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

## Theorem

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 $\AA$ be such that no two of its elements are disjoint:

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

Then $\ds \bigcup \AA$ is itself connected.

### Corollary

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 $A := \ds \bigcup \AA$.

Let $D = \set {0, 1}$, with the discrete topology.

Let $f: A \to D$ be a continuous mapping.

To show that $A$ is connected, we need to show that $f$ is not a surjection.

From Connected Set in Subspace, each $C \in \AA$ is connected in $A$.

From Restriction of Continuous Mapping is Continuous, the restriction $f \restriction_C$ is continuous for all $C \in \AA$.

From Continuous Image of Connected Space is Connected, $f \sqbrk C$ is connected in $D$ for all $C \in \AA$.

From Discrete Space is Totally Disconnected, $f \sqbrk C$ is a singleton for all $C \in \AA$:

$f \sqbrk C = \set {\map \epsilon C}$

where $\map \epsilon C = 0$ or $1$.

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

Hence:

$\forall B, C \in \AA: \map \epsilon B = \map \epsilon C$

Thus $f$ is constant on $A$ as required.

$\blacksquare$