Elementary bounds on ideal cardinal invariants

Finite unions. If $A_0,\dots ,A_m\in \mathcal{I}$ with $\mathcal{I}\in \{\mathcal{N},\mathcal{M}\}$, then ${\bigcup }_{i\le m}{A}_i\in \mathcal{I}$: extend the finite list to the sequence $A_n:=\mathrm{\varnothing }$ for $n>m$, the empty set being in both ideals, and apply the countable-union clause of [F2] respectively [F3].

$\mathrm{add}\le \mathrm{cov}$. If $\mathcal{A}\subseteq \mathcal{I}$ with $\bigcup \mathcal{A}=\mathbb{R}$, then $\mathbb{R}\mathrm{\notin }\mathcal{I}$ gives $\bigcup \mathcal{A}\mathrm{\notin }\mathcal{I}$, so $\mathcal{A}$ is a candidate in the minimum defining the additivity and $\mathrm{add}(\mathcal{I})\le \mid \mathcal{A}\mid$; minimizing over covers of $\mathbb{R}$ by members of $\mathcal{I}$ gives $\mathrm{add}(\mathcal{I})\le \mathrm{cov}(\mathcal{I})$.

$\mathrm{add}\le \mathrm{non}$. If $Y\subseteq \mathbb{R}$ with $Y\mathrm{\notin }\mathcal{I}$, then $Y={\bigcup }_{y\in Y}\{y\}$ is the union of the family $\{\{y\}:y\in Y\}$ of members of $\mathcal{I}$, of cardinality $\mid Y\mid$; this family is a candidate in the minimum defining the additivity, so $\mathrm{add}(\mathcal{I})\le \mid Y\mid$, and minimizing over $Y\mathrm{\notin }\mathcal{I}$ gives $\mathrm{add}(\mathcal{I})\le \mathrm{non}(\mathcal{I})$.

$\mathrm{cov}\le \mathrm{cof}$. Let $\mathcal{A}\subseteq \mathcal{I}$ be inclusion-cofinal in $\mathcal{I}$ with $\mid \mathcal{A}\mid =\mathrm{cof}(\mathcal{I})$ (attained by [F1]). For each $x\in \mathbb{R}$ the singleton $\{x\}$ is a member of $\mathcal{I}$, so the set $\{A\in \mathcal{A}:x\in A\}$ is nonempty, and the Axiom of Choice selects a member $A_x\in \mathcal{A}$ containing $x$. Every real therefore lies in some member of $\mathcal{A}$, that is, $\bigcup \mathcal{A}=\mathbb{R}$, so $\mathcal{A}$ is a cover of $\mathbb{R}$ by members of $\mathcal{I}$ and $\mathrm{cov}(\mathcal{I})\le \mid \mathcal{A}\mid =\mathrm{cof}(\mathcal{I})$.

$\mathrm{non}\le \mathrm{cof}$. Let $\mathcal{A}\subseteq \mathcal{I}$ again be inclusion-cofinal with $\mid \mathcal{A}\mid =\mathrm{cof}(\mathcal{I})$. Since $\mathbb{R}\mathrm{\notin }\mathcal{I}$, no member $A\in \mathcal{A}$ equals $\mathbb{R}$, so each set $\mathbb{R}\setminus A$ is nonempty and the Axiom of Choice selects a point $x_A\in \mathbb{R}\setminus A$ for every $A\in \mathcal{A}$. Put $Y:=\{x_A:A\in \mathcal{A}\}$; then $Y\subseteq \mathbb{R}$ and $\mid Y\mid \le \mid \mathcal{A}\mid$ because $Y$ is the image of $\mathcal{A}$ under $A\mapsto x_A$. If $Y$ were a member of $\mathcal{I}$, cofinality would give $A_{\ast }\in \mathcal{A}$ with $Y\subseteq A_{\ast }$, and then $x_{A_{\ast }}\in Y\subseteq A_{\ast }$ would contradict $x_{A_{\ast }}\in \mathbb{R}\setminus A_{\ast }$; hence $Y\mathrm{\notin }\mathcal{I}$ and $\mathrm{non}(\mathcal{I})\le \mid Y\mid \le \mathrm{cof}(\mathcal{I})$.

$\mathrm{cof}(\mathcal{N})\le \mathfrak{c}$. Let $E\in \mathcal{N}$. By [F4] there is a $G_{\delta }$ set $G$ with $E\subseteq G$ and ${\lambda }^{\ast }(G)={\lambda }^{\ast }(E)$; here ${\lambda }^{\ast }(E)=\lambda (E)=0$ because $E$ is measurable with $\lambda (E)=0$, so ${\lambda }^{\ast }(G)=0$, and $G$ is a Borel set, hence a member of $\mathcal{N}$, containing $E$. Therefore the family ${\mathcal{B}}_{\mathcal{N}}:=\{B\in \mathcal{B}(\mathbb{R}):B\in \mathcal{N}\}$ consists of members of $\mathcal{N}$ and is inclusion-cofinal in it, so $\mathrm{cof}(\mathcal{N})\le \mid {\mathcal{B}}_{\mathcal{N}}\mid \le \mid \mathcal{B}(\mathbb{R})\mid =\mathfrak{c}$ by [F1] and [F5].

$\mathrm{cof}(\mathcal{M})\le \mathfrak{c}$. Let $E\in \mathcal{M}$ and, by [F1], let $(N_n{)}_{n\in \mathbb{N}}$ be a sequence of nowhere dense sets with $E\subseteq {\bigcup }_n{N}_n$. Put $B:={\bigcup }_n\overline{N_n}$. Each $\overline{N_n}$ is closed by [F7], hence $\overline{N_n}=\overline{\stackrel{‾}{N_n}}$ by [F7], so its interior $\mathrm{int}(\overline{\stackrel{‾}{N_n}})=\mathrm{int}(\overline{N_n})=\mathrm{\varnothing }$ is empty, that is, $\overline{N_n}$ is nowhere dense; thus $B$ is the union of a sequence of nowhere dense sets and is meagre, and $B$ is $F_{\sigma }$, hence Borel and a member of $\mathcal{M}$, with $E\subseteq B$. Therefore ${\mathcal{B}}_{\mathcal{M}}:=\{B\in \mathcal{B}(\mathbb{R}):B\in \mathcal{M}\}$ is inclusion-cofinal in $\mathcal{M}$, and $\mathrm{cof}(\mathcal{M})\le \mid {\mathcal{B}}_{\mathcal{M}}\mid \le \mid \mathcal{B}(\mathbb{R})\mid =\mathfrak{c}$.

Countable families never witness. If $\mathcal{A}\subseteq \mathcal{I}$ is at most countable, then $\bigcup \mathcal{A}\in \mathcal{I}$: a finite family is handled by step 1.1, and a countably infinite family can be listed as a sequence and is handled by the countable-union clause of [F2] for $\mathcal{N}$ and of [F3] for $\mathcal{M}$.

${\mathrm{\aleph }}_1\le \mathrm{add}(\mathcal{I})$. By [F1] the minimum defining the additivity is attained, so there is $\mathcal{A}\subseteq \mathcal{I}$ with $\mid \mathcal{A}\mid =\mathrm{add}(\mathcal{I})$ and $\bigcup \mathcal{A}\mathrm{\notin }\mathcal{I}$. Step 2.1 shows that such an $\mathcal{A}$ is not at most countable, so $\mathrm{add}(\mathcal{I})>{\mathrm{\aleph }}_0$, and since $\mathrm{add}(\mathcal{I})$ is a cardinal and ${\mathrm{\aleph }}_1={\mathrm{\aleph }}_0^{+}$ is the least cardinal strictly above ${\mathrm{\aleph }}_0$, it follows that ${\mathrm{\aleph }}_1\le \mathrm{add}(\mathcal{I})$.

Combining step 3.1 with steps 1.2 and 1.3 gives ${\mathrm{\aleph }}_1\le \mathrm{add}(\mathcal{I})\le \min (\mathrm{cov}(\mathcal{I}),\mathrm{non}(\mathcal{I}))$; steps 1.4 and 1.5 give $\max (\mathrm{cov}(\mathcal{I}),\mathrm{non}(\mathcal{I}))\le \mathrm{cof}(\mathcal{I})$; steps 1.6 and 1.7 give $\mathrm{cof}(\mathcal{I})\le \mathfrak{c}$ for the two ideals; and $\min \le \max$ of two cardinals is immediate. This is the displayed chain. ∎