Proof. Let $\mathbb {P}= \bigcup _{i\in \omega } \mathcal {P}_i$ , where the $\mathcal {P}_i$ are centered. Fix $i \in \omega $ . Note that for each $n \in \omega $ , there is $y \in \mathcal {P}(\{0, ..., n-1\})$ such that no $p\in \mathcal {P}_i$ forces

$$\begin{align*}\dot{x} \cap n \neq y. \end{align*}$$

Otherwise, for every $y \in \mathcal {P}(\{0, ..., n-1\})$ , there is a condition $p_y\in \mathcal {P}_i$ such that $p_y \Vdash \dot {x} \cap n \neq y$ . Since $\mathcal {P}_i$ is centered, there exists $p\in \mathbb {P}$ such that $p \leq p_y$ for all $y \in \mathcal {P}(\{0, ..., n-1\})$ , then $p \Vdash \dot {x} \cap n \neq y$ for all $y \in \mathcal {P}(\{0, ..., n-1\})$ , which is a contradiction. By König’s lemma, there is $X_i \subset \omega $ such that for all $n \in \omega $ no $p\in \mathcal {P}_i$ forces

(1) $$ \begin{align} \dot{x} \cap n \neq X_i \cap n. \end{align} $$

Let A be an infinite subset of $\omega $ , and suppose that for all $i\in \omega $ , there are infinitely many $n \in \omega $ such that

$$\begin{align*}\frac{d_n(X_i \cap A)}{d_n(A)} \not\in (1/2 - \epsilon, \epsilon + 1/2). \end{align*}$$

Fix $m \in \omega $ and a condition $p \in \mathbb {P}$ . There is $\mathcal {P}_i$ such that $p \in \mathcal {P}_i$ . Also, there is $n>m$ such that

$$\begin{align*}\frac{d_n(X_i \cap A)}{d_n(A)} \not\in (1/2 - \epsilon, \epsilon + 1/2). \end{align*}$$

By (1), there is $q\leq p$ such that

$$\begin{align*}q \Vdash \dot{x} \cap n = X_i \cap n. \end{align*}$$

Therefore,

$$\begin{align*}q \Vdash \frac{d_n(\dot{x} \cap A)}{d_n(A)} = \frac{d_n(X_i \cap A)}{d_n(A)} \not\in (1/2 - \epsilon, \epsilon + 1/2). \end{align*}$$

By the previous paragraph, the set $D_m=\{ q\in \mathbb {P} \, | \, $ . There is $n>m$ such that $ q \Vdash \frac {d_n(\dot {x} \cap A)}{d_n(A)} \not \in (1/2 - \epsilon , \epsilon + 1/2) \}$ is dense. Then

$$\begin{align*}\Vdash \exists^\infty n \in \omega \left(\frac{d_n(\dot{x} \cap A)}{d_n(A)} \not\in (1/2 - \epsilon, \epsilon + 1/2) \right).\\[-41pt] \end{align*}$$

Proof. No new reals arise in limit stages of uncountable cofinality so that the lemma vacuously holds if cf $(\delta )>\omega $ . Hence assume cf $(\delta )=\omega $ . To simplify notation, suppose $\delta =\omega $ .

Let $\dot {x}$ be a $\mathbb {P}_\omega $ -name for an element of $[\omega ]^\omega $ . Fix $m \in \omega $ . Work in $V^{\mathbb {P}_m}$ for the moment. There is a decreasing sequence of conditions $p_k=p_{m,k}$ in the remainder forcing $\mathbb {P}_\omega \setminus \mathbb {P}_m$ such that $p_k$ decides the k-th element of $\dot {x}$ . Say

$$\begin{align*}p_k \Vdash "l_k \text{ is the } k\text{-th element of }\dot{x}\text{"}. \end{align*}$$

Let $X_m=\{l_k \,| \, k\in \omega \}$ . Work in the ground model V. We have a $\mathbb {P}_m$ -name $\dot {x}_m$ for $X_m$ . Since $\mathbb {P}_m$ has property $(\star )$ , we can find $X_{m,i}$ , $i\in \omega $ which satisfy what property $(\star )$ said for $\dot {x}_m$ . Now, fix $A\in [\omega ]^\omega $ . We claim that if for all $i,m\in \omega $ , there are infinitely many $n \in \omega $ such that

$$\begin{align*}\frac{d_n(X_{m,i} \cap A)}{d_n(A)} \not\in (1/2 - \epsilon, \epsilon + 1/2) \end{align*}$$

then

$$\begin{align*}\Vdash \exists^\infty n \in \omega \left(\frac{d_n(\dot{x} \cap A)}{d_n(A)} \not\in (1/2 - \epsilon, \epsilon + 1/2) \right). \end{align*}$$

To see this, let $p \in \mathbb {P}_\omega $ and $n\in \omega $ . Fix m such that $p \in \mathbb {P}_m$ and work in $V^{\mathbb {P}_m}$ . We know that there is $j> n$ such that

$$\begin{align*}p_{m,j} \Vdash \frac{d_{j}(\dot{x} \cap A)}{d_{j}(A)} = \frac{d_{j}(X_m \cap A)}{d_{j}(A)} \not\in (1/2 - \epsilon, \epsilon + 1/2). \end{align*}$$

Now work in the ground model V. By strengthening $p\in \mathbb {P}_n$ , if necessary, we may assume p decides $p_{m,j}$ . Then

$$\begin{align*}q= p^\frown p_{m,j} \Vdash \frac{d_{j}(\dot{x} \cap A)}{d_{j}(A)} \not\in (1/2 - \epsilon, \epsilon + 1/2).\\[-41pt] \end{align*}$$

Proof. Fix $\kappa> \aleph _1$ . Let $\mathbb {C}_\kappa $ be an fsi of Cohen forcing of length $\kappa $ . Then $V^{\mathbb {C}_\kappa } \models $ cov $(\mathcal {M}) = \mathfrak {s}_{1/2\pm \epsilon } = \kappa $ . Denote $M= V^{\mathbb {C}_\kappa }$ . Now, let $\mathbb {D}_{\aleph _1}$ be an fsi of Hechler forcing of length $\aleph _1$ over M. Then $M^{\mathbb {D}_{\aleph _1}} \models \mathfrak {d} = \aleph _1$ because $\aleph _1$ dominating reals are added. By Lemmas 2.3 and 2.2, $\mathbb {D}_{\aleph _1}$ has property $(\star )$ , so $M^{\mathbb {D}_{\aleph _1}} \models \mathfrak {s}_{1/2\pm \epsilon }=\kappa $ .

Proof. Fix $\kappa> \aleph _1$ and assume CH in the ground model. Let $\mathbb {D}_{\kappa }$ be an fsi of Hechler forcing of length $\kappa $ . It is known that $V^{\mathbb {D}_\kappa } \models \mathfrak {b}=\kappa $ . Besides, the set $V \cap [\omega ]^\omega $ remains an $\epsilon $ -almost reaping family in $V^{\mathbb {D}_\kappa }$ by Corollary 2.4. Thus $V^{\mathbb {D}_\kappa } \models \mathfrak {r}_{1/2\pm \epsilon }= \aleph _1$ .

Proof. Denote as $\mathcal {I}^\star $ the set of finite Boolean combinations of $\mathcal {I}$ . Consider a filter $\mathcal {F}$ over $\omega $ such that

$$\begin{align*}Y \in \mathcal{F} \rightarrow \text{ for all } I\in \mathcal{I}^\star, \text{ we have that } Y \cap I \text{ is infinite } \end{align*}$$

and $\mathcal {F}$ is maximal with this property. Let $\dot {m}$ be an $\mathbb {M}_{\mathcal {F}}$ -name of the generic real. Consider $I \in \mathcal {I}^\star $ . For each $n \in \omega $ , we define

$$\begin{align*}B_n &:= \{ (s,A) \, | \, \text{ there is }m>n \text{ such that }m \in s \cap I \} \\C_n&:= \{ (s,A) \, | \, \text{ there is }m>n \text{ such that }m \in I \text{ but }m \not\in s \cup A \}. \end{align*}$$

It is easy to check that $B_n$ and $C_n$ are dense sets. Therefore, $\Vdash _{\mathbb {M}_{\mathcal {F}}} \text {"} I \cap \dot {m} \text { is infinite"}$ and $\Vdash _{\mathbb {M}_{\mathcal {F}}} \text {"} I \cap \omega \setminus \dot {m} \text { is infinite"}$ . Now, consider $x \subseteq \omega $ such that $x \not \in \mathcal {I}$ . If $\{x\}\cup \mathcal {I}$ is not an independent family, then there is nothing to do. Thus, suppose that $\{x\}\cup \mathcal {I}$ is an independent family. If $x \in \mathcal {F}$ , then $\Vdash _{\mathbb {M}_{\mathcal {F}}} \dot {m} \subseteq ^{\star } x$ . If $x \not \in \mathcal {F}$ , then there is $y \in \mathcal {F}$ and $I \in \mathcal {I}^\star $ such that $y \cap x \cap I$ is finite. Since $\Vdash _{\mathbb {M}_{\mathcal {F}}} \dot {m} \subseteq ^{\star } y$ , then $\Vdash _{\mathbb {M}_{\mathcal {F}}} \text {"}\dot {m} \cap x \cap I \text { is finite"}$ .

Proof. Fix $\kappa> \aleph _1$ . Let $\mathbb {C}_\kappa $ be an fsi of Cohen forcing of length $\kappa $ . Then $V^{\mathbb {C}_\kappa } \models $ cov $(\mathcal {M}) = \mathfrak {s}_{1/2\pm \epsilon } = \kappa $ . Denote $M= V^{\mathbb {C}_\kappa }$ . Let $\mathcal {I}_0 \in M$ be a countable independent family. Build in M an iterated forcing construction $\mathbb {P}=(\mathbb {P}_\alpha , \dot {\mathbb {Q}}_\alpha : \alpha < \omega _1)$ such that $\Vdash _\alpha \text {"}\mathcal {I}_0 \cup \{\dot {m}_\beta : \beta < \alpha \} \text { is an independent family"}$ , where $\{\dot {m}_\beta : \beta < \alpha \} $ are the $\mathbb {P}_\alpha $ -names of generic reals added by $\mathbb {P}_\alpha $ , and $\Vdash _\alpha \text {"}\dot {\mathbb {Q}}_\alpha $ is a Mathias forcing with $\dot {\mathcal {F}}_\alpha $ ”, where $\dot {\mathcal {F}}_\alpha $ is a $\mathbb {P}_\alpha $ -name of the filter given by Proposition 2.9 with the family $\mathcal {I}_0 \cup \{\dot {m}_\beta : \beta < \alpha \}$ . Work in $M^{\mathbb {P}}$ . Denote as $\mathcal {I}$ the family $\mathcal {I}_0 \cup \{m_\beta : \beta < \omega _1\}$ . By Proposition 2.9, $\mathcal {I}$ is a maximal independent family. Then, $M^{\mathbb {P}} \models \mathfrak {i}=\aleph _1$ . Note that $\mathbb {P}$ has property ( $\star $ ) because of Lemmas 2.2 and 2.3. Therefore, $M^{\mathbb {P}} \models \mathfrak {s}_{1/2\pm \epsilon } = \kappa $ .

Proof. Let $\{x_i\}_{i\in \omega }$ be an increasing enumeration of X. For each $m\in \omega $ consider the set $ B_{2m} := \{ B \subseteq \{x_1, ..., x_{2m} \} \, | \, |B|=m \}$ , and for each $B\in B_{2m}$ consider the function $f_B: \{x_1, ..., x_{2m} \} \rightarrow \{0,1\}$ where $f_B(x_i)=1$ if and only if $x_i\in B$ . Denote as $[f_B]$ the open set generated by $f_B$ . Define $A_{2m}:= \bigcup _{B \in B_{2m}} [f_B]$ . Note that Y cofinally bisects X if and only if there are infinitely many m such that $Y\in A_{2m}$ . Thus,

$$\begin{align*}B(X)= \bigcap_{n \in \omega} \bigcup_{m>n} A_{2m}. \end{align*}$$

Therefore $[\omega ]^\omega \setminus B(X)$ is a countable union of closed sets. Furthermore, it is already shown in [Reference Brendle, Halbeisen, Klausner, Lischka and Shelah5] that $[\omega ]^\omega \setminus B(X)$ is a set of measure zero, so $[\omega ]^\omega \setminus B(X)$ is a countable union of closed sets of measure zero.

Proof. Assume that B splits $\mathcal {A}$ . To see that B cofinally bisects A it suffices to show that both $\{n\in \omega \, | \, \frac {d_n(B\cap A)}{d_n(A)}\geq 1/2 \}$ and $\{n\in \omega \, | \, \frac {d_n(B\cap A)}{d_n(A)}\leq 1/2 \}$ are infinite. Fix $n\in \omega $ . There are $m_1,m_2>n$ such that $A_{m_1} \subseteq B$ and $A_{m_2} \cap B = \emptyset $ . Letting $K_1:=a_{(\sum _{i \leq m_1+1} 2^i )+1}$ and $K_2:=a_{(\sum _{i \leq m_2+1} 2^i )+1}$ , we have

$$\begin{align*}\kern-12pt \frac{d_{K_1}(B\cap A)}{d_{K_1}(A)} \geq \frac{2^{m_1+1}}{\sum_{i \leq m_1+1} 2^i } \geq \frac{1}{2} \text{ and } \frac{d_{K_2}(B\cap A)}{d_{K_2}(A)} \leq \frac{\sum_{i \leq m_2} 2^i}{\sum_{i \leq m_2+1} 2^i } \leq \frac{1}{2}.\\[-41pt] \end{align*}$$

Proof. Let $p\in \mathbb {D}$ . There are $r \leq p $ and $f\in \omega ^\omega $ a recursive function such that $r \Vdash |\dot {A}_n|=f(n)$ . Since we are going to work below r, to simplify the notation, we can assume

$$\begin{align*}\Vdash |\dot{A}_n|=f(n). \end{align*}$$

For $s\in \omega ^{<\omega }$ , $i\in \omega $ , and $0<n\leq f(i)$ , say that s favors that k is the n-th element of $\dot {A}_i$ if there is no condition with first coordinate s which forces that “k is not the n-th element of $\dot {A}_i$ ”. Define the rank rk $_n^i(s)$ by recursion on the ordinals, as follows:

• • rk $_n^i(s)=0$ if for some $k \in \omega $ , s favors that k is the n-th element of $\dot {A}_i$ .

• • for $\alpha>0$ : rk $_n^i(s)=\alpha $ if there is no $\beta < \alpha $ such that rk $_n^i(s)=\beta $ and there are infinitely many l such that rk $_n^i(s^\frown l)<\alpha $ .

Clearly, rk $_n^i(s)$ must be a countable ordinal or undefined (in which case we write rk $_n^i(s) = \infty $ ). We show the latter never happens.

We continue the proof of the lemma. Consider the function $F(i)= \sum _{j< i} f(j)$ . Let $s\in \omega ^{<\omega }$ . We make a recursive construction on the rank to get the block sequences we need. Given $s\in \omega ^{<\omega }$ , there are three cases we need to consider.

1. Case 0. There infinitely many $(n,i)$ such that rk $_n^i(s)=0$ . It is easy to check that this actually implies rk $_n^i(s)=0$ for all $(n,i)$ .

2. Case 1. There is $(n,i) \in \omega \times \omega $ such that rk $_n^i(s)=1$ . This constitutes our base case for the recursive construction.

3. Case 2. There is some $1<\alpha <\omega _1$ and $(n,i) \in \omega \times \omega $ such that rk $_n^i(s)=\alpha $ .

Let us begin the construction.

1. Case 0. Let $s\in \omega ^{<\omega }$ be such that rk $_n^i(s)=0$ for infinitely many $(n,i)$ , or equivalently, rk $_n^i(s)=0$ for all $(n,i)$ . For all $m \in \omega $ , we can find a finite sequence of finite sets $\mathcal {B}_m=(B_i^m : i<m )$ such that s favors that $\mathcal {B}_m$ are the first m blocks of $\dot {\mathcal {A}}$ . Fix $i<\omega $ . Choose one block of the collection $\{B_i^m\}_{m> i}$ such that its maximum element coincides with $\min \{ \max (B_i^m) \, | \, m \in \omega \}$ , and denote this block as $B_i$ . We denote the collection $(B_i : i< \omega )$ as $\mathcal {A}_s$ . Note that $\mathcal {A}_s$ is a good block sequence.

2. Case 1. Let $(n,i) \in \omega \times \omega $ and $s\in \omega ^{<\omega }$ such that rk $_n^i(s)=1$ . There are infinitely many $l\in \omega $ such that rk $_n^i(s^\frown l)=0$ . For each such l, we may find ${A_l \in [\omega ]^{F(i)+n}}$ such that $s ^\frown l$ favors that $A_l$ are the first $(F(i)+n)$ -th elements of $\bigcup \dot {\mathcal {A}}$ . Define:

   $$\begin{align*}i_s:= \max \{j \in \omega \, | \, \{a_j : a_j \text{ is the }j\text{-th element of some }A_l \} \text{ is finite} \}. \end{align*}$$

   Then there is $A \in [\omega ]^{i_s}$ such that the set $\{ l \, | \, A \text { are the first }i_s\text {-th elements} \text{of } A_l \}$ is infinite. We denote this set as $A_{s,n}^i$ . We can construct a block sequence $\mathcal {A}_{s,n}^i$ such that

   • • $\mathcal {A}_{s,n}^i \subseteq [\omega ]^{F(i)+n-i_s}$ , so it is a good block sequence. Note that we must have $i_s < F(i)+n$ because rk $_n^i(s)>0$ .

   • • For each $l\in \omega $ , there is $l^\prime> l$ and $D \in \mathcal {A}_{s,n}^i$ such that $s ^\frown l^\prime $ favors that $A_{s,n}^i \cup D$ are the first $(F(i)+n)$ -th elements of $\bigcup \dot {\mathcal {A}}$ .

   Note that s favors that $A_{s,n}^i$ are the first $i_s$ -th elements of $\bigcup \dot {\mathcal {A}}$ .

3. Case 2. Let $(n,i) \in \omega \times \omega $ and $s\in \omega ^{<\omega }$ such that rk $_n^i(s)=\alpha>1$ . There are infinitely many $l\in \omega $ such that $0<$ rk $_n^i(s^\frown l)<\alpha $ . For each $s^\frown l$ , there exist a set $A_{s^\frown l,n}^i$ and a natural number $i_{s^\frown l}$ such that $s^\frown l$ favors that $A_{s^\frown l,n}^i$ are the first $i_{s^\frown l}$ elements of $\bigcup \dot {\mathcal {A}}$ . Also, there is a block sequence $\mathcal {A}_{s^\frown l,n}^i$ . Define:

   $$\begin{align*}i_s:= \max \{j \in \omega \, | \, \{a_i : a_i \text{ is the} j\text{-th element of some }A_{s^\frown l,n}^i \} \text{ is finite} \}. \end{align*}$$
   Then there is $A \in [\omega ]^{i_s}$ such that the set $\{ l \, | \, A \text { are the first} i_s\text {-th elements} \text{of } A_{s^\frown l,n}^i \}$ is infinite. We denote this set as $A_{s,n}^i$ . Also, choose a number $m\leq F(i)+n$ such that the set $\{ l \, | \, A_{s,n}^i \text { are the first }i_s\text {-th elements of} A_{s^\frown l,n}^i \text {and } m=|A_{s^\frown l,n}^i \setminus A_{s,n}^i| \}$ is infinite. We can construct a block sequence $\mathcal {A}_{s,n}^i$ such that

   • • $\mathcal {A}_{s,n}^i \subseteq [\omega ]^{m}$ , so it is a good block sequence. Note that $m>0$ since rk $_n^i(s)>0$ and the maximality of $i_s$ .

   • • For each $l\in \omega $ , there is $l^\prime> l$ and $D \in \mathcal {A}_{s,n}^i$ such that $D = A_{s^\frown l^\prime ,n}^i \setminus A_{s,n}^i$ .

   Note that s favors that $A_{s,n}^i$ are the first $i_s$ -th elements of $\bigcup \dot {\mathcal {A}}$ .

We claim that if B splits all $\mathcal {A}_s$ and all $\mathcal {A}_{s,n}^i$ , then $\Vdash B \text { splits } \dot {\mathcal {A}}$ . Let $(s, \phi )$ be a condition and let $m\in \omega $ . We need to find $(t,\psi ) \leq (s, \phi )$ and $m_1,m_2\geq m$ such that $(t,\psi )$ forces that $\dot {A}_{m_1} \subseteq B$ and $\dot {A}_{m_2} \cap B = \emptyset $ . Since the construction of $m_1$ and $m_2$ is analogous, it suffices to produce the former.

1. Case 0. There are infinitely many $(n,i)$ such that rk $_n^i(s)=0$ . Since B splits $\mathcal {A}_s$ , there is $n>m$ such that $B_n \in \mathcal {A}_s $ and $B_n \subseteq B$ . By definition of $\mathcal {A}_s$ , s favors that $B_n$ is the n-th block of $\dot {\mathcal {A}}$ . Therefore, there is $(t,\psi ) \leq (s, \phi ) $ such that $(t,\psi ) \Vdash \dot {A}_n = B_n \subseteq B$ .

Now, suppose that rk $_n^i(s)>0$ for all but finitely many $(n,i)$ . Choose $i\geq m$ such that rk $_n^i(s)>0$ for all $0<n\leq f(i)$ . There are two cases.

1. Case 1. Suppose that rk $_{f(i)}^i(s) =1 $ . Since B splits $\mathcal {A}_{s,f(i)}^i$ , we can find $l\geq \phi (s)$ and $D\in \mathcal {A}_{s,f(i)}^i$ such that

   • • $D \subseteq B$ and $s ^\frown l$ favors that $A_{s,f(i)}^i \cup D$ are the first $(F(i)+f(i))$ -th elements of $\bigcup \dot {\mathcal {A}}$ .

   Note that $|A_{s,f(i)}^i|=i_s \leq F(i)$ because rk $_n^i(s)>0$ for all $0<n\leq f(i)$ , thus $s ^\frown l$ favors that D contains the i-th block of $\dot {\mathcal {A}}$ . Therefore, there is $(r,\psi ) \leq (s, \phi ) $ such that

   $$\begin{align*}(r,\psi) \Vdash \dot{A}_i \subseteq D \subseteq B. \end{align*}$$

2. Case 2. Suppose that rk $_{f(i)}^i(s) = \alpha> 1$ . Since B splits $\mathcal {A}_{s,f(i)}^i$ , we can find $l_1\geq \phi (s)$ and $D_1\in \mathcal {A}_{s,f(i)}^i$ such that

   • • $1\leq $ rk $_{f(i)}^i(s^\frown l_1) <$ rk $_{f(i)}^i(s)$

   • • $D_1 \subseteq B$ and $D_1 = A_{s^\frown l_1,f(i)}^i \setminus A_{s,f(i)}^i$ .

   Repeating this argument, we can extend s to $s^\frown l_1 ^{\frown ... \frown } l_k=t$ and find finite sets $D_1, ..., D_k$ such that

   • • $t(i)\geq \phi (t\upharpoonright i)$ for all $i \in |t| \setminus |s|$ , and rk $_{f(i)}^i(t)=1$ .

   • • $D_j \subseteq B$ and $D_j = A_{s^\frown l_1 ^{\frown ... \frown } l_j,f(i)}^i \setminus A_{l_1 ^{\frown ... \frown } l_{j-1},f(i)}^i$ .

   Finally, since rk $_{f(i)}^i(t)=1$ and B splits $\mathcal {A}_{t,f(i)}^i$ , we can find $l\geq \phi (t)$ and $D\in \mathcal {A}_{t,f(i)}^i$ such that

   • • $D \subseteq B$ and $t ^\frown l$ favors that $A_{t,f(i)}^i \cup D$ are the first $(F(i)+f(i))$ -th elements of $\bigcup \dot {\mathcal {A}}$ .

   Note that $|A_{s,f(i)}^i|=i_s \leq F(i)$ because rk $_n^i(s)>0$ for all $0<n\leq f(i)$ . Also, by construction, $A_{t,f(i)}^i = D_1 \cup ... \cup D_k \cup A_{s,f(i)}^i $ , thus $t ^\frown l$ favors that $D_1 \cup ... \cup D_k \cup D$ contains the i-th block of $\dot {\mathcal {A}}$ . Therefore, there is $(r,\psi ) \leq (s, \phi ) $ such that

   $$\begin{align*}(r,\psi) \Vdash \dot{A}_i \subseteq D_1 \cup ... \cup D_k \cup D \subseteq B.\\[-31pt] \end{align*}$$

Proof. No new reals arise in limit stages of uncountable cofinality so that the lemma vacuously holds if cf $(\delta )>\omega $ . Hence assume cf $(\delta )=\omega $ . To simplify notation, suppose $\delta =\omega $ .

Let $\dot {\mathcal {A}}=(\dot {A}_n : n \in \omega )$ be a $\mathbb {P}_\omega $ -name of a good block sequence. Fix $m \in \omega $ . Work in $V^{\mathbb {P}_m}$ for the moment. There is a decreasing sequence of conditions $p_k=p_{m,k}$ in the remainder forcing $\mathbb {P}_\omega \setminus \mathbb {P}_m$ such that $p_k$ decides the k-th block of $\dot {\mathcal {A}}$ . Say $p_k \Vdash B_k= \dot {A}_k$ . Let $\mathcal {A}_m=(B_n: n \in \omega )$ .

Work in the ground model V. We have a $\mathbb {P}_m$ -name $\dot {\mathcal {A}}_m$ for $\mathcal {A}_m$ . Since $\mathbb {P}_m$ has property $(\star \star )$ , there are good block sequences $\mathcal {A}_{m,i}$ , $i\in \omega $ , such that whenever $B \in [\omega ]^\omega $ splits all $\mathcal {A}_{m,i}$ , then $\Vdash B \text { splits } \dot {\mathcal {A}}_m$ .

Now, fix $B\in [\omega ]^\omega $ . We claim that if $B \in [\omega ]^\omega $ splits all $\mathcal {A}_{m,i}$ for all $i,m\in \omega $ , then $\Vdash B \text { splits } \dot {\mathcal {A}}$ .

To see this, let $p \in \mathbb {P}_\omega $ and $n\in \omega $ . We need to find $q \leq p$ and $m_1,m_2\geq n$ such that q forces that $\dot {A}_{m_1} \subseteq B$ and $\dot {A}_{m_2} \cap B = \emptyset $ . Since the construction of $m_1$ and $m_2$ is analogous, it suffices to produce the former. Fix m such that $p \in \mathbb {P}_m$ and work in $V^{\mathbb {P}_m}$ . We know that B splits $\mathcal {A}_m$ . Thus there is $k> n$ such that $B_k\subseteq B$ and

$$\begin{align*}p_{m,k} \Vdash B_k= \dot{A}_k. \end{align*}$$

Now work in the ground model V. By strengthening $p\in \mathbb {P}_n$ , if necessary, we may assume p decides $p_k$ and $B_k$ . Then

$$\begin{align*}q= p^\frown p_{m,k} \Vdash \dot{A}_k \subseteq B.\\[-31pt] \end{align*}$$

Proof. Let $c_\alpha $ , $\alpha < \omega _1$ denote the Cohen reals added by the first $\aleph _1$ many Hechler reals. It is easy to see that every countable family of good block sequences in $V^{\mathbb {D}_{\aleph _1}}$ is split by some $c_\alpha $ . By the previous lemmas, $\mathbb {D}_\kappa $ has property $(\star \star )$ , so every good block sequence in $V^{\mathbb {D}_{\kappa }}$ is split by $\{c_\alpha \}_{\alpha <\omega _1}$ . Therefore, by Corollary 3.6, we get $V^{\mathbb {D}_\kappa }\models \aleph _1 = \mathfrak {fs}_{g} =\mathfrak {s}_{1/2}^{\infty }$ . Besides, since $\mathfrak {fs} = \max \{ \mathfrak {b}, \mathfrak {s} \}$ , we have $V^{\mathbb {D}_\kappa }\models \mathfrak {b}=\mathfrak {fs} = \kappa $ .

Proof. See [Reference Bartoszynski and Judah1, Theorem 2.6.9].

Proof. Let $\kappa>\aleph _1$ . By Corollary 3.10, $V^{\mathbb {D}_\kappa }\models \mathfrak {s}_{1/2}^{\infty }= \aleph _1$ . Furthermore, it is known that $V^{\mathbb {D}_\kappa }\models $ non $(\mathcal {M}) = \mathfrak {b} =$ non $(\mathcal {N}) = \kappa $ , then $V^{\mathbb {D}_\kappa }\models $ non $(\mathcal {E})= \kappa $ by Theorem 3.11.

Proof. Fix $\kappa> \aleph _1$ . Let $\mathbb {C}_\kappa $ be an fsi of Cohen forcing of length $\kappa $ . Because of $\mathfrak {fr}=\min \{\mathfrak {d},\mathfrak {r}\}$ and Corollary 3.6, we have $V^{\mathbb {C}_\kappa } \models \kappa =$ cov $(\mathcal {M}) = \mathfrak {d}=\mathfrak {r} = \mathfrak {fr} = \mathfrak {fr}_g = \mathfrak {r}_{1/2}^\infty $ . Denote $M= V^{\mathbb {C}_\kappa }$ . Now, let $\mathbb {D}_{\aleph _1}$ be an fsi of Hechler forcing of length $\aleph _1$ over M. Then $M^{\mathbb {D}_{\aleph _1}} \models \mathfrak {d} =$ cov $(\mathcal {M}) =$ cov $(\mathcal {N}) = \aleph _1$ because $\aleph _1$ dominating reals are added. Hence $M^{\mathbb {D}_{\aleph _1}} \models \aleph _1 =$ cov $(\mathcal {E})$ by Theorem 3.11. Also, $M^{\mathbb {D}_{\aleph _1}} \models \kappa = \mathfrak {fr}_g = \mathfrak {r}_{1/2}^{\infty } $ because $\mathbb {D}_{\aleph _1}$ has property $(\star \star )$ .

Proof. First suppose that $l < \frac {2m}{\epsilon }$ . Note that

$$\begin{align*}\frac{|(R \cap m) \cup (S\cap [m,l) )|}{l} &= \frac{|R \cap m |}{l} + \frac{|S\cap [m,l )|}{l } \\&\leq \frac{m}{l}(1-\epsilon) + \frac{l-m}{l} \\&= 1- \epsilon\frac{m}{l} \\&<1 - \frac{\epsilon^2}{2}. \end{align*}$$

For the lower bound, we get

$$\begin{align*}\frac{|(R \cap m) \cup (S\cap [m,l) )|}{l} &= \frac{|R \cap m |}{l} + \frac{|S\cap [m,l )|}{l } \\ &\geq \frac{m}{l}(\epsilon) \\ &> \frac{\epsilon^2}{2}. \end{align*}$$

Now, suppose this were false for some $l^{\star } \geq \frac {2m}{\epsilon }$ . Then, without loss of generality,

$$\begin{align*}\frac{|(R \cap m) \cup (S\cap [m,l^{\star}) )|}{l^{\star}} \geq 1-\frac{\epsilon^2}{2}. \end{align*}$$

Since

$$\begin{align*}\frac{|R \cap m |}{m} < 1-\epsilon, \end{align*}$$

we get

$$\begin{align*}\frac{|S\cap [m,l^{\star})|}{l^{\star}} \geq 1-\frac{\epsilon^2}{2} - \frac{m}{l^{\star}}(1-\epsilon). \end{align*}$$

But then

$$\begin{align*}\frac{|S \cap m|}{m}> \epsilon \end{align*}$$

implies

$$\begin{align*}\frac{|S \cap l^{\star}|}{l^{\star}} = \frac{|(S \cap m) \cup (S\cap [m,l^{\star}) ) |}{l^{\star}} & \displaystyle> \frac{m}{l^{\star}}(\epsilon) + 1-\frac{\epsilon^2}{2} - \frac{m}{l^{\star}}(1-\epsilon) \\&= 1-\frac{\epsilon^2}{2} + \frac{2m}{l^{\star}}(\epsilon) - \frac{m}{l^{\star}} \\& > 1-\frac{\epsilon^2}{2} - \frac{m}{l^{\star}} \\&\geq 1- \epsilon, \end{align*}$$

which is a contradiction.

Proof. Same proof as [Reference Brendle, Halbeisen, Klausner, Lischka and Shelah5, Lemma 3.14].

Proof. Let $\delta>0$ be such that $\epsilon _0 < \delta \leq \underline {d}(W) \leq \bar {d}(W) \leq 1- \delta < 1-\epsilon _0$ . First, notice that

$$\begin{align*}\begin{array}{c} \displaystyle\limsup_{n \to \infty} \frac{|W \cap n|}{n} = \limsup_{n \to \infty} \frac{|W \cap [k,n)|}{n-k} \\ \displaystyle\liminf_{n \to \infty} \frac{|W \cap n|}{n} = \liminf_{n \to \infty} \frac{|W \cap [k,n)|}{n-k} \end{array} \end{align*}$$

then, there is $N>0$ such that for all $l>N$ , we have

$$\begin{align*}\delta \leq \frac{|W \cap [k,l)|}{l-k}\leq 1 - \delta. \end{align*}$$

Thus, for any $l>N$ , we can apply Lemma 4.2 to $R:= k$ , $S:=[k,l)$ , $A:= Z$ , $B:=W\cap [k,l)$ , $q_0:= \delta $ , and $q_1:=1-\delta $ to obtain

$$\begin{align*}\delta - \frac{1}{c} \leq \frac{|Z_l|}{l} \leq 1-\delta + \frac{1}{c}, \end{align*}$$

where $c=(l-k)/k$ . Choose $k^\prime $ large enough to guarantee $\epsilon _0 <\delta -\frac {1}{c}$ .

Proof. Same proof as Corollary 4.3.

Proof. Fix $E \subseteq n$ . For any $k \leq l \leq k^\prime $ , we have

$$\begin{align*}\frac{d_{l}(\bigcap_{i \in E} B_i \cap Z^\prime)}{\prod_{i\in E}d_{l}(B_i)d_{l}(Z^\prime)} &= \frac{1}{\prod_{i\in E}d_{l}(B_i)}\frac{|\bigcap_{i \in E} B_i \cap Z\cap k|+|\bigcap_{i \in E} B_i \cap W \cap [k,l)|}{|Z\cap k|+|W\cap [k,l)|} \\&\leq \left(1+ \epsilon \right)\left( \frac{ |W \cap l |}{ |\bigcap_{i \in E} B_i \cap W \cap l|} \right)\\& \quad \times \left( \frac{|\bigcap_{i \in E} B_i \cap Z \cap k|+|\bigcap_{i \in E} B_i \cap W \cap [k ,l)|}{|Z\cap k|+|W\cap [k,l)|} \right) \\&= \left(1+ \epsilon\right)\left(\frac{|W\cap l|}{ |Z\cap k|+|W\cap [k,l)|} \right)\\& \quad \times \left( \frac{|\bigcap_{i \in E} B_i \cap Z \cap k|+|\bigcap_{i \in E} B_i \cap W \cap [k,l)|}{|\bigcap_{i \in E} B_i \cap W \cap l|} \right) \\&= \left(1+ \epsilon\right)\left(\frac{|W\cap k|+ l^\prime}{ |Z\cap k|+l^\prime} \right) \left( \frac{|\bigcap_{i \in E} B_i \cap Z \cap k|+l^{\prime\prime}}{|\bigcap_{i \in E} B_i \cap W \cap k|+l^{\prime\prime}} \right), \end{align*}$$

where $l^\prime =|W\cap [k,l)|$ and $l^{\prime \prime }=|\bigcap _{i \in E} B_i \cap W \cap [k,l)|$ . We have four cases:

$|W\cap k|\leq |Z\cap k|$ and $|\bigcap _{i \in E} B_i \cap W \cap k| \geq |\bigcap _{i \in E} B_i \cap Z \cap k|$ . In this case, we get

$$\begin{align*}\left(\frac{|W\cap k|+ l^\prime}{ |Z\cap k|+l^\prime} \right) \left( \frac{|\bigcap_{i \in E} B_i \cap Z \cap k|+l^{\prime\prime}}{|\bigcap_{i \in E} B_i \cap W \cap k|+l^{\prime\prime}} \right) \leq 1. \end{align*}$$

$|W\cap k|\geq |Z\cap k|$ and $|\bigcap _{i \in E} B_i \cap W \cap k|\leq |\bigcap _{i \in E} B_i \cap Z \cap k|$ . Note that

(2) $$ \begin{align} \displaystyle \frac{|W\cap k |}{ |Z \cap k|} \frac{|\bigcap_{i \in E} B_i \cap Z \cap k |}{|\bigcap_{i \in E} B_i \cap W \cap k |} = \frac{\prod_{i\in E}d_{k}(B_i) |W\cap k|}{|\bigcap_{i \in E} B_i \cap W \cap k| } \frac{|\bigcap_{i \in E} B_i \cap Z \cap k|}{\prod_{i\in E}d_{k}(B_i) |Z\cap k|} \leq \left(1+ \epsilon \right)^2. \end{align} $$

Therefore, in this case, we get

$$\begin{align*}&\left(\frac{|W\cap k|+ l^\prime}{ |Z\cap k|+l^\prime} \right) \left( \frac{|\bigcap_{i \in E} B_i \cap Z \cap k|+l^{\prime\prime}}{|\bigcap_{i \in E} B_i \cap W \cap k|+l^{\prime\prime}} \right)\\& \quad \leq \left(\frac{|W\cap k|}{ |Z\cap k|} \right) \left( \frac{|\bigcap_{i \in E} B_i \cap Z \cap k|}{|\bigcap_{i \in E} B_i \cap W \cap k|} \right) \leq \left(1+ \epsilon \right)^2. \end{align*}$$

$|W\cap k|\geq |Z\cap k|$ and $|\bigcap _{i \in E} B_i \cap W \cap k|\geq |\bigcap _{i \in E} B_i \cap Z \cap k|$ . In this case, we have

$$\begin{align*}\left(\frac{|W\cap k|+ l^\prime}{ |Z\cap k|+l^\prime} \right) \left( \frac{|\bigcap_{i \in E} B_i \cap Z \cap k|+l^{\prime\prime}}{|\bigcap_{i \in E} B_i \cap W \cap k|+l^{\prime\prime}} \right) \leq \frac{|W\cap k|}{ |Z\cap k|} \leq 1+ \epsilon. \end{align*}$$

$|W\cap k|\leq |Z\cap k|$ and $|\bigcap _{i \in E} B_i \cap W \cap k|\leq |\bigcap _{i \in E} B_i \cap Z \cap k|$ . Note that in (2), we also get

$$\begin{align*}\frac{|\bigcap_{i \in E} B_i \cap Z \cap k|}{|\bigcap_{i \in E} B_i \cap W \cap k| } \leq (1+\epsilon)^2 \frac{|Z \cap k|}{|W \cap k|} \leq (1+ \epsilon)^3. \end{align*}$$

Thus, in this case, we have

$$\begin{align*}\left(\frac{|W\cap k|+ l^\prime}{ |Z\cap k|+l^\prime} \right) \left( \frac{|\bigcap_{i \in E} B_i \cap Z \cap k|+l^{\prime\prime}}{|\bigcap_{i \in E} B_i \cap W \cap k|+l^{\prime\prime}} \right) \leq \frac{|\bigcap_{i \in E} B_i \cap Z \cap k|}{|\bigcap_{i \in E} B_i \cap W \cap k| } \leq (1+ \epsilon)^3. \end{align*}$$

To summarize, we get

$$\begin{align*}\displaystyle \frac{d_{l}(\bigcap_{i \in E} B_i \cap Z^\prime)}{\prod_{i\in E}d_{l}(B_i)d_{l}(Z^\prime)} \leq (1+\epsilon)^4 \leq 1+ 15\epsilon. \end{align*}$$

In a similar way, we have the lower bound.

Proof. First, note that

$$\begin{align*}\displaystyle \lim_{l \to \infty} \frac{d_{l}(\bigcap_{i \in E} B_i \cap W)}{\prod_{i\in E}d_{l}(B_i)d_{l}(W)} = \displaystyle \lim_{l \to \infty} \frac{ |\bigcap_{i \in E} B_i \cap W \cap [k, l)| }{\prod_{i\in E}d_{l}(B_i) |W \cap [k, l)|} \end{align*}$$

for any $E\subseteq n$ . Therefore, there is $N\in \omega $ such that for all $E\subseteq n$ and $l>N$ , we have

$$\begin{align*}\displaystyle \frac{|\bigcap_{i \in E} B_i \cap W \cap [k,l)|}{\prod_{i\in E}d_{l}(B_i) |W \cap [k, l)|} \in \left( 1-\frac{\epsilon}{2}, 1+ \frac{\epsilon}{2} \right). \end{align*}$$

Choose some $k^\prime> N$ . For the upper bound, we have

$$\begin{align*}\begin{array}{ll} \displaystyle \frac{d_{k^\prime}(\bigcap_{i \in E} B_i \cap Z_{k^\prime})}{\prod_{i\in E}d_{k^\prime}(B_i)d_{k^\prime}(Z_{k^\prime})} & \leq \displaystyle\frac{ k + |\bigcap_{i \in E} B_i \cap W \cap [k,k^\prime)| }{\prod_{i\in E}d_{k^\prime}(B_i) |W \cap [k, k^\prime)|} \\ \\ & \leq \displaystyle \frac{k(1+ \frac{\epsilon}{2})}{|\bigcap_{i \in E} B_i \cap W \cap [k, k^\prime)|} + 1+ \frac{\epsilon}{2}. \end{array} \end{align*}$$

For the lower bound, we have

$$\begin{align*}\begin{array}{ll} \displaystyle \frac{d_{k^\prime}(\bigcap_{i \in E} B_i \cap Z_{k^\prime})}{\prod_{i\in E}d_{k^\prime}(B_i)d_{k^\prime}(Z_{k^\prime})} & \geq \displaystyle\frac{ |\bigcap_{i \in E} B_i \cap W \cap [k, k^\prime)| }{\prod_{i\in E}d_{k^\prime}(B_i) (k+|W \cap [k, k^\prime)|)} \\ \\ & \geq \displaystyle\frac{ (1- \frac{\epsilon}{2}) \prod_{i\in E}d_{k^\prime}(B_i) ||W \cap [k, k^\prime)| }{\prod_{i\in E}d_{k^\prime}(B_i) (k+|W \cap [k, k^\prime)|)} \\ \\ &= \displaystyle \left(1-\frac{\epsilon}{2}\right) \frac{1}{1+\frac{k}{|W \cap [k, k^\prime)|}}. \end{array} \end{align*}$$

Since the collection of subsets of $ \{0, ..., n-1\}$ is finite, we can choose $k^\prime $ sufficiently large as to guarantee $(1-\frac {\epsilon }{2}) \frac {1}{1+\frac {k}{|W \cap [k, k^\prime )|}}\geq 1-\epsilon $ and $\frac {k(1+ \epsilon /2)}{|\bigcap _{i \in E} B_i \cap W \cap [k, k^\prime )|} \leq \frac {\epsilon }{2}$ for all $E \subseteq n$ .

Proof. First, for each function $f:n \rightarrow \{-1,1\}$ , we define

$$\begin{align*}\mathcal{B}_f := \bigcap_{i<n} B_i^{f(i)}, \end{align*}$$

where $B_i^1:=B_i$ and $B_i^{-1}:=\omega \setminus B_i$ . Note that the collection $\{ \mathcal {B}_f \, | \, f \in \{-1,1\}^n \}$ is a partition of $\omega $ . Fix a function $f \in \{-1,1\}^n$ . Let $\{d_j\}_{j\in \omega }$ be an increasing enumeration of $\mathcal {B}_f$ . Let $\frac {a}{b}$ be the irreducible fraction of q, and let A be a set containing exactly a residue classes modulo b. Define

$$\begin{align*}X_f := \{ d_j \, | \, j \in \bigcup A \}. \end{align*}$$

It is easy to show that $\displaystyle \lim _{l \to \infty } \frac {|X_f \cap l|}{|\mathcal {B}_f \cap l |} = \frac {a}{b}$ . Now, define

$$\begin{align*}X_q:= \bigcup_{ f \in \{-1,1\}^n } X_f. \end{align*}$$

Let us see that $X_q$ satisfies the conditions of the lemma. It suffices to show that for any $E \subseteq n$ , we have

$$\begin{align*}\lim_{l \to \infty}\frac{|\bigcap_{i \in E} B_i \cap X_q \cap l |}{|\bigcap_{i \in E} B_i \cap l |} = \frac{a}{b}. \end{align*}$$

Let E be a subset of n. Consider the set $G:=\{ f \in \{-1,1\}^n \,| \, f(i)=1 \text { for all} i \in E \} $ . Note that

$$\begin{align*}\bigcap_{i \in E} B_i = \bigcup_{f \in G} \mathcal{B}_f. \end{align*}$$

Also, given $\epsilon>0$ , there exists $N\in \omega $ such that for all $l>N$ , we have

$$\begin{align*}\frac{|X_f \cap l|}{|\mathcal{B}_f \cap l |} \in \left( \frac{a}{b}- \epsilon, \frac{a}{b}+\epsilon \right) \end{align*}$$

for all $f\in G$ . Thus, for $l>N$ , we get

$$\begin{align*}\begin{array}{ll} \displaystyle\frac{|\bigcap_{i \in E} B_i \cap X_q \cap l |}{|\bigcap_{i \in E} B_i \cap l |} &\displaystyle= \sum_{f \in G} \frac{|\mathcal{B}_f \cap X_q \cap l|}{ |\bigcap_{i \in E} B_i \cap l |} \\ \\ &\displaystyle= \sum_{f \in G} \frac{|X_f \cap l|}{|\bigcap_{i \in E} B_i \cap l |} = \sum_{f \in G} \frac{|X_f \cap l|}{|\mathcal{B}_f \cap l |}\frac{|\mathcal{B}_f \cap l |}{|\bigcap_{i \in E} B_i \cap l |} \\ \\ &\displaystyle\leq \left( \frac{a}{b} + \epsilon\right)\sum_{f \in G}\frac{|\mathcal{B}_f \cap l |}{|\bigcap_{i \in E} B_i \cap l |} = \left( \frac{a}{b} + \epsilon\right). \end{array} \end{align*}$$

In a similar way, we get $\left ( \frac {a}{b} - \epsilon \right ) \leq \frac {|\bigcap _{i \in E} B_i \cap X_q \cap l |}{|\bigcap _{i \in E} B_i \cap l |}$ .

Proof. First, note that there exists some rational $\delta>0$ such that for any q rational number such that $\epsilon _0 \leq q \leq 1- \epsilon _0,$ we have

$$\begin{align*}1-\epsilon<\frac{q-\delta}{q+\delta},\frac{q+\delta}{q-\delta} < 1 + \epsilon. \end{align*}$$

In fact, it is easy to prove that any $\delta < \frac {\epsilon _0 \epsilon }{2+\epsilon }$ satisfies the previous inequality. Now, consider a finite collection of intervals with rational center $\mathcal {I}_0= \{( q_j - \delta , q_j + \delta ) \}_{j < J}$ such that $\epsilon _0 < q_j < 1-\epsilon _0$ and $[\epsilon _0,1-\epsilon _0] \subseteq \bigcup \mathcal {I}_0 $ . Let $X_{q_j}$ be the sets which are given by Lemma 4.7. Let $k_0 \in \omega $ such that for any $E \subseteq n$ , $ j < J$ and for all $l \geq k_0,$ we have

• • $ \displaystyle \frac {d_{l}(\bigcap _{i \in E} B_i \cap X_{q_j})}{\prod _{i\in E}d_{l}(B_i)d_{l}(X_{q_j})}, \frac {\prod _{i\in E}d_{l}(B_i)d_{l}(X_{q_j})}{d_{l}(\bigcap _{i \in E} B_i \cap X_{q_j})} \in \left ( 1- \epsilon , 1+ \epsilon \right ) $

• • $\displaystyle \frac {|X_{q_j}\cap l|}{l} \in \left ( q_j - \delta , q_j + \delta \right ) $

• • $\displaystyle \epsilon _0 < \frac {|X_{q_j}\cap l|}{l} < 1-\epsilon _0. $

Besides, there is a finite set $Z_{k_0}\subseteq k_0$ such that

• • $Z_{k_0} \cap k = Z$

• • $\epsilon _0 \leq \frac {|Z_{k_0} \cap l|}{l} \leq 1- \epsilon _0 $ for all $k \leq l \leq k_0$ .

• • $\displaystyle \frac {d_{l}(\bigcap _{i \in E} B_i \cap Z_{k_0} )}{\prod _{i\in E}d_{l}(B_i)d_{l}(Z_{k_0} )} \in \left ( 1-\epsilon , 1+ \epsilon \right )$ for all $k \leq l \leq k_0 $ .

Also, there is some $j< J$ such that $\frac {|Z_{k_0} \cap k_0 |}{k_0} \in \left (q_j-\delta , q_j + \delta \right )$ . Denote this rational as $r_1$ . Then, we get

$$\begin{align*}1 - \epsilon \leq \frac{r_1-\delta}{r_1+\delta} \leq \frac{|X_{r_1}\cap k_0 |}{ |Z_{k_0} \cap k_0 | } , \frac{|Z_{k_0} \cap k_0 |}{|X_{r_1}\cap k_0 |} \leq \frac{r_1+\delta}{r_1-\delta} \leq 1 + \epsilon. \end{align*}$$

Denote $Z_{k_0} \cup (X_{r_1} \cap [k_0, l))$ by $Z_l^1$ . Therefore, we can apply Lemmas 4.1, 4.5, and 4.6 and Corollaries 4.3 and 4.4 with $Z_{k_0}$ and $X_{r_1}$ to find $k_1$ such that

• • $\displaystyle \frac {d_{l}(\bigcap _{i \in E} B_i \cap Z_{l}^1)}{\prod _{i\in E}d_{l}(B_i)d_{l}(Z_{l}^1)} \in \left ( 1-15\epsilon , 1+ 15\epsilon \right )$ for all $k_0 \leq l < k_1$ .

• • $\displaystyle \frac {\epsilon _0^2}{2} \leq \frac {|Z_l^1 \cap l|}{l} \leq 1- \frac {\epsilon _0^2}{2} $ for all $k \leq l < k_1$

• • $\displaystyle \frac {d_{l}(\bigcap _{i \in E} B_i \cap Z_{l}^1)}{\prod _{i\in E}d_{l}(B_i)d_{l}(Z_{l}^1)} \in \left ( 1- \epsilon , 1+ \epsilon \right )$ for all $l \geq k_1$

• • $\displaystyle \epsilon _0 \leq \frac {|Z_l^1 \cap l|}{l} \leq 1- \epsilon _0 $ for all $l \geq k_1$

• • $\displaystyle \frac {|Z_{l}^1 \cap l|}{l} \in \left (r_1 -\delta , r_1 + \delta \right )$ for all $l \geq k_1.$

If there is $l \geq k_1$ such that $\frac {|W \cap l|}{l}\in \left (r_1 -\delta , r_1 + \delta \right )$ , then we have the result with $Z^\prime := Z_l^1$ and $k^\prime =l$ . Otherwise, there is $k_1^\prime>k_1$ such that

$$\begin{align*}\frac{|W \cap l|}{l} < r_1 -\delta \text{ or } \frac{|W \cap l|}{l}> r_1 +\delta \end{align*}$$

for all $l\geq k_1^\prime $ . Without loss of generality, we can assume that $\frac {|W \cap l|}{l} < r_1 -\delta $ for all $l>k_1^\prime $ . Now, define $r_2:= r_1 - \frac {\delta }{2}$ . Note that $r_2> \epsilon _0$ since $ \frac {|W \cap k_1^\prime |}{k_1^\prime } \geq \epsilon _0$ . Consider the set $X_{r_2}$ given by Lemma 4.7. Since there is $\delta ^\prime $ such that $(r_1- \delta ^\prime , r_1 + \delta ^\prime ) \subseteq \left (r_2 -\delta , r_2 + \delta \right )$ , by Corollary 4.4, there is some $k_1^{\prime \prime } \geq k_1^\prime $ such that for all $l \geq k_1^{\prime \prime }$

$$\begin{align*}\frac{|Z_{l}^1 \cap l|}{l} \in (r_1- \delta^\prime, r_1 + \delta^\prime) \subseteq \left(r_2 -\delta, r_2 + \delta \right). \end{align*}$$

Besides, if $k_1^{\prime \prime }$ is large enough, we can ask that for any $E \subseteq n$ and for all $l \geq k_1^{\prime \prime }$

• • $ \displaystyle \frac {d_{l}(\bigcap _{i \in E} B_i \cap X_{r_2})}{\prod _{i\in E}d_{l}(B_i)d_{l}(X_{r_2})}, \frac {\prod _{i\in E}d_{l}(B_i)d_{l}(X_{r_2})}{d_{l}(\bigcap _{i \in E} B_i \cap X_{r_2})} \in \left ( 1- \epsilon , 1+ \epsilon \right ) $

• • $\displaystyle \frac {|X_{r_2}\cap l|}{l} \in \left ( r_2 - \delta , r_2 + \delta \right ) $

• • $\displaystyle \epsilon _0 < \frac {|X_{r_2}\cap l|}{l} < 1-\epsilon _0. $

Then, we get

$$\begin{align*}1 - \epsilon \leq \frac{r_2-\delta}{r_2+\delta} \leq \frac{|X_{r_2}\cap k_1^{\prime\prime} |}{ |Z_{k_1^{\prime\prime}}^1 \cap k_1^{\prime\prime} | } , \frac{|Z_{k_1^{\prime\prime}}^1 \cap k_1^{\prime\prime} |}{|X_{r_2}\cap k_1^{\prime\prime} |} \leq \frac{r_2+\delta}{r_2-\delta} \leq 1 + \epsilon. \end{align*}$$

Denote $Z_{k_1^{\prime \prime }}^1 \cup (X_{r_2} \cap [k_1^{\prime \prime }, l))$ by $Z_l^2$ . Therefore, we can apply again Lemmas 4.1, 4.5, and 4.6 and Corollaries 4.3 and 4.4 with $Z_{k_1^{\prime \prime }}^1 $ and $X_{r_2}$ to find $k_2$ such that

• • $\displaystyle \frac {d_{l}(\bigcap _{i \in E} B_i \cap Z_{l}^2)}{\prod _{i\in E}d_{l}(B_i)d_{l}(Z_{l}^2)} \in \left ( 1-15\epsilon , 1+ 15\epsilon \right )$ for all $k_1^{\prime \prime } \leq l < k_2$ .

• • $\displaystyle \frac {\epsilon _0^2}{2} \leq \frac {|Z_l^2 \cap l|}{l} \leq 1- \frac {\epsilon _0^2}{2} $ for all $k_1^{\prime \prime } \leq l < k_2$

• • $\displaystyle \frac {d_{l}(\bigcap _{i \in E} B_i \cap Z_{l}^2)}{\prod _{i\in E}d_{l}(B_i)d_{l}(Z_{l}^2)} \in \left ( 1- \epsilon , 1+ \epsilon \right )$ for all $l \geq k_2$

• • $\displaystyle \epsilon _0 \leq \frac {|Z_l^2 \cap l|}{l} \leq 1- \epsilon _0 $ for all $l \geq k_2$

• • $\displaystyle \frac {|Z_{l}^2 \cap l|}{l} \in \left (r_2 -\delta , r_2 + \delta \right )$ for all $l \geq k_2$ .

If there is $l \geq k_2$ such that $\frac {|W \cap l|}{l}\in \left (r_2 -\delta , r_2 + \delta \right )$ , then we have the result with ${Z^\prime := Z_l^2}$ and $k^\prime =l$ . Otherwise, $\frac {|W \cap l|}{l} < r_2 -\delta $ for all $l \geq k_2$ . Then, define ${r_3= r_2 - \frac {\delta }{2}}$ and repeat the previous step. This construction has an ending because ${\epsilon _0 \leq \frac {|W \cap l|}{l} \leq 1- \epsilon _0}$ for all $l>k$ .

Proof. It suffices to consider what happens when forcing with just $\mathbb {C}=$ Fn $(2,\omega )$ because we are going to build a maximal $*$ -independent family of size $\aleph _1$ that is preserved by Cohen forcing, and a new real added by $\mathbb {C}_{\kappa }$ is already added by a countable fragment which is isomorphic to Cohen forcing. Let $\mathcal {A}_0=\{A_n \subseteq [\omega ]^\omega \, | \, n \in \omega \}$ be an $*$ -independent family. Fix an enumeration $\{(p_\alpha , \dot {X}_{\alpha }) \, | \, \omega \leq \alpha \leq \omega _1 \}$ of all pairs $(p, \dot {X})$ such that $p\in \mathbb {C}$ and $\dot {X}$ is a nice name for a subset of $\omega $ . Note that since $V\models $ CH, there are just $\aleph _1$ many nice names for subsets of $\omega $ in V.

We now construct a maximal $*$ -independent family $\mathcal {A}$ from $\mathcal {A}_0$ as follows. Let $\omega \leq \alpha < \omega _1$ and assume we have already defined set $A_\beta $ for all $\beta <\alpha $ . Below, we will construct $A_\alpha \subseteq \omega $ such that the following two properties hold:

1. (i) The family $\{A_\beta \, | \, \beta \leq \alpha \}$ is $*$ -independent.

2. (ii) If there is a rational number $\epsilon _0>0$ and $N\in \omega $ in the ground model such that

   $$\begin{align*} &p_\alpha \Vdash \text{"}\epsilon_0 < d_l(\dot{X}_\alpha) < 1-\epsilon_0 \text{ for all }l\geq N,\\ & \quad \text{and } \{A_\beta \, | \, \beta < \alpha \} \cup \{ \dot{X}_\alpha \}\text{ is } *\text{-independent",} \end{align*}$$
   then there exists $\epsilon ^\prime < \epsilon _0$ and a sequence of intervals of $\omega $ , $\mathcal {I}_\alpha =\{ I_j \}_{j \in \omega }$ , with $\max I_j < \min I_{j+1}$ , such that

   1. (a) $|I_n \cap A_\alpha |> (1-\epsilon ^\prime )| (\max I_n +1) \cap A_\alpha |$ .

   2. (b) For all $m\in \omega $ , the set

      $$\begin{align*}D^\alpha_{m}:=\{ q \in \mathbb{C} \, | \, \exists n \geq m : q \Vdash A_\alpha \cap I_n = \dot{X}_{\alpha} \cap I_n \} \end{align*}$$
      is dense below $p_\alpha $ .

We first show that the family $\mathcal {A}=\{A_\beta \, | \, \beta < \omega _1 \}$ constructed this way is a maximal $*$ -independent family in $V^{\mathbb {C}}$ . Clearly, $\mathcal {A}$ is an $*$ -independent family. Suppose it were not maximal. Then, there is a condition p and a nice name $\dot {X}$ for a subset of $\omega $ such that

$$\begin{align*}p \Vdash \text{"}\mathcal{A} \cup \{\dot{X}\} \text{ is } *\text{-independent"}. \end{align*}$$

By strengthening p, we can assume that there is a rational number $\epsilon _0$ and $N\in \omega $ such that

$$\begin{align*}p \Vdash \epsilon_0 < d_l(\dot{X}) < 1-\epsilon_0 \end{align*}$$

for $l \geq N$ . This is possible since $\Vdash $ “ $\dot {X}$ is moderate”. Let $\alpha $ be such that $(p,\dot {X})=(p_\alpha , \dot {X}_\alpha )$ , so

$$\begin{align*}p_\alpha \Vdash \text{"} \{ A_\alpha, \dot{X}_\alpha\} \text{ is } *\text{-independent"}. \end{align*}$$

By property (ii), there exist $\epsilon ^\prime < \epsilon _0$ and a sequence of intervals of $\omega $ , $\mathcal {I}_\alpha =\{ I_n \}_{n \in \omega }$ , such that $|I_n \cap A_\alpha |> (1-\epsilon ^\prime )| (\max I_n +1) \cap A_\alpha |$ . Let $\epsilon = \frac {1-\epsilon ^\prime }{1-\epsilon _0} - 1$ . Find a condition $q\leq p_\alpha $ and $M \in \omega $ such that for all $n\geq M$

(3) $$ \begin{align} q\Vdash \frac{d_n(A_\alpha \cap \dot{X}_\alpha) }{d_n(A_\alpha) \cdot d_n(\dot{X}_\alpha) } \in ( 1- \epsilon, 1+ \epsilon ). \end{align} $$

Consider an interval $I_m$ such that $\min I_m> N, M$ . Now by the density of $D^\alpha _{m}$ , we can find $r \leq q$ and some $n\geq m$ such that $r \Vdash A_\alpha \cap I_n = \dot {X}_{\alpha } \cap I_n$ . Denote j as $\max I_n +1$ . We have

$$\begin{align*} r \Vdash \frac{d_{j}(A_\alpha \cap \dot{X}_\alpha) }{ d_{j}(\dot{X}_\alpha) \cdot d_{j}(A_\alpha) } &= \frac{|A_\alpha \cap \dot{X}_\alpha \cap j|}{ \frac{|\dot{X}_\alpha \cap j | }{j} \cdot |A_\alpha \cap j | } \\ &\geq \frac{|A_\alpha \cap \dot{X}_\alpha \cap j|}{ (1-\epsilon_0) \cdot |A_\alpha \cap j | } \\ & \geq \frac{|A_\alpha \cap \dot{X}_\alpha \cap I_n|}{ (1-\epsilon_0) \cdot |A_\alpha \cap j | } = \frac{|A_\alpha \cap I_n|}{ (1-\epsilon_0) \cdot |A_\alpha \cap j | }> \frac{1-\epsilon^\prime}{1- \epsilon_0} = 1 + \epsilon, \end{align*}$$

which contradicts Equation (3).

Now, we are going to show that we can construct such an $A_\alpha $ satisfying the properties (i) and (ii) for $\omega \leq \alpha < \omega _1$ . We only have to consider those $\alpha $ such that $\dot {X}_\alpha $ satisfies the assumption in property (ii), since finding an $A_\alpha $ with property (i) is straightforward. Therefore, assume that there is a rational number $\epsilon _0>0$ and $N\in \omega $ in the ground model such that

$$\begin{align*}p_\alpha \Vdash \text{"}\epsilon_0 < d_l(\dot{X}_\alpha) < 1-\epsilon_0 \text{ for all } l\geq N,\text{ and } \{A_\beta \, | \, \beta < \alpha \} \cup \{ \dot{X}_\alpha \} \text{ is *-independent",} \end{align*}$$

and consider $\epsilon ^\prime < \epsilon _0$ . Let $\{ B_n \, | \, n<\omega \}$ be an enumeration of $\{A_\beta \, | \, \beta < \alpha \}$ . We further pick some strictly decreasing sequence of real numbers $\{\delta _n\}_{n\in \omega }$ with $\delta _0<1$ and $\lim _{n \to \infty } \delta _n =0$ , and let $\{q_n \, | \, n \in \omega \}$ be some sequence enumerating all conditions below $p_\alpha $ infinitely often. By induction on $n<\omega $ , we will construct conditions $r_n \leq q_n^\prime \leq q_n$ , a strictly increasing of natural numbers $\{k_n\}_{n \in \omega }$ and initial segments $Z_n=A_\alpha \cap k_n$ of $A_\alpha $ such that the following statementsFootnote ¹ will hold:

$*$ -Independence:

1. C1. $\displaystyle \frac {d_{k_n}(\bigcap _{i \in E} B_i \cap Z_n)}{\prod _{i\in E}d_{k_n}(B_i)d_{k_n}(Z_n)} \in \left ( 1-\frac {\delta _n}{15}, 1+ \frac {\delta _n}{15} \right )$ for all $E \subseteq \{0, ..., n \}$ .

2. C2. $\displaystyle q_n^\prime \Vdash \displaystyle \frac {d_{l}(\bigcap _{i \in E} B_i \cap \dot {X}_\alpha )}{\prod _{i\in E}d_{l}(B_i)d_{l}(\dot {X}_\alpha )} \in \left ( 1-\frac {\delta _n}{15}, 1+ \frac {\delta _n}{15} \right )$ and $\displaystyle q_n^\prime \Vdash \displaystyle \frac {\prod _{i\in E}d_{l}(B_i)d_{l}(\dot {X}_\alpha )}{d_{l}(\bigcap _{i \in E} B_i \cap \dot {X}_\alpha )} \in \left ( 1-\frac {\delta _n}{15}, 1+ \frac {\delta _n}{15} \right )$ for all $l\geq k_n$ and $E \subseteq \{0, ..., n \}$ .

3. C3. $\displaystyle \frac {d_{l}(\bigcap _{i \in E} B_i \cap Z_{n+1})}{\prod _{i\in E}d_{l}(B_i)d_{l}(Z_{n+1})} \in \left ( 1-\delta _n, 1+ \delta _n \right )$ for all $k_n \leq l \leq k_{n+1}$ and $E \subseteq \{0, ..., n \}$ .

Moderate:

1. C4. $ \epsilon ^\prime \leq \displaystyle d_{k_n}(Z_n) \leq 1-\epsilon ^\prime $ .

2. C5. $ \frac {(\epsilon ^\prime )^2}{2} \leq \displaystyle d_{l}(Z_{n+1}) \leq 1-\frac {(\epsilon ^\prime )^2}{2}$ for all $k_n \leq l \leq k_{n+1}$ with $n>0$ .

For the construction:

1. C6. For every $m \geq k_n$ , there is a finite set $Y_m \subseteq m$ such that

   • • $Y_m \cap k_n = Z_n$

   • • $\epsilon ^\prime \leq \frac {|Y_m \cap l|}{l} \leq 1- \epsilon ^\prime $ for all $k_n \leq l \leq m$ .

   • • $\displaystyle \frac {d_{l}(\bigcap _{i \in E} B_i \cap Y_m)}{\prod _{i\in E}d_{l}(B_i)d_{l}(Y_m)} \in \left ( 1-\frac {\delta _n}{15}, 1+ \frac {\delta _n}{15} \right )$ for all $k_n \leq l \leq m$ .

Intervals: There is $k_n^\prime \geq k_n$ such that

1. C7. $\displaystyle |Z_{n+1} \cap [k_n^\prime , k_{n+1}) |> (1-\epsilon ^\prime )|Z_{n+1} \cap k_{n+1} |$

2. C8. $\displaystyle r_n \Vdash Z_{n+1} \cap [k_n^\prime , k_{n+1}) = \dot {X}_\alpha \cap [k_n^\prime , k_{n+1}) $ .

It is clear that (C1)–(C8) taken together for all $n<\omega $ imply that $A_\alpha := \bigcup _{n<\omega } Z_n$ is moderate and satisfies the properties (i) and (ii) with the intervals $I_n:=[k_n^\prime , k_{n+1})$ . Now, we are going to begin our construction.

For $n=0$ , choose any $k_0$ , $q^\prime _0 \leq q_0$ and $Z_0$ such that (C1), (C2), (C4), and (C6) hold. There is nothing to show yet for (C3), (C5), (C7), and (C8).

Now assume that we have obtained $k_n \in \omega $ , $q_n^\prime \leq q_n$ and $Z_n$ such that (C1), (C2), (C4), and (C6) hold for n. We will construct $r_n \leq q_n^\prime $ , $k_{n+1} \in \omega $ , $q_{n+1}^\prime \leq q_{n+1}$ and $Z_{n+1}$ such that (C3), (C5), (C7), and (C8) hold for n and (C1), (C2), (C4), and (C6) hold for $n+1$ . First, note that $Z_{n}$ , $\frac {\delta _n}{15}$ , and $\epsilon ^\prime $ satisfy the conditions of Lemma 4.8 since (C1) and (C6) hold for n. Then, we can find $k_n^\prime \geq k_n$ , a condition $q_n^{\prime \prime } \leq q_n^\prime $ , and a finite set $Z_n^\prime \subseteq k_n^\prime $ such that

• • $Z^\prime _n \cap k_n = Z_n$ .

• • $\displaystyle \frac {(\epsilon ^\prime )^2}{2} \leq \frac {|Z^\prime _n \cap l|}{l} \leq 1- \frac {(\epsilon ^\prime )^2}{2} $ for all $k_n \leq l < k^\prime _n$ .

• • $\displaystyle \frac {d_{l}(\bigcap _{i \in E} B_i \cap Z^\prime _n)}{\prod _{i\in E}d_{l}(B_i)d_{l}(Z^\prime _n)} \in \left ( 1- \delta _n, 1+ \delta _n \right )$ for all $k_n \leq l < k^\prime _n$ and for any $E \subseteq n$ .

• • $\displaystyle \epsilon ^\prime \leq \frac {|Z^\prime _n \cap k^\prime _n|}{k^\prime _n} \leq 1- \epsilon ^\prime $

• • $\displaystyle \frac {d_{k^\prime _n}(\bigcap _{i \in E} B_i \cap Z^\prime _n)}{\prod _{i\in E}d_{k^\prime _n}(B_i)d_{k^\prime }(Z^\prime _n)} \in \left ( 1- \frac {\delta _n}{15}, 1+ \frac {\delta _n}{15} \right )$ for any $E \subseteq n$ .

• • $q_n^{\prime \prime } \Vdash \displaystyle \frac {|Z^\prime _n \cap k^\prime _n|}{|\dot {X}_\alpha \cap k^\prime _n|}, \frac {|\dot {X}_\alpha \cap k^\prime _n|}{|Z^\prime _n \cap k^\prime _n|} \in \left (1 -\frac {\delta _n}{15}, 1 + \frac {\delta _n}{15}\right )$ .

Now, find a condition $q_{n+1}^\prime \leq q_{n+1}$ and $k_n^{\prime \prime } \geq k_n^\prime $ such that for all $E \subseteq \{0, ..., n+1\}$ , we have that

$$\begin{align*}\displaystyle q_{n+1}^\prime \Vdash \frac{d_{l}(\bigcap_{i \in E} B_i \cap \dot{X}_\alpha)}{\prod_{i\in E}d_{l}(B_i)d_{l}(\dot{X}_\alpha)} \in \left( 1-\frac{\delta_{n+1}}{15}, 1+ \frac{\delta_{n+1}}{15} \right) \end{align*}$$

and

$$\begin{align*}\displaystyle q_{n+1}^\prime \Vdash \frac{\prod_{i\in E}d_{l}(B_i)d_{l}(\dot{X}_\alpha)}{d_{l}(\bigcap_{i \in E} B_i \cap \dot{X}_\alpha)} \in \left( 1-\frac{\delta_{n+1}}{15}, 1+ \frac{\delta_{n+1}}{15} \right) \end{align*}$$

for all $l\geq k_n^{\prime \prime }$ . This is possible since the assumption in property (ii) is true. This implies that (C2) holds for $n+1$ . Next, we find $r_n \leq q_n^{\prime \prime }$ and a sufficiently large $k_{n+1}> k_n^{\prime \prime }$ such that for all $E \subseteq \{0, ..., n+1\}$ , we have

(4) $$ \begin{align} \displaystyle r_n \Vdash \displaystyle \frac{d_{k_{n+1}}(\bigcap_{i \in E} B_i \cap (Z_n^\prime \cup (\dot{X}_\alpha \cap [k_n^\prime, k_{n+1}))) )}{\prod_{i\in E}d_{k_{n+1}}(B_i)d_{k_{n+1}}(Z_n^\prime \cup (\dot{X}_\alpha \cap [k_n^\prime, k_{n+1})))} \in \left( 1- \frac{\delta_{n+1}}{15}, 1+ \frac{\delta_{n+1}}{15} \right) \end{align} $$

(5) $$ \begin{align} \displaystyle r_n \Vdash \epsilon_0 \leq \frac{\dot{X}_\alpha \cap [k_n^\prime, k_{n+1})}{k_{n+1}-k_n^\prime} \leq 1- \epsilon_0 \end{align} $$

and

$$\begin{align*}\displaystyle r_n \Vdash |\dot{X}_\alpha \cap [k_n^\prime, k_{n+1})|> (1-\epsilon^\prime)|\dot{X}_\alpha \cap [k_n^\prime, k_{n+1}) | + (1-\epsilon^\prime) |Z_{n}^\prime |. \end{align*}$$

Besides, we can assume that $r_n$ decides $\dot {X}_\alpha \cap k_{n+1}$ . Specifically, there is $X_n \subseteq [k_n^\prime , k_{n+1})$ such that $r_n \Vdash X_n=\dot {X}_\alpha \cap [k_n^\prime , k_{n+1})$ . All this is also possible because of the assumption in property (ii), Lemma 4.6, and the following is forced:

$$\begin{align*}\begin{array}{rl} \displaystyle \limsup\limits_{l\rightarrow \infty} \frac{|\dot{X}_\alpha \cap l|}{l} &= \displaystyle \limsup\limits_{l\rightarrow \infty} \frac{|\dot{X}_\alpha \cap [k_n^\prime, l)|}{l-k_n^\prime} \\ \displaystyle \liminf\limits_{l\rightarrow \infty} \frac{|\dot{X}_\alpha \cap l|}{l} &= \displaystyle \liminf\limits_{l\rightarrow \infty} \frac{|\dot{X}_\alpha \cap [k_n^\prime, l)|}{l-k_n^\prime}. \end{array} \end{align*}$$

Let $Z_{n+1}:=Z_n^\prime \cup X_n$ . Note that (C7) and (C8) hold for n, and (C1) holds for $n+1$ by definition of $Z_{n+1}$ . Let us see that the remaining conditions hold. Consider W such that $r_n \Vdash W=\dot {X}_\alpha \cap k_{n+1}$ .

• Condition (C5). By construction $ \frac {(\epsilon ^\prime )^2}{2} \leq \frac {|Z_n^\prime \cap l|}{l} \leq 1- \frac {(\epsilon ^\prime )^2}{2} $ for $k_n \leq l < k_n^\prime $ and $\displaystyle \epsilon ^\prime \leq \frac {|Z^\prime _n \cap k^\prime _n|}{k^\prime _n} \leq 1- \epsilon ^\prime $ . Since

  $$\begin{align*}\epsilon^\prime \leq \frac{|W \cap k_n^\prime|}{k_n^\prime}, \frac{|W \cap l|}{l} \leq 1-\epsilon^\prime \end{align*}$$
  for any l with $k_n^\prime \leq l \leq k_{n+1}$ , by Lemma 4.1, we get
  $$\begin{align*}\frac{(\epsilon^\prime)^2}{2} \leq \displaystyle d_{l}(Z_{n+1}) \leq 1- \frac{(\epsilon^\prime)^2}{2}. \end{align*}$$
  Thus (C5) holds for n.

• Condition (C4). Note that by Equation (5) and (C8) for n, we have

  $$\begin{align*}\epsilon_0 \leq \frac{X_n}{k_{n+1}-k_n^\prime} \leq 1-\epsilon_0. \end{align*}$$
  Thus, we can apply Lemma 4.2 to $R:= k_n^\prime $ , $S:=[k_n^\prime , k_{n+1})$ , $A=Z_n^\prime $ , and $B=X_n$ to obtain
  $$\begin{align*}\epsilon_0 - \frac{1}{c} \leq \frac{Z_{n}^\prime\cup X_n}{k_n^\prime \cup [k_n^\prime, k_{n+1})} = d_{k_{n+1}}(Z_{n+1}) \leq 1-\epsilon_0 + \frac{1}{c}, \end{align*}$$
  where $c=\frac {k_{n+1}-k_n^\prime }{k_n^\prime }$ . Note that we can choose $k_{n+1}$ sufficiently large as to guarantee $\epsilon ^\prime < \epsilon _0 - \frac {1}{c}$ , so (C4) holds for $n+1$ .

• Condition (C6). Let $l \geq k_{n+1}$ . Let $X_l$ be any finite set that is a witness of $\dot {X}_\alpha \cap [k_n^\prime , l)$ , that is, there is a condition $r \leq r_n$ such that $r \Vdash X_l = \dot {X}_\alpha \cap [k_n^\prime , l)$ . Then, by the previous arguments, $Y_l:= Z_n^\prime \cup X_l = Z_{n+1} \cup X_l \setminus X_n$ satisfies the inequalities of the condition (C6). Thus (C6) holds for $n+1$ .

• Condition (C3). By construction, we already have

  $$\begin{align*}\displaystyle \frac{d_{l}(\bigcap_{i \in E} B_i \cap Z_{n+1})}{\prod_{i\in E}d_{l}(B_i)d_{l}(Z_{n+1})} \in \left( 1- \delta_n, 1+ \delta_n \right) \end{align*}$$
  for all $k_n \leq l < k^\prime _n$ and $E \subseteq \{0, ..., n\}$ . Besides, by (C2) and (C8) for n, we have

  • • $\displaystyle \frac {d_{k_n^\prime }(\bigcap _{i \in E} B_i \cap Z_n^\prime )}{\prod _{i\in E}d_{k_n^\prime }(B_i)d_{k_n^\prime }(Z_n^\prime )} \in \left ( 1-\frac {\delta _n}{15}, 1+ \frac {\delta _n}{15} \right )$ for any $E \subseteq \{0, ..., n\}$ .

  • • $\displaystyle \frac {d_{l}(\bigcap _{i \in E} B_i \cap W)}{\prod _{i\in E}d_{l}(B_i)d_{l}(W)}, \frac {\prod _{i\in E}d_{l}(B_i)d_{l}(W)}{d_{l}(\bigcap _{i \in E} B_i \cap W)} \in \left ( 1-\frac {\delta _n}{15}, 1+ \frac {\delta _n}{15} \right ) $ for $l \geq k_n^\prime $ and $E \subseteq \{0, ..., n\}$ .

  • • $\displaystyle \frac {|Z^\prime _n \cap k^\prime _n|}{|W \cap k^\prime _n|}, \frac {|W \cap k^\prime _n|}{|Z^\prime _n \cap k^\prime _n|} \in \left (1 -\frac {\delta _n}{15}, 1 + \frac {\delta _n}{15}\right )$ .

  Thus, apply Lemma 4.5 to $Z_n^\prime $ and W to get
  $$\begin{align*}\displaystyle \frac{d_{l}(\bigcap_{i \in E} B_i \cap Z_{n+1})}{\prod_{i\in E}d_{l}(B_i)d_{l}(Z_{n+1})} \in \left( 1- \delta_n, 1+ \delta_n \right) \end{align*}$$
  for $k_n^\prime \leq l \leq k_{n+1} $ and $E \subseteq \{0, ..., n\}$ . Therefore, (C3) holds for $(n)$ .