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

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 $H$.

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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