Splitting and reaping comparisons with b and d

Interval partitions. Call $P=(i_n^P{)}_{n\in \mathbb{N}}$ a partition when $i_0^P=0$ and $i_n^P<i_{n+1}^P$ for all $n$; it is identified with the partition of $\omega$ into the finite intervals $[i_n^P,i_{n+1}^P)=\{m:i_n^P\le m<i_{n+1}^P\}$. For partitions $P,Q$ say that $Q$ almost dominates $P$ when

The functions ${\mathrm{func}}_P$ and the partition ${\mathrm{part}}_g$. For a partition $P$ and $x\in \omega$ let ${\mathrm{func}}_P(x)=i_{n+2}^P-1$, where $n$ is the unique index with $x\in [i_n^P,i_{n+1}^P)$, so that ${\mathrm{func}}_P\in {}^{\omega }\omega$. For $g\in {}^{\omega }\omega$ define $Q={\mathrm{part}}_g$ by $i_0^Q=0$ and, given $i_k^Q$, let $i_{k+1}^Q$ be the least $j>i_k^Q$ such that $g(x)<j$ for every $x\le i_k^Q$; the finite set $\{g(0),\dots ,g(i_k^Q)\}$ has a greatest element $m$ by [F5], and $j=\max \{i_k^Q,m\}+1$ qualifies. Then $Q$ is a partition, and its defining property is

Countable lower bound for $s$. Let $\{Y_i:i\in \mathbb{N}\}\subseteq [\omega {]}^{\omega }$ be a countable family; we construct $Z\in [\omega {]}^{\omega }$ that no $Y_i$ splits. Write $Y_i^0=Y_i$ and $Y_i^1=\omega \setminus Y_i$. Recursively choose $\epsilon (i)\in \{0,1\}$ so that $C_i:={\bigcap }_{j\le i}{Y}_j^{\epsilon (j)}$ is infinite: at stage $i=0$, one of $Y_0$ and its complement is infinite; at each later stage, the infinite set $C_{i-1}$ is the union $(C_{i-1}\cap Y_i)\cup (C_{i-1}\cap Y_i^1)$, so at least one part is infinite. After choosing $C_i$, let $m_i$ be its least element outside the finite set $\{m_0,\dots ,m_{i-1}\}$. Then the $m_i$ are pairwise distinct and $Z=\{m_i:i\in \mathbb{N}\}\in [\omega {]}^{\omega }$. For fixed $i$ and every $j\ge i$ one has $m_j\in C_j\subseteq Y_i^{\epsilon (i)}$, so $Z\setminus Y_i^{\epsilon (i)}\subseteq \{m_0,\dots ,m_{i-1}\}$ is finite. If $\epsilon (i)=0$, then $Z\setminus Y_i$ is finite; if $\epsilon (i)=1$, then $Z\cap Y_i$ is finite. In neither case are both $Z\cap Y_i$ and $Z\setminus Y_i$ infinite, so $Y_i$ does not split $Z$. Thus no countable family is a splitting family, and since $s$ is a cardinal, ${\mathrm{\aleph }}_1\le s$.

The partition $\psi (X)$ of a set. For $X\in [\omega {]}^{\omega }$ define $Q=\psi (X)$ by $i_0^Q=0$ and, given $i_n^Q$, let $i_{n+1}^Q$ be the least $j>i_n^Q$ with $[i_n^Q,j)\cap X\mathrm{\ne }\mathrm{\varnothing }$. Such a $j$ exists because $X$ is infinite, so there is $x\in X$ with $x\ge i_n^Q$, and $j=\max \{i_n^Q,x\}+1$ qualifies; the least one is determined by [F5]. Then $Q$ is a partition and by construction $[i_n^Q,i_{n+1}^Q)\cap X\mathrm{\ne }\mathrm{\varnothing }$ for every $n$.

The even-block set $\phi (P)$. For a partition $P$ put $\phi (P)={\bigcup }_{n\in \mathbb{N}}[i_{2n}^P,i_{2n+1}^P)$. Each interval $[i_{2n}^P,i_{2n+1}^P)$ is nonempty because $i_{2n}^P<i_{2n+1}^P$, these intervals are pairwise disjoint, and they are infinitely many, so $\phi (P)\in [\omega {]}^{\omega }$.

First coding lemma. If $P$ is a partition, $g\in {}^{\omega }\omega$ and ${\mathrm{func}}_P{\le }^{\ast }g$, then ${\mathrm{part}}_g$ almost dominates $P$. Let $Q={\mathrm{part}}_g$ and choose $p$ with ${\mathrm{func}}_P(n)\le g(n)$ for all $n\ge p$. Given $n\ge p$, choose $k$ with $i_n^Q\in [i_k^P,i_{k+1}^P)$ and let $x\in [i_{k+1}^P,i_{k+2}^P)$; then $p\le n\le i_n^Q<i_{k+1}^P\le x\le i_{k+2}^P-1={\mathrm{func}}_P(i_n^Q)\le g(i_n^Q)<i_{n+1}^Q$, using the defining property of ${\mathrm{part}}_g$ at $x=i_n^Q\le i_n^Q$. Hence $[i_{k+1}^P,i_{k+2}^P)\subseteq [i_n^Q,i_{n+1}^Q)$, and $n\ge p$ was arbitrary, so $Q$ almost dominates $P$.

Second coding lemma. If $P$ is a partition, $g\in {}^{\omega }\omega$ and $P$ almost dominates ${\mathrm{part}}_g$, then $g{\le }^{\ast }{\mathrm{func}}_P$. Let $Q={\mathrm{part}}_g$ and choose $m$ so that for every $n\ge m$ there is $k$ with $[i_k^Q,i_{k+1}^Q)\subseteq [i_n^P,i_{n+1}^P)$. Let $x\ge i_m^P$ and let $n$ be the index with $x\in [i_n^P,i_{n+1}^P)$, so $n\ge m$ and hence also $n+1\ge m$. Choose $k$ with $[i_k^Q,i_{k+1}^Q)\subseteq [i_{n+1}^P,i_{n+2}^P)$; then $x<i_{n+1}^P\le i_k^Q$, so $g(x)<i_{k+1}^Q\le i_{n+2}^P$, that is $g(x)\le i_{n+2}^P-1={\mathrm{func}}_P(x)$. Hence $g{\le }^{\ast }{\mathrm{func}}_P$.

Splitter lemma. If a partition $P$ almost dominates $\psi (X)$ for some $X\in [\omega {]}^{\omega }$, then $\phi (P)$ splits $X$. Let $Q=\psi (X)$ and choose $m$ so that for all $n\ge m$ there is $k$ with $[i_k^Q,i_{k+1}^Q)\subseteq [i_n^P,i_{n+1}^P)$. Since every block of $Q$ meets $X$ by step 2.1, $X\cap [i_n^P,i_{n+1}^P)\mathrm{\ne }\mathrm{\varnothing }$ for every $n\ge m$. The intervals $[i_n^P,i_{n+1}^P)$ are pairwise disjoint, so the sets $X\cap [i_{2n}^P,i_{2n+1}^P)$ for even $2n\ge m$ are pairwise disjoint nonempty subsets of $X\cap \phi (P)$, and the sets $X\cap [i_{2n+1}^P,i_{2n+2}^P)$ for odd $2n+1\ge m$ are pairwise disjoint nonempty subsets of $X\setminus \phi (P)$. Both families are infinite, so $X\cap \phi (P)$ and $X\setminus \phi (P)$ are infinite and $\phi (P)\in [\omega {]}^{\omega }$ splits $X$ by step 2.2.

$s\le d$. Let $D\subseteq {}^{\omega }\omega$ be a ${\le }^{\ast }$-dominating family with $\mid D\mid =d$, and fix $X\in [\omega {]}^{\omega }$. The partition $\psi (X)$ of step 2.1 is a partition, and ${\mathrm{func}}_{\psi (X)}\in {}^{\omega }\omega$ is dominated by some $g\in D$. By step 2.3 the partition ${\mathrm{part}}_g$ almost dominates $\psi (X)$, so by step 3.1 the set $\phi ({\mathrm{part}}_g)$ splits $X$. Hence $\{\phi ({\mathrm{part}}_g):g\in D\}$ is a splitting family: it is contained in $[\omega {]}^{\omega }$ by step 2.2 and its size is at most $\mid D\mid =d$. Therefore $s\le d$.

$b\le r$. Let $\mathcal{R}=\{X_{\alpha }:\alpha <r\}$ be an unreaped family of infinite sets, of size $r$. Consider the family $\mathrm{\Psi }=\{\psi (X_{\alpha }):\alpha <r\}$ of partitions. No partition $P$ almost dominates every member of $\mathrm{\Psi }$: otherwise $P$ almost dominates $\psi (X_{\alpha })$ for every $\alpha <r$, so by step 3.1 the set $\phi (P)$ splits every $X_{\alpha }$, contradicting that $\mathcal{R}$ is unreaped. Consequently the family of functions $\{{\mathrm{func}}_{\psi (X_{\alpha })}:\alpha <r\}$ is ${\le }^{\ast }$-unbounded: if some $g$ dominated all of them, step 2.3 would make ${\mathrm{part}}_g$ a partition almost dominating every member of $\mathrm{\Psi }$, contrary to what was just shown. An unbounded family has size at least $b$, and $\mid \{{\mathrm{func}}_{\psi (X_{\alpha })}:\alpha <r\}\mid \le r$, so $b\le r$.

Steps 1.3, 4.1 and [F3] give ${\mathrm{\aleph }}_1\le s\le d\le \mathfrak{c}$, and steps 4.2 and [F1] with [F3] give ${\mathrm{\aleph }}_1\le b\le r\le \mathfrak{c}$. This is the statement. ∎