Excluded Point Space is not Arc-Connected

Theorem
Let $S$ be a set with at least two distinct elements.

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

Then $T$ is not arc-connected.

Proof
Let $q \in S$ be such that $q \ne p$.

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


 * $f \left({0}\right) = p$
 * $f \left({1}\right) = q$

Because $f$ is an injection, it must be that:


 * $f^{-1} \left[{\left\{{q}\right\}}\right] = \left\{{1}\right\}$

where $f^{-1}$ denotes the preimage under $f$.

Now since $p \notin \left\{{q}\right\}$, $\left\{{q}\right\}$ is open in $\tau_{\bar p}$, by definition of the excluded point topology.

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

Thus we have exhibited an open set whose preimage under $f$ is not open.

Hence $f$ is not continuous.

Since $f$ was an arbitrary injection, no arc between $p$ and $q$ can exist.

Hence $T$ is not arc-connected.