Components are Open iff Union of Open Connected Sets/Space is Union of Open Connected Sets implies Components are Open
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Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let $S$ be a union of open connected sets of $T$.
Then:
- The components of $T$ are open sets.
Proof
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.
$\blacksquare$