Must a closed set with empty interior in a separable compact space of cardinality at most $2^{\aleph_{0}}$ be separable?

Question

Consider a separable compact Hausdorff space $X$ with $|X| \leq 2^{\aleph_{0}}$.

If $A\subseteq X$ is a closed set with empty interior, must $A$ be separable?

Each example I've been able to come up with in this setting is either countable or can be proven to be separable.

EDIT: Edited question to reflect the question at base.

Answer 1

Here is a counterexample in ZFC. Let $X\subset\{0,1\}^\mathbb{R}$ be the set of characteristic functions of (possibly degenerate or empty) intervals. Then $X$ is closed in $\{0,1\}^\mathbb{R}$ (if $f\in \{0,1\}^\mathbb{R}\setminus X$ then this is witnessed by points $x<y<z$ such that $f(x)=f(z)=1$ and $f(y)=0$ so the complement is open) so $X$ is compact. Also, $X$ is separable, since the characteristic functions of intervals with rational endpoints are dense in $X$, and clearly $|X|=2^{\aleph_0}$.

Now let $A\subset X$ be the set of characteristic functions of sets with at most one element. It is easy to see that $A$ is closed in $X$ and has empty interior. However, $A$ is not separable. Indeed, $A$ is the 1-point compactification of a discrete space of cardinality $2^{\aleph_0}$ (the characteristic functions of singletons form a discrete space, and the zero function is the point at infinity).

Here is another closely related example. Let $Y=[0,1]\times \{0,1\}$ with the lexicographic order and let $X=Y\times Y$. Let $D\subset X$ be the set of points of the form $((x,0),(x,1))$. Then $D$ is discrete, since $(-\infty,(x,1))\times((x,0),\infty)$ isolates $((x,0),(x,1))$ in $D$. So the closure $A$ of $D$ is not separable, but can easily be seen to have empty interior.

(To see how this is related to the first example, identify $((x,i),(y,j))$ with the interval from $x$ to $y$ in $\mathbb R$ where the second coordinates $i$ and $j$ determine whether the endpoints are included.)

Answer 2

Here is an example under the assumption that $\beta \mathbb N \setminus \mathbb N$ has P-points:

In this paper of Jayachandran and Rajagopalan, a construction of a topological space $X$ is referenced, which is a compactification of $\mathbb N \cup \{p\}$, where $p$ is a P-point of $\beta \mathbb N \setminus \mathbb N$, such that $A := X \setminus \mathbb N$ is homeomorphic to $\omega_1+1$ (see Remark 1.6a).

Hence, $X$ and $A$ are as required.

Answer 3

Assuming Martin's Axiom + $\neg\mathrm{CH}$, we have a counterexample given by $X = D^{\aleph_1}$ (where $D$ is the unit disk of the complex plane) and $A$ being the unit ball of $\ell^p(\aleph_1)$ for any $1 \leq p < \infty$. $X$ being separable follows from the Hewitt-Marczewski-Pondiczery theorem. Martin's Axiom + $\neg\mathrm{CH}$ implies $2^{\aleph_1} = \mathfrak{c}$, so $|X| = (2^{\aleph_0})^{\aleph_1} = 2^{\aleph_0\times\aleph_1} = 2^{\aleph_1} = \mathfrak{c}$. $A$ is compact under the weak$^*$ topology. It is easy to verify that, in $\ell^p(\aleph_1)$, the weak$^*$ topology and the topology of pointwise convergence coincide on bounded sets, so $A$ is compact and therefore closed in $X$. It is easy to see that $A$ has no interior. $A$ is however not separable, as any element of $A$ can only take nonzero values on countably many coordinates, so countably many elements of $A$ must leave $\aleph_1$ many coordinates all zero, whence these elements cannot be dense in $A$.