# Connected Subset of Union of Disjoint Open Sets

## 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 $V' = \O$.

Then

 $\displaystyle A$ $=$ $\displaystyle U' \cup V'$ $\displaystyle$ $=$ $\displaystyle U' \cup \O$ $\displaystyle$ $=$ $\displaystyle U'$ Union with Empty Set $\displaystyle$ $=$ $\displaystyle A \cap U$ $\displaystyle \leadsto \ \$ $\displaystyle A$ $\subseteq$ $\displaystyle U$ Intersection with Subset is Subset

$\blacksquare$