Separated Sets are Clopen in Union
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Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $A$ and $B$ be separated sets in $T$.
Let $H = A \cup B$ be given the subspace topology.
Then $A$ and $B$ are each both open and closed in $H$.
Proof
By hypothesis, $A$ and $B$ are separated:
- $A \cap B^- = A^- \cap B = \O$
Then:
\(\ds H \cap B^-\) | \(=\) | \(\ds \paren {A \cup B} \cap B^-\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {A \cap B^-} \cup \paren {B \cap B^-}\) | Intersection Absorbs Union | |||||||||||
\(\ds \) | \(=\) | \(\ds \O \cup B\) | Set is Subset of its Topological Closure and Intersection with Subset is Subset‎ | |||||||||||
\(\ds \) | \(=\) | \(\ds B\) | Union with Empty Set |
Since the intersection of a closed set with a subspace is closed in the subspace, $B$ is closed in $H$.
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Since $A = H \setminus B$ and $B$ is closed in $H$, $A$ is open in $H$.
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By the same argument with the roles of $A$ and $B$ reversed, $A$ is closed in $H$ and $B$ is open in $H$.
Hence the result.
$\blacksquare$