Path Connectedness¶

Definition¶

In the metric space $(X, d_{X})\text{,}$ a nonempty subset $E$ is termed path-connected if, for any two distinct points $a, b \in E\text{,}$ a path $p\colon [0,1] \to E$ exists such that $p(0) = a$ and $p(1) = b\text{.}$

Interpretation¶

You can walk from any point to any other point in a path-connected set without ever leaving the set: there is a path inside $E$ that joins them. It is a constructive way of saying that a set is in one piece.

The path must lie in $E$ for every $t\in[0,1]$ $[0, 1]$ , not only at its endpoints; a straight segment between two points of $E$ that leaves $E$ on the way does not count. Restricting the definition to distinct points loses nothing, since a constant path joins a point to itself.

To show that a set is path-connected, one writes down paths. In a set that contains the straight segment between any two of its points, such as a disc, a box or $\mathbb{R}^n$ itself, the segment $p(t)=(1-t)a+tb$ always works. Other sets need bent paths: the annulus $\{\boldsymbol{x}\in\mathbb{R}^2\colon 1\le\|\boldsymbol{x}\|\le2\}$ is path-connected, because any two of its points can be joined by an arc of a circle followed by a radial segment.

Every path-connected set is connected, as the Key Results state, and path-connectedness is usually the easiest way to prove connectedness. In $\mathbb{R}$ the two notions coincide, since both describe the intervals, and for open subsets of $\mathbb{R}^n$ they coincide as well.

A common slip is to assume that the converse holds in general. The two Examples are connected sets with a point that no path inside the set reaches: one oscillates and the other crowds towards that point, so the set stays in one piece without being traversable.

Key Results¶

In the metric space $(X, d_{X})\text{,}$ any path-connected subset of $X$ is connected.

Examples¶

The topologist's sine curve is connected but not path-connected:

$\{(0,0)\}\cup\{(x,\sin(1/x))\colon x>0\}.$

As $x\to0^+$ the graph oscillates ever faster between $-1$ and $1$ ; the set is connected, but no path inside it reaches $(0,0)$ .
The collection of line segments joining the origin to points $(1, 1/n)$ for all positive integers $n$ , together with the point $(1,0)$ is also connected but not path-connected.

Segments from the origin to $(1,1/n)$ crowd towards the missing segment to $(1,0)$ ; only its endpoint $(1,0)$ belongs to the set, and no path inside the set reaches it.