Path Component is not necessarily Arc Component

Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let $P$ be a path component of $T$.

Then it is not necessarily the case that $P$ is also an arc component of $T$.

Proof

Let $T = \struct {S, \tau_p}$ be a finite particular point space.

From Particular Point Space is Path-Connected, $T$ is path-connected.

Therefore $S$ is a path component in $T$.

But from Particular Point Space is not Arc-Connected, $T$ is not arc-connected.

Therefore $S$ is not an arc component in $T$.

Hence the result.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $4$: Connectedness