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

From ProofWiki
Jump to navigation Jump to search

Theorem

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


The following statements 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 $S = \ds \bigcup \set {U \subseteq S : U \in \tau \text { and } U \text { is path-connected} }$.

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


Lemma

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

Then:

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

$\Box$


Then:

\(\ds C\) \(=\) \(\ds C \cap S\) Intersection with Subset is Subset
\(\ds \) \(=\) \(\ds C \cap \bigcup \set {U \subseteq S : U \in \tau \text { and } U \text { is path-connected} }\)
\(\ds \) \(=\) \(\ds \bigcup \set {C \cap U : U \in \tau \text { and } U \text { is path-connected} }\) Intersection Distributes over Union
\(\ds \) \(=\) \(\ds \bigcup \set {C \cap U : U \in \tau, U \cap C \ne \O \text { and } U \text { is path-connected} }\) Union with Empty Set
\(\ds \) \(=\) \(\ds \bigcup \set {U \subseteq C : U \in \tau, U \ne \O \text { and } U \text { is path-connected} }\) Lemma


Hence $C$ is the union of open sets.

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

The result follows.

$\blacksquare$


Also see