# Clopen Set contains Components of All its Points

## Theorem

Let $T = \left({S, \tau}\right)$ be a topological space.

Let $H \subseteq S$ be both closed and open in $T$.

Then $H$ contains the components of all of its points.

## Proof

Let $H$ be a clopen set in $T$.

By definition, $H$ is open and so $H \in \tau$.

But as $H$ is also closed, by definition $\complement_S \left({H}\right) \in \tau$ where $\complement_S$ denotes complement relative to $S$.

Thus $H$ and $\complement_S \left({H}\right)$ are both open such that:

- $H \cap \complement_S \left({H}\right) = \varnothing$ from Intersection with Relative Complement is Empty
- $H \cup \complement_S \left({H}\right) = S$ from Union with Relative Complement

and so forming a partition of $T$.

As $H$ and $\complement_S \left({H}\right)$ are both closed, it follows from Closed Set Equals its Closure that:

- $H \cap \left({\complement_S \left({H}\right)}\right)^- = \varnothing = H^- \cap \complement_S \left({H}\right)$

and so by definition $H$ and $\complement_S \left({H}\right)$ are separated.

Aiming for a contradiction, suppose that $H$ does not contain the components of all of its points.

That is:

- $\exists x \in H: \operatorname{Comp}_x \left({T}\right) \nsubseteq H$

where $\operatorname{Comp}_x \left({T}\right)$ is the component of $x$ in $T$.

Then:

- $\exists z \in \operatorname{Comp}_x \left({T}\right): z \in H \cup \complement_S \left({H}\right)$

This means that $H \cup \complement_S \left({H}\right)$ is connected.

This contradicts the fact that $H$ and $\complement_S \left({H}\right)$ are separated.

The result follows by Proof by Contradiction.

$\blacksquare$

## Sources

- 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*... (previous) ... (next): $\text{I}: \ \S 4$