Complement of Interior equals 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 $\map \complement H$ be the complement of $H$ in $T$:


 * $\map \complement H = T \setminus H$

Then:
 * $\map \complement {H^\circ} = \paren {\map \complement H}^-$

and similarly:


 * $\paren {\map \complement H}^\circ = \map \complement {H^-}$

These can alternatively be written:


 * $T \setminus H^\circ = \paren {T \setminus H}^-$


 * $\paren {T \setminus H}^\circ = T \setminus H^-$

which, it can be argued, is easier to follow.

Proof
Let $\tau$ be the topology on $T$.

Let $\mathbb K = \set {K \in \tau: K \subseteq H}$.

Then:

By the definition of closed set, $K$ is open in $T$ $T \setminus K$ is closed in $T$.

Also, from Set Complement inverts Subsets we have that $T \setminus K \supseteq T \setminus H$.

Now consider the set $\mathbb K'$ defined as:
 * $\mathbb K' := \set {K' \subseteq T: \paren {T \setminus H} \subseteq K', K' \text { closed in } T}$

From the above we see that:
 * $K \in \mathbb K \iff T \setminus K \in \mathbb K'$

Thus:

Then we note that:

and so: