Excluded Point Space is Connected/Proof 2
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Theorem
Let $T = \left({S, \tau_{\bar p}}\right)$ be an excluded point space.
Then $T^*_{\bar p}$ is a connected space.
Proof
The only open set of $T$ which contains $p$ is $S$.
Therefore it is impossible to set up a separation of $T$, as $S$ will always need to be an element of such a separation.
$\blacksquare$
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: $3$