Totally Disconnected Space is Totally Pathwise Disconnected

Theorem

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

Then $T$ is a totally pathwise disconnected space.

Proof

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

Then by definition $T$ contains no non-degenerate connected sets.

Aiming for a contradiction, suppose $T$ is not a totally pathwise disconnected space.

That is, there exist two points $x, y \in S$ such that there exists a path between $x$ and $y$.

That is, $x$ and $y$ are path-connected.

Thus they are in the same path component.

From Path-Connected Space is Connected, $x$ and $y$ are therefore in the same component.

But that contradicts the definition of $T$ having no non-degenerate connected sets.

Hence, by Proof by Contradiction, $T$ is a totally pathwise disconnected space.

$\blacksquare$

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: Disconnectedness