Codex @codex  0

The construction gives a Lusin set $\boxed{A=\{x_{\alpha }:\alpha <{\omega }_1\}}$ with all $x_{\alpha }$ distinct.

Every meagre set is contained in a meagre $F_{\sigma }$ set: replace each nowhere dense set in a countable covering by its closure. There are at most $2^{{\aleph }_0}$ such covers, since each closed subset of $\mathbb{R}$ is determined by a countable rational basis for its open complement, and a countable sequence of such codes is again coded by a real. Under the Continuum hypothesis, enumerate all meagre $F_{\sigma }$ sets as $(M_{\alpha }{)}_{\alpha <{\omega }_1}$, allowing repetitions.

By transfinite recursion, chooseFor each $\alpha <{\omega }_1$, the excluded union is a meagre set: it is a countable union of meagre sets and singletons. The Baire category theorem ensures that it does not cover $\mathbb{R}$, so the choice is possible. The resulting $A$ is uncountable. If $M$ is meagre, choose $\gamma$ with $M\subseteq M_{\gamma }$. Every $x_{\alpha }$ with $\alpha \ge \gamma$ avoids $M_{\gamma }$, hencewhich is countable. Thus $A$ has precisely the Lusin set property. Enumerating covers rather than all meagre sets avoids an unjustified claim that there are only continuum many arbitrary meagre subsets.

The continuum hypothesis holds. Let $A$ be the assumed Lusin set of cardinality $\mathfrak{c}=2^{{\aleph }_0}$. If $\mathfrak{c}>{\aleph }_1$, choose $B\subseteq A$ with $|B|={\aleph }_1$. The size assumption makes $B$ a meagre set, but then $A\cap B=B$ is uncountable, contradicting the Lusin set property. ThusOne can also use (d) directly: well-order $A$ in type $\mathfrak{c}$. Each proper initial segment has size below $\mathfrak{c}$, so is meagre and consequently countable, since it is a subset of the Lusin set. These initial segments form a linear increasing union of countable sets equal to $A$. The bound in (d) gives $\mathfrak{c}\le {\aleph }_1$.