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