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

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

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

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

Then:
 * $U \cap C \ne \O$ $U \ne \O$ and $U \subseteq C$

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

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

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

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

Hence $C = U \cup C$.

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

Then:

The result follows.

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

Then:

The result follows.