Interior equals Complement of Closure of Complement

Theorem

Let $T$ be a topological space.

Let $H \subseteq T$.

Let $H^-$ denote the closure of $H$ and $H^\circ$ denote the interior of $H$.

Let $H^\prime$ denote the complement of $H$ in $T$:

$H^\prime = T \setminus H$

Then:

$H^\circ = H^{\prime \, - \, \prime}$

Proof

\(\ds H^{\circ \, \prime}\)

\(=\)

\(\ds H^{\prime \, -}\)

Complement of Interior equals Closure of Complement

\(\ds \leadsto \ \ \)

\(\ds \paren {H^{\circ \, \prime} }^\prime\)

\(=\)

\(\ds \paren {H^{\prime \, -} }^\prime\)

\(\ds \leadsto \ \ \)

\(\ds H^{\circ \, \prime \, \prime}\)

\(=\)

\(\ds H^{\prime \, - \, \prime}\)

Composition of Mappings is Associative

\(\ds \leadsto \ \ \)

\(\ds H^\circ\)

\(=\)

\(\ds H^{\prime \, - \, \prime}\)

Relative Complement of Relative Complement

$\blacksquare$