Definition 4.5.2 (Closure). Let $X$ be a topological space, $A \subset X$, $x \in X$, and $\fB \subset \cn(x)$ be a fundamental system of neighbourhoods, then the following are equivalent:
For every $B \supset A$ closed, $x \in B$.
For every $U \in \cn(x)$, $U \cap A \ne \emptyset$.
For every $U \in \fB$, $U \cap A \ne \emptyset$.
There exists a filter $\fF \subset 2^{A}$ that converges to $\fF$.
The set $\ol{A}$ of all points satisfying the above is the closure of $A$ in $X$. By (1), it is the smallest closed set containing $A$.
Proof. $\neg (2) \Rightarrow \neg (1)$: Let $U \in \cn(x)$ such that $U \cap A = \emptyset$ and $V \in \cn^{o}(x)$ with $V \subset U$. Since $x \in V$ by (V1), $V^{c} \supset A$ is closed with $x \not\in V^{c}$.
$(2) \Rightarrow (3)$: $\cn(x) \supset \fB$.
$(3) \Rightarrow (4)$: Let $\fF = \bracs{V \cap A| V \in \cn(x)}$. For any $V \cap A \in \fF$, there exists $W \in \fB$ such that $W \subset V$. Since $W \cap A \ne \emptyset$, $V \cap A \ne \emptyset$ as well. Thus $\emptyset \not\in \fF$, and $\fF$ is a filter in $A$, and $\fF$ converges to $x$ by definition.
$\neg (1) \Rightarrow \neg (4)$: Let $B \supset A$ such that $x\not\in B$, then $B^{c} \in \cn^{o}(x)$ with $B^{c} \cap A = \emptyset$. Thus there exists no filter on $A$ containing $B^{c} \cap A$, and no filter on $A$ converging to $x$.$\square$