Set with Dispersion Point is Biconnected
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Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be a connected set in $T$.
Let $p \in H$ be a dispersion point of $H$.
Then $H$ is biconnected.
Proof
Aiming for a contradiction, suppose $H$ is not biconnected.
Then by definition there exist disjoint non-degenerate connected sets $U, V$ such that $H = U \cup V$.
Without loss of generality, let $p \in U$.
Then $V \subset H \setminus \set p$.
As $p$ is a dispersion point of $H$, $H \setminus \set p$ is totally disconnected.
Thus, by definition, $H \setminus \set p$ contains no non-degenerate connected sets.
But this contradicts our definition of $V$, as $V \subset H \setminus \set p$.
It follows by Proof by Contradiction that $H$ is biconnected.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $4$: Connectedness: Biconnectedness and Continua