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

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

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

Without loss of generality assume that $V’ = \empty$.

Then

\(\displaystyle A\) \(=\) \(\displaystyle U’ \cup V’\)
\(\displaystyle \) \(=\) \(\displaystyle U’ \cup \empty\)
\(\displaystyle \) \(=\) \(\displaystyle U’\) Union with Empty Set
\(\displaystyle \) \(=\) \(\displaystyle A \cap U\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle A\) \(\subseteq\) \(\displaystyle U\) Intersection with Subset is Subset

$\blacksquare$

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