Union of Connected Sets with Non-Empty Intersections is Connected

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Theorem

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

Let $I$ be an indexing set.

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

Let $\mathcal A$ be such that no two of its elements are disjoint:

$\forall B, C \in \mathcal A: B \cap C \ne \O$


Then $\displaystyle \bigcup \mathcal A$ is itself connected.


Corollary

Let $T = \left({S, \tau}\right)$ be a topological space.

Let $I$ be an indexing set.

Let $\mathcal A = \left \langle{A_\alpha}\right \rangle_{\alpha \mathop \in I}$ be an indexed family of subsets of $S$, all connected in $T$.

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

$\forall C \in \mathcal A: B \cap C \ne \varnothing$


Then $\displaystyle B \cup \bigcup \mathcal A$ is connected.


Proof

Let $A := \displaystyle \bigcup \mathcal A$.

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

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

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


Since each $C \in \mathcal A$ is connected and the restriction $f \restriction_C$ is continuous:

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

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

But, for all $B, C \in \mathcal A$:

$B \cap C \ne \O$

Hence $\map \epsilon B = \map \epsilon C$.

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

$\blacksquare$


Sources