If $int\overline{X} = int\overline{Y} = \emptyset$ then $int(\overline{X\cup Y}) = \emptyset$.

Question

Let $X$ and $Y$ be subsets of a metric space. If $int\overline{X} = int\overline{Y} = \emptyset$ then $int(\overline{X\cup Y}) = \emptyset$.

I know that $int(\overline{X\cup Y}) = int(\overline{X}\cup \overline{Y})$ and so I'm trying to prove the more general statement $$ int(X) = int(Y) = \emptyset \implies int(X\cup Y) = \emptyset $$ which seems true to me but I not getting the proof.

What I tried so far is to consider an $x\in int(X\cup Y)$, so there is a open ball $B_x$ that is contained in $X\cup Y$ but it cannot be contained in $X$ or $Y$ because these have empty interior. But I don't know what to do next.

Thanks in advance for any help.

Answer 1

$X$ is called nowhere dense iff the interior of its closure is empty iff the complement of $X$ contains a dense and open subset iff every non-empty open subset $O$ contains a non-empty open subset $O'$ that misses $X$.

As the finite intersection of dense and open subsets is dense and open, by de Morgan, the finite union of nowhere dense sets is nowhere dense. Or using the final characterisation: first pick $O'$ to avoid $X$ and an even smaller one inside $O'$ to also avoid $Y$.

Answer 2

Unfortunately, the statement you were expecting to hold does not, as Daniel Schepler illustratively pointed out. There is however a series of elementary propositions valid over any topological space $(X, \mathscr{T})$:

Proposition 1. For any $M, N \subseteq X$ it holds that $\overline{M \cup N}=\overline{M} \cup \overline{N}$ (finite additivity of the closure operator).

Proposition 2. For any subset $M \subseteq X$ and open subset $U \in \mathscr{T}$ it holds that $U \cap \overline{M} \subseteq \overline{U \cap M}$.

Corollary 1. If $T \subseteq X$ is a dense subset (i.e. $\overline{T}=X$), then $\overline{U}=\overline{U \cap T}$.

Corollary 2. The intersection of two dense open subsets is also dense and open.

If $F, G$ are two closed sets of empty interior, then their complementaries are open dense subsets and subsequently so is their intersection (corollary 2). Hence $F \cup G$ has empty interior.