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 $x \in \operatorname{Fr} A$ :
 * for every $U \in \mathcal B$:
 * if $x \in U$
 * then $A \cap U \ne \varnothing$ and $\complement_S \left({A}\right) \cap U \ne \varnothing$

where:


 * $\complement_S \left({A}\right) = S \setminus A$ denotes the complement of $A$ in $S$


 * $\operatorname{Fr} A$ denotes the boundary of $A$ in $T$.

Sufficient Condition
Let $x \in \operatorname{Fr} A$.

Let $U \in \mathcal B$.

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

Thus from Characterization of Boundary by Open Sets:
 * if $x \in U$
 * then $A \cap U \ne \varnothing$ and $\complement_S \left({A}\right) \cap U \ne \varnothing$.

Necessary Condition
Let $x$ be such that for every $U \in \mathcal B$:
 * if $x \in U$
 * then $A \cap U \ne \varnothing$ and $\complement_S \left({A}\right) \cap U \ne \varnothing$.

By Characterization of Boundary by Open Sets, 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 $A \cap U \ne \varnothing$ and $\complement_S \left({A}\right) \cap U \ne \varnothing$.

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

and:
 * $\complement_S \left({A}\right) \cap V \ne \varnothing$

From the corollary to Set Intersection Preserves Subsets:


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

and:
 * $\complement_S \left({A}\right) \cap V \subseteq \complement_S \left({A}\right) \cap U$

So:
 * $A \cap U \ne \varnothing$ and $\complement_S \left({A}\right) \cap U \ne \varnothing$

and hence the result.