# Boundary is Intersection of Closure with Closure of Complement

## Theorem

Let $T$ be a topological space.

Let $X \subseteq T$.

Let $\partial X$ denote the boundary of $X$, defined as:

$\partial X = X^- \setminus X^\circ$

Let $\overline X = T \setminus X$ denote the complement of $X$ in $T$.

Let $X^-$ denote the closure of $X$.

Then:

$\partial X = X^- \cap \paren {\overline X}^-$

## Proof

 $\ds \partial X$ $=$ $\ds X^- \setminus X^\circ$ Definition 1 of Boundary (Topology): $X^\circ$ is the interior of $X$ $\ds$ $=$ $\ds X^- \cap \overline {\paren {X^\circ} }$ Set Difference as Intersection with Relative Complement $\ds$ $=$ $\ds X^- \cap \paren {\overline X}^-$ Complement of Interior equals Closure of Complement

$\blacksquare$