[proofplan] Let $\mathfrak c := 2^{\aleph_0}$, and let $\mathcal M$ denote the ideal of meagre subsets of $\mathbb R$. We prove the two inequalities separately. The upper bound follows by covering $\mathbb R$ with its singleton subsets, each of which is nowhere dense. For the lower bound, we take an arbitrary family of fewer than $\mathfrak c$ many meagre sets, refine it to fewer than $\mathfrak c$ many nowhere dense sets, and use Martin's axiom on a countable chain condition poset of rational open intervals to build a real outside all of them. [/proofplan] [step:Cover $\mathbb R$ by $\mathfrak c$ many nowhere dense singletons] For each $x \in \mathbb R$, the singleton $\{x\}$ is nowhere dense in $\mathbb R$: its closure is $\{x\}$, and no nonempty open interval in $\mathbb R$ is contained in $\{x\}$. Hence $\{x\} \in \mathcal M$ for every $x \in \mathbb R$. The family \begin{align*} \mathcal A := \{\{x\} : x \in \mathbb R\} \end{align*} has cardinality $|\mathcal A| = |\mathbb R| = \mathfrak c$ and satisfies $\bigcup \mathcal A = \mathbb R$. Therefore \begin{align*} \operatorname{cov}(\mathcal M) \leq \mathfrak c. \end{align*} [/step] [step:Refine any smaller meagre family to fewer than $\mathfrak c$ many nowhere dense sets] Let $\lambda < \mathfrak c$ be a cardinal, let $I$ be a set with $|I| = \lambda$, and let $(M_i)_{i \in I}$ be a family of meagre subsets of $\mathbb R$. By definition, a subset of $\mathbb R$ is meagre if it is contained in a countable union of nowhere dense subsets of $\mathbb R$. Hence, for each $i \in I$, choose nowhere dense sets $N_{i,n} \subset \mathbb R$, indexed by $n \in \mathbb N$, such that \begin{align*} M_i \subseteq \bigcup_{n \in \mathbb N} N_{i,n}. \end{align*} Define \begin{align*} J := I \times \mathbb N. \end{align*} Then \begin{align*} |J| = |I \times \mathbb N| \leq \max\{\lambda,\aleph_0\} < \mathfrak c. \end{align*} Thus the proposed cover is contained in the union of fewer than $\mathfrak c$ many nowhere dense sets: \begin{align*} \bigcup_{i \in I} M_i \subseteq \bigcup_{(i,n) \in J} N_{i,n}. \end{align*} [/step] [step:Use Martin's axiom to choose nested intervals avoiding every nowhere dense set] Let $\mathbb P$ be the set of all nonempty open intervals in $\mathbb R$ with rational endpoints. Order $\mathbb P$ by declaring $q \leq p$ when either $q = p$ or $\overline q \subset p$. This relation is reflexive by the equality alternative, transitive because strict containment of closures inside intervals composes, and antisymmetric because $q \leq p$ and $p \leq q$ force $q = p$. Thus $\mathbb P$ is a poset. The poset $\mathbb P$ is countable, so every antichain is countable; hence $\mathbb P$ satisfies the countable chain condition. For each $(i,n) \in J$, define \begin{align*} D_{i,n} := \{p \in \mathbb P : \overline p \cap N_{i,n} = \varnothing\}. \end{align*} Each $D_{i,n}$ is dense in $\mathbb P$. Indeed, if $p \in \mathbb P$, then $p \nsubseteq \overline{N_{i,n}}$ because $N_{i,n}$ is nowhere dense. Hence $p \setminus \overline{N_{i,n}}$ contains a nonempty open interval, and we may choose $q \in \mathbb P$ with $\overline q \subset p \setminus \overline{N_{i,n}}$. Then $q \leq p$ and $q \in D_{i,n}$. For each $m \in \mathbb N$, define \begin{align*} E_m := \{p \in \mathbb P : \text{the length of } p \text{ is less than } 1/m\}. \end{align*} Each $E_m$ is dense in $\mathbb P$, since every nonempty rational open interval contains a smaller rational open interval whose closure is still contained in the original interval. The family of dense subsets \begin{align*} \mathcal D := \{D_{i,n} : (i,n) \in J\} \cup \{E_m : m \in \mathbb N\} \end{align*} has cardinality less than $\mathfrak c$. Define the cardinal $\kappa := |\mathcal D|$. Then $\kappa < \mathfrak c$, so the hypothesis gives $MA_\kappa$. Applying $MA_\kappa$ to the countable chain condition poset $\mathbb P$ and to the family $\mathcal D$ of at most $\kappa$ dense subsets, there is a filter $G \subseteq \mathbb P$ such that \begin{align*} G \cap D_{i,n} \neq \varnothing \end{align*} for every $(i,n) \in J$, and \begin{align*} G \cap E_m \neq \varnothing \end{align*} for every $m \in \mathbb N$. [guided] The purpose of the forcing construction is to build one real number while imposing fewer than $\mathfrak c$ many avoidance requirements. The conditions are rational open intervals, and a stronger condition is a smaller interval whose closure lies inside the old one. Formally, let $\mathbb P$ be the set of all nonempty open intervals in $\mathbb R$ with rational endpoints, ordered by $q \leq p$ iff either $q = p$ or $\overline q \subset p$. We first verify that Martin's axiom can be applied. The equality alternative makes the order reflexive. Transitivity follows because if $r \leq q$ and $q \leq p$, then either one of the relations is equality or $\overline r \subset q \subset \overline q \subset p$, so $r \leq p$. Antisymmetry follows because strict closure containment in both directions is impossible, so $q \leq p$ and $p \leq q$ imply $q = p$. Thus $\mathbb P$ is a poset. It is countable because rational endpoints form a [countable set](/page/Countable%20Set) of pairs. Therefore every antichain in $\mathbb P$ is countable, so $\mathbb P$ has the countable chain condition. Now fix one [nowhere dense set](/page/Nowhere%20Dense%20Set) $N_{i,n}$. We want a dense set of conditions that already avoid $N_{i,n}$. Define \begin{align*} D_{i,n} := \{p \in \mathbb P : \overline p \cap N_{i,n} = \varnothing\}. \end{align*} This is dense. To prove density, take any $p \in \mathbb P$. Since $N_{i,n}$ is nowhere dense, the [closed set](/page/Closed%20Set) $\overline{N_{i,n}}$ has empty interior. Hence $p$ cannot be contained in $\overline{N_{i,n}}$, so $p \setminus \overline{N_{i,n}}$ contains a nonempty open interval. Choose a rational open interval $q$ whose closure is contained in $p \setminus \overline{N_{i,n}}$. Then $q \leq p$ and $\overline q \cap N_{i,n} = \varnothing$, so $q \in D_{i,n}$. We also need the intervals chosen by the filter to shrink to a single point. For each $m \in \mathbb N$, define \begin{align*} E_m := \{p \in \mathbb P : \text{the length of } p \text{ is less than } 1/m\}. \end{align*} This set is dense because every nonempty rational open interval contains a rational open subinterval of length less than $1/m$ whose closure lies inside the original interval. The number of dense requirements is still less than $\mathfrak c$: \begin{align*} |\{D_{i,n} : (i,n) \in J\} \cup \{E_m : m \in \mathbb N\}| \leq |J| + \aleph_0 < \mathfrak c. \end{align*} Define $\kappa := |\mathcal D|$. Since $\kappa < \mathfrak c$, the hypothesis gives $MA_\kappa$. Applying $MA_\kappa$ to the countable chain condition poset $\mathbb P$ and this family of at most $\kappa$ dense subsets gives a filter $G \subseteq \mathbb P$ meeting every $D_{i,n}$ and every $E_m$. [/guided] [/step] [step:Extract a real from the generic filter] Choose $p_0 \in G$. Since $G$ meets every $E_m$ and is directed downward, we may recursively choose a sequence $(p_m)_{m \in \mathbb N}$ in $G$ such that \begin{align*} p_{m+1} \leq p_m \end{align*} and $p_m \in E_m$ for every $m \in \mathbb N$. Thus the compact intervals $\overline{p_m}$ are nested and their lengths tend to $0$. Each $\overline{p_m}$ is a nonempty compact interval in $\mathbb R$, the containment $p_{m+1} \leq p_m$ gives $\overline{p_{m+1}} \subset p_m \subset \overline{p_m}$, and the lengths of the intervals tend to $0$ because $p_m \in E_m$. By the nested interval form of completeness for $\mathbb R$, applied to this decreasing sequence of nonempty compact intervals with lengths tending to $0$, there is a unique point $x \in \mathbb R$ such that \begin{align*} \{x\} = \bigcap_{m \in \mathbb N} \overline{p_m}. \end{align*} We claim that $x \in \overline p$ for every $p \in G$. Fix $p \in G$. For each $m \in \mathbb N$, directedness of $G$ gives $q_m \in G$ with $q_m \leq p$ and $q_m \leq p_m$. Hence $q_m \subset p \cap p_m$, so choose $y_m \in q_m$. Then $y_m \in p_m$. Since both $y_m$ and $x$ lie in $\overline{p_m}$, their distance is bounded by the length of $p_m$, which tends to $0$. Hence $(y_m)_{m \in \mathbb N}$ converges to $x$. Also $y_m \in p$ for every $m$, so $x \in \overline p$. [guided] The filter $G$ gives many rational intervals, but it does not arrive as a sequence. The dense sets $E_m$ force intervals of arbitrarily small length, and directedness lets us arrange them into one nested sequence. Choose $p_0 \in G$. Having chosen $p_m \in G$, use that $G$ meets $E_{m+1}$ to choose $r_{m+1} \in G \cap E_{m+1}$, and then use directedness of $G$ to choose $p_{m+1} \in G$ such that $p_{m+1} \leq p_m$ and $p_{m+1} \leq r_{m+1}$. Since $p_{m+1} \leq r_{m+1}$ means either $p_{m+1} = r_{m+1}$ or $\overline{p_{m+1}} \subset r_{m+1}$, the length of $p_{m+1}$ is at most the length of $r_{m+1}$ and is therefore less than $1/(m+1)$. Thus $p_{m+1} \in E_{m+1}$. Thus $(\overline{p_m})_{m \in \mathbb N}$ is a decreasing sequence of nonempty compact intervals in $\mathbb R$, and the length of $p_m$ is less than $1/m$. The nested interval form of completeness for $\mathbb R$ applies: each interval is nonempty and compact, the intervals are nested, and their lengths tend to $0$. Hence there is a unique point $x \in \mathbb R$ such that \begin{align*} \{x\} = \bigcap_{m \in \mathbb N} \overline{p_m}. \end{align*} It remains to connect this particular point to every condition in the filter, not only to the chosen sequence. Fix $p \in G$. For each $m \in \mathbb N$, directedness of $G$ gives $q_m \in G$ with $q_m \leq p$ and $q_m \leq p_m$. The order definition gives $q_m \subset p \cap p_m$, so choose $y_m \in q_m$. Since $y_m \in p_m$ and $x \in \overline{p_m}$, both points lie in the compact interval $\overline{p_m}$. Therefore $|y_m - x|$ is bounded by the length of $p_m$, and this length tends to $0$. Hence $(y_m)_{m \in \mathbb N}$ converges to $x$. Since $y_m \in p$ for every $m$, this convergence gives $x \in \overline p$. [/guided] [/step] [step:Show the extracted real avoids the proposed cover] Fix $(i,n) \in J$. Since $G$ meets $D_{i,n}$, choose $p_{i,n} \in G \cap D_{i,n}$. By the preceding step, $x \in \overline{p_{i,n}}$. Since $p_{i,n} \in D_{i,n}$, we have \begin{align*} \overline{p_{i,n}} \cap N_{i,n} = \varnothing. \end{align*} Therefore $x \notin N_{i,n}$. Since this holds for every $(i,n) \in J$, \begin{align*} x \notin \bigcup_{(i,n) \in J} N_{i,n}. \end{align*} Using the refinement from the second step, \begin{align*} x \notin \bigcup_{i \in I} M_i. \end{align*} Thus no family of fewer than $\mathfrak c$ many meagre subsets of $\mathbb R$ covers $\mathbb R$, so \begin{align*} \operatorname{cov}(\mathcal M) \geq \mathfrak c. \end{align*} Together with $\operatorname{cov}(\mathcal M) \leq \mathfrak c$, this proves \begin{align*} \operatorname{cov}(\mathcal M) = \mathfrak c. \end{align*} [/step]