Characterization of Boundary by Open Sets

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

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

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

Then $x \in \operatorname{fr} A$ iff for every open subset $U$ of $T$ if $x \in U$, then $A \cap U \neq \varnothing$ and $A` \cap U \neq \varnothing$

where $A` = S \setminus A$ stands for complement of $A$.

Proof
To prove first implication assume $x \in \operatorname{fr} A$.

Then $x \in \operatorname{Cl} A`$ and $x \in \operatorname{Cl} A$ by Boundary is Intersection of Closure with Closure of Complement.

Hence $A \cap U \neq \varnothing$ and $A` \cap U \neq \varnothing$ for every open set $U$ of $T$ if $x \in U$ by Condition for Point being in Closure.

Assume for every open set $U$ of $T$ if $x \in U$, then $A \cap U \neq \varnothing$ and $A` \cap U \neq \varnothing$.

Then $x \in \operatorname{Cl} A`$ and $x \in \operatorname{Cl} A$ by Condition for Point being in Closure.

Hence $x \in \operatorname{fr} A$ by Boundary is Intersection of Closure with Closure of Complement.