Definition 4.5.5 (Dense). Let $X$ be a topological space and $A \subset X$, then the following are equivalent:
$\ol{A}= X$.
For every $\emptyset \ne U \subset X$ open, $A \cap U \ne \emptyset$.
For every $U \subset X$ open, $\overline{A \cap U}\supset U$.
If the above holds, then $A$ is a dense subset of $X$.
Proof. $(1) \Rightarrow (2)$: Since $U \ne \emptyset$, there exists $x \in U$, so $U \in \cn(x)$. By (2) of Definition 4.5.2, $U \cap A \ne \emptyset$.
$(2) \Rightarrow (3)$: Let $\emptyset \ne U \subset X$ open. For each $x \in U$ and $V \in \cn(x)$, $U \cap V \in \cn(x)$ by (F2). Thus there exists $y \in A \cap U \cap V$. By (3) of Definition 4.5.2, $x \in \overline{A \cap U}$.
$(3) \Rightarrow (1)$: $X$ is open.$\square$