Characterization of Boundary by Basis

Theorem
Let $T = \struct {S, \tau}$ be a topological space.

Let $\BB \subseteq \tau$ be a basis.

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

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

Then $x \in \partial A$ :
 * for every $U \in \BB$:
 * if $x \in U$
 * then $A \cap U \ne \O$ and $\relcomp S A \cap U \ne \O$

where:


 * $\relcomp S A = S \setminus A$ denotes the complement of $A$ in $S$


 * $\partial A$ denotes the boundary of $A$ in $T$.

Sufficient Condition
Let $x \in \partial A$.

Let $U \in \BB$.

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

Thus from Characterization of Boundary by Open Sets:
 * if $x \in U$
 * then $A \cap U \ne \O$ and $\relcomp S A \cap U \ne \O$.

Necessary Condition
Let $x$ be such that for every $U \in \BB$:
 * if $x \in U$
 * then $A \cap U \ne \O$ and $\relcomp S A \cap U \ne \O$.

By Characterization of Boundary by Open Sets, to prove that $x \in \partial A$ it is enough to prove that:
 * for every open set $U$ of $T$:
 * if $x \in U$ then $A \cap U \ne \O$ and $\relcomp S A \cap U \ne \O$.

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

Let $x \in U$.

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

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

and:
 * $\relcomp S A \cap V \ne \O$

From the corollary to Set Intersection Preserves Subsets:


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

and:
 * $\relcomp S A \cap V \subseteq \relcomp S A \cap U$

So:
 * $A \cap U \ne \O$ and $\relcomp S A \cap U \ne \O$

and hence the result.