Characterization of Closure by Basis

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Let $\mathcal B \subseteq \tau$ be a basis.

Let $A$ be a subset of $T$.

Let $x$ be a point of $T$.

Then $x \in A^-$ :
 * for every $U \in \mathcal B$:
 * if $x \in U$ then $A \cap U \ne \varnothing$

where:


 * $A^-$ denotes the closure of $A$

Sufficient Condition
Let $x \in A^-$.

Let $U \in \mathcal B$.

By definition of basis, $U$ is an open set of $T$.

Thus from Condition for Point being in Closure:
 * if $x \in U$ then $A \cap U \ne \varnothing$.

Necessary Condition
Let $x$ be such that for every $U \in \mathcal B$:
 * if $x \in U$
 * then $A \cap U \ne \varnothing$.

By Condition for Point being in Closure, to prove that $x \in \operatorname{Fr} A$ it is enough to prove that:
 * for every open set $U$ of $T$:
 * if $x \in U$ then $A \cap U \ne \varnothing$.

Let $U$ be an open set of $T$.

Let $x \in U$.

By definition of (analytic) basis, there exists $V \in \mathcal B$ such that:
 * $x \in V \subseteq U$

By assumption:
 * $A \cap V \ne \varnothing$

From the corollary to Set Intersection Preserves Subsets:


 * $A \cap V \subseteq A \cap U$

So:
 * $A \cap U \ne \varnothing$

and hence the result.