[proofplan] We refine each meagre set into countably many closed nowhere dense sets, obtaining fewer than continuum many closed nowhere dense obstacles. The proof uses Martin's Axiom; the additional hypothesis $\neg CH$ is part of the stated assumption but is not needed for this stronger conclusion. We then use the standard interval forcing poset: conditions are nonempty open intervals with rational endpoints, ordered by reverse inclusion. For each closed nowhere dense obstacle, the set of intervals whose closure avoids it is dense, and Martin's Axiom gives a filter meeting all these dense sets. We also add dense requirements forcing arbitrarily small intervals; the resulting filter yields a nested sequence of closed intervals with diameters tending to $0$. Completeness of $\mathbb{R}$ then gives a unique real number lying in all their closures, and compatibility with each avoidance condition places that real outside every obstacle, hence outside every $M_i$. [/proofplan]

[step:Replace the meagre family by fewer than continuum many closed nowhere dense sets]Let $\mathbb{N} := \{1,2,3,\dots\}$ denote the set of positive natural numbers, and let $\mathbb{Q} \subseteq \mathbb{R}$ denote the set of rational numbers. For each $i < \lambda$, since $M_i \subseteq \mathbb{R}$ is meagre, choose a sequence $(A_{i,n})_{n \in \mathbb{N}}$ of nowhere dense subsets of $\mathbb{R}$ such that \begin{align*} M_i \subseteq \bigcup_{n \in \mathbb{N}} A_{i,n}. \end{align*} For each pair $(i,n) \in \lambda \times \mathbb{N}$, define \begin{align*} F_{i,n} := \overline{A_{i,n}}, \end{align*} where the closure is taken in the usual topology on $\mathbb{R}$. Since $A_{i,n}$ is nowhere dense, $F_{i,n}$ is closed and nowhere dense. Also \begin{align*} M_i \subseteq \bigcup_{n \in \mathbb{N}} F_{i,n}. \end{align*} Let \begin{align*} I := \lambda \times \mathbb{N}. \end{align*} Since $\lambda < 2^{\aleph_0}$ and $\mathbb{N}$ is countable, we have \begin{align*} |I| = |\lambda \times \mathbb{N}| < 2^{\aleph_0}. \end{align*} Thus it suffices to find a real number outside every $F_{i,n}$.[/step]

[guided]The reason for passing from meagre sets to closed nowhere dense sets is that closed nowhere dense sets can be avoided by shrinking intervals. We use $\mathbb{N} := \{1,2,3,\dots\}$ for the positive natural numbers and $\mathbb{Q} \subseteq \mathbb{R}$ for the rational numbers. For each $i < \lambda$, the hypothesis that $M_i$ is meagre means precisely that there is a countable sequence $(A_{i,n})_{n \in \mathbb{N}}$ of nowhere dense subsets of $\mathbb{R}$ satisfying \begin{align*} M_i \subseteq \bigcup_{n \in \mathbb{N}} A_{i,n}. \end{align*} We now replace each $A_{i,n}$ by its closure. Define \begin{align*} F_{i,n} := \overline{A_{i,n}}. \end{align*} Because $A_{i,n}$ is nowhere dense, its closure has empty interior. Hence $F_{i,n}$ is closed and nowhere dense. Since $A_{i,n} \subseteq F_{i,n}$, we still have \begin{align*} M_i \subseteq \bigcup_{n \in \mathbb{N}} F_{i,n}. \end{align*} The total collection of closed nowhere dense sets is indexed by \begin{align*} I := \lambda \times \mathbb{N}. \end{align*} Because $\lambda < 2^{\aleph_0}$ and multiplying an infinite cardinal below the continuum by $\aleph_0$ leaves it below the continuum, while finite $\lambda$ gives a countable product, we get \begin{align*} |I| < 2^{\aleph_0}. \end{align*} Therefore, if we construct a point $x \in \mathbb{R}$ with $x \notin F_{i,n}$ for every $(i,n) \in I$, then automatically \begin{align*} x \notin \bigcup_{i < \lambda} M_i. \end{align*}[/guided]

[step:Define the interval forcing poset and verify the countable chain condition] Let $\mathbb{P}$ be the set of all nonempty bounded open intervals in $\mathbb{R}$ with rational endpoints. Define the order $\leq_{\mathbb{P}}$ on $\mathbb{P}$ by \begin{align*} p \leq_{\mathbb{P}} q \quad \Longleftrightarrow \quad p \subseteq q. \end{align*} Thus stronger conditions are smaller intervals. The set $\mathbb{P}$ is countable, since it is indexed by ordered pairs $(a,b) \in \mathbb{Q}^2$ with $a < b$. Hence every antichain in $\mathbb{P}$ is countable, so $\mathbb{P}$ satisfies the countable chain condition. [/step]

[step:Build dense sets forcing avoidance of each closed nowhere dense obstacle] For each $(i,n) \in I$, define \begin{align*} D_{i,n} := \{p \in \mathbb{P} : \overline{p} \cap F_{i,n} = \varnothing\}. \end{align*} Here $\overline{p}$ denotes the closure of the interval $p$ in $\mathbb{R}$. We show that $D_{i,n}$ is dense in $\mathbb{P}$. Fix $q \in \mathbb{P}$. Since $F_{i,n}$ is nowhere dense, the nonempty open interval $q$ is not contained in $\overline{F_{i,n}} = F_{i,n}$. Hence there exists $x \in q \setminus F_{i,n}$. Because $\mathbb{R} \setminus F_{i,n}$ is open, there exists $\varepsilon > 0$ such that \begin{align*} (x-\varepsilon,x+\varepsilon) \subseteq q \setminus F_{i,n}. \end{align*} Choose rational numbers $a,b \in \mathbb{Q}$ with \begin{align*} x-\varepsilon < a < x < b < x+\varepsilon. \end{align*} Set \begin{align*} p := (a,b). \end{align*} Then $p \in \mathbb{P}$, $p \subseteq q$, and $\overline{p} = [a,b] \subseteq (x-\varepsilon,x+\varepsilon)$, so $\overline{p} \cap F_{i,n} = \varnothing$. Therefore $p \leq_{\mathbb{P}} q$ and $p \in D_{i,n}$, proving that $D_{i,n}$ is dense. [/step]

[step:Apply Martin's Axiom to meet avoidance and shrinking requirements]For each $m \in \mathbb{N}$, define the diameter requirement \begin{align*} H_m := \{p \in \mathbb{P} : \operatorname{diam}(p) < 1/m\}, \end{align*} where for an interval $p = (a,b)$ the diameter is $\operatorname{diam}(p) := b-a$. Each $H_m$ is dense in $\mathbb{P}$. Indeed, if $q = (a,b) \in \mathbb{P}$, choose rational numbers $c,d \in \mathbb{Q}$ with $a < c < d < b$ and $d-c < 1/m$, using the density of $\mathbb{Q}$ in $\mathbb{R}$. Then $p := (c,d)$ belongs to $H_m$ and satisfies $p \leq_{\mathbb{P}} q$. Define the family of dense subsets \begin{align*} \mathcal{D} := \{D_{i,n} : (i,n) \in I\} \cup \{H_m : m \in \mathbb{N}\}. \end{align*} Its cardinality is at most $|I| + \aleph_0 < 2^{\aleph_0}$. Since $\mathbb{P}$ satisfies the countable chain condition, Martin's Axiom applies to $\mathbb{P}$ and $\mathcal{D}$. Hence there exists a filter $G \subseteq \mathbb{P}$ such that \begin{align*} G \cap D \neq \varnothing \end{align*} for every $D \in \mathcal{D}$.[/step]

[guided]The filter must do two jobs. First, it must meet each $D_{i,n}$, because that gives an interval whose closure avoids the closed [nowhere dense set](/page/Nowhere%20Dense%20Set) $F_{i,n}$. Second, it must meet arbitrarily small interval requirements, because a point cannot be obtained from an arbitrary filter of intervals unless the intervals are forced to converge. For each $m \in \mathbb{N}$, define \begin{align*} H_m := \{p \in \mathbb{P} : \operatorname{diam}(p) < 1/m\}, \end{align*} where $\operatorname{diam}(p) := b-a$ when $p = (a,b)$. We verify density. Let $q = (a,b) \in \mathbb{P}$. Since rational numbers are dense in $\mathbb{R}$, choose $c,d \in \mathbb{Q}$ such that $a < c < d < b$ and $d-c < 1/m$. Then $p := (c,d)$ is a nonempty bounded open interval with rational endpoints, so $p \in \mathbb{P}$. Also $p \subseteq q$, hence $p \leq_{\mathbb{P}} q$, and $\operatorname{diam}(p) < 1/m$. Thus $H_m$ is dense. Now set \begin{align*} \mathcal{D} := \{D_{i,n} : (i,n) \in I\} \cup \{H_m : m \in \mathbb{N}\}. \end{align*} Every member of $\mathcal{D}$ is dense in $\mathbb{P}$, the poset $\mathbb{P}$ has the countable chain condition, and \begin{align*} |\mathcal{D}| \leq |I| + \aleph_0 < 2^{\aleph_0}. \end{align*} These are precisely the hypotheses needed for Martin's Axiom. Therefore there exists a filter $G \subseteq \mathbb{P}$ meeting every dense set in $\mathcal{D}$, meaning \begin{align*} G \cap D \neq \varnothing \end{align*} for every $D \in \mathcal{D}$.[/guided]

[step:Construct the real determined by the filter and prove it avoids every meagre set]For each $m \in \mathbb{N}$, choose $h_m \in G \cap H_m$. We construct a sequence $(q_m)_{m \in \mathbb{N}}$ in $G$ recursively. Set $q_1 := h_1$. Having chosen $q_m \in G$, use the directedness of the filter $G$ to choose $q_{m+1} \in G$ such that \begin{align*} q_{m+1} \leq_{\mathbb{P}} q_m \end{align*} and \begin{align*} q_{m+1} \leq_{\mathbb{P}} h_{m+1}. \end{align*} By the order definition, $q_{m+1} \subseteq q_m$ and $q_{m+1} \subseteq h_{m+1}$. Hence the closed intervals $(\overline{q_m})_{m \in \mathbb{N}}$ are nested, nonempty, and satisfy \begin{align*} \operatorname{diam}(\overline{q_m}) \leq \operatorname{diam}(h_m) < 1/m. \end{align*} By completeness of $\mathbb{R}$, or equivalently the nested interval theorem, there is a unique real number $x \in \mathbb{R}$ such that \begin{align*} x \in \bigcap_{m \in \mathbb{N}} \overline{q_m}. \end{align*} We claim that $x \in \overline{p}$ for every $p \in G$. Fix $p \in G$. Since $G$ is directed, for each $m \in \mathbb{N}$ choose $s_m \in G$ with $s_m \leq_{\mathbb{P}} p$ and $s_m \leq_{\mathbb{P}} q_m$. Thus $p \cap q_m$ is nonempty for every $m$. Since $q_m$ is an interval and $x \in \overline{q_m}$ with $\operatorname{diam}(q_m) < 1/m$, the distance from $x$ to $p$ is at most $1/m$ for every $m \in \mathbb{N}$. Therefore $x \in \overline{p}$. Now fix $(i,n) \in I$. Choose $p_{i,n} \in G \cap D_{i,n}$. The preceding paragraph gives $x \in \overline{p_{i,n}}$, while $p_{i,n} \in D_{i,n}$ gives \begin{align*} \overline{p_{i,n}} \cap F_{i,n} = \varnothing. \end{align*} Hence $x \notin F_{i,n}$. Since this holds for every $(i,n) \in I$, \begin{align*} x \notin \bigcup_{(i,n) \in I} F_{i,n}. \end{align*} Because every $M_i$ is contained in $\bigcup_{n \in \mathbb{N}}F_{i,n}$, it follows that \begin{align*} x \notin \bigcup_{i < \lambda} M_i. \end{align*} Therefore \begin{align*} \bigcup_{i < \lambda} M_i \neq \mathbb{R}. \end{align*}[/step]

[guided]The filter gives many intervals, not a single interval chosen once and for all. To extract a point from it, we use the shrinking dense sets $H_m$. For each $m \in \mathbb{N}$, choose $h_m \in G \cap H_m$, so $h_m$ is an interval in the filter with diameter less than $1/m$. We now turn these small intervals into a nested sequence. Define $q_1 := h_1$. Suppose $q_m \in G$ has already been chosen. Since $G$ is a filter, it is directed downward: for the two conditions $q_m,h_{m+1} \in G$, there exists $q_{m+1} \in G$ such that $q_{m+1} \leq_{\mathbb{P}} q_m$ and $q_{m+1} \leq_{\mathbb{P}} h_{m+1}$. Because the order is reverse inclusion, this means $q_{m+1} \subseteq q_m$ and $q_{m+1} \subseteq h_{m+1}$. Thus the closures $\overline{q_m}$ form a nested sequence of nonempty compact intervals, since each $q_m$ is a bounded open interval in $\mathbb{R}$. Moreover, \begin{align*} \operatorname{diam}(\overline{q_m}) \leq \operatorname{diam}(h_m) < 1/m. \end{align*} The nested interval theorem, which follows from completeness of $\mathbb{R}$, gives at least one real number in the intersection of these nested compact intervals. The diameter bound gives uniqueness: if two distinct points lay in every $\overline{q_m}$, their positive distance would be less than $1/m$ for every $m \in \mathbb{N}$. Hence there is a unique real number $x \in \mathbb{R}$ satisfying \begin{align*} x \in \bigcap_{m \in \mathbb{N}} \overline{q_m}. \end{align*} The uniqueness comes from the diameter bound: two distinct points in all the closures would have positive distance apart, contradicting $\operatorname{diam}(\overline{q_m}) < 1/m$ for sufficiently large $m$. We still need to connect this point $x$ to every interval in the filter, not just to the chosen nested sequence. Fix $p \in G$. For each $m \in \mathbb{N}$, the directedness of $G$ gives $s_m \in G$ with $s_m \leq_{\mathbb{P}} p$ and $s_m \leq_{\mathbb{P}} q_m$. Hence $s_m \subseteq p \cap q_m$, so $p \cap q_m$ is nonempty. Since $x \in \overline{q_m}$ and the interval $q_m$ has diameter less than $1/m$, some point of $p$ lies within distance at most $1/m$ of $x$. Therefore the distance from $x$ to $p$ is at most $1/m$ for every $m$, and so $x \in \overline{p}$. Finally fix one obstacle $F_{i,n}$. Because $G$ meets $D_{i,n}$, choose $p_{i,n} \in G \cap D_{i,n}$. From the preceding paragraph, $x \in \overline{p_{i,n}}$. From the definition of $D_{i,n}$, \begin{align*} \overline{p_{i,n}} \cap F_{i,n} = \varnothing. \end{align*} Therefore $x \notin F_{i,n}$. Since $(i,n)$ was arbitrary, \begin{align*} x \notin \bigcup_{(i,n) \in I} F_{i,n}. \end{align*} The earlier reduction gave $M_i \subseteq \bigcup_{n \in \mathbb{N}} F_{i,n}$ for every $i < \lambda$, so \begin{align*} x \notin \bigcup_{i < \lambda} M_i. \end{align*} This produces a real number outside the union, and hence \begin{align*} \bigcup_{i < \lambda} M_i \neq \mathbb{R}. \end{align*}[/guided]