Excluded Point Space is Connected/Proof 2

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: $3$
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