Proof Let $\mathbb {P} = \bigcup _{n\in \omega } C_n$ , where each $C_n$ is centred, let $\dot {B}$ be a $\mathbb {P}$ -name for an element of $[\omega ]^{\omega }$ , and assume towards a contradiction that

$$ \begin{align*} p \Vdash \text{"}\dot{B}\; \varepsilon\text{-almost bisects every } X\in[\omega]^{\omega}\cap V\text{"} \end{align*} $$

for some $p\in \mathbb {P}$ and , that is,

$$ \begin{align*} p\Vdash \text{"}\frac{1}{2}-\varepsilon < \frac{|\dot{B}\cap X\cap n|}{|X\cap n|} < \frac{1}{2}+\varepsilon \text{ for almost all } n\text{"} \end{align*} $$

for every $X\in [\omega ]^{\omega }\cap V$ . Fix an interval partition $(I_n)$ of $\omega $ such that $|I_0|\geq 2$ and $|I_n|> 2^{n}|I_{<n}|$ for every n (where $I_{<n}=\bigcup _{k<n}I_k$ and $I_{<0}=\varnothing $ ). First of all, we show that

$$ \begin{align*} p\Vdash \text{"}\frac{1}{2}-\varepsilon' < \frac{|\dot{B}\cap I_n|}{|I_n|} < \frac{1}{2}+\varepsilon' \text{ for almost all } n\text{"} \end{align*} $$

holds for every . To see this, fix such an $\varepsilon '$ ; then, as $p\Vdash $ “ $\dot {B} \ \varepsilon $ -almost bisects $\omega $ ,” p forces that for every sufficiently large n,

$$ \begin{align*} \frac{|\dot{B}\cap I_n|}{|I_n|} &\leq \frac{(1+2^{-n})|\dot{B}\cap I_{\leq n}|}{ (1+2^{-n})|I_n|}< \frac{(1+2^{-n})|\dot{B}\cap I_{\leq n}|}{ |I_{\leq n}|} \\ &<(1+2^{-n})\left(\frac{1}{2}+\varepsilon\right)<\frac{1}{2}+\varepsilon'\text{,} \end{align*} $$

where the second inequality follows from $|I_n|+2^{-n}|I_n|>|I_n|+|I_{<n}|$ . The lower bound can be established similarly: p forces that for every sufficiently large n,

$$ \begin{align*} \frac{|\dot{B}\cap I_n|}{|I_n|}> \frac{|\dot{B}\cap I_n|}{|I_n|}+\frac{|I_{<n}|}{|I_n|}-2^{-n}\geq\frac{|\dot{B}\cap I_{\leq n}|}{|I_{\leq n}|}-2^{-n}>\frac{1}{2}-\varepsilon-2^{-n}>\frac{1}{2}-\varepsilon'\text{.} \end{align*} $$

Fix an $\varepsilon '$ as above. Then there is a $p'\leq p$ and an $N\in \omega $ such that

$$ \begin{align*} p'\Vdash\text{"}\frac{1}{2}-\varepsilon' < \frac{|\dot{B}\cap I_n|}{|I_n|} < \frac{1}{2}+\varepsilon'\text{ for every } n\geq N \text{."} \end{align*} $$

By modifying $\dot {B}$ on $I_{<N}$ (which does not affect the property of being $\varepsilon $ -almost bisecting), we can assume that $p'$ forces these inequalities for every n; in particular, $\dot {B}\cap I_n\ne \varnothing $ for every n (as $\varepsilon '$ can be very small, we assume that $|I_n|$ is even for every n). In the rest of the proof we assume that $p'=1_{\mathbb {P}}$ , that is,

(*1) $$ \begin{align} \Vdash\text{"}\frac{1}{2}-\varepsilon' < \frac{|\dot{B}\cap I_n|}{|I_n|} < \frac{1}{2}+\varepsilon'\text{ for every } n\text{."} \end{align} $$

Let $\mathcal {E}_n=\{E\subseteq I_n \mid \forall \, p\in C_n\; \exists \, q\leq p\colon q\Vdash \dot {B}\cap I_n=E\}$ . Note that $\mathcal {E}_n\neq \varnothing $ : Otherwise, for every $E\subseteq I_n$ we can fix $p_E\in C_n$ such that $p_E\Vdash \dot {B}\cap I_n\neq E$ , but these $p_E$ have a common extension, which is a contradiction. Fix an arbitrary sequence $(E_n)_{n \in \omega }\in V$ such that $E_n\in \mathcal {E}_n$ for every n (we know that $E_n\ne \varnothing $ ), and let $X=\bigcup _{n\in \omega }E_n\in [\omega ]^{\omega }\cap V$ .

Since $\dot {B} \ \varepsilon $ -almost bisects X, we can fix some $p\in \mathbb {P}$ and $M \in \omega $ such that

(*2)

Now fix a k such that

As each condition has incompatible extensions, there are extensions of p in infinitely many $C_n$ ; hence for some $n \geq k$ , we can fix a $p'\in C_n$ below p. By the definition of $\mathcal {E}_n$ , there is a $q\leq p'$ which forces that $\dot {B}\cap I_n=E_n=X\cap I_n$ ; in particular, q forces that

where the second inequality follows from Equation (*1) and the third inequality follows from $|I_{<n}|<2^{-n}|I_n|$ . This contradicts Equation (*2), because $I_{\leq n}=\min (I_{n+1})\geq \min (I_{k+1})\geq M$ .

Proof Assume towards a contradiction that a $p\in \mathbb {P}$ forces that $\dot {B}\in V^{\mathbb {P}} \ \varepsilon $ -almost bisects every $X\in [\omega ]^{\omega }\cap V$ . As in the proof of Lemma 3.2, we fix an interval partition $(I_n)$ in V such that $|I_0|\geq 2$ and $|I_n|>2^n|I_{<n}|$ for every n as well as an

, and just like above, we can assume that

(*3) $$ \begin{align} p\Vdash\text{"}\frac{1}{2}-\varepsilon' < \frac{|\dot{B}\cap I_n|}{|I_n|} < \frac{1}{2}+\varepsilon'\text{ for every } n \text{."} \end{align} $$

Let $Q_0=I_1$ , $Q_1=I_2\cup I_3$ , …, $Q_m=\bigcup _{m'<2^m} I_{2^{m}+m'}$ be the union of the next $2^m$ many intervals in the partition. Applying the Laver property, there are an $S\in \prod _{m\in \omega }[\mathcal {P}(Q_m)]^{2^m}$ in V, $S(m)=\{S^m_{m'} \mid m'<2^m\}$ , a $p' \leq p$ , and a $\mathbb {P}$ -name $\dot {b}$ for an element of $\prod _{m\in \omega }2^m$ such that $p'\Vdash $ “ $\dot {B}\cap Q_m=S^m_{\dot {b}(m)}\in S(m)$ for every m.” Define

$$ \begin{align*} X=\bigcup_{m\in\omega}\bigcup_{m'< 2^m} S^m_{m'}\cap I_{2^m+m'}\in V\text{.} \end{align*} $$

Then X is infinite because $X\cap I_{2^m+\dot {b}(m)}= S^m_{\dot {b}(m)}\cap I_{2^m+\dot {b}(m)}=\dot {B}\cap I_{2^m+\dot {b}(m)}\neq \varnothing $ for every m. We claim that $p'\Vdash $ “ $\dot {B}$ does not $\varepsilon $ -almost bisect X.” To see this, let m be sufficiently large such that

Then $p'$ forces that

where we used Equation (*3) in the second inequality and $|I_{<n}|/|I_n|<2^{-n}$ in the third one.

Proof The inequalities follow by dualising the ones in Theorem 2.3. To show the consistency of the strict inequalities and that there are no further inequalities in $\mathrm {ZFC}$ , keeping in mind the consistent cuts of the Cichoń’s diagram, it is enough to show that the following are consistent: $\mathfrak {r}<\mathrm {non}(\mathcal {M})$ and (separately, of course).

$\mathfrak {r}<\mathrm {non}(\mathcal {M})$ holds in the model presented in [Reference Brendle, Halbeisen, Klausner, Lischka and Shelah3, Section 5], because it is a model of $\mathfrak {u}<\mathfrak {s}$ and we know that $\mathfrak {r}\leq \mathfrak {u}$ and $\mathfrak {s}\leq \mathrm {non}(\mathcal {M})$ (for more details on the ultrafilter number $\mathfrak {u}$ , see [Reference Blass and Shelah2, Section 9]).

We show that holds after the $\mathfrak {c}^+$ stage finite support iteration $\mathbb {D}_{\mathfrak {c}^+}$ of the Hechler forcing $\mathbb {D}$ . We know that if $\kappa \geq \mathfrak {c}$ is regular, then $V^{\mathbb {D}_{\kappa }}\vDash \mathrm {add}(\mathcal {M})=\min \{\mathfrak {b},\mathrm {cov}(\mathcal {M})\}=\kappa =\mathfrak {c}$ because $\mathbb {D}$ adds both dominating and Cohen reals. Recall (folklore, see [Reference Goldstern and Judah7]) that if $\delta \leq \mathfrak {c}$ (hence also if $\delta <\mathfrak {c}^+$ ) and $(\mathbb {P}_{\alpha },\dot {\mathbb {Q}}_{\beta })_{\alpha \leq \delta , \beta <\delta }$ is a finite support iteration of $\sigma $ -centred forcing notions (that is, $\Vdash _{\alpha }$ “ $\dot {\mathbb {Q}}_{\alpha }$ is $\sigma $ -centred” for every $\alpha <\delta $ ), then $\mathbb {P}_{\delta }$ is also $\sigma $ -centred. Applying Lemma 3.2, it follows that if $\delta \leq \mathfrak {c}^+$ and $(\mathbb {P}_{\alpha },\dot {\mathbb {Q}}_{\beta })_{\alpha \leq \delta ,\beta <\delta }$ is a finite support iteration of $\sigma $ -centred forcing notions, then

$$ \begin{align*} \Vdash_{\delta}\text{"no }S\in [\omega]^{\omega}\text{ can }\varepsilon\text{-almost bisect all elements of }[\omega]^{\omega}\cap V\text{";} \end{align*} $$

in particular, .

Question C. Is consistent?

Proof Fix an $S\in [\omega ]^{\omega }$ , an , and a $p\in \mathbb {P}$ . Pick an n in $\omega \smallsetminus \mathrm {dom}(p)$ such that

$$ \begin{align*} 2^{-n+1}\leq \frac{1}{2}-\varepsilon \qquad \text{and hence} \qquad \frac{1}{2^{-n+1}+1}\geq \frac{1}{2}+\varepsilon\text{.} \end{align*} $$

We distinguish two cases:

Case 1: $|S\cap I_n|>|I_n|/2$ . Define $q_n\in \mathbb {P}$ , $\mathrm {dom}(q_n)=\mathrm {dom}(p)\cup \{n\}$ , $q_n{\upharpoonright }_{\mathrm {dom}(p)}=p$ , and $q_n(n)=S\cap I_n$ . Then $q_n\leq p$ and $q_n$ forces that

$$ \begin{align*} \frac{|S\cap \dot{X}\cap I_{\leq n}|}{|\dot{X}\cap I_{\leq n}|}&\geq \frac{|S\cap I_n|}{|I_{<n}|+|S\cap I_n|}= \frac{1}{\frac{|I_{<n}|}{|S\cap I_n|}+1}\\ &>\frac{1}{\frac{|I_{<n}|}{|I_n|/2}+1}>\frac{1}{2^{-n+1}+1}\geq \frac{1}{2}+\varepsilon\text{.} \end{align*} $$

As n can be arbitrarily large, p cannot force that $S \ \varepsilon $ -almost bisects $\dot {X}$ .

Case 2: $|S\cap I_n|\leq |I_n|/2$ . Define $r_n\in \mathbb {P}$ , $\mathrm {dom}(r_n)=\mathrm {dom}(p)\cup \{n\}$ , $r_n{\upharpoonright }_{\mathrm {dom}(p)}=p$ , and $r_n(n)=I_n\smallsetminus S$ . Then $r_n\leq p$ and $r_n$ forces that

$$ \begin{align*} \frac{|S\cap \dot{X}\cap I_{\leq n}|}{|\dot{X}\cap I_{\leq n}|}\leq \frac{|I_{<n}|}{|I_n\smallsetminus S|}\leq \frac{|I_{<n}|}{|I_n|/2}<2^{-n+1}\leq \frac{1}{2}-\varepsilon\text{.} \end{align*} $$

As n can be arbitrarily large, p cannot force that $S \ \varepsilon $ -almost bisects $\dot {X}$ .

Proof (1): First let $X\in [\omega ]^{\omega }$ . If there is an infinite $H\subseteq \omega $ such that for every $k\in H$ , then $X\sqsubset _1 (H,(I_k\smallsetminus X)_{k \in \omega })$ . If there is no such H, then for every $k\geq K$ for some $K\in \omega $ , and if we choose some $E_k\subseteq I_k$ with for every $k\geq K(\geq 2)$ , then

$$ \begin{align*} |X\cap E_k|\leq |E_k| <\frac{3}{8}|I_k| < \Big(\frac{1}{2}+\varepsilon\Big)|X\cap I_k|\text{ for every } k\geq K\text{,} \end{align*} $$

hence $X\sqsubset _0(\omega \smallsetminus K,(E_k)_{k \in \omega })$ .

Now let $(H,(E_k)_{k \in \omega })\in \mathcal {O}$ . If $X=\omega \smallsetminus \bigcup _{k\in H}E_k$ is infinite, then $X\sqsubset _1 (H,(E_k))$ . If this set is finite, however, then there is a K such that $\omega \smallsetminus K\subseteq H$ and $E_k=I_k$ for every $k\geq K$ . If $Y\in [\omega ]^{\omega }$ such that $|Y\cap I_k|=1$ for every k, then

$$ \begin{align*} |Y\cap E_k|\leq 1 < \Big(\frac{1}{2}+\varepsilon\Big)\big(1+|I_{<k}|\big)\text{ holds for every } k\in H \smallsetminus 1 \text{,} \end{align*} $$

and so $Y\sqsubset _1 (H,(E_k))$ .

(2): If $X\not \sqsubset _n (H, (E_k))$ , then this is witnessed by a $k_0\in H\smallsetminus n$ , that is,

$$ \begin{align*} |X\cap E_{k_0}|\geq \Big(\frac{1}{2}+\varepsilon\Big)\big(|X\cap I_{k_0}|+|I_{<k_0}|\big)\text{.} \end{align*} $$

Then $X'\not \sqsubset _n (H',(E^{\prime }_k))$ for all $X'\in [\omega ]^{\omega }$ and $(H',(E^{\prime }_k))\in \mathcal {O}$ such that $X'\cap I_{k_0}=X\cap I_{k_0}$ , $k_0\in H'$ , and $E^{\prime }_{k_0}=E_{k_0}$ ; these pairs $(X',(H',(E^{\prime }_k)))\in [\omega ]^{\omega }\times \mathcal {O}$ form an open neighbourhood of $(X,(H,(E_k)))$ in $[\omega ]^{\omega }\times \mathcal {O}$ .

(3): Fix an $X\in [\omega ]^{\omega }$ with $X\sqsubset _n(H,(E_k))$ and a basic open neighbourhood $U_m=\{Y\in [\omega ]^{\omega } \mid Y\cap m = X\cap m\}$ of X (for an $m\in \omega $ ). If $k\in H\smallsetminus n$ such that $\min (I_k)\geq m$ (and k is sufficiently large, see below), and $Y\in U_m$ such that $Y\cap I_k=E_k$ , then k witnesses $Y\not \sqsubset _n (H,(E_k))$ , that is, because

$$ \begin{align*} \frac{|E_k|}{|I_k|}>\Big(\frac{1}{2}+\varepsilon\Big)\Big(\frac{|E_k|}{|I_k|}+2^{-k}\Big) \end{align*} $$

holds if k is sufficiently large.

(4): $\mathrm {cov}(\mathcal {M})\leq \mathfrak {d}(\sqsubset )$ follows from (1) and (3). Now let $S\in [\omega ]^{\omega }$ , define $S'=S$ if for infinitely many k and $S'=\omega \smallsetminus S$ otherwise, and let . For $k\in H_S$ , let $E^S_k=S'\cap I_k$ , and for $k\in \omega \smallsetminus H_S$ , let $E_k=I_k$ . It is enough to show that if $S \ \varepsilon $ -almost bisects X, then $X\sqsubset (H_S,(E^S_k))$ , i.e., that is a Tukey connection. Clearly, iff ; therefore if $k\in H_S$ is sufficiently large, then $X\cap I_{\leq k}\neq \varnothing $ and

$$ \begin{align*} \frac{|X\cap E^{S}_k|}{|X\cap I_k|+|I_{<k}|} \leq \frac{|S'\cap X\cap I_{\leq k}|}{|X\cap I_{\leq k}|} < \frac{1}{2}+\varepsilon\text{,} \end{align*} $$

finishing the proof.

Proof Let $(\dot {H},(\dot {E}_k))$ be a $\mathbb {D}$ -name for an element of $\mathcal {O}$ . We will construct a countable family $\mathcal {O}'\subseteq \mathcal {O}$ such that whenever $X\in [\omega ]^{\omega }\cap V$ and $X\not \sqsubset \mathcal {O}'$ , then $\Vdash X\not \sqsubset (\dot {H},(\dot {E}_k))$ .

Let $\dot {H}=\{\dot {k}_0<\dot {k}_1<\cdots \}$ be an enumeration in $V^{\mathbb {D}}$ . Recall that we denote the generic real of $\mathbb {D}$ by $\dot {d}$ . By thinning out $\dot {H}$ , we can assume that $\dot {d}(n) < \dot {k}_n$ for every n. (Note that if $\Vdash X\not \sqsubset (\dot {J},(\dot {E}_k))$ for some infinite $\dot {J}\subseteq \dot {H}$ , then this holds for $\dot {H}$ as well.)

We define a rank function $\rho _n$ on $\omega ^{<\uparrow \omega }$ for every fixed $n\in \omega $ as follows: We set $\rho _n(s)=0$ if there are $k_{n,s}\in \omega $ and $E_{n,s}\subseteq I_{k_{n,s}}$ such that whenever $T\in \mathbb {D}$ and $\mathrm {stem}(T)=s$ , then there is a $T'\leq T$ which forces that “ $\dot {k}_n=k_{n,s}$ and $\dot {E}_{k_{n,s}}=E_{n,s}$ .” Then we proceed by recursion: At the $\alpha $ th stage, after already having defined $\{s\in \omega ^{<\omega } \mid \rho _n(s)=\beta \}$ for every $\beta <\alpha $ , we set $\rho _n(s)=\alpha $ if

$$ \begin{align*} Y_{n,s}=\left\{m \;\middle|\; \rho_n\big(s^{\frown}(m)\big)<\alpha\right\}\text{ is infinite.} \end{align*} $$

We show that $\mathrm {dom}(\rho _n)=\omega ^{<\uparrow \omega }$ for every n. Assume towards a contradiction that $\rho _n(s)$ is not defined. Then $\{m \mid \rho _n(s^{\frown }(m))\text { is not defined}\}$ is cofinite; hence we can construct a $T\in \mathbb {D}$ with stem s such that $\rho _n(t)$ is not defined for every $t\in T$ above s. There are a $T'\leq T$ , $k\in \omega $ , and $E\subseteq I_k$ such that $T'\Vdash \text {"}\dot {k}_n=k$ and $\dot {E}_k=E \text {"}$ ; in particular, k and E witness that $\rho _n(\mathrm {stem}(T'))=0$ , a contradiction.

Also, we will need the fact that if $n\geq |s|$ , then $\rho _n(s)>0$ . To see this, let (for $k\in \omega $ ) $T_k\in \mathbb {D}$ such that $\mathrm {stem}(T_k)=s$ and $\mathrm {ext}_{T_k}(s)=\omega \smallsetminus k$ . Then $T_k\Vdash k\leq \dot {d}(|s|)\leq \dot {d}(n)<\dot {k}_n$ . If $\rho _n(s)=0$ , then we could pick $k_{n,s}$ such that $T\Vdash \dot {k}_n=k_{n,s}$ whenever $T\in \mathbb {D}$ with stem s; but in that case, $T=T_{k_{n,s}}\Vdash k_{n,s}<\dot {k}_n$ , a contradiction.

Now, if $\rho _n(s)=1$ and $m\in Y_{n,s}$ , then $\rho _n(s^{\frown }(m))=0$ and hence we have defined $k_{n,s^{\frown }(m)}$ and $E_{n,s^{\frown }(m)}$ . Note that

$$ \begin{align*} \left\{m\in Y_{n,s} \;\middle|\; k_{n,s^{\frown}(m)}=k\right\}\text{is finite} \end{align*} $$

for each k. Otherwise, there are $k \in \omega $ and $E\subseteq I_k$ such that $X=\{m\in Y_{n,s} \mid k_{n,s^{\frown }(m)}=k \text { and } E_{n,s^{\frown }(m)}=E\}$ is also infinite, and hence if $T\in \mathbb {D}$ with $\mathrm {stem}(T)=s$ , then there is an $m\in X\cap \mathrm {ext}_T(s)$ , and so there is a $T'\leq T {\upharpoonright }_{s^{\frown }(m)} \leq T$ such that $T'\Vdash \text {"} \dot {k}_n=k \text { and } \dot {E}_{k}=E \text {,"}$ in other words, $\rho _n(s)=0$ , a contradiction.

For such n and s we can thus fix an infinite $Z_{n,s}\subseteq Y_{n,s}$ such that

$$ \begin{align*} \text{if }m,m'\in Z_{n,s}\text{ and }m<m'\text{, then }k_{n,s^{\frown} (m)}<k_{n,s^{\frown}(m')}\text{.} \end{align*} $$

We let $K_{n,s} = \{k_{n,s^{\frown }(m)} \mid m\in Z_{n,s}\}$ , $E^{n,s}_k=E_{n,s^{\frown }(m)}$ if $m\in Z_{n,s}$ and $k_{n,s^{\frown }(m)}=k$ and $E^{n,s}_k=I_k$ otherwise, and

$$ \begin{align*} \mathcal{O}' = \left\{(K_{n,s},(E^{n,s}_k)_{k\in\omega}) \;\middle|\; n\in\omega,\;s\in\omega^{\uparrow<\omega},\;\rho_n(s)=1\right\} \subseteq\mathcal{O}\text{.} \end{align*} $$

To finish the proof, fix an $X \in [\omega ]^{\omega }\cap V$ and assume that $X \not \sqsubset \mathcal {O}'$ , i.e., $X\not \sqsubset (K_{n,s},(E^{n,s}_k))$ whenever $\rho _n(s)=1$ , or more explicitly, for infinitely many $k\in K_{n,s}$

(• _{k,X,(E k ^n,s))} $$\begin{align} |X\cap E^{n,s}_k|\geq \Big(\frac{1}{2} + \varepsilon\Big)\big(|X\cap I_k|+|I_{<k}|\big)\text{.} \end{align}$$

To show that $\Vdash X\not \sqsubset (\dot {H},(\dot {E}_k))$ , we fix $T\in \mathbb {D}$ and $n\in \omega $ . We will find a $T'\leq T$ and a $k\geq n$ such that $T' \Vdash \text {"} k\in \dot {H} \text { and } (\bullet _{k,X,(\dot {E}_k)})\text {."}$ We can assume that $n\geq |\mathrm {stem}(T)|$ and hence $\rho _n(\mathrm {stem}(T))>0$ . By induction on this rank, one can easily show that there is an $s\in T$ above the stem such that $\rho _n(s)=1$ . Pick a $k=k_{n,s^{\frown }(m)}\in K_{n,s}\cap \mathrm {ext}_T(s)\smallsetminus n$ such that $(\bullet _{k,X,(E^{n,s}_k)})$ . By the definition of $\rho _n(s^{\frown }(m))=0$ , there is a $T' \leq T{\upharpoonright }_{s^{\frown }(m)} \leq T$ which forces that $\dot {k}_n=k$ and $\dot {E}_k=E_{n,s^{\frown }(m)}=E^{n,s}_k$ ; in particular, $T'$ also forces that $k\in \dot {H}\smallsetminus n$ and $(\bullet _{k,X,(\dot {E}_k)})$ .

Fact 5.1. Let $\rho _0,\rho _1\in (0,1)$ and $A,B,X\in [\omega ]^{\omega }$ . Then the following statements hold:

1. (1) If $A\mid _{\rho _0} X$ and $B\mid _{\rho _1} A\cap X$ , then $A\cap B\mid _{\rho _0\rho _1} X$ ; hence $\max \{\mathfrak {s}_{\rho _0},\mathfrak {s}_{\rho _1}\}\geq \mathfrak {s}_{\rho _0\rho _1}$ and $\min \{\mathfrak {r}_{\rho _0},\mathfrak {r}_{\rho _1}\}\leq \mathfrak {r}_{\rho _0\rho _1}$ .

2. (2) If $A\mid _{\rho _0} X$ and $B\mid _{\rho _1} X\smallsetminus A$ , then $A\cup B\mid _{\rho _0+\rho _1-\rho _0\rho _1} X$ ; hence $\max \{\mathfrak {s}_{\rho _0},\mathfrak {s}_{\rho _1}\}\geq \mathfrak {s}_{\rho _0+\rho _1-\rho _0\rho _1}$ and $\min \{\mathfrak {r}_{\rho _0},\mathfrak {r}_{\rho _1}\}\leq \mathfrak {r}_{\rho _0+\rho _1-\rho _0\rho _1}$ .

Proof (1) follows from

$$ \begin{align*} \frac{|(A\cap B)\cap X\cap n|}{|X\cap n|}=\frac{|B\cap (A\cap X)\cap n|}{|(A\cap X)\cap n|}\cdot\frac{|A\cap X\cap n|}{|X\cap n|}\text{,} \end{align*} $$

and (2) follows from

$$ \begin{align*} \frac{|(A\cup B)\cap X\cap n|}{|X\cap n|}&=\frac{|A\cap X\cap n|+ |B\cap (X\smallsetminus A)\cap n|}{|X\cap n|}\\ &=\frac{|A\cap X\cap n|}{|X\cap n|}+\frac{|B\cap (X\smallsetminus A)\cap n|}{|(X\smallsetminus A)\cap n|}\cdot\frac{|(X\smallsetminus A)\cap n|}{|X\cap n|}\text{.}\\[-42pt] \end{align*} $$

Fact 5.2 (Non-integer bases).

Let $b>1$ be a real number. Then every $x>0$ can be written as $x=\sum _{n=-\infty }^Nc_nb^n$ where $N\geq 0$ is an integer and $0\leq c_n< b$ are also integers for all n.

Proof Fix a . We show that .

To show that , let $\mathcal {R} \subseteq [\omega ]^{\omega }$ such that . We will construct an $S \in [\omega ]^{\omega }$ such that $S \mid _{\rho } \mathcal {R}$ (that is, such that $S \mid _{\rho } X$ for every $X\in \mathcal {R}$ ).

By recursion on $n\in \omega $ , one can easily construct $S_n\in [\omega ]^{\omega }$ and $\mathcal {R}_n \subseteq [\omega ]^{\omega }$ such that $S_0=\omega $ , $\mathcal {R}_0=\mathcal {R}$ , and

$$ \begin{align*} \mathcal{R}_{n+1} = \left\{S_{n+1} \cap X \;\middle|\; X \in \mathcal{R}_n\right\}\!\text{.} \end{align*} $$

For $m \geq 1$ , define

$$ \begin{align*} I_m=\bigcap_{n\leq m}S_n\qquad \text{and} \qquad D_m=I_{m-1} \smallsetminus I_m =I_{m-1}\smallsetminus S_m\text{.} \end{align*} $$

It follows that $\mathcal {R}_m=\{I_m\cap X \mid X\in \mathcal {R}\}$ and hence that $I_m,D_m\in [\omega ]^{\omega }$ for every $m\geq 1$ . First of all, we show that and for every $m\geq 1$ . This holds for $m=1$ by the definitions above. We proceed by induction on m; assume that the claim holds for a fixed m. If $X\in \mathcal {R}$ , then $I_m\cap X \in \mathcal {R}_m$ and thus . Since holds as well, we can apply Fact 5.1(1) with , , $A=I_m$ , $B=S_{m+1}$ , and X, and obtain that . It follows that (because ).

Now let $P \subseteq \omega \smallsetminus \{0\}$ be such that $\sum _{m \in P}{2^{-m}} = \rho $ (a representation of $\rho <1$ in base $2$ ) and $S= \bigcup _{m\in P}{D_m}$ . We show that $S \mid _{\rho } \mathcal {R}$ . Fix an $X\in \mathcal {R}$ and define the lower and upper relative density of $S\in [\omega ]^{\omega }$ in $X\in [\omega ]^{\omega }$ as follows:

$$ \begin{align*} \overline{d}_X(S)&=\limsup_{n\to\infty}\frac{|S\cap X\cap n|}{|X\cap n|}\text{,}\\ \underline{d}_X(S)&=\liminf_{n\to\infty}\frac{|S\cap X\cap n|}{|X\cap n|}\text{.} \end{align*} $$

Clearly, $\overline {d}_X$ and $\underline {d}_X$ are monotone, $\overline {d}_X(\omega \smallsetminus S)=1-\underline {d}_X(S)$ and $\underline {d}_X(\omega \smallsetminus S)=1-\overline {d}_X(S)$ , and $S \mid _{\rho } X$ iff $\overline {d}_X(S)=\underline {d}_X(S)=\rho $ ; in this case we write $d_X(S)=\rho $ . Also, notice that $d_X$ is finitely additive. It follows that it is enough to show that $\underline {d}_X(S)\geq \rho $ and $\underline {d}_X(\omega \smallsetminus S) \geq 1-\rho $ . For $k\in \omega $ , let $A_k=\bigcup _{m\in P\cap k}D_m$ (and $\bigcup \varnothing =\varnothing $ ), then $\underline {d}_X(A_k)\leq \underline {d}_X(S)$ , and at the same time $\underline {d}_X(A_k)=d_X(A_k)=\sum _{m\in P\cap k}d_X(D_m)=\sum _{m\in P\cap k}2^{-m}\xrightarrow {k\to \infty }\rho $ , hence $\rho \leq \underline {d}_X(S)$ . To show that $\underline {d}_X(\omega \smallsetminus S)\geq 1-\rho $ , write $\omega \smallsetminus S$ as $\bigcup _{m\in \omega \smallsetminus P}D_m$ . (Since finite modifications of $S_n$ do not affect the above, we can assume without loss of generality that $\bigcap _{n\in \omega }S_n=\varnothing $ and hence that $\omega =\bigcup _{m\geq 1}D_m$ is a partition.)

The proof of the converse inequality is similar but with a little twist: We start with a family $\mathcal {R}'\subseteq [\omega ]^{\omega }$ such that $|\mathcal {R}'| < \mathfrak {r}_{\rho }$ and have to construct an $S' \in [\omega ]^{\omega }$ such that . By recursion on $n\in \omega $ , we define $S^{\prime }_n\in [\omega ]^{\omega }$ and $\mathcal {R}^{\prime }_n \subseteq [\omega ]^{\omega }$ such that $S^{\prime }_0=\omega $ , $\mathcal {R}^{\prime }_0=\mathcal {R}'$ , $S^{\prime }_{n+1}\mid _{\rho }\mathcal {R}^{\prime }_n$ , and $\mathcal {R}^{\prime }_{n+1} = \{S^{\prime }_{n+1} \cap X \mid X \in \mathcal {R}^{\prime }_n\}$ . For $m \geq 1$ , define $I^{\prime }_m=\bigcap _{n\leq m}S_n$ and $D^{\prime }_m=I_{m-1} \smallsetminus I_m$ . It follows (by induction) that $I^{\prime }_m \mid _{\rho ^m} \mathcal {R}'$ and $D^{\prime }_m \mid _{\rho ^{m-1}(1-\rho )} \mathcal {R}'$ for every $m\geq 1$ . If we can find a $P' \subseteq \omega \smallsetminus \{0\}$ such that , then, just like above, . The problem is that when writing in the non-integer base $1<b=\rho ^{-1}<2$ , i.e., , we may have to use non-zero coefficients for at least some $0 \leq n \leq N$ . Therefore, we have two cases:

Case 1: If , i.e., , then $N<0$ and we can pick an appropriate $P'$ and hence conclude that .

Case 2: If then, applying Fact 5.1(1), $\mathfrak {r}_{\rho }=\min \{\mathfrak {r}_{\rho },\mathfrak {r}_{\rho }\}\leq \mathfrak {r}_{\rho ^2}=\min \{\mathfrak {r}_{\rho ^2},\mathfrak {r}_{\rho ^2}\}\leq \mathfrak {r}_{\rho ^4}\leq \cdots $ ; in other words, $\mathfrak {r}_{\rho }\leq \mathfrak {r}_{\rho ^2}\leq \mathfrak {r}_{\rho ^4}\leq \mathfrak {r}_{\rho ^8}\leq \cdots $ . There is an n such that (because ), and hence .

Question D. Is it consistent that $\mathfrak {s}_{\rho }\ne \mathfrak {s}_{\tau }$ for some $\rho ,\tau \in (0,1)$ ?

Question E. Does $\mathbf {Cov}(\mathcal {N})\preccurlyeq \mathbf {Reap}_{\rho }$ , or at least $\mathrm {non}(\mathcal {N})\geq \mathfrak {s}_{\rho }$ and $\mathrm {cov}(\mathcal {N})\leq \mathfrak {r}_{\rho }$ , hold?

Fact 5.4. $\mathbf {Dom}^{\perp }\preccurlyeq \mathbf {Reap}_0$ and hence $\mathfrak {d}\geq \mathfrak {s}_0$ and $\mathfrak {b}\leq \mathfrak {r}_0$ .

Proof Instead of $\omega ^{\omega }$ , we work with $\mathcal {X}=\{x\in \omega ^{\uparrow \omega } \mid |\omega \smallsetminus \mathrm {ran}(x)|=\omega \}$ . It is trivial to show that $(\mathcal {X},\leq ^*,\mathcal {X})\equiv (\omega ^{\omega },\leq ^*,\omega ^{\omega })=\mathbf {Dom}$ . We define a Tukey connection $(F,G)\colon \mathbf {Reap}_0^{\perp }\to \mathbf {Dom}$ , that is, an

$$ \begin{align*} (F,G)\colon\big([\omega]^{\omega},\text{"is } 0\text{-split by,"}\left\{S\in [\omega]^{\omega} \;\middle|\; |\omega\smallsetminus S|=\omega\right\}\big)\to (\mathcal{X},\leq^*,\mathcal{X}) \end{align*} $$

as follows: Let $F\colon [\omega ]^{\omega }\to \mathcal {X}$ be defined by $F(R)(n)=r_{2^n}$ where $R=\{r_0<r_1<\cdots \}\in [\omega ]^{\omega }$ and let $G\colon \mathcal {X}\to \{S\in [\omega ]^{\omega } \mid |\omega \smallsetminus S|=\omega \}$ be defined by $G(x)=\mathrm {ran}(x)$ . If $F(R)\leq ^* x$ , $r_{2^n}\leq x(n)$ for every $n\geq N$ for some N and $r_{2^n}<k\leq r_{2^{n+1}}$ , then

$$ \begin{align*} \frac{|\mathrm{ran}(x)\cap R\cap k|}{|R\cap k|}\leq \frac{N+n}{2^n}\xrightarrow{n\to\infty} 0\text{,} \end{align*} $$

and hence $G(x)=\mathrm {ran}(x) \ 0$ -splits R.

Question F. Does $\mathbf {Dom}^{\perp }\equiv \mathbf {Reap}_0$ , and hence $\mathfrak {d}=\mathfrak {s}_0$ and $\mathfrak {b}=\mathfrak {r}_0$ , hold? Or do at least $\mathfrak {s}_0\geq \mathfrak {s}$ and $\mathfrak {r}_0\leq \mathfrak {r}$ hold?