Either-Or Topology is not Locally Arc-Connected

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Theorem

Let $T = \left({S, \tau}\right)$ be the either-or space.


Then $T$ is not a locally arc-connected space.


Proof

Let $\mathcal B$ be a basis for $\tau$.

Suppose that $0 \in B$, where $B \in \mathcal B$.

Then by definition of the either-or topology, $\left({-1 \,.\,.\, 1}\right) \subseteq B$.

In particular, $\dfrac 1 2 \in B$.


Let $f: \left[{0 \,.\,.\, 1}\right] \to B$ be any injection such that:

$f \left({0}\right) = 0$
$f \left({1}\right) = \dfrac 1 2$

Since $0 \notin \left\{{\dfrac 1 2}\right\}$, it follows that $\left\{{\dfrac 1 2}\right\}$ is open in $\tau$.

As $f$ is an injection, it follows that:

$f^{-1} \left({\left\{{\dfrac 1 2}\right\}}\right) = \left\{{1}\right\}$

By Closed Real Interval is not Open Set, $\left\{{1}\right\} = \left[{1 \,.\,.\, 1}\right]$ is not open in $\left[{0 \,.\,.\, 1}\right]$.


Thus, $f$ cannot be continuous.

Hence, no arc from $0$ to $\dfrac 1 2$ can exist.

It follows that $B$ is not arc-connected, and hence $T$ is not locally arc-connected.

$\blacksquare$


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