= Solution
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>, choose
$$
x_\alpha\in\mathbb R\setminus\left(\bigcup_{\beta\leq\alpha}M_\beta\ \cup\ \{x_\beta:\beta<\alpha\}\right).
$$
For 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 $\alpha\geq\gamma$ avoids $M_\gamma$, hence
$$
A\cap M\subseteq\{x_\alpha:\alpha<\gamma\},
$$
which 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.
Back to article page
