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