Is $U\cap C$ dense in $C$?

Question

Suppose $X$ is a topological space, $C$ is a closed subset of $X$, $U$ is an open subset of $X$ and $U$ is dense in $X$, is $U\cap C$ dense in $C$?

Answer 1

No, take $C=\mathbb{Z}$ (closed in $X=\mathbb{R}$ (usual topology)) and $U = \mathbb{R}\setminus{Z}$ which is open and dense. $C \cap U=\emptyset$ is not dense in $C$.

Any space with a closed set $C$ with empty interior (and $U$ its complement) will do.

Answer 2

This is true if and only if $C$ is the closure of its interior. Such sets are called regular closed sets.

Indeed, if $C=\overline{V}$ for some open $V$, then $U\cap V$ is dense in $V$, and hence in $C$, so $U\cap C\supseteq U\cap V$ is dense in $C$. (Note that we do not really use the assumption that $U$ is open --- only density matters.)

Otherwise, if $C$ is not the closure of its interior, $U=(X\setminus C)\cup \operatorname{int}(C)$ is open and dense, and $U\cap C=\operatorname{int}(C)$, so $U\cap C$ is not dense in $C$.