Union of Interiors and Boundary Equals Whole Space

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

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

Then:
 * $S = \operatorname{Int} A \cup \operatorname{Fr} A \cup \operatorname{Int} A'$

where:
 * $A' = S \setminus A$ denotes the complement of $A$ relative to $S$
 * $\operatorname{Int} A$ denotes the interior of $A$
 * $\operatorname{Fr} A$ denotes the boundary of $A$.