Connected Subset of Union of Disjoint Open Sets

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Theorem

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

Let $A$ be a connected set of $T$.

Let $U, V$ be disjoint open sets.

Let $A \subseteq U \cup V$.


Then

either $A \subseteq U$ or $A \subseteq V$.


Proof

Let $U' = A \cap U$ and $V' = A \cap V$.

By definition $U'$ and $V'$ are open sets in the subspace $\struct{A, \tau_A}$.

From Intersection is Empty Implies Intersection of Subsets is Empty $U'$ and $V'$ are disjoint.

Hence $U'$ and $V'$ are separated sets by definition.


Now

\(\ds A\) \(=\) \(\ds A \cap \paren {U \cup V}\) Intersection with Subset is Subset
\(\ds \) \(=\) \(\ds \paren {A \cap U} \cup \paren {A \cap V}\) Intersection Distributes over Union
\(\ds \) \(=\) \(\ds U' \cup V'\)

Since $A$ is connected then one of $U'$ or $V'$ is empty.

Without loss of generality assume that $V' = \O$.

Then

\(\ds A\) \(=\) \(\ds U' \cup V'\)
\(\ds \) \(=\) \(\ds U' \cup \O\)
\(\ds \) \(=\) \(\ds U'\) Union with Empty Set
\(\ds \) \(=\) \(\ds A \cap U\)
\(\ds \leadsto \ \ \) \(\ds A\) \(\subseteq\) \(\ds U\) Intersection with Subset is Subset

$\blacksquare$


Sources

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