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.

$\blacksquare$


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