Path-Connected Space is Connected
Theorem
Let $T$ be a topological space which is path-connected.
Then $T$ is connected.
Proof 1
Suppose $X \mid Y$ is a separation of $T$.
As $X$ and $Y$ are non-empty, we can find $x \in X$ and $y \in Y$.
As $T$ is path-connected, there exists a path $f : \closedint 0 1 \to T$ with initial point $x$ and final point $y$.
Subset of Real Numbers is Interval iff Connected shows that $\closedint 0 1$ is connected.
Continuous Image of Connected Space is Connected shows that $\Img f$ is connected.
Connected Subspace Lie in One Component of Separation shows that either $\Img f \cap X = \O$, or $\Img f \cap Y = \O$.
As we have $x, y \in \Img f$, this is a contradiction.
It follows that there can be no separation of $T$, so $T$ is connected.
$\blacksquare$
Proof 2
Let $D$ be the discrete space $\set {0, 1}$.
Let $T$ be path-connected.
Let $f: T \to D$ be a continuous surjection.
Let $x, y \in T: \map f x = 0, \map f y = 1$.
Let $I \subset \R$ be the closed real interval $\closedint 0 1$.
Let $g: I \to T$ be a path from $x$ to $y$.
Then by Composite of Continuous Mappings is Continuous 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.
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Hence the result.
$\blacksquare$
Also see
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
- 2000: James R. Munkres: Topology (2nd ed.): $3$: Connectedness and Compactness: $\S 24$: Connected Subspaces of the Real Line
- 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): connected (of a space)