Clopen Set contains Components of All its Points

Theorem
Let $T = \struct {S, \tau}$ 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 $\relcomp S H \in \tau$ where $\complement_S$ denotes complement relative to $S$.

Thus $H$ and $\relcomp S H$ are both open such that:
 * $H \cap \relcomp S H = \O$ from Intersection with Relative Complement is Empty
 * $H \cup \relcomp S H = S$ from Union with Relative Complement

and so forming a partition of $T$.

As $H$ and $\relcomp S H$ are both closed, it follows from Closed Set Equals its Closure that:
 * $H \cap \paren {\relcomp S H}^- = \O = H^- \cap \relcomp S H$

and so by definition $H$ and $\relcomp S H$ are separated.

that $H$ does not contain the components of all of its points.

That is:
 * $\exists x \in H: \map {\operatorname {Comp}_x} T \nsubseteq H$

where $\map {\operatorname {Comp}_x} T$ is the component of $x$ in $T$.

Then:
 * $\exists z \in \map {\operatorname {Comp}_x} T: z \in H \cup \relcomp S H$

This means that $H \cup \relcomp S H$ is connected.

This contradicts the fact that $H$ and $\relcomp S H$ are separated.

The result follows by Proof by Contradiction.