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 $\complement \left({H}\right)$ be the complement of $H$ in $T$:


 * $\complement \left({H}\right) = T \setminus H$

Then:
 * $\complement \left({ H^\circ}\right) = \left({\complement \left({H}\right)}\right)^-$

and similarly:


 * $\left({\complement \left({ H}\right)}\right)^\circ = \complement \left({H^-}\right)$

These can alternatively be written:


 * $T \setminus H^\circ = \left({T \setminus H}\right)^-$


 * $\left({T \setminus H}\right)^\circ = T \setminus H^-$

which is arguably easier to follow.

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

Let $\mathbb K = \left\{{K \in \vartheta: K \subseteq H}\right\}$.

Then:

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

Also, from Complements Invert Subsets we have that $T \setminus K \supseteq T \setminus H$.

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

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

Thus:
 * $\displaystyle T \setminus H^\circ = \bigcap_{K \in \mathbb K'} \left({T \setminus K}\right)$

The other result is demonstrated similarly.

Alternatively, we note that:

and so: