ProofWiki:Sandbox

Theorem
Let $T$ 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 set $U$ of $T$ if $x \in U$, then $A \cap U \neq \emptyset$ and $A` \cap U \neq \emptyset$.

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 $A \cap U \neq \emptyset$ and $A` \cap U \neq \emptyset$ for every open set $U$ of $T$ if $x \in U$ by Characterization of Closure by Open Sets.

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

Then $x \in \operatorname{Cl} A`$ and $x \in \operatorname{Cl} A$ by Characterization of Closure by Open Sets.

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