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

## Theorem

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

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

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

Then:

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

## Proof

### Necessary Condition

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

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:

 $\displaystyle U$ $=$ $\displaystyle U \cap C$ Intersection with Subset is Subset $\displaystyle$ $\neq$ $\displaystyle \O$ Assumption

The result follows.

$\Box$

### Sufficient Condition

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

Then:

 $\displaystyle U \cap C$ $=$ $\displaystyle U$ Intersection with Subset is Subset $\displaystyle$ $\neq$ $\displaystyle \O$ Assumption

The result follows.

$\blacksquare$