Union of Connected Sets with Common Point is Connected
Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let $\family {B_\alpha}_{\alpha \mathop \in A}$ be a family of connected sets of $T$.
Let $\exists x \in \ds \bigcap \family {B_\alpha}_{\alpha \mathop \in A}$.
Then
- $\ds \bigcup \family {B_\alpha}_{\alpha \mathop \in A}$ is a connected set of $T$.
Proof 1
Let $B = \ds \bigcup_{\alpha \mathop \in A} B_\alpha$.
From definition 3 of a connected set, $B$ is connected in $T$ if and only if the subspace $\struct {B, \tau_B}$ is a connected space.
From definition 4 of a connected space, $\struct {B, \tau_B}$ is connected if and only if the only clopen sets in $\struct {B, \tau_B}$ are $B$ and $\O$.
Let $U$ be any clopen set of the subspace of $\struct {B, \tau_B}$.
Let $V = B \setminus U$.
From Complement of Clopen Set is Clopen, $V$ is also clopen.
Hence $U, V$ are disjoint clopen sets such that $B = U \cup V$.
Without loss of generality assume $x \in U$.
From Set is Subset of Union:
- $\ds \forall \alpha \in A : B_\alpha \subseteq B = U \cup V$
From Connected Subset of Union of Disjoint Open Sets:
- $\forall \alpha \in A : B_\alpha \subseteq U$
From Union is Smallest Superset:
- $B \subseteq U$
Since $U \subseteq B$ then $U = B$.
From Set Difference with Self is Empty Set,
- $V = B \setminus U = B \setminus B = \O$
Hence the only clopen sets in $\struct {B, \tau_B}$ are $B$ and $\O$.
The result follows.
$\blacksquare$
Proof 2
Follows immediately from Union of Connected Sets with Non-Empty Intersections is Connected.
$\blacksquare$
Sources
- 2000: John M. Lee: Introduction to Topological Manifolds: $\S 4$ Connectedness and Compactness, Proposition $4.9 \ \text {(d)}$