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

## 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 1

Let $C \in \AA$.

From Union of Connected Sets with Non-Empty Intersections is Connected applied to $B$ and $C$, the union $B \cup C$ is connected.

Thus the set $\tilde \AA = \set {B \cup C: C \in \AA}$ satisfies the conditions of the theorem.

Hence the result.

$\blacksquare$

## Proof 2

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

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

That is, that $H$ is disconnected.

We have that:

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

Thus:

- $\exists x \in H: x \in B \cap C$

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

So:

- $\forall s \in S: s \in U \lor s \in V$

Aiming for a contradiction, suppose $s \in V$.

Then $H, V$ serve as separated sets whose union is a cover for $\family {A_\alpha}_{\alpha \mathop \in I}$.

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Note that the disconnection is valid as:

- $\exists x \in U \cap \family {A_\alpha}_{\alpha \mathop \in I}$

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As $s \in V$, it follows that the $\family {A_\alpha}_{\alpha \mathop \in I}$ are disconnected.

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From this contradiction it follows that $s \notin V$.

Hence $s \in U$.

This holds for arbitrary $s \in H$.

Hence:

- $H \subseteq U$

But as $U$ and $V$ are separated, $U \cap H$ is empty, as required per the definition of a disconnected set.

This contradicts our deduction that $H \subseteq U$.

Hence the sets $U, V$ are not separated sets in $H$.

Thus $H$ is connected in $T$.

$\blacksquare$

## Proof 3

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 $T$.

$\blacksquare$