# Path-Connected Space is Connected

## Contents

## Theorem

Let $T$ be a topological space which is path-connected.

Then $T$ is connected.

## Proof

Let $D$ be the discrete space $\left\{{0, 1}\right\}$.

Let $T$ be path-connected.

Let $f: T \to D$ be a continuous surjection.

Let $x, y \in T: f \left({x}\right) = 0, f \left({y}\right) = 1$.

Let $I \subset \R$ be the closed real interval $\left[{0 \,.\,.\, 1}\right]$.

Let $g: I \to T$ be a path from $x$ to $y$.

Then by Continuity of Composite Mapping it follows that $f \circ g: I \to D$ is a continuous surjection.

This contradicts the connectedness of $I$ as proved in Subset of Real Numbers is Interval iff Connected.

Hence the result.

$\blacksquare$

## Also see

- Closed Topologist's Sine Curve is not Path-Connected for a proof that the converse does not apply.

## Sources

- 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*... (previous) ... (next): $\text{I}: \ \S 4$ - 1975: W.A. Sutherland:
*Introduction to Metric and Topological Spaces*... (previous) ... (next): $6.4$: Comparison of Definitions: Proposition $6.4.1$