Excluded Point Space is Ultraconnected/Proof 2

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Theorem

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


Then $T$ is ultraconnected.


Proof

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.

$\blacksquare$


Sources