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.
Aiming for a contradiction, suppose 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.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $4$: Connectedness