# 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:

\(\displaystyle H \cap B^-\) | \(=\) | \(\displaystyle \paren {A \cup B} \cap B^-\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {A \cap B^-} \cup \paren {B \cap B^-}\) | Intersection Absorbs Union | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \O \cup B\) | Set is Subset of its Topological Closure and Intersection with Subset is Subset | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle B\) | Union with Empty Set |

Since the intersection of a closed set with a subspace is closed in the subspace, $B$ is closed in $H$.

Since $A = H \setminus B$ and $B$ is closed in $H$, $A$ is open in $H$.

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.

$\blacksquare$