Path-Connected Space is Connected

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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.



Hence the result.

$\blacksquare$


Also see


Sources