Components are Open iff Union of Open Connected Sets

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Theorem

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


The following 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 $S = \bigcup \{ U \subseteq S : U \in \tau \text { and } U \text { is connected} \}$.

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


Lemma

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


Then:

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


Hence $C$ is the union of open sets.

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

The result follows.

$\blacksquare$

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