$$

If $(p',F',\eta ')$ is an other fusion condition, we say that $(p',F',\eta ')\le ^1 (p,F,\eta )$, if the following assertions are satisfied:

• (a) $p'\le _{F,\eta } p$;

• (b) $F\subseteq F'$ and $\left| {F'{\setminus } F}\right| \le 1$;

• (c) if $F'\setminus F = \left\{ {\beta }\right\}$, then ${\eta }{\upharpoonright }{F} = {\eta '}{\upharpoonright }{F}$ and $\eta '(\beta ) = 0$;

• (d) if $F' = F$, there exists precisely one $\beta \in F$ such that ${\eta }{\upharpoonright }{(F{\setminus }\left\{ {\beta }\right\} )}={\eta '}{\upharpoonright }{(F{\setminus }\left\{ {\beta }\right\} )}$ and $\eta '(\beta ) = \eta (\beta ) + 1$.

If $(p',F',\eta ')\le ^1 (p,F,\eta )$ and $\beta$ is the unique element from (c) or (d), then we call $(p',F',\eta ')$ a $\beta$-extension of $(p,F,\eta )$. A sequence $\left\langle {{(p_n,F_n,\eta _n)} : {n<\omega }}\right\rangle$ of fusion conditions is called a fusion sequence if

• (a) $\left| {F_0}\right| = 1$;

• (b) $(p_{n+1},F_{n+1},\eta _{n+1})\le ^1 (p_n,F_n,\eta _n)$;

• (c) $\bigcup _{n<\omega }F_n = \bigcup _{n<\omega }\textrm{supp}(p_n)$;

• (d) $0\in \bigcup _{n<\omega }F_n$;

• (e) For all $\beta \in \textrm{supp}(p_n)$ there exists $m<\omega$ such that $\eta _m(\beta ) \ge n$.

For every $1\le n<\omega$, we call the unique $\beta \in F_n$ from (c) or (d) of the definition of $\le ^1$ the n-th active coordinate. The 0-th active coordinate is the unique element of $F_0$. Clearly a fusion sequence uniquely determines the sequence $\left\langle {{\beta _n} : {n<\omega }}\right\rangle$ of its active coordinates, and $\left\langle {{(p_n,\beta _n)} : {n<\omega }}\right\rangle$ determines the corresponding fusion sequence.

Put $F_\omega := \bigcup _{n<\omega } F_n$ and define $p_\omega \in \mathbb {P}_\alpha$ such that for every $\beta \in \alpha {\setminus } F_\omega$, ${p_\omega }{\upharpoonright }{\beta }\Vdash _\beta p_\omega (\beta ) = {\textbf{1}}_\mathbb {P}$ and for every $\beta \in F_\omega$,

$$\begin{aligned}{p_\omega }{\upharpoonright }{\beta }\Vdash _\beta p_\omega (\beta ) = \bigcap _{n<\omega } p_n(\beta ).\end{aligned}$$

We call $p_\omega$ the fusion limit of $\left\langle {{(p_n,F_n,\eta _n)} : {n<\omega }}\right\rangle$.

Let $(p,F,\eta )$ be a fusion condition and $\sigma \in \prod _{\beta \in F} {^{(\eta (\beta )+1)}}{2}{}$. We define $p*\sigma \in \mathbb {P}_\alpha$ such that

$$\begin{aligned}{(p*\sigma )}{\upharpoonright }{\beta }\Vdash (p*\sigma )(\beta ) ={\left\{ \begin{array}{ll} p(\beta )*{\sigma (\beta )} &{}\beta \in F\\ p(\beta ) &{}\beta \notin F. \end{array}\right. } \end{aligned}$$

We will fix some notation: If $\sigma : F\rightarrow {^{<\omega }}{2}{}$ for some $F\subseteq \omega _2$, $\beta < \omega _2$ and $j<2$, we define $\sigma ^{(\beta ,j)}: F\cup \left\{ {\beta }\right\} \rightarrow {^{<\omega }}{2}{}$ with ${\sigma ^{(\beta ,j)}}{\upharpoonright }{(F{\setminus }\left\{ {\beta }\right\} )} = {\sigma }{\upharpoonright }{(F{\setminus }\left\{ {\beta }\right\} )}$ and

$$\begin{aligned}\sigma ^{(\beta ,j)}(\beta ):= {\left\{ \begin{array}{ll} \sigma (\beta )^\frown j &{}\beta \in F\\ \left\langle {j}\right\rangle &{}\beta \notin F. \end{array}\right. } \end{aligned}$$

For $f: \omega \rightarrow \omega _2$ we define $\sigma _s^f: f[\left| {s}\right| ]\rightarrow {^{<\omega }}{2}{}$ for every $s\in {^{<\omega }}{2}{}$ by recursion on s:

$$\begin{aligned} \sigma _\emptyset ^f&:= \emptyset \nonumber \\ \sigma _{s^\frown j}^f&:= (\sigma _s^f)^{(f(\left| {s}\right| ),j)}. \end{aligned}$$

(2)

Lemma 2.3

Let $\mathbb {P}\in \left\{ {\mathbb {V},\mathbb {U}, \mathbb {S}}\right\}$ and $\alpha \le \omega _2$. Let $p\in \mathbb {P}_\alpha$ and $f: \omega \rightarrow \textrm{supp}(p)$ such that $f^{-1}[\left\{ {\beta }\right\} ]$ is infinite for every $\beta \in \textrm{supp}(p)$. Let $\left\langle {{\sigma _s^f} : {s\in {^{<\omega }}{2}{}}}\right\rangle$ be as in (2). For every $s\in {^{<\omega }}{2}{}$, put $a_s:= p*\sigma _s^f$.

• (1) For every $s\in {^{<\omega }}{2}{}$ and $\left| {s}\right| \le n < \omega$, $\left\langle {{a_t} : {s\subseteq t\in {^{n}}{2}{}}}\right\rangle$ is a maximal antichain below $a_s$.

• (2) For every $s\in {^{<\omega }}{2}{}$, $\beta \in \textrm{supp}(p)$, and $p'\le a_s$ there exists $t\in {^{<\omega }}{2}{}$ such that $f(\left| {t}\right| ) = \beta$, $s\subseteq t$, and $p'$ is compatible with both $a_{t^\frown 0}$ and $a_{t^\frown 1}$.

Proof

(1). It suffices to consider $n = \left| {s}\right| + 1$. Clearly, $a_{s^\frown 0}$ and $a_{s^\frown 1}$ are incompatible. Let $p\le a_s$ and $\beta := f(\left| {s}\right| )$. Since $p\le a_s$,

$$\begin{aligned}{p}{\upharpoonright }{\beta }\Vdash p(\beta )\le a_s(\beta ).\end{aligned}$$

Since

$$\begin{aligned}{p}{\upharpoonright }{\beta }\Vdash \{a_{s^\frown 0}(\beta ), a_{s^\frown 1}(\beta )\}\text { is a maximal antichain below }a_{s}(\beta ),\end{aligned}$$

there exist a $\mathbb {P}_\beta$-name $\dot{r}$, $j < 2$, and $q\le {p}{\upharpoonright }{\beta }$ such that

$$\begin{aligned}q\Vdash \dot{r}\le p(\beta ), a_{s^\frown j}(\beta ).\end{aligned}$$

Then $q^\frown \dot{r}^\frown ({p}{\upharpoonright }{[\beta +1,\alpha )})\le p, a_{s^\frown j}$.

(2). We can assume without loss of generality that ${p'}{\upharpoonright }{\beta }$ decides $\textrm{stem}(p'(\beta ))$ and that ${a_s}{\upharpoonright }{(f(\left| {s}\right| ))}$ decides $\textrm{stem}(a_s(f(\left| {s}\right| )))$ for every $s\in {^{<\omega }}{2}{}$.

Construct recursively $t_0,t_1,\dots$ and $p_0,p_1,\dots$ with the following properties:

• (a) $p_n \le a_{t_n}$;

• (b) $p_{n+1}\le p_n$;

• (c) $t_{n+1} = t_n^\frown j$ for some $j<2$;

• (d) ${p_n}{\upharpoonright }{\beta }\Vdash \textrm{stem}(a_{t_n}(\beta ))\subseteq \textrm{stem}(p_n(\beta )) = \textrm{stem}(p'(\beta ))$.

Note that Property (d) guarantees that the construction will finish after finitely many steps.

Put $p_0 = p'$ and $t_0 = s$. Clearly, Properties (a), (b), and (c) are satisfied. For Property (d), note that

$$\begin{aligned}{p'}{\upharpoonright }{\beta }\Vdash p'(\beta )\le a_s(\beta ),\end{aligned}$$

and hence,

$$\begin{aligned}{p'}{\upharpoonright }{\beta }\Vdash \textrm{stem}(a_s(\beta ))\subseteq \textrm{stem}(p'(\beta )).\end{aligned}$$

Now, assume that $p_n$ and $t_n$ have been constructed for some $n<\omega$.

Since $\left\{ {a_{t_n^\frown 0}, a_{t_n^\frown 1}}\right\}$ is a maximal antichain below $a_{t_n}$, there exists $j< 2$ such that $p_n$ is compatible with $a_{t_n^\frown j}$. If $\beta _n:= f(\left| {t_n}\right| )\ne \beta$, choose any such j and put $t_{n+1}:= t_n^\frown j$. Define $p_{n+1}$ such that ${p_{n+1}}{\upharpoonright }{(\alpha {\setminus }\left\{ {\beta _n}\right\} )} = {p_n}{\upharpoonright }{(\alpha {\setminus }\left\{ {\beta _n}\right\} )}$ and

$$\begin{aligned}{p_{n+1}}{\upharpoonright }{\beta _n}\Vdash p_{n+1}(\beta _n) = p_n(\beta _n)\cap a_{t_{n+1}}(\beta _n).\end{aligned}$$

Since

$$\begin{aligned}{p_n}{\upharpoonright }{\beta _n}\Vdash (p_n(\beta _n)\le a_{t_n}(\beta _n)\wedge (p_n(\beta _n)\parallel a_{t_{n+1}}(\beta _n))),\end{aligned}$$

it follows that $p_{n+1}$ satisfies Properties (a) and (b). Since ${p_n}{\upharpoonright }{\beta }$ decides both $\textrm{stem}(a_{t_n}(\beta ))$ and $\textrm{stem}(p_n(\beta ))$ and ${p_n}{\upharpoonright }{\beta }\Vdash \textrm{stem}(a_{t_{n+1}}(\beta )) = \textrm{stem}(a_{t_n}(\beta ))$, Property (d) is satisfied in this case.

If $f(\left| {t_n}\right| ) = \beta$, distinguish two cases. If

$$\begin{aligned}{p_n}{\upharpoonright }{\beta }\Vdash \textrm{stem}(a_{t_n}(\beta ))\subsetneq \textrm{stem}(p'(\beta )).\end{aligned}$$

Since ${p_n}{\upharpoonright }{\beta }$ decides both $\textrm{stem}(a_{t_n}(\beta ))$ and $\textrm{stem}(p'(\beta ))$, there is a unique $j< 2$ such that $\textrm{stem}(a_{t_n}(\beta ))^\frown j\subseteq \textrm{stem}(p'(\beta ))$. Put $t_{n+1}:= t_n^\frown j$ and define $p_{n+1}$ as in the previous case. It is clear that all properties are satisfied. If

$$\begin{aligned}{p_n}{\upharpoonright }{\beta }\Vdash \textrm{stem}(a_{t_n}(\beta )) = \textrm{stem}(p'(\beta )),\end{aligned}$$

the constriction is finished. It is easy to see that $p'$ is compatible with both $a_{t_n^\frown 0}$ and $a_{t_n^\frown 1}$. $\square$

Building on the proof of the previous lemma we obtain the following result, which only works, if one considers iterations of uniform or Silver trees.

Lemma 2.4

Let $\mathbb {P}\in \left\{ {\mathbb {V},\mathbb {U}}\right\}$ and $\alpha \le \omega _2$. Let $p\in \mathbb {P}_\alpha$ and $f: \omega \rightarrow \textrm{supp}(p)$ such that $f^{-1}[\left\{ {\beta }\right\} ]$ is infinite for every $\beta \in \textrm{supp}(p)$. Let $\left\langle {{\sigma _s^f} : {s\in {^{<\omega }}{2}{}}}\right\rangle$ be as in (2) and put $a_s:= p*\sigma _s^f$ for every $s\in {^{<\omega }}{2}{}$. Let $s\in {^{<\omega }}{2}{}$ with $f(\left| {s}\right| ) = 0$ and $p'\le a_s$ such that $p'$ is compatible with both $a_{s^\frown 0}$ and $a_{s^\frown 1}$. Then there exist $t_0,t_1\in {^{<\omega }}{2}{}$ such that $s^\frown j\subseteq t_j$, $\left| {t_0}\right| = \left| {t_1}\right|$, $f(\left| {t_0}\right| ) = 0$ and $p'$ is compatible with $a_{t_j^\frown 0}$ and $a_{t_{j}^\frown 1}$ for $j<2$.

Proof

The only thing to prove is that $t_0,t_1$ can be found such that $\left| {t_0}\right| = \left| {t_1}\right|$. We can assume without loss of generality that $p'\le a_s$. Since $p'(0)$ and $a_{s}(0)$ are uniform trees with $p'(0)\le a_s(0)$ such that $p'(0)$ is compatible with both $a_{s^\frown 0}(0)$ and $a_{s^\frown 1}(0)$, we know that $\textrm{stem}(p'(0)) = \textrm{stem}(a_s(0))$. Furthermore, there exists $m < \omega$ such that $\textrm{split}_1(p'(0))\subseteq \textrm{split}_m(a_s(0))$. In the proof of the previous lemma with $\beta = 0$, the length of $t_0$ and $t_1$ is completely determined by m. In particular, $\left| {t_0}\right| = \left| {t_1}\right|$. $\square$

Definition 2.5

Let $\mathbb {P}\in \left\{ {\mathbb {V},\mathbb {U},\mathbb {S}}\right\}$. For $p\in \mathbb {P}_\alpha$, $f:\omega \rightarrow \textrm{supp}(p)$ such that $f^{-1}[\left\{ {\beta }\right\} ]$ is infinite for every $\beta \in \textrm{supp}(p)$, and $\left\langle {{\sigma _s^f} : {s\in {^{<\omega }}{2}{}}}\right\rangle$ as in (2), we call $\left\langle {{a_s} : {s\in {^{<\omega }}{2}{}}}\right\rangle$ with $a_s = p*\sigma _s^f$ the antichain refinement associated with (p, f). We write $\bar{A}(p,f)$ for this family.

3 $\textrm{cov}({s^{0}})$ in $V^{\mathbb {U}_{\omega _2}}$ and $V^{\mathbb {V}_{\omega _2}}$

In this section we will show that $\textrm{cov}({s^{0}})$ remains small in the Silver and the uniform Sacks model. To obtain a family of ${s^{0}}$-sets (in the extension models) that covers the reals, we will construct a family of dense subsets $D\subseteq \mathbb {S}$. Note that sets of the form

$$\begin{aligned}\mathcal {X}(D):= {^{\omega }}{2}{}\setminus \bigcup \left\{ {[q]: q\in D}\right\} ,\end{aligned}$$

where $D\subseteq \mathbb {S}$ is pre-dense form a base of ${s^{0}}$.

The following definition yields a convenient way to describe the continuous reading of names that we need in this section and the next.

Definition 3.1

Let $\mathbb {P}\in \left\{ {\mathbb {V},\mathbb {U},\mathbb {S}}\right\}$ and $\alpha < \omega _2$. An $\mathbb {P}_\alpha$-coding triple is a triple $(\dot{x}, p, f)$, where $\dot{x}$ is a $\mathbb {P}_\alpha$-name, $p\in \mathbb {P}_\alpha$, and $f: \omega \rightarrow \textrm{supp}(p)$ such that

• (a) $p\Vdash _{\mathbb {P}_\alpha } \dot{x} \in {^{\omega }}{2}{}\setminus \bigcup _{\beta < \alpha } V^{\mathbb {P}_\beta }$;

• (b) $f^{-1}[\left\{ {\beta }\right\} ]$ is infinite for every $\beta \in \textrm{supp}(p)$;

• (c) for every $s\in {^{<\omega }}{2}{}$, ${(p*\sigma _s^f)}{\upharpoonright }{f(\left| {s}\right| )}$ decides $\textrm{stem}((p*\sigma _s^f)(f(\left| {s}\right| )))$;

• (d) the associated antichain refinement $\bar{A}(p,f) = \left\langle {{a_s} : {s\in {^{<\omega }}{2}{}}}\right\rangle$ satisfies

  $$\begin{aligned}\forall s,t\in {^{<\omega }}{2}{} (s\perp t\implies \dot{x}[a_s]\perp \dot{x}[a_t]).\end{aligned}$$

For a $\mathbb {P}_\alpha$-coding triple $(\dot{x}, p, f)$ with $\bar{A} (p,f) = \left\langle {{a_s} : {s\in {^{<\omega }}{2}{}}}\right\rangle$ define

$$\begin{aligned}\bar{T}(\dot{x}, p, f) = \left\langle {{t_s} : {s\in {^{<\omega }}{2}{}}}\right\rangle \end{aligned}$$

such that $t_s$ is the longest common initial segment of $\dot{x}[a_{s^\frown 0}]$ and $\dot{x}[a_{s^\frown 1}]$ for every $s\in {^{<\omega }}{2}{}$. Furthermore put

$$\begin{aligned}a(\dot{x}, p, f):= {\left\{ \begin{array}{ll} f^{-1}[\left\{ {\beta }\right\} ] &{} \beta = \max (\textrm{supp}(p))\\ \emptyset &{} \textrm{supp}(p)\text { has no maximal element.} \end{array}\right. }\end{aligned}$$

A splitting family is a strictly increasing family $\bar{T} = \left\langle {{t_s} : {s\in {^{<\omega }}{2}{}}}\right\rangle$ such that for every $s\in {^{<\omega }}{2}{}$, $t_{s^\frown 0}$ and $t_{s^\frown 1}$ are incompatible and $t_s$ is their longest common initial segment. It is clear that if T is the tree generated by such a family $\bar{T}$, then T is perfect, $\bar{T}$ enumerates $\textrm{split}(T)$ and ${\bar{T}}{\upharpoonright }{{^{n}}{2}{}}: {^{n}}{2}{}\rightarrow \textrm{split}_n(T)$ is a bijection for every $n<\omega$.

The following proposition is a reformulation of continuous reading of names for the respective iterations. For $\mathbb {V}$, a stronger version of this proposition is proved by the second author in [5, Theorem 3.1], for $\mathbb {U}$ the proposition was proved by the authors in [2, Section 7] and for $\mathbb {S}$ a stronger version of the proposition can be found below (see Proposition 4.1).

Proposition 3.2

Let $\mathbb {P}\in \left\{ {\mathbb {V},\mathbb {U},\mathbb {S}}\right\}$ and $\alpha \le \omega _2$. Let $\dot{x}$ be a $\mathbb {P}_\alpha$-name and $p'\in \mathbb {P}_\alpha$ such that $p'\Vdash _\alpha \dot{x} \in {^{\omega }}{2}{}{\setminus }\bigcup _{\beta < \alpha } V^{\mathbb {P}_\beta }$. Then there exist $p\le p'$ and $f: \omega \rightarrow \textrm{supp}(p)$ such that $(\dot{x}, p, f)$ is an $\mathbb {P}_\alpha$-coding triple.

Fix $\mathbb {P}\in \left\{ {\mathbb {V},\mathbb {U}}\right\}$. Let $\Omega$ be the set of infinite, co-infinite subsets of $\omega$. For every splitting family $\bar{T}: {^{<\omega }}{2}{}\rightarrow {^{<\omega }}{2}{}$ and every $a\in \Omega \cup \left\{ {\emptyset ,\omega }\right\}$ in the ground model we will construct a family $\langle {\dot{D}_\gamma (\bar{T}, a)} : {\gamma <\omega _2} \rangle$ such that $\dot{D}_\gamma (\bar{T}, a)$ is a $\mathbb {P}_{\gamma +1}$-name for an open-dense subset of $\mathbb {S}^{V^{\mathbb {P}_{\gamma +1}}}$ such that for every $\mathbb {P}_\alpha$-coding triple $(\dot{x}, p, f)$ (for some $\alpha < \omega _2$) with $\bar{T} = \bar{T}(\dot{x}, p, f)$ and $a = a(\dot{x}, p, f)$ we have

$$\begin{aligned}p\Vdash _{\mathbb {P}_{\omega _2}} \dot{x}\in \mathcal {X}(D_\gamma (\bar{T}, a)).\end{aligned}$$

Clearly, there are only $\aleph _1$-many of these pairs. This ensures that $\textrm{cov}({s^{0}}) = \aleph _1$ in $V^{\mathbb {V}_{\omega _2}}$ and $V^{\mathbb {U}_{\omega _2}}$.

Remark 3.3

In the next section we will have to invoke a notion of isomorphism for coding triples (see Definition 4.9). There we will see that reals that come from isomorphic coding triples are captured by the same ideal set. The fact that there are only $\aleph _1$-many isomorphism types shows that the respective covering remains small. In this situation, however, it suffices to consider $\bar{T}(\dot{x},p,f)$ and $a(\dot{x}, p, f)$. Both are invariant with respect to the notion of isomorphism.

Before we define the coding systems, we need one more definition.

Definition 3.4

Given a tree $T\subseteq {^{<\omega }}{\omega }{}$, we write [T] for the set of infinite branches of T, i.e.

$$\begin{aligned} =\left\{ {x\in {^{\omega }}{\omega }{}:\forall n <\omega ({x}{\upharpoonright }{n}\in T) }\right\} .\end{aligned}$$

Let $\bar{T} = \left\langle {{t_s} : {s\in {^{<\omega }}{2}{}}}\right\rangle$ be a splitting family. Let T be the perfect tree generated by $\bar{T}$. For every $x\in [T]$ there exists a unique real $c_x \in {^{\omega }}{2}{}$ such that

$$\begin{aligned}x = \bigcup \left\{ {t_{{c_x}{\upharpoonright }{n}}: n<\omega }\right\} .\end{aligned}$$

Let $a\subseteq \omega$ and $g:\textrm{dom}(g)\rightarrow 2$ with $a\subseteq \textrm{dom}(g)\subseteq \omega$. We call x ($\bar{T}$-)incompatible with g on a, if ${c_x}{\upharpoonright }{a}\ne {g}{\upharpoonright }{a}$. We say that a perfect subtree S of T is incompatible with g on a, if every $x\in [S]$ is incompatible with g on a. We say that $S\le _\mathbb {S}T$ is incompatible with $A\subseteq {^{\omega }}{2}{}$ on a, if S is incompatible with every $g\in A$ on a.

Definition 3.5

Let $\bar{T} = \left\langle {{t_s} : {s\in {^{<\omega }}{2}{}}}\right\rangle$ be a splitting family that generates the perfect tree T. Let $a\in \Omega \cup \left\{ {\emptyset ,\omega }\right\}$ and, assuming $a\ne \emptyset$, let $\varphi : a\rightarrow \omega$ be the order isomorphism. Let $\gamma < \omega _2$ and let $x_\gamma$ be a $\mathbb {P}_{\gamma +1}$-name for the generic real that is adjoined in the $\gamma$-th step of the iteration. Let $\dot{D}_\gamma = \dot{D}_\gamma (\bar{T}, a)$ be a $\mathbb {P}_{\gamma +1}$-name with the following properties:

• (a) $\Vdash _{\gamma +1} \dot{D}_\gamma \text { is an open-dense subset of }\mathbb {S}$;

• (b) $\Vdash _{\gamma +1}\forall S\in \dot{D}_\gamma \;(S\not \subseteq T\implies \textrm{stem}(S)\notin T)$;

• (c) $\Vdash _{\gamma +1}\forall S\in \dot{D}_\gamma \;(S\subseteq T\implies (\textrm{split}(S)\subseteq \bigcup _{n\in a}\textrm{split}_n(T)\vee \textrm{split}(S)\subseteq \bigcup _{n\in \omega \setminus a}\textrm{split}_n(T)))$;

• (d) $\Vdash _{\gamma +1}\forall S\in \dot{D}_\gamma \; (S\subseteq T\implies \forall n<\omega (\left| {\textrm{split}(S)\cap \textrm{split}_n(T)}\right| \le 1))$;

• (e) $\Vdash _{\gamma +1} \forall S\in \dot{D}_\gamma \;(\textrm{split}(S)\subseteq \bigcup _{n\in \omega \setminus a}\textrm{split}_n(T)$ $\implies S\text { is }\bar{T}\text {-incompatible with}{^{\omega }}{2}{}\cap V^{\mathbb {P}_\gamma }\text { on }\omega \setminus a)$;

• (f) $\Vdash _{\gamma +1}\forall S\in \dot{D}_\gamma \; (\textrm{split}(S)\subseteq \bigcup _{n\in a}\textrm{split}_n(T) \implies S\text { is } \bar{T}\text {-incompatible with }x_\gamma \circ \varphi \text { on }a)$;

• (g) $\Vdash _{\gamma +1} \forall S\in \dot{D}_\gamma \; (S\subseteq T\implies [S]\cap V^{\mathbb {P}_\gamma } = \emptyset )$.

We quickly deal with the fact that the $\dot{D}_\gamma (\bar{T}, a)$’s from the previous definition exist: Since we are only concerned with names for successor steps of the iteration, the analysis can be reduced to a single step of $\mathbb {P}$. Since the sets of all trees with the respective properties are clearly open, it suffices to prove that they are dense. $\textrm{ZFC}$ proves that trees with properties (b), (c), and (d) are open-dense. The remaining properties are dealt with in the following lemma.

Lemma 3.6

Let $a\in [\omega ]^\omega$, let $\bar{T} = \left\langle {{t_s} : {s\in {^{<\omega }}{2}{}}}\right\rangle$ be a splitting family and let T be the tree generated by $\bar{T}$. Let $\mathbb {P}\in \left\{ {\mathbb {V},\mathbb {U},\mathbb {S}}\right\}$, $\mathbb {Q}\in \left\{ {\mathbb {U},\mathbb {S}}\right\}$ and let $\dot{S}$ be a $\mathbb {P}$-name such that

$$\begin{aligned}\Vdash _{\mathbb {P}}\dot{S}\in \mathbb {Q}\wedge \dot{S}\subseteq T\wedge \textrm{split}(\dot{S})\subseteq \bigcup _{n\in a}\textrm{split}_n(T).\end{aligned}$$

Then there exists a $\mathbb {P}$-name $\dot{R}$ such that

$$\begin{aligned}\Vdash _{\mathbb {P}}\dot{R}\le _{\mathbb {Q}}\dot{S}\wedge (\dot{R}\text { is }\bar{T}\text { -incompatible with }{^{\omega }}{2}{}\cap V\text { on }a).\end{aligned}$$

Proof

Let $\dot{I}$ be a $\mathbb {P}$-name such that $\Vdash \dot{I}\in \mathbb {S}$ and

$$\begin{aligned}\Vdash \textrm{split}(\dot{S}) = \lbrace {t_{s}: s\in \textrm{split}(\dot{I})}\rbrace .\end{aligned}$$

We consider the case $\mathbb {Q}= \mathbb {S}$ first, and deal with $\mathbb {Q}= \mathbb {U}$ in a remark afterwards. It suffices to find a $\mathbb {P}$-name $\dot{J}$ such that

$$\begin{aligned}\Vdash _{\mathbb {P}}\dot{J}\le _{\mathbb {S}}\dot{I}\wedge \forall x\in [\dot{J}] ({x}{\upharpoonright }{a}\notin V).\end{aligned}$$

Let $p\in \mathbb {P}$. The proof of the following claim is a simple fusion below p and is left to the reader.

Claim 3.6.1

There exists $q\le p$ and a family $\left\langle {{S_t} : {t\in {^{<\omega }}{2}{}}}\right\rangle$ with the following properties:

• (a) $S_t\subseteq {^{<\omega }}{2}{}$;

• (b) $q*t\Vdash S_t\subseteq \textrm{split}_{2\left| {t}\right| }(\dot{I})$;

• (c) for every $t\in {^{<\omega }}{2}{}$, $j<2$, $\sigma \in S_t$ there exist $\sigma _0,\sigma _{1}\in S_{t^\frown j}$ such that $\sigma ^\frown j\subseteq \sigma _0,\sigma _1$ and $\sigma _0\perp \sigma _1$;

• (d) $\left| {S_{t}}\right| = 2^{\left| {t}\right| }$.

Let q and $\left\langle {{S_t} : {t\in {^{<\omega }}{2}{}}}\right\rangle$ be as in Claim 3.6.1.

Claim 3.6.2

For every $y\in {^{a}}{2}{}$ and every $t\in {^{<\omega }}{2}{}$,

$$\begin{aligned}\left| {\left\{ {\sigma \in S_{t}: {\sigma }{\upharpoonright }{a}\subseteq y}\right\} }\right| \le 1.\end{aligned}$$

Proof

By induction on $\left| {t}\right|$. For $t = \emptyset$, this is easy, because $\left| {S_\emptyset }\right| = 2^0 = 1$. Assume that the assertion is true for t. Let $j<2$. If there is no $\sigma \in S_t$ such that ${\sigma }{\upharpoonright }{a}\subseteq y$, there is no such element in $S_{t^\frown j}$, since every element of $S_{t^\frown j}$ extends an element from $S_t$. So, let $\sigma \in S_t$ be the unique element with ${\sigma }{\upharpoonright }{a}\subseteq y$. Since $q*t\Vdash \sigma \in \textrm{split}(\dot{S})$, $\left| {\sigma }\right| \in a$. Let $\sigma _0,\sigma _1\in S_{t^\frown j}$ be the two elements extending $\sigma ^\frown j$. Clearly, no other element of $S_{t^\frown j}$ can have the desired property. If $j = 1 - y(\left| {\sigma }\right| )$, neither of the two extension is compatible with y. If $j = y(\left| {\sigma }\right| )$, there can exist $k < 2$ such that ${\sigma _k}{\upharpoonright }{a}\subseteq y$. Since $\sigma _0\perp \sigma _1$ and the split nodes of $\dot{S}$ have lengths in a, ${\sigma _0}{\upharpoonright }{a}\ne {\sigma _1}{\upharpoonright }{a}$, so there is at most one such k. $\square$

Let $\dot{J}$ be the $\mathbb {P}$-name generated by $\bigcup \left\{ {S_{t}: t\subseteq \dot{g}}\right\}$.

Claim 3.6.3

$q\Vdash \forall x\in [\dot{J}]({x}{\upharpoonright }{a}\notin V)$.

Proof

Let $y\in {^{a}}{2}{}$ and $q'\le q$. Let $t\in {^{<\omega }}{2}{}$ be maximal with $q'\le q*t$. By Claim 2, there exists at most one $\sigma \in S_t$ such that ${\sigma }{\upharpoonright }{s}\subseteq y$. Since $\left| {\sigma }\right| \in a$ for every $\sigma \in S_t$, there exists $j< 2$ such that for every $\sigma \in S_t$, ${(\sigma ^\frown j)}{\upharpoonright }{a}\not \subseteq y$. Let $q'':= q'\cap (q*(t^\frown j))$ and let $\dot{x}$ be a $\mathbb {P}$-name with $q''\Vdash \dot{x}\in [\dot{J}]$. Then there exists a unique $\sigma \in S_{t^\frown j}$ such that $q''\Vdash \sigma \subseteq \dot{x}$. But ${\sigma }{\upharpoonright }{a}\not \subseteq y$, and hence, $q''\Vdash {\dot{x}}{\upharpoonright }{a}\ne y$. $\square$

This concludes the proof for $\mathbb {Q}= \mathbb {S}$. In case $\mathbb {Q}= \mathbb {U}$, let $\left\langle {{\dot{\sigma }_s} : {s\in {^{<\omega }}{2}{}}}\right\rangle$ be a $\mathbb {P}$-name for the canonical enumeration of $\textrm{split}(\dot{I})$. Note that there exists a $\mathbb {P}$-name $\dot{u}$ for a uniform tree such that

$$\begin{aligned}\Vdash _{\mathbb {P}}\dot{J} = \left\{ {\sigma _{s}: s\in \textrm{split}(\dot{u})}\right\} .\end{aligned}$$

In particular, if $\dot{S} = \lbrace {t_{s}: s\in \textrm{split}(\dot{I})}\rbrace$ is a uniform subtree of T, then so is $\dot{R} = \lbrace {t_{s}: s\in \textrm{split}(\dot{J})}\rbrace$. $\square$

The following lemmas show that for all coding triples $(\dot{x}, p, f)$, p forces that $\dot{x}$ is not in a branch set of one of the open-dense sets defined in Definition 3.5. We will deal with the consequences of Property (d) first. This is the only part of the proof that utilizes the fact that we deal with iterations of uniform trees.

Lemma 3.7

Let $\mathbb {P}\in \left\{ {\mathbb {V},\mathbb {U}}\right\}$ and $\alpha \le \omega _2$. Let $(\dot{x}, p, f)$ be an $\mathbb {P}_\alpha$-coding triple, $\bar{T}(\dot{x}, p, f) = \left\langle {{t_s} : {s\in {^{<\omega }}{2}{}}}\right\rangle$, and T the tree generated by $\bar{T}$. Let $S\in \mathbb {S}$ such that

$$\begin{aligned}\forall n<\omega (\left| {\textrm{split}(S)\cap \textrm{split}_n(T)}\right| \le 1).\end{aligned}$$

Then $p\Vdash \dot{x}\notin [S]$.

Proof

Let $\bar{A}(p,f) = \left\langle {{a_s} : {s\in {^{<\omega }}{2}{}}}\right\rangle$ be the antichain refinement associated with $(\dot{x}, p, f)$. If $\textrm{stem}(S)\notin T$, we are done. So, assume that $\textrm{stem}(S)\in T$. Then $\textrm{stem}(S) = t_s$ for some $s\in {^{<\omega }}{2}{}$. Let $p'\le p$. If $p'$ is incompatible with $a_s$, $p'\Vdash t_s\not \subseteq \dot{x}$, and hence, $p'\Vdash \dot{x}\notin [S]$. So, assume that $p'$ is compatible with $a_s$. Without loss of generality, $p'\le a_s$. By Lemma 2.3(2), there exists $s\subseteq s'\in {^{<\omega }}{2}{}$ such that $f(\left| {s'}\right| ) = 0$ and $p'$ is compatible with both $a_{{s'}^\frown 0}$ and $a_{{s'}^\frown 1}$. By Lemma 2.4 there exist $s_0,s_1\in {^{<\omega }}{2}{}$ such that $s'^\frown j\subseteq s_j$, $\left| {s_0}\right| = \left| {s_1}\right|$, $f(\left| {s_j}\right| ) = 0$, and $p'$ is compatible with $a_{s_j^\frown 0}$ and $a_{s_j^\frown 1}$ ($j<2$). By choice of S, there exists $j<2$ such that $t_{s_j}\notin \textrm{split}(S)$. Hence, there exists $k < 2$ such that $t_{s_j}^\frown k\notin S$. Let $\hat{s}\in {^{\left| {s_j}\right| +1}}{2}{}$ such that $t_{s_j}^\frown k\subseteq t_{\hat{s}}$. Then, $a_{\hat{s}}\Vdash t_{\hat{s}}\subseteq \dot{x}$, and thus, $a_{\hat{s}}\Vdash \dot{x}\notin [S]$. By choice of $s_j$, $a_{\hat{s}}$ is compatible with $p'$. This concludes the proof of the lemma. $\square$

The proof of following lemma is a straightforward application of the definition of $\bar{T}$-compatibility and is therefore left to the reader.

Lemma 3.8

Let $\mathbb {P}\in \left\{ {\mathbb {V},\mathbb {U}}\right\}$, $\alpha \le \omega _2$, and let $(\dot{x}, p, f)$ be an $\mathbb {P}_\alpha$-coding triple, $\bar{T} = \bar{T}(\dot{x}, p, f)$, and $a = a(\dot{x}, p, f)$. Assume that $a\ne \emptyset$, i.e. $\textrm{supp}(p)$ has a maximal element $\beta$, $\alpha = \beta + 1$ and $a = f^{-1}[\left\{ {\beta }\right\} ]$. Let $\dot{\varphi }$ be a $\mathbb {P}_\beta$-name such that

$$\begin{aligned}{p}{\upharpoonright }{\beta }\Vdash \dot{\varphi }: a\rightarrow a^{p(\beta )}\text { is an order isomorphism}.\end{aligned}$$

Let $x_\beta$ a $\mathbb {P}_\alpha$-name for the generic real adjoined in the $\beta$-th step of the iteration. Let $\dot{S}$ be an $\mathbb {P}_\alpha$-name for an element of $\mathbb {S}$ and let $p'\le p$ such that either

$$\begin{aligned}p'\Vdash _{\mathbb {P}_\alpha } \dot{S}\text { is incompatible with }x_\beta \circ \dot{\varphi }\text { on }a\end{aligned}$$

or

$$\begin{aligned}p'\Vdash _{\mathbb {P}_\alpha } \dot{S}\text { is incompatible with }{^{\omega }}{2}{}\cap V\text { on }\omega \setminus a.\end{aligned}$$

Then $p'\Vdash \dot{x}\notin [\dot{S}]$.

The following lemma is the crucial part of the proof that $\textrm{cov}({s^{0}})$ remains small in $V^{\mathbb {V}_{\omega _2}}$ and $V^{\mathbb {U}_{\omega _2}}$.

Lemma 3.9

Let $\mathbb {P}\in \left\{ {\mathbb {V},\mathbb {U}}\right\}$, $\alpha <\omega _2$, $(\dot{x}, p, f)$ a $\mathbb {P}_\alpha$-coding triple, $\bar{T} = \bar{T}(\dot{x}, p, f) = \left\langle {{t_s} : {s\in {^{<\omega }}{2}{}}}\right\rangle$, and $a = a(\dot{x}, p, f)$. Let $\gamma < \omega _2$, then

$$\begin{aligned}p\Vdash _{\omega _2} \dot{x}\in \mathcal {X}(\dot{D}_\gamma (\bar{T}, a)).\end{aligned}$$

Proof

Let T be the tree generated by $\bar{T}$. Put $\dot{D}_\gamma := \dot{D}_\gamma (\bar{T}, a)$. For every $\beta < \omega _2$ let $x_\beta$ be a $\mathbb {P}_{\beta +1}$-name for the generic real adjoined in the $\beta$-th step of the iteration.

Let $\dot{S}$ be a $\mathbb {P}_{\gamma +1}$-name with $p\Vdash _{\omega _2}\dot{S}\in \dot{D}_\gamma$. Let $p'\le p$. We can assume that one of the following statements is true. Either, $p'\Vdash \dot{S}\subseteq T$ or $p'\Vdash \textrm{stem}(\dot{S})\notin T$. In the latter case, $p'\Vdash \dot{x}\notin [\dot{S}]$. So, assume that $p'\Vdash \dot{S}\subseteq T$. Now, distinguish three cases:

Case 1. $\gamma + 1 < \alpha$. Let $\beta = \min (\textrm{supp}(p){\setminus } (\gamma +1))$. Let $G_{\beta }\subseteq \mathbb {P}_{\beta }$ be a V-generic filter with ${p}{\upharpoonright }{\beta }\in G_{\beta }$. Let $b:= f^{-1}[\textrm{supp}(p)\cap (\gamma + 1)]$. For every $n\in b$, there exists $k_n < \omega$ such that n is the $k_n$-th element of $f^{-1}[\left\{ {f(n)}\right\} ]$. Let

$$\begin{aligned}\bar{T}_{\beta } = \left\langle {{t_s} : {s\in {^{<\omega }}{2}{}, \forall n\in \left| {s}\right| \cap b (s(n) = x_{f(n)}(k_n))}}\right\rangle .\end{aligned}$$

Since $\beta < \alpha$, $\dot{x}$ has not been decided in $V[G_{\beta }]$, and hence, $\bar{T}_{\beta }$ generates a perfect tree $T_{\beta }$ with

$$\begin{aligned}V[G_{\beta }]\models ({p}{\upharpoonright }{[\beta , \omega _2)}\Vdash \dot{x}\in [T_{\beta }]).\end{aligned}$$

By Property (d) and the fact that $T_\beta$ is a subtree of T, we know that

$$\begin{aligned}\forall n<\omega (\vert {\textrm{split}(\dot{S}[G_{\beta }])\cap \textrm{split}_n(T_{\beta })}\vert \le 1).\end{aligned}$$

We can apply Lemma 3.7 to this situation to obtain, ${p}{\upharpoonright }{[\beta ,\omega _2)}\Vdash \dot{x} \notin \dot{q}[G_{\beta }]$. In particular, $p\Vdash \dot{x}\notin [\dot{q}]$.

Case 2. $\gamma + 1 = \alpha$. Then $\gamma$ is the maximal element of $\textrm{supp}(p)$ and $a = f^{-1}[\left\{ {\gamma }\right\} ]$. By Property (c),

$$\begin{aligned}\Vdash _{\omega _2} \textrm{split}(\dot{S})\subseteq \bigcup _{n\in a}\textrm{split}_n(T)\vee \textrm{split}(\dot{S})\subseteq \bigcup _{n\in \omega \setminus a}\textrm{split}_n(T).\end{aligned}$$

Without loss of generality, we can assume that $p'\Vdash _{\omega _2} \textrm{split}(\dot{S})\subseteq \bigcup _{n\in a}\textrm{split}_n(T)$ or $p'\Vdash _{\omega _2} \textrm{split}(\dot{S})\subseteq \bigcup _{n\in \omega {\setminus } a}\textrm{split}_n(T)$.

First, if $p'\Vdash _{\omega _2} \textrm{split}(\dot{S})\subseteq \bigcup _{n\in a}\textrm{split}_n(T)$, then, by Property (f),

$$\begin{aligned}p'\Vdash _{\gamma + 1}\dot{S} \text { is incompatible with }x_\gamma \circ \dot{\varphi }\text { on }a,\end{aligned}$$

where $\dot{\varphi }$ is a $\mathbb {P}_\gamma$-name such that ${p}{\upharpoonright }{\gamma }$ forces that $\dot{\varphi }$ is the order isomorphism $a\rightarrow a^{p(\gamma )}$. By Lemma 3.8, $p'\Vdash \dot{x}\notin [\dot{S}]$. So, assume $p'\Vdash _{\omega _2} \textrm{split}(\dot{S})\subseteq \bigcup _{n\in \omega {\setminus } a}\textrm{split}_n(T)$. By Property (e),

$$\begin{aligned}p'\Vdash \dot{S}\text { is incompatible with }{^{\omega }}{2}{}\cap V^{\mathbb {P}_\gamma }\text { on }\omega \setminus a.\end{aligned}$$

Then, Lemma 3.8 implies $p'\Vdash \dot{x} \notin [\dot{S}]$. It follows that $p\Vdash \dot{x}\notin [\dot{S}]$.

Case 3. $\gamma + 1 > \alpha$. By Property (g), $\Vdash _{\omega _2} [\dot{S}]\cap V^{\mathbb {P}_\gamma } = \emptyset$. Since $p\Vdash \dot{x}\in V^{\mathbb {P}_\gamma }$, this implies $p\Vdash \dot{x}\notin [\dot{S}]$. $\square$

The following theorem is the main result of this section. It shows that $\textrm{cov}({s^{0}})$ remains small in $V^{\mathbb {V}_{\omega _2}}$ and $V^{\mathbb {U}_{\omega _2}}$. The fact that both forcings increase the covering number of their respective ideals, is a Löwenheim–Skolem argument which can be found in [4, Theorem 1.2] or [2, Claim 5.5.1].

Theorem 3.10

Let $\mathbb {P}\in \left\{ {\mathbb {V},\mathbb {U}}\right\}$ and let $j^0$ be the associated ideal. Let $\mathbb {P}_{\omega _2}$ be the countable support iteration of $\mathbb {P}$ over a model V for $\textrm{ZFC}+ \textrm{CH}$. For every V-generic filter $G\subseteq \mathbb {P}_{\omega _2}$,

$$\begin{aligned}V[G]\models \textrm{cov}({s^{0}}) = \aleph _1 < \aleph _2 = {2^{\aleph _0}}= \textrm{cov}(j^0).\end{aligned}$$

Proof

For every splitting family $\bar{T} = \left\langle {{t_s} : {s\in {^{<\omega }}{2}{}}}\right\rangle$ and $a\in \Omega \cup \left\{ {\emptyset ,\omega }\right\}$ in the ground model we have the family

$$\begin{aligned}\langle {\dot{D}_\gamma (\bar{T}, a)} : {\gamma < \omega _2} \rangle \end{aligned}$$

as in Definition 3.5. In V[G], put

$$\begin{aligned}D(\bar{T}, a):= \bigcup _{\gamma < \omega _2} (\dot{D}_\gamma (\bar{T}, a))[G].\end{aligned}$$

Since every perfect tree in V[G] is added by some initial segment of the iteration, $D(\bar{T}, a)$ is a dense subset of $\mathbb {S}$ (in V[G]).

For every real x in V[G], there exist $\alpha < \omega _2$ and a $\mathbb {P}_\alpha$-coding triple $(\dot{x}, p, f)$ with $p\in G$ and $\dot{x}[G] = x$. Let $\bar{T} = \bar{T}(\dot{x}, p,f)$ and $a = a(\dot{x}, p, f)$. Then

$$\begin{aligned}p\Vdash \dot{x} \in \bigcap _{\gamma < \omega _2}\mathcal {X}(\dot{D}_\gamma (\bar{T}, a)),\end{aligned}$$

by Lemma 3.9. In particular, $x = \dot{x}[G] \in \mathcal {X}(D(\bar{T}, a))$. $\square$

4 $\textrm{cov}({u^{0}})$ and $\textrm{cov}({\mathfrak {C}_{2}})$ in $V^{\mathbb {S}_{\omega _2}}$

The proof that $\textrm{cov}({\mathfrak {C}_{2}})$ remains small in $V^{\mathbb {S}_{\omega _2}}$ is very similar to the first author’s proof of this fact in $V^{\mathbb {V}_{\omega _2}}$. We refer the reader to [5] for more motivation on the definition of our coding systems below.

We start with a version of continuous reading of names for Sacks forcing. The analogue theorem for Silver forcing, which can be found in [5, Theorem 3.1], is a little weaker ($f(n)\le f(\left| {s}\right| )$ has to be replaced with $f(n) < f(\left| {s}\right| )$). This stronger version can easily be proved using the same methods. Sacks forcing allows for this stronger version, since it admits to more liberty in the construction of fusion sequences.

Proposition 4.1

Let $\alpha \le \omega _2$ and let $\dot{x}$ be a $\mathbb {S}_\alpha$-name and $p'\in \mathbb {S}_\alpha$ such that $p'\Vdash \dot{x} \in {^{\omega }}{2}{}{\setminus }\bigcup _{\beta < \alpha }V^{\mathbb {S}_\beta }$. Then there exists an $\mathbb {S}_\alpha$-coding triple $(\dot{x}, p, f)$ such that $p\le p'$ and $\bar{T}(\dot{x}, p, f) = \left\langle {{t_s} : {s\in {^{<\omega }}{2}{}}}\right\rangle$ satisfies the following assertions:

• (a) $\forall s,s'\in {^{<\omega }}{2}{}(\left| {s}\right|< \left| {s'}\right| \implies \left| {t_{s}}\right| < \left| {t_{s'}}\right| )$;

• (b) For all $s,s'\in {^{<\omega }}{2}{}$ with $\left| {s}\right| = \left| {s'}\right|$ we have

  $$\begin{aligned}\left\{ {n < \left| {s}\right| : s(n)\ne s'(n)\;\wedge \; f(n) \le f(\left| {s}\right| )}\right\} \ne \emptyset \implies \left| {t_s}\right| \ne \left| {t_{s'}}\right| .\end{aligned}$$

The proof of the proposition is a fusion. It requires some lemmas:

Lemma 4.2

Let $\alpha \le \omega _2$ and let $\dot{x}$ be an $\mathbb {S}_\alpha$-name and $(p,F,\eta )$ an $\mathbb {S}_\alpha$-fusion condition such that $p\Vdash \dot{x} \in {^{\omega }}{2}{}$. For every $N<\omega$ there exists $q\le _{F,\eta }$ such that $(q,F,\eta )$ weakly decides ${\dot{x}}{\upharpoonright }{N}$.

Proof

Left to the reader. $\square$

Remark 4.3

Let $\dot{x}$ be an $\mathbb {S}_\alpha$-name, let $p\in \mathbb {S}_\alpha$ such that $p\Vdash \dot{x}\in {^{\omega }}{\omega }{}{\setminus }\bigcup _{\beta <\alpha }V^{\mathbb {S}_\beta }$. Let $\beta < \alpha$. Put $F = \left\{ {\beta }\right\}$ and $\eta = \left\{ {(\beta ,0)}\right\}$. Then there exists $q\le _{F,\eta }$ such that $(q,F,\eta )$ splits $\dot{x}$ at $\beta$.

Lemma 4.4

Let $\alpha \le \omega _2$ and let $\dot{x}$ be an $\mathbb {S}_\alpha$-name, let $(p,F,\eta )$ be an $\mathbb {S}_\alpha$-fusion condition such that $p\Vdash \dot{x}\in {^{\omega }}{\omega }{}{\setminus }\bigcup _{\beta <\alpha }V^{\mathbb {S}_\beta }$. Let $\beta \in \textrm{supp}(p)$ such that $\beta < \gamma$ for every $\gamma \in F$ and $N < \omega$. Then there exists $q\le _{F,\eta } p$ such that the $\beta$-extension $(q,F',\eta ')$ of $(p,F,\eta )$ splits $\dot{x}$ at $\beta$ and for every $\sigma \in \prod _{\gamma \in F}{^{(\eta (\gamma )+1)}}{2}{}$,

$$\begin{aligned} \left| {\dot{x}[q*\sigma ^{(\beta ,0)}]\wedge \dot{x}[q*\sigma ^{(\beta ,1)}]}\right| >N. \end{aligned}$$

Proof

We prove the claim by induction on $\left| {F}\right|$. For $F = \emptyset$, let $p'\le p$ such that $p'$ decides ${\dot{x}}{\upharpoonright }{(N+1)}$. Now, find $q\le _{F',\eta '}p'$ such that $(q,F',\eta ')$ splits $\dot{x}$ at $\beta$.

Now, let $F\ne \emptyset$. We can assume without loss of generality, that $(p,F,\eta )$ weakly decides ${\dot{x}}{\upharpoonright }{(N+1)}$. Let $\delta$ be the minimal element of F and let $\left\langle {{s_l} : {l<d}}\right\rangle$ enumerate . Put $q_{0}:= p$. Let $l < d$ and assume that $q_{l}\le _{F',\eta '} p$ has been constructed such that for every $j<l$, $(q_l*\left\{ {(\delta , s_{j})}\right\} , F'\setminus \left\{ {\delta }\right\} , \eta ')$ splits $\dot{x}$ at $\beta$. By the induction hypothesis, there exists $r_l\le _{F'{\setminus }\left\{ {\delta }\right\} ,\eta '} q_l*\left\{ {(\delta ,s_l)}\right\}$ such $(r_l,F'\setminus \left\{ {\delta }\right\} ,\eta ')$ splits $\dot{x}$ at $\beta$. Define $q_{l+1}\le _{F',\eta '} q_l$ as follows:

• ${q_{l+1}}{\upharpoonright }{\delta } = {r_l}{\upharpoonright }{\delta }$;

• ${q_{l+1}}{\upharpoonright }{\delta }\Vdash (q_{l+1}*\left\{ {(\delta , s_j)}\right\} )(\delta ) = (q_{l}*\left\{ {(\delta , s_j)}\right\} )(\delta )$ for every $j\ne l$;

• ${q_{l+1}}{\upharpoonright }{\delta }\Vdash (q_{l+1}*\left\{ {(\delta , s_l)}\right\} )(\delta ) = (r_{l})(\delta )$;

• ${(q_{l+1}*\left\{ {(\delta ,s_l)}\right\} )}{\upharpoonright }{(\delta +1)}\Vdash {q_{l+1}}{\upharpoonright }{(\delta ,\alpha )} = {r_l}{\upharpoonright }{(\delta ,\alpha )}$;

• ${(q_{l+1}*\left\{ {(\delta ,s_j)}\right\} )}{\upharpoonright }{(\delta +1)}\Vdash {q_{l+1}}{\upharpoonright }{(\delta ,\alpha )} = {q_l}{\upharpoonright }{(\delta ,\alpha )}$ for every $j \ne l$.

Finally, $q = q_d$ is the desired element. $\square$

Lemma 4.5

Let $\alpha \le \omega _2$ and let $\dot{x}$ be an $\mathbb {S}_\alpha$-name, let $(p,F,\eta )$ be an $\mathbb {S}_\alpha$-fusion condition such that $p\Vdash \dot{x}\in {^{\omega }}{\omega }{}{\setminus }\bigcup _{\beta <\alpha }V^{\mathbb {S}_\beta }$. Let $\beta = \min (F)$ and $N<\omega$. Then there exists $q\le _{F,\eta } p$ such that the $\beta$-extension $(q,F,\eta ')$ of $(p,F,\eta )$ splits $\dot{x}$ at $\beta$ and

• (a) for every $\sigma \in \prod _{\gamma \in F}{^{(\eta (\gamma )+1)}}{2}{}$, $\left| {\dot{x}[q*\sigma ^{(\beta ,0)}]\wedge \dot{x}[q*\sigma ^{(\beta ,1)}]}\right| >N$;

• (b) for every $\sigma _0,\sigma _1\in \prod _{\gamma \in F}{^{(\eta (\gamma )+1)}}{2}{}$ with $\sigma _0(\beta )\ne \sigma _1(\beta )$,

  $$\begin{aligned} \left| {\dot{x}[q*\sigma _0^{(\beta ,0)}]\wedge \dot{x}[q*\sigma _0^{(\beta ,1)}]}\right| \ne \left| {\dot{x}[q*\sigma _1^{(\beta ,0)}]\wedge \dot{x}[q*\sigma _1^{(\beta ,1)}]}\right| . \end{aligned}$$

Proof

We can assume without loss of generality that $(p,F,\eta )$ weakly decides ${\dot{x}}{\upharpoonright }{(N+1)}$. Let $\left\langle {{s_l} : {l<d}}\right\rangle$ enumerate ${^{(\eta (\beta )+1)}}{2}{}$. Put $q_{0}:= p$. Let $l < d$ and assume that $q_{l}\le _{F,\eta }p$ has been constructed such that for every $j_{0}< j_{1}< l$, $\sigma _0,\sigma _1\in \prod _{\gamma \in F}{^{(\eta (\gamma )+1)}}{2}{}$ with $\sigma _{0}(\beta )= s_{j_0}$ and $\sigma _{1}(\beta )= s_{j_1}$

$$\begin{aligned} \left| {\dot{x}[q_l*\sigma _0^{(\beta ,0)}]\wedge \dot{x}[q_l*\sigma _0^{(\beta ,1)}]}\right| \ne \left| {\dot{x}[q_l*\sigma _1^{(\beta ,0)}]\wedge \dot{x}[q_l*\sigma _1^{(\beta ,1)}]}\right| . \end{aligned}$$

By the previous lemma, there exists $r_l\le _{F{\setminus }\left\{ {\beta }\right\} ,\eta } q_l*\left\{ {(\beta ,s_l)}\right\}$ such that, for every $\sigma \in \prod _{\gamma \in F{\setminus }\left\{ {\beta }\right\} }{^{(\eta (\gamma )+1)}}{2}{}$, $\left| {\dot{x}[r_l*\sigma ^{(\beta ,0)}]\wedge \dot{x}[r_l*\sigma ^{(\beta ,1)}]}\right|$ is longer than any $\left| {\dot{x}[q_{l}*\hat{\sigma }^{(\beta ,0)}]\wedge \dot{x}[q_{l}*\hat{\sigma }^{(\beta ,1)}]}\right|$ for $\hat{\sigma }\in \prod _{\gamma \in F}{^{(\eta (\gamma )+1)}}{2}{}$ with $\hat{\sigma }(\beta ) = s_j$ for some $j<l$. Define $q_{l+1}\le _{F,\eta }q_l$ as follows:

• ${q_{l+1}}{\upharpoonright }{\beta } = {r_l}{\upharpoonright }{\beta }$,

• ${q_{l+1}}{\upharpoonright }{\beta }\Vdash (q_{l+1}*\left\{ {(\beta , s_j)}\right\} )(\beta ) = (q_{l}*\left\{ {(\beta , s_j)}\right\} )(\beta )$ for every $j\ne l$;

• ${q_{l+1}}{\upharpoonright }{\beta }\Vdash (q_{l+1}*\left\{ {(\beta , s_l)}\right\} )(\beta ) = (r_{l})(\beta )$;

• ${(q_{l+1}*\left\{ {(\beta ,s_j)}\right\} )}{\upharpoonright }{(\beta +1)}\Vdash {q_{l+1}}{\upharpoonright }{(\beta ,\alpha )} = {r_l}{\upharpoonright }{(\beta ,\alpha )}$ for every $j\le l$;

• ${(q_{l+1}*\left\{ {(\beta ,s_j)}\right\} )}{\upharpoonright }{(\beta +1)}\Vdash {q_{l+1}}{\upharpoonright }{(\beta ,\alpha )} = {q_l}{\upharpoonright }{(\beta ,\alpha )}$ for every $j > l$.

Finally $q = q_d$ is the desired element. $\square$

Lemma 4.6

Let $\alpha \le \omega _2$ and let $\dot{x}$ be an $\mathbb {S}_\alpha$-name and $(p,F,\eta )$ and an $\mathbb {S}_\alpha$-fusion condition such that $p\Vdash \dot{x}\in {^{\omega }}{\omega }{}{\setminus }\bigcup _{\beta <\alpha }V^{\mathbb {S}_\beta }$. Let $\beta < \alpha$ and $N<\omega$. Then there exists $q\le _{F,\eta }p$ such that the $\beta$-extension $(q,F',\eta ')$ of $(p,F,\eta )$ splits $\dot{x}$ at $\beta$ and for every $\sigma _0,\sigma _1\in \prod _{\gamma \in F}{^{(\eta (\gamma )+1)}}{2}{}$ with

• (a) for every $\sigma \in \prod _{\gamma \in F}{^{(\eta (\gamma )+1)}}{2}{}$, $\left| {\dot{x}[q*\sigma ^{(\beta ,0)}]\wedge \dot{x}[q*\sigma ^{(\beta ,1)}]}\right| >N$;

• (b) for every $\sigma _0,\sigma _1\in \prod _{\gamma \in F}{^{(\eta (\gamma )+1)}}{2}{}$ with ${\sigma _0}{\upharpoonright }{(\beta +1)}\ne {\sigma _1}{\upharpoonright }{(\beta +1)}$,

  $$\begin{aligned} \left| {\dot{x}[q*\sigma _0^{(\beta ,0)}]\wedge \dot{x}[q*\sigma _0^{(\beta ,1)}]}\right| \ne \left| {\dot{x}[q*\sigma _1^{(\beta ,0)}]\wedge \dot{x}[q*\sigma _1^{(\beta ,1)}]}\right| . \end{aligned}$$

Proof

We can assume without loss of generality, that $(p,F,\eta )$ weakly decides ${\dot{x}}{\upharpoonright }{(N+1)}$. Let $\left\langle {\delta _{0},\dots ,\delta _n}\right\rangle$ increasingly enumerate $\left\{ {\delta \in F\cup \left\{ {\beta }\right\} : \delta \le \beta }\right\}$. We prove the claim by induction on n. For $n = 0$ this one of the previous lemmas. So, assume $n > 0$. Let $\left\langle {{s_j} : {j<d}}\right\rangle$ enumerate . Put $q_{0}:=p$. Let $l < d$ and assume that $q_l\le _{F,\eta }p$ has been constructed such that for every $\sigma _0,\sigma _1\in \prod _{\gamma \in F}{^{(\eta (\gamma )+1)}}{2}{}$ such that $\sigma _0(\delta _{0})= s_{j_0}$ and $\sigma _1(\delta _0)=s_{j_1}$ for some $j_0,j_{1}<l$ and ${\sigma _0}{\upharpoonright }{(\beta +1)}\ne {\sigma _1}{\upharpoonright }{(\beta +1)}$, then

$$\begin{aligned} \left| {\dot{x}[q*\sigma _0^{(\beta ,0)}]\wedge \dot{x}[q*\sigma _0^{(\beta ,1)}]}\right| \ne \left| {\dot{x}[q*\sigma _1^{(\beta ,0)}]\wedge \dot{x}[q*\sigma _1^{(\beta ,1)}]}\right| . \end{aligned}$$

By the induction hypothesis, there exists $r_l\le _{F\setminus \left\{ {\delta _0}\right\} ,\eta }q_l*\left\{ {(\delta _0,s_l)}\right\}$ such that for every $\sigma \in \prod _{\gamma \in F{\setminus }\left\{ {\delta _0}\right\} }{^{(\eta (\gamma )+1)}}{2}{}$, $\left| {\dot{x}[r_l*\sigma ^{(\beta ,0)}]\wedge \dot{x}[r_l*\sigma ^{(\beta ,1)}]}\right|$ is longer than any $\left| {\dot{x}[q_{l}*\hat{\sigma }^{(\beta ,0)}]\wedge \dot{x}[q_{l}*\hat{\sigma }^{(\beta ,1)}]}\right|$ for $\hat{\sigma }\in \prod _{\gamma \in F}{^{(\eta (\gamma )+1)}}{2}{}$ with $\hat{\sigma }(\delta _0) = s_j$ for some $j<l$. Define $q_{l+1}\le _{F,\eta }q_l$ as follows:

• ${q_{l+1}}{\upharpoonright }{\delta _0} = {r_l}{\upharpoonright }{\delta _0}$,

• ${q_{l+1}}{\upharpoonright }{\delta _0}\Vdash (q_{l+1}*\left\{ {(\delta _0, s_j)}\right\} )(\delta _0) = (q_{l}*\left\{ {(\delta _0, s_j)}\right\} )(\delta _0)$ for every $j\ne l$;

• ${q_{l+1}}{\upharpoonright }{\delta _0}\Vdash (q_{l+1}*\left\{ {(\delta _0, s_l)}\right\} )(\delta _0) = (r_{l})(\delta _0)$;

• ${(q_{l+1}*\left\{ {(\delta _0,s_l)}\right\} )}{\upharpoonright }{(\delta _0 +1)}\Vdash {q_{l+1}}{\upharpoonright }{(\delta _0,\alpha )} = {r_l}{\upharpoonright }{(\delta _0,\alpha )}$;

• ${(q_{l+1}*\left\{ {(\beta ,s_j)}\right\} )}{\upharpoonright }{(\beta +1)}\Vdash {q_{l+1}}{\upharpoonright }{(\beta ,\alpha )} = {q_l}{\upharpoonright }{(\beta ,\alpha )}$ for every $j \ne l$.

Finally $q = q_d$ is the desired element. $\square$

Proof of Proposition 4.1

Using the previous lemmas one may construct a fusion sequence $\left\langle {{(p_n,F_n,\eta _n)} : {n<\omega }}\right\rangle$ below $p'$ with active coordinates $\left\langle {{\beta _n} : {n<\omega }}\right\rangle$ and $\left\langle {{\sigma _s} : {s\in {^{<\omega }}{2}{}}}\right\rangle$ as in (2) with the following properties:

($\alpha$):

$(p_n,F_n,\eta _n)$ splits $\dot{x}$ at $\beta _n$;

($\beta$):

For every $s\in {^{n}}{2}{}$, $s'\in {^{n+1}}{2}{}$,

$$\begin{aligned} \left| {\dot{x}[p_n*\sigma _{s^\frown 0}]\wedge \dot{x}[p_n*\sigma _{s^\frown 1}]}\right| < \left| {\dot{x}[p_{n+1}*\sigma _{s'^\frown 0}]\wedge \dot{x}[p_{n+1}*\sigma _{s'^\frown 1}]}\right| ; \end{aligned}$$

($\gamma$):

For every $s,s'\in {^{<\omega }}{2}{}$ with $\left| {s}\right| = \left| {s'}\right| =: n$ and ${\sigma _s}{\upharpoonright }{(\beta _{n}+1)}\ne {\sigma _{s'}}{\upharpoonright }{(\beta _{n}+1)}$,

$$\begin{aligned} \left| {\dot{x}[p_{n}*\sigma _{s^\frown 0}]\wedge \dot{x}[p_{n}*\sigma _{s^\frown 1}]}\right| \ne \left| {\dot{x}[p_{n}*\sigma _{s'^\frown 0}]\wedge \dot{x}[p_{n}*\sigma _{s'^\frown 1}]}\right| . \end{aligned}$$

Let $p_\omega$ be the fusion limit and $f = \left\langle {{\beta _n} : {n<\omega }}\right\rangle$. Then $(\dot{x},p_\omega ,f)$ is an $\mathbb {S}_\alpha$-coding triple. Furthermore,

$$\begin{aligned} t_{s}= \dot{x}[p_\omega *\sigma _{s^\frown 0}]\wedge \dot{x}[p_\omega *\sigma _{s^\frown 1}] =\dot{x}[p_{\left| {s}\right| }*\sigma _{s^\frown 0}]\wedge \dot{x}[p_{\left| {s}\right| }*\sigma _{s^\frown 1}], \end{aligned}$$

and hence, Properties (a) and (b) of the assertion follow from Properties $(\beta )$ and ($\gamma$) of the fusion. $\square$

The following lemma is pivotal in the construction of our coding systems below. It has the same role in this proof as Lemma 3.7 in the proof of the previous section.

Lemma 4.7

Let $\alpha \le \omega _2$ and let $(\dot{x}, p, f)$ be an $\mathbb {S}_\alpha$-coding triple such that the associated splitting family $\bar{T}(\dot{x},p, f) = \left\langle {{t_s} : {s\in {^{<\omega }}{2}{}}}\right\rangle$ satisfies the assertions of Proposition 4.1. Let $s, s_0,s_1\in {^{<\omega }}{2}{}$ such that $f(\left| {s}\right| ) = f(\left| {s_0}\right| ) = f(\left| {s_1}\right| )$, $\left| {t_{s_0}}\right| = \left| {t_{s_1}}\right|$, and $t_{s_0}, t_{s_1}$ extend $t_s$. Then $s_0(\left| {s}\right| ) = s_1(\left| {s}\right| )$.

Proof

Since $t_{s_0}$ and $t_{s_1}$ both extend $t_s$, $s_0$ and $s_1$ both extend s. Since $\left| {t_{s_0}}\right| = \left| {t_{s_1}}\right|$, Proposition 4.1(1) implies $\left| {s_0}\right| = \left| {s_1}\right|$. Then, Proposition 4.1(2) implies $\left\{ {n < \left| {s_0}\right| : s_0(n)\ne s_1(n)\;\wedge \; f(n)\le f(\left| {s_0}\right| )}\right\} = \emptyset$. Since $f(\left| {s}\right| ) = f(\left| {s_0}\right| )$, $s_0(\left| {s}\right| ) = s_1(\left| {s}\right| )$. $\square$

Before we come to the definition of the coding systems, we need one final lemma. Its proof is the same as for the analogous lemma for Silver forcing.

Lemma 4.8

(cf. [5, Lemma 5.1]) Let $\dot{b}$ be a $\mathbb {S}$-name and $p\in \mathbb {S}$ with $p\Vdash \dot{b}\in [\omega ]^\omega \setminus V$. Then there exist $\mathbb {S}$-names $\dot{c}, \dot{y}$ and $q\le p$ such that

$$\begin{aligned}q\Vdash \dot{c}\in [\dot{b}]^\omega \wedge \dot{y}\in {^{\dot{c}}}{2}{}\setminus \left\{ {{x}{\upharpoonright }{\dot{c}}:x\in {^{\omega }}{2}{}\cap V}\right\} .\end{aligned}$$

Definition 4.9

Let $(\dot{x}, p, f)$ and $(\dot{y}, q, g)$ be two coding triples. We say that $(\dot{x}, p, f)$ and $(\dot{y}, q, g)$ are isomorphic, if there is an order isomorphism $\pi : \textrm{supp}(p)\rightarrow \textrm{supp}(q)$ that satisfies $g = \pi \circ f$ and $\bar{T}(\dot{x},p, f) = \bar{T}(\dot{y}, q, g)$.

It is easy to see that there are only $\aleph _1$ isomorphism types of coding triples.

Definition 4.10

Let $\alpha \le \omega _2$ and let $(\dot{x}, p, f)$ be an $\mathbb {S}_\alpha$-coding triple with associated splitting family $\bar{T}(\dot{x}, p, f) = \left\langle {{t_s} : {s\in {^{<\omega }}{2}{}}}\right\rangle$. For every $\beta \in \textrm{supp}(p)$ put

$$\begin{aligned}L(\beta ):= \left\{ {\left| {t_s}\right| :f(\left| {s}\right| ) = \beta }\right\} .\end{aligned}$$

Let $\Omega (\dot{x}, p,f)$ be the set of all $b\in [\omega ]^\omega$ that satisfy one of the following assertions:

• (a) There exists some $\beta \in \textrm{supp}(p)$ such that $b\subseteq L(\beta )$;

• (b) $b\subseteq \bigcup \left\{ {L(\beta ): \beta \in \textrm{supp}(p)}\right\}$ and $\left| {b\cap L(\beta )}\right| \le 1$ for every $\beta \in \textrm{supp}(p)$;

• (c) $b\cap \bigcup \left\{ {L(\beta ):\beta \in \textrm{supp}(p)}\right\} = \emptyset$.

Clearly, $\Omega (\dot{x},p,f)$ is a dense subset of $[\omega ]^\omega$. We come to the definition of the coding systems. These will be extended at successor stages of the iteration. Let $\gamma < \omega _2$ we work in $V^{\mathbb {S}_{\gamma +1}}$. Let

$$\begin{aligned}\Omega _\gamma (\dot{x}, p,f) = \left\{ {b\in \Omega (\dot{x}, p, f): \lnot \exists b_0\in \Omega (\dot{x}, p, f)\cap V^{\mathbb {S}_\gamma }(b_0\subseteq b)}\right\} .\end{aligned}$$

For every $b\in \Omega _\gamma (\dot{x}, p, f)$ we define $u^b\in \mathbb {U}$ depending on b:

• (0) For all cases:

  • (a) $\emptyset \in u^b$;

  • (b) if $\sigma \in u^b$ and $\left| {\sigma }\right| \notin b$, then $\sigma ^\frown 0,\sigma ^\frown 1\in u^b$;

  • (c) if $\sigma \in u^b{\setminus }\textrm{split}(T)$ and $\left| {\sigma }\right| \in b$, then let $\sigma ^\frown j\in u^b$ and $\sigma ^\frown (1-j)\notin u^b$, where $j<2$ is minimal such that $\sigma ^\frown j\notin T$.

• (1) $b\subseteq L(\beta )$ for some $\beta \in \textrm{supp}(p)$. Split b into three infinite sets $b_0,b_1,b_2$. Use Lemma 4.8 to obtain a $y\in {^{b_2}}{2}{}$ such that for every $x\in {^{\omega }}{2}{}\cap V^{\mathbb {S}_\gamma }$, ${x}{\upharpoonright }{b_2}\ne y$. Define $u^b$ recursively.

  • (a) If $\sigma \in u^b$, $\left| {\sigma }\right| \in b_0$, $\sigma \in \textrm{split}(T)$, let n be the minimal element of $b_0\setminus (\left| {\sigma }\right| +1)$. If $S:=\textrm{split}(T)\cap T(n)\cap T_\sigma \ne \emptyset$, there exists $j<2$ such that every element of S extends $\sigma ^\frown j$ (see Lemma 4.7). Let $\sigma ^\frown (1-j)\in u^b$ and $\sigma ^\frown j\notin u^b$. If S is empty, let $\sigma ^\frown 0\in u^b$ and $\sigma ^\frown 1\notin u^b$.

  • (b) If $\sigma \in u^b$, $\left| {\sigma }\right| \in b_1$, and $\sigma \in \textrm{split}(T)$, then $\sigma = t_s$ for some $s\in {^{<\omega }}{2}{}$ with $f(\left| {s}\right| ) = \beta$. Let $k < \omega$ such that $\left| {s}\right|$ is the k-th element of $f^{-1}[\left\{ {\beta }\right\} ]$. Let $\sigma ^\frown (1-t_{s^\frown g_\gamma (k)}(\left| {\sigma }\right| ))\in u^b$ and $\sigma ^\frown (t_{s^\frown g_\gamma (k)}(\left| {\sigma }\right| ))\notin u^b$.

  • (c) For $\sigma \in u^b$ with $\left| {\sigma }\right| \in b_2$ and $\sigma \in \textrm{split}(T)$, i.e. $\sigma = t_s$ for some s, let $\sigma ^\frown (t_{s^\frown y(\left| {\sigma }\right| )}(\left| {\sigma }\right| ))\in u^b$ and $\sigma ^\frown (1- t_{s^\frown y(\left| {\sigma }\right| )}(\left| {\sigma }\right| ))\notin u^b$.

• (2) $b\subseteq \bigcup \left\{ {L(\beta ): \beta \in \textrm{supp}(p)}\right\}$ and $\left| {b\cap L(\beta )}\right| \le 1$ for every $\beta \in \textrm{supp}(p)$. For $b'\subseteq b$ put

  $$\begin{aligned}Y(b'):=\left\{ {\beta \in \textrm{supp}(p): L(\beta )\cap b'\ne \emptyset }\right\} .\end{aligned}$$

  Let $\delta$ be the minimal ordinal such that $Y(b)\setminus \delta$ is finite. Let $b_0,b_1$ be two disjoint infinite subsets of b such that $\sup (Y(b_0)) = \sup (Y(b_1)) = \delta$ and $\delta \notin Y(b_0),Y(b_1)$. Use Lemma 4.8 to obtain a $y\in {^{b_1}}{2}{}$ such that for every $x\in {^{\omega }}{2}{}\cap V^{\mathbb {S}_\gamma }$, ${x}{\upharpoonright }{b_1}\ne y$. Define $u^b$ recursively:

  • (a) if $\sigma \in u^b$, $\left| {\sigma }\right| \in b_0$, and $\sigma \in \textrm{split}(T)$, let $\sigma ^\frown 0\in u^b$ and $\sigma ^\frown 1\notin u^b$;

  • (b) For $\sigma \in u^b$ with $\left| {\sigma }\right| \in b_1$ and $\sigma \in \textrm{split}(T)$, i.e. $\sigma = t_s$ for some s, let $\sigma ^\frown (t_{s^\frown y(\left| {\sigma }\right| )}(\left| {\sigma }\right| ))\in u^b$ and $\sigma ^\frown (1- t_{s^\frown y(\left| {\sigma }\right| )}(\left| {\sigma }\right| ))\notin u^b$.

• (3) If $b\cap \bigcup \left\{ {L(\beta ): \beta \in \textrm{supp}(p)}\right\} = \emptyset$, $u^b$ has been completely defined by Case (0).

Proposition 4.11

Let $\alpha ,\alpha '\le \omega _2$ and let $(\dot{x}, p, f)$ be an $\mathbb {S}_\alpha$-coding triple. Let $\gamma < \omega _2$ and $b\in \Omega _\gamma (\dot{x},p, f)$. Let $u^b$ be as in Definition 4.10. Then for every $\mathbb {S}_{\alpha '}$-coding triple $(\dot{y}, q, g)\cong (\dot{x}, p, f)$, $q\Vdash _{\omega _2} \dot{y}\notin [u^b]$.

Proof

Let $\pi : \textrm{supp}(p) \rightarrow \textrm{supp}(q)$ be the order isomorphism. Let $q'\le q$.

Case 1. $b\subseteq L(\beta )$ for some $\beta \in \textrm{supp}(p)$. Let $b = b_0{\dot{\cup }}b_1{\dot{\cup }}b_2$ as in Definition 4.10. Now, distinguish three cases:

Case 1.1. $\pi (\beta ) > \gamma$. Without loss of generality, $q'$ decides ${\dot{y}}{\upharpoonright }{l}$ as $\sigma$ for some $l\in b_0$. If $\sigma \notin \textrm{split}(T)$, there exists $j<2$ such that $q'\Vdash \dot{y}(l) = j$ and $\sigma ^\frown j\notin u^b$, by Property (0)(c). So, assume that $\sigma$ is a split node of T. Let n be the minimal element of $b_0\setminus (\left| {\sigma }\right| +1)$. By Property (1)(a), there are no split nodes of T in $T_\sigma (n)\cap u^b$. By Property (0)(c), $q'\Vdash \dot{y}\notin [u^b]$.

Case 1.2. $\pi (\beta ) = \gamma$. We can assume without loss of generality that $q'$ decides ${\dot{y}}{\upharpoonright }{l}$ as $\sigma$ for some $l\in b_1$. Again, one may assume that $\sigma \in \textrm{split}(T)$. Then $\sigma = t_s$ for some $s\in {^{<\omega }}{2}{}$. Let $k < \omega$ such that $\left| {s}\right|$ is the k-th element of $f^{-1}[\left\{ {\beta }\right\} ]$. Then $\left| {s}\right|$ is the k-th element of $g^{-1}[\left\{ {\gamma }\right\} ]$. In particular,

$$\begin{aligned}q'\Vdash {\dot{y}}{\upharpoonright }{(l+1)} = \sigma ^\frown (t_{s^\frown g_\gamma (k)}(\left| {\sigma }\right| )).\end{aligned}$$

By Property (1)(b), $\sigma ^\frown (t_{s^\frown g_\gamma (k)}(\left| {\sigma }\right| ))\notin u^b$, and hence, $q'\Vdash \dot{y}\notin [u^b]$.

Case 1.3. $\pi (\beta ) < \gamma$. Let $G_\gamma \subseteq \mathbb {S}_\gamma$ be a V-generic filter with ${q'}{\upharpoonright }{\gamma }\in G_\gamma$. In $V[G_\gamma ]$, define $F: L(\beta )\rightarrow 2$ as follows: For $l\in L(\beta )$, there exists $s\in {^{<\omega }}{2}{}$ with $f(\left| {s}\right| ) = \beta$ and $\left| {t_s}\right| = l$. There exists a unique $k < \omega$ such that $\left| {s}\right|$ is the k-th element of $f^{-1}[\left\{ {\beta }\right\} ]$. By Proposition 4.1(1), k only depends on l and not on s. Define $F(l):= g_{\pi (\beta )}(k)$. Note that

$$\begin{aligned}(q'\Vdash {\dot{y}}{\upharpoonright }{l} = t_s)\implies (q'\Vdash {\dot{y}}{\upharpoonright }{(l+1)} \subseteq t_{s^\frown g_{\pi (\beta )}(k)}).\end{aligned}$$

By choice of y, $y\ne {F}{\upharpoonright }{b_2}$. Hence, there exists $l\in b_2$ such that $F(l)\ne y(l)$. We can assume that $q'$ decides ${\dot{y}}{\upharpoonright }{l}$ as $\sigma \in {^{l}}{2}{}$. The interesting case is, when $\sigma \in \textrm{split}(T)$, i.e. $\sigma = t_s$ for some s. Let k be as in the definition of F, then, by Property (1)(c),

$$\begin{aligned}q'\Vdash {\dot{y}}{\upharpoonright }{(l+1)}\notin u^b.\end{aligned}$$

Case 2. Assume that $b\subseteq \bigcup \left\{ {L(\beta ):\beta \in \textrm{supp}(p)}\right\}$ and $\left| {L(\beta )\cap b}\right| \le 1$ for every $\beta \in \textrm{supp}(b)$. Let $b_0,b_1$ be the partition of b as in Definition 4.10(2).

Case 2.1. $\sup (\pi [Y(b_0)]) > \gamma$. Let $\beta \in Y(b_0)$ such that $\pi (\beta ) > \gamma$. Using Lemma 2.3, find $\tilde{s}\in {^{<\omega }}{2}{}$ such that $f(\left| {\tilde{s}}\right| ) = \beta$ and $q'$ is compatible with $q*\sigma _{\tilde{s}^\frown 0}^g$ and $q*\sigma _{\tilde{s}^\frown 1}^g$. Since $\left\{ {\delta \in Y(b_0): \pi (\delta ) > \gamma }\right\}$ is infinite, there exists $s,s_0,s_1\in {^{<\omega }}{2}{}$ such that $\tilde{s}^\frown j\subseteq s_j$, $\left| {s}\right| = \left| {s_0}\right| = \left| {s_1}\right|$, $\left| {t_s}\right| \in b_0\cap L(f(\left| {s}\right| ))$, and $\beta \le f(\left| {s}\right| )$. By construction,

$$\begin{aligned}\left\{ {n < \left| {s}\right| : s_0(n)\ne s_1(n)\wedge \pi (f(n)) \in [\gamma , \pi (f(\left| {s}\right| )))}\right\} \ne \emptyset .\end{aligned}$$

By Proposition 4.1(b), there is at most one $j < 2$ such that $\left| {t_{s_j}}\right| = l$. Hence, one may choose $j < 2$ and $q''\le q$ such that $\left| {t_{s_j}}\right| \ne l$. We can assume that $q''$ decides ${\dot{y}}{\upharpoonright }{l}$, say as $\sigma$. By Proposition 4.1(a), $\sigma \notin \textrm{split}(T)$. By Property (0)(c), $q''\Vdash \dot{y}\notin [u^b]$.

Case 2.2. Assume that $\sup (\pi [Y(b_0)])\le \gamma$. The restricted tree $T_\gamma$ in the model $V[G_\gamma ]$ already decides $\dot{y}$ on coordinates associated with $\beta \in \textrm{supp}(q)\cap \gamma$. A similar construction as in Case 1.3. yields the claim.

Case 3. If $b\cap \bigcup \left\{ {L(\beta ):\beta \in \textrm{supp}(p)}\right\} = \emptyset$, no split node of T has length in b. Property (0) yields $q'\Vdash \dot{y}\notin [u^b]$.

This concludes the proof that $\left\{ {q'\le q: q'\Vdash \dot{y}\notin [u^b]}\right\}$ is dense below q. It follows that $q\Vdash \dot{y}\notin [u^b]$. $\square$

Theorem 4.12

$V^{\mathbb {S}_{\omega _2}}\models \textrm{cov}({\mathfrak {C}_{2}}) = \aleph _1$.

Since $\textrm{cov}({u^{0}})\le \textrm{cov}({\mathfrak {C}_{2}})$ is provable in $\textrm{ZFC}$, one obtains the following corollary:

Corollary 4.13

$V^{\mathbb {S}_{\omega _2}}\models \textrm{cov}({u^{0}}) = \aleph _1$.