Example for hollow sets whose complement is not dense in $\mathbb{R}$. A set is hollow if it has empty interior. A set is no where dense (closure is hollow) if and only if i̶t̶s̶ ̶c̶o̶m̶p̶l̶e̶m̶e̶n̶t̶ ̶i̶s̶ ̶d̶e̶n̶s̶e̶ .̶ **its complement contains a dense open set.** Therefore the complement of closed hollow set is dense, and there has to be some hollow set whose complement is not dense. Working in $\mathbb{R}$ with the usual topology, I am trying to come up with an example of not-closed hollow set whose complement is not dense. But it seems for each open set $O\subset \mathbb{R}$, $O$ has to contain elements in the complement of the hollow set, or else $O$ would be in the interior of the hollow set. Edit: I think my definition was wrong.

Your last paragraph is basically already the answer to your question. Suppose $A$ is hollow. Then for any (nonempty) open set $O$, $O$ must contain some $x\in \mathbb{R}\setminus A$ (why? otherwise $O$ would be a subset of the interior of $A$, which would make $A$ not hollow).

But this is _exactly_ the statement that the complement of $A$ is dense!

So indeed, the complement of a hollow set is always dense.

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Note that there's nothing special about $\mathbb{R}$, here - this is a general fact about topological spaces.