Path Components are Open iff Union of Open Path-Connected Sets

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Theorem

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


The following are equivalent::

(1) $\quad$ The path components of $T$ are open.
(2) $\quad S$ is a union of open path-connected sets of $T$.


Proof

Condition (1) implies Condition (2)

Let the path components of $T$ be open sets.

By definition, the path components of $T$ are a partition of $S$.

Hence $S$ is the union of the open path components of $T$.

Since a path component is a maximal path-connected set by definition, then $S$ is a union of open path-connected sets of $T$

$\Box$

Condition (2) implies Condition (1)

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:
$U \cap C \neq \O$ if and only if $U \neq \O$ 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.

$\blacksquare$

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