$C$ closed with nonempty interior $\Rightarrow$ $\operatorname{int}C$ is dense Let $X$ be a topological space. Let's consider $C$ a closed set in $X$, such that $\operatorname{int}C\neq\emptyset$. I should prove that...--prophetes.ai

$C$ closed with nonempty interior $\Rightarrow$ $\operatorname{int}C$ is dense Let $X$ be a topological space. Let's consider $C$ a closed set in $X$, such that $\operatorname{int}C\neq\emptyset$. I should prove that $\operatorname{int}C$ is dense". The book doesn't specify _where_ should it be dense, but I think it means "$\operatorname{int}C$ is dense in $C$". How can I prove this? Mant thanks

The statement is false: consider the set $C=[0,1]\cup\\{2\\}$ in the space $\Bbb R$ with the usual topology. The interior of $C$ is $(0,1)$, which is not dense in $C$.