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 $\complement \left({H}\right)$ 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, it can be argued, is easier to follow.


Proof

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

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


Then:

\(\displaystyle T \setminus H^\circ\) \(=\) \(\displaystyle T \setminus \bigcup_{K \mathop \in \mathbb K} K\) Definition of Interior (Topology) of $H$
\(\displaystyle \) \(=\) \(\displaystyle \bigcap_{K \mathop \in \mathbb K} \left({T \setminus K}\right)\) 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:

$\mathbb K' := \left\{{K' \subseteq T: \left( T \setminus H \right) \subseteq K', K' \text { closed in } T}\right\}$.

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

Thus:

\(\displaystyle T \setminus H^\circ\) \(=\) \(\displaystyle \bigcap_{K' \mathop \in \mathbb K'} K'\)
\(\displaystyle \) \(=\) \(\displaystyle \left( T \setminus H \right)^-\) Definition of Closure of $T \setminus H$

$\Box$


Then we note that:

\(\displaystyle H^\circ\) \(=\) \(\displaystyle T \setminus \left({T \setminus H^\circ}\right)\) Relative Complement of Relative Complement
\(\displaystyle \) \(=\) \(\displaystyle T \setminus \left({\left({T \setminus H}\right)^-}\right)\) from above

and so:

\(\displaystyle \left({T \setminus H}\right)^\circ\) \(=\) \(\displaystyle T \setminus \left({\left({T \setminus \left({T \setminus H}\right)}\right)^-}\right)\) from above
\(\displaystyle \) \(=\) \(\displaystyle T \setminus H^-\) Relative Complement of Relative Complement

$\blacksquare$


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