Union of Connected Sets with Common Point is Connected

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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