Excluded Point Space is not Locally Arc-Connected

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Theorem

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


Then $T$ is not locally arc-connected.


Proof

Let $\mathcal B \subseteq \tau_{\bar p}$ be a basis for $\tau_{\bar p}$.

Since $\mathcal B$ covers $S$, there must be an open set $B \in \mathcal B$ such that $p \in B$.

By definition of the excluded point topology, the only open set containing $p$ is $S$ itself.

Hence necessarily $S \in \mathcal B$.


But by Excluded Point Space is not Arc-Connected, $S$ is not arc-connected.

Hence $\mathcal B$ does not consist only of arc-connected sets.


Because $\mathcal B$ was arbitrary, there cannot exist a basis for $\tau_{\bar p}$ comprising only arc-connected sets.

Hence, by definition, $T$ is not locally arc-connected.

$\blacksquare$


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