Component is not necessarily Path Component
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Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let $C$ be a component of $T$.
Then it is not necessarily the case that $C$ is also an path component of $T$.
Proof
Let $C$ be the closed topologist's sine curve embedded in the real Euclidean plane.
From Closed Topologist's Sine Curve is Connected, $C$ is connected in $T$
Therefore $C$ is a component in the subspace of $T$ induced by $C$.
From Closed Topologist's Sine Curve is not Path-Connected, $C$ is not path-connected.
Therefore $C$ is not a path component in the subspace of $T$ induced by $C$.
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