# Path Components are Open iff Union of Open Path-Connected Sets/Space is Union of Open Path-Connected Sets implies Path Components are Open

## Theorem

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

Let $S$ be the union of open path-connected sets of $T$.

Then:

The path components of $T$ are open sets.

## Proof

Let $S = \bigcup \{ U \subseteq S : U \in \tau \text { and } U \text { is path-connected} \}$.

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

#### Lemma

For any path-connected set $U$ then:
$U \cap C \ne \O$ if and only if $U \ne \O$ and $U \subseteq C$

Then:

 $\displaystyle C$ $=$ $\displaystyle C \cap S$ Intersection with Subset is Subset $\displaystyle$ $=$ $\displaystyle C \cap \bigcup \{ U \subseteq S : U \in \tau \text { and } U \text { is path-connected} \}$ $\displaystyle$ $=$ $\displaystyle \bigcup \{ C \cap U : U \in \tau \text { and } U \text { is path-connected} \}$ Intersection Distributes over Union $\displaystyle$ $=$ $\displaystyle \bigcup \{ C \cap U : U \in \tau, U \cap C \neq \O \text { and } U \text { is path-connected} \}$ Union with Empty Set $\displaystyle$ $=$ $\displaystyle \bigcup \{ U \subseteq C : U \in \tau, U \neq \O \text { and } U \text { is path-connected} \}$ Lemma

Hence $C$ is the union of open sets.

By definition of a topology then $C$ is an open set.

The result follows.

$\blacksquare$