Cross-ideal and bounding inequalities in Cichoń's diagram

Enumerate the rationals as $(q_i{)}_i$. For each $n$, choose open intervals around $q_i$ whose total length is $<2^{-n}$, and let $O_n$ be their union. Each $O_n$ is dense open and has measure $<2^{-n}$, so $B={\bigcap }_n{O}_n$ is dense $G_{\delta }$ and null. Its complement $C={\bigcup }_n(\mathbb{R}\setminus O_n)$ is meagre. Every translate of $B$ is null and comeagre; every translate of $C$ is meagre and conull.

For $g\in {\omega }^{\omega }$ put $B_g=\{x\in {\omega }^{\omega }:x{\le }^{\ast }g\}$. It is meagre: it is the union over $m$ of the closed nowhere-dense sets ${\bigcap }_{n\ge m}\{x:x(n)\le g(n)\}$. A family of size below $\mathfrak{b}$ is bounded by some $g$, so its image in Baire space is meagre. By [F2], a nonmeagre subset of $\mathbb{R}$ has a nonmeagre intersection with the irrationals, since the rationals are countable meagre. Consequently $\mathfrak{b}\le \mathrm{non}(\mathcal{M})$. A dominating family $D$ of size $\mathfrak{d}$ gives the meagre cover $(B_g{)}_{g\in D}$ of Baire space; transport it to the irrationals and add the rational set to one cover member. Each transported member is meagre in $\mathbb{R}$, because the irrationals are a dense $G_{\delta }$ subspace with countable complement. Hence $\mathrm{cov}(\mathcal{M})\le \mathfrak{d}$.

In Cantor space, for each $n$ list every clopen interval cylinder $S_m^n=\{x:x\upharpoonright [n,k)=s\},k>n,s\in 2^{[n,k)}\mathrm{.}$ For any dense open $D$, some listed $S_m^n$ lies in $D$: successively extend a common suffix while processing the finitely many length-$n$ prefixes, so all their concatenations land in $D$. Every listed interval cylinder meets every length-$n$ prefix cylinder. It follows that for any $f\in {\omega }^{\omega }$, each tail union ${\bigcup }_{n\ge r}{S}_{f(n)}^n$ is dense open, and $M_f=2^{\omega }\setminus {\mathrm{lim\hspace{0.17em}sup}}_n{S}_{f(n)}^n$ is meagre. This family is inclusion-cofinal: if $A\subseteq {\bigcup }_j{F}_j$ with $F_j$ closed nowhere dense, choose $S_{f(n)}^n$ inside the dense open complement of ${\bigcup }_{j\le n}{F}_j$, making $A\subseteq M_f$.

If $X$ is nonmeagre and $y\in \mathbb{R}$, then $X$ cannot be contained in the meagre complement of $y-B$, so $y=x+b$ for some $x\in X,b\in B$. Thus the null translates $(x+B{)}_{x\in X}$ cover $\mathbb{R}$, giving $\mathrm{cov}(\mathcal{N})\le \mathrm{\mid }X\mathrm{\mid }$. Take $\mathrm{\mid }X\mathrm{\mid }=\mathrm{non}(\mathcal{M})$. Likewise, if $X$ is nonnull, it meets the conull translate $y-C$ for every $y$, so the meagre translates $(x+C{)}_{x\in X}$ cover $\mathbb{R}$ and give $\mathrm{cov}(\mathcal{M})\le \mathrm{non}(\mathcal{N})$.

Let $k_f(n)>n$ be the right endpoint of $S_{f(n)}^n$. For a strictly increasing $g$ with $g(n)>n$, put $E_g=\{x:\text{for all sufficiently large }n,\text{some }i\in [n,g(n))\text{ has }x(i)=1\}\mathrm{.}$ This is meagre: for every $r$, the union over $n\ge r$ of the zero-block cylinders $\{x:x\upharpoonright [n,g(n))=0\}$ is dense open, so its limsup is comeagre and its complement is $E_g$.

We claim $E_g\subseteq M_f\Rightarrow g{\le }^{\ast }{k}_f$. If $g(n)>k_f(n)$ infinitely often, choose increasing $n_j$ from those indices with $g(n_j)<n_{j+1}$. Define $x$ to agree with the prescribed pattern $S_{f(n_j)}^{n_j}$ on $[n_j,k_f(n_j))$ and to equal $1$ elsewhere. The blocks are disjoint, so $x$ belongs to infinitely many $S_{f(n_j)}^{n_j}$ and hence $x\mathrm{\notin }{M}_f$. Yet $x\in E_g$: outside the prescribed blocks $x(n)=1$; when $n$ lies inside the block starting at $n_j$, monotonicity gives $g(n)\ge g(n_j)>k_f(n_j)$, and the coordinate $k_f(n_j)$ is outside all prescribed blocks and has value $1$. Thus $E_g\subseteq ̸M_f$, proving the claim.

Choose an unbounded family $G$ of strictly increasing functions of size $\mathfrak{b}$; replacing an arbitrary witness by its strictly increasing running majorants preserves unboundedness. If ${\bigcup }_{g\in G}{E}_g$ were meagre, step 1.3 would put it in one $M_f$, and step 3.1 would make $k_f$ dominate every $g\in G$, a contradiction. Therefore $\mathrm{add}({\mathcal{M}}_{2^{\omega }})\le \mathfrak{b}$. If $(A_i{)}_{i<\mathrm{cof}({\mathcal{M}}_{2^{\omega }})}$ is an inclusion-cofinal meagre family, use [F4] to select $f_i$ with $A_i\subseteq M_{f_i}$. Every $E_g$ lies in some $A_i$, so step 3.1 says $g{\le }^{\ast }{k}_{f_i}$. The family $(k_{f_i}{)}_i$ dominates, whence $\mathfrak{d}\le \mathrm{cof}({\mathcal{M}}_{2^{\omega }})$. Transfer these two inequalities to $\mathbb{R}$ by [F3]. AC is used exactly for the cardinal witness families and indexed choices of $f_i$; the interval construction itself uses finite searches. ∎