Proposition

Show that a subset $A \subset X$ is dense if and only if every nonempty open subset of $X$ contains a point of $A$.

Solution

Suppose $A \subset X$ is dense. Let $U \subset X$ be a nonempty open subset. Suppose that $U$ contains no point of $A$. Then $A \subset U^c$. Since $U^c$ is a closed set containing $A$, $\overline{A} \subset U^c$. Since $U$ is nonempty, $U^c \subsetneq X$. This implies $\overline{A} \ne X$, which is a contradiction.

On the other hand, suppose that every nonempty open subset of $X$ contains a point of $A$. Since $\overline{A}$ is a closed set containing $A$, $\overline{A}^c$ is an open subset of $X$ that does not contain a point of $A$. Therefore, $\overline{A}^c$ must be closed. In other words, $\overline{A} = X$, so $A$ is dense.