# Union of Connected Sets with Non-Empty Intersections is Connected/Corollary/Proof 3

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## Corollary to Union of Connected Sets with Non-Empty Intersections is Connected

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

Let $I$ be an indexing set.

Let $\AA = \family {A_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of $S$, all connected in $T$.

Let $B$ be a connected set of $T$ such that:

- $\forall C \in \AA: B \cap C \ne \O$

Then $\ds B \cup \bigcup \AA$ is connected.

## Proof

Let $\ds H = B \cup \bigcup \AA$

Aiming for a contradiction, suppose that $H$ is not connected in $T$.

That is, that $H$ is disconnected.

From the definition of disconnected, there exist separated sets $U, V$ whose union is $H$.

This needs considerable tedious hard slog to complete it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Finish}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

It follows that $B$ is disconnected.

This contradicts our condition that $B$ is a connected set in $T$.

It follows by Proof by Contradiction that $H$ is connected in $T$.

$\blacksquare$

## Sources

- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*(2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $4$: Connectedness: Functions and Products