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
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $13 \text { - } 15$. Excluded Point Topology: $3$