# 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 $V' = \O$.

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 |

$\blacksquare$

## Sources

- 2000: John M. Lee:
*Introduction to Topological Manifolds*: $\S 4$ Connectedness and Compactness, Proposition $4.9 \ \text {(a)}$