# Complement of Interior equals Closure of Complement

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## 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:

 $\displaystyle T \setminus H^\circ$ $=$ $\displaystyle T \setminus \bigcup_{K \mathop \in \mathbb K} K$ Definition of Interior (Topology) of $H$ $\displaystyle$ $=$ $\displaystyle \bigcap_{K \mathop \in \mathbb K} \paren {T \setminus K}$ De Morgan's Laws: Difference with Union

By the definition of closed set, $K$ is open in $T$ if and only if $T \setminus K$ 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:

 $\displaystyle T \setminus H^\circ$ $=$ $\displaystyle \bigcap_{K' \mathop \in \mathbb K'} K'$ $\displaystyle$ $=$ $\displaystyle \paren {T \setminus H}^-$ Definition 2 of Closure of $T \setminus H$

$\Box$

Then we note that:

 $\displaystyle H^\circ$ $=$ $\displaystyle T \setminus \paren {T \setminus H^\circ}$ Relative Complement of Relative Complement $\displaystyle$ $=$ $\displaystyle T \setminus \paren {\paren {T \setminus H}^-}$ from above

and so:

 $\displaystyle \paren {T \setminus H}^\circ$ $=$ $\displaystyle T \setminus \paren {\paren {T \setminus \paren {T \setminus H} }^-}$ from above $\displaystyle$ $=$ $\displaystyle T \setminus H^-$ Relative Complement of Relative Complement

$\blacksquare$