Components are Open iff Union of Open Connected Sets

Theorem

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

The following statements are equivalent:

$(1): \quad$ The components of $T$ are open.

$(2): \quad S$ is a union of open connected sets of $T$.

Proof

Condition (1) implies Condition (2)

Let the components of $T$ be open.

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

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

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

$\Box$

Condition (2) implies Condition (1)

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

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

Lemma

For any connected set $U$ 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

\(\ds \)

\(=\)

\(\ds C \cap \bigcup \set { U \subseteq S : U \in \tau \text { and } U \text { is connected} }\)

\(\ds \)

\(=\)

\(\ds \bigcup \set {C \cap U : U \in \tau \text { and } U \text { is connected} }\)

Intersection Distributes over Union

\(\ds \)

\(=\)

\(\ds \bigcup \set {C \cap U : U \in \tau, U \cap C \ne \empty \text { and } U \text { is connected} }\)

Union with Empty Set

\(\ds \)

\(=\)

\(\ds \bigcup \set {U \subseteq C : U \in \tau \text { and } U \text { is connected} }\)

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

• Path Components are Open iff Union of Open Path-Connected Sets, an analogous result for path components