Interior equals Complement of 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 $H^\prime$ denote the complement of $H$ in $T$:
- $H^\prime = T \setminus H$
Then:
- $H^\circ = H^{\prime \, - \, \prime}$
Proof
\(\ds H^{\circ \, \prime}\) | \(=\) | \(\ds H^{\prime \, -}\) | Complement of Interior equals Closure of Complement | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {H^{\circ \, \prime} }^\prime\) | \(=\) | \(\ds \paren {H^{\prime \, -} }^\prime\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds H^{\circ \, \prime \, \prime}\) | \(=\) | \(\ds H^{\prime \, - \, \prime}\) | Composition of Mappings is Associative | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds H^\circ\) | \(=\) | \(\ds H^{\prime \, - \, \prime}\) | Relative Complement of Relative Complement |
$\blacksquare$