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 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