Characterization of Boundary by Open Sets

Theorem

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

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

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

Then $x \in \partial A$ if and only if:

for every open set $U$ of $T$:

if $x \in U$

then $A \cap U \ne \O$ and $\relcomp S A \cap U \ne \O$

where:

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

$\partial A$ denotes the boundary of $A$.

Proof

Sufficient Condition

Let $x \in \partial A$.

Then by Boundary is Intersection of Closure with Closure of Complement:

$x \in \paren {\relcomp S A}^-$ and $x \in A^-$

where $A^-$ denotes the closure of $A$.

Hence by Condition for Point being in Closure, for every open set $U$ of $T$:

$x \in U \implies A \cap U \ne \O$

and:

$x \in U \implies \relcomp S A \cap U \ne \O$

Necessary Condition

Let $x$ be such that for every open set $U$ of $T$:

if $x \in U$

then $A \cap U \ne \O$ and $\relcomp S A \cap U \ne \O$.

Then by Condition for Point being in Closure:

$x \in \paren {\relcomp S A}^-$ and $x \in A^-$.

Hence by Boundary is Intersection of Closure with Closure of Complement:

$x \in \partial A$

$\blacksquare$

Sources

• 1989: Ryszard Engelking: General Topology (revised and completed ed.)

• Mizar article TOPS_1:28