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 $T$:

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

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:

	\(\displaystyle \)	\(\displaystyle T \setminus H^\circ\)	\(=\)	\(\displaystyle T \setminus \bigcup_{K \in \mathbb K} K\)	\(\displaystyle \)	Definition of interior of $H$
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \bigcap_{K \in \mathbb K} \left({T \setminus K}\right)\)	\(\displaystyle \)	De Morgan's Laws

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:

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

$\blacksquare$

The other result is demonstrated similarly.

Alternatively, we note that:

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

and so:

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

$\blacksquare$

Sources

• Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (1970)... (previous)... (next): $\text{I}: \ \S 1$: Closures and Interiors