Excluded Point Space is Ultraconnected

Theorem

Let $T = \left({S, \tau_{\bar p}}\right)$ be an excluded point space.

Then $T$ is ultraconnected.

Proof 1

Excluded Point Topology is Open Extension Topology of Discrete Topology

Open Extension Space is Ultraconnected

$\blacksquare$

Proof 2

Apart from $S$, every open set of $T$ does not contain $p$, by definition of excluded point space.

So, apart from $\O$, every closed set of $T$ does contain $p$, by definition of closed set.

So every pair of closed sets of $T$ has an intersection which contains at least $p$.

So there are no non-empty disjoint closed sets of $T$.

Hence the result, by definition of ultraconnected.

$\blacksquare$