Components are Open iff Union of Open Connected Sets/Lemma 1

Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let $C$ be a component of $T$.

Let $U$ be a connected set of $T$.

Then:

$U \cap C \ne \O$ if and only if $U \ne \O$ and $U \subseteq C$

Proof

Necessary Condition

Let $U \cap C \ne \O$.

From Union of Connected Sets with Common Point is Connected, $U \cup C$ is a connected set of $T$.

From Set is Subset of Union, $C \subseteq U \cup C$.

By definition of a component, $C$ is a maximal connected set.

Hence $C = U \cup C$.

From Union with Superset is Superset, $U \subseteq C$.

Then:

\(\ds U\)

\(=\)

\(\ds U \cap C\)

Intersection with Subset is Subset

\(\ds \)

\(\ne\)

\(\ds \O\)

Assumption

The result follows.

$\Box$

Sufficient Condition

Let $U \ne \O$ and $U \subseteq C$.

Then:

\(\ds U \cap C\)

\(=\)

\(\ds U\)

Intersection with Subset is Subset

\(\ds \)

\(\ne\)

\(\ds \O\)

Assumption

The result follows.

$\blacksquare$