Characterization of Boundary 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 $x \in \operatorname{Fr} A$ iff for every subset $U \in \mathcal B$ if $x \in U$, then $A \cap U \neq \varnothing$ and $A' \cap U \neq \varnothing$

where
 * $A' = S \setminus A$ denotes the complement of $A$,
 * $\operatorname{Fr} A$ denotes the boundary of $A$.

Proof
First implication follows from Characterization of Boundary by Open Sets because an element of basis is open.

Assume for every subset $U \in \mathcal B$ if $x \in U$, then $A \cap U \neq \varnothing$ and $A' \cap U \neq \varnothing$.

To prove that $x \in \operatorname{Fr} A$ according to Characterization of Boundary by Open Sets it is enough to prove that for every open subset $U$ of $T$ if $x \in U$ then $A \cap U \neq \varnothing$.

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

Assume that $x \in U$. Then there exists $V \in \mathcal B$ such that $x \in V \subseteq U$ by Definition:Basis (Topology)/Analytic Basis.

By assuption $A \cap V \neq \varnothing$ and $A' \cap V \neq \varnothing$.

$\varnothing \neq A \cap V \subseteq A \cap U$. $\varnothing \neq A' \cap V \subseteq A' \cap U$.