Path-Connectedness

In this lesson you’ll define paths and path-connectedness, see that it implies connectedness but not conversely, work through the topologist’s sine curve, and meet local path-connectedness, which is the condition that makes the two notions agree.

A path in $X$ from $a$ to $b$ is a continuous map $\gamma:[0,1]\to X$ with $\gamma(0)=a$ and $\gamma(1)=b$.

$X$ is path-connected if every pair of points is joined by a path.

This is closer to intuition than connectedness: it says you can walk from anywhere to anywhere. Note that a path is a map, not just its image, and that it may retrace itself, stop, or be nowhere injective.

Paths compose. If $\gamma_1$ runs $a\to b$ and $\gamma_2$ runs $b\to c$, then the concatenation

$$\gamma(t) = \begin{cases}\gamma_1(2t) & 0\le t\le\tfrac12\\ \gamma_2(2t-1) & \tfrac12\le t\le1\end{cases}$$

is continuous by the pasting lemma, and runs $a\to c$. That gives transitivity, and with the constant path and the reversal $t\mapsto\gamma(1-t)$, being joinable by a path is an equivalence relation. Its classes are the path components.

Path-connected $\implies$ connected.

Proof. Suppose $X$ is path-connected and $X = U\sqcup V$ were a separation. Pick $a\in U$ and $b\in V$ and a path $\gamma$ joining them. Then $\gamma^{-1}(U)$ and $\gamma^{-1}(V)$ separate $[0,1]$, contradicting connectedness of the interval. ∎

The converse fails, and the counterexample is the standard one in all of topology.

Let

$$S = \left\{\left(x,\sin\tfrac1x\right) : 0<x\le1\right\}, \qquad T = S\cup\big(\{0\}\times[-1,1]\big)$$

$T$ is connected. $S$ is a continuous image of $(0,1]$, hence connected. As $x\to0^+$ the function $\sin\frac1x$ passes through every value in $[-1,1]$ infinitely often, so every point of the vertical segment is a limit point of $S$. Therefore $S\subseteq T\subseteq\bar S$, and anything between a connected set and its closure is connected.

$T$ is not path-connected. Suppose a path $\gamma$ started on the segment at $(0,y_0)$ and reached a point of $S$. Consider the last time $t^*$ at which $\gamma$ is on the segment. Just after $t^*$ the path is in $S$ with $x$-coordinate positive but arbitrarily small, and on any such interval the second coordinate oscillates through the whole of $[-1,1]$. So $\gamma$ cannot be continuous at $t^*$: its values fail to settle near $\gamma(t^*)$.

The figure makes the obstruction concrete. Every window $(0,d]$ contains infinitely many zeros of $\sin\frac1x$, at the points $x = \frac{1}{k\pi}$, so there is no scale at which the oscillation calms down. A path has to cross all of them in finite time, and continuity forbids it.

$$T \text{ has 2 path components and 1 connected component}$$

Those two counts differing is exactly the point. They are different invariants, and both are worth having.

$X$ is locally path-connected if every point has arbitrarily small path-connected neighbourhoods.

Theorem. A connected and locally path-connected space is path-connected.

This is why the distinction almost never matters in practice. Open subsets of $\mathbb{R}^n$, manifolds, and CW-complexes are all locally path-connected, so for them connected and path-connected coincide. The topologist’s sine curve fails local path-connectedness at every point of the segment, which is precisely where it goes wrong.

A useful special case:

An open connected subset of $\mathbb{R}^n$ is path-connected, and in fact any two points can be joined by a polygonal path.

Convex sets are trivially path-connected: use the straight segment. Slightly weaker and often enough:

$X\subseteq\mathbb{R}^n$ is star-shaped about $c$ if the segment from $c$ to any point of $X$ stays in $X$.

Star-shaped implies path-connected, and it also implies the stronger property of being simply connected, which lesson 17 needs. The chain to remember is

$$\text{convex} \implies \text{star-shaped} \implies \text{path-connected} \implies \text{connected}$$

with every implication strict.

Example 1: Show $\mathbb{R}^n$ is path-connected.

Solution. For $a,b\in\mathbb{R}^n$ the straight-line path

$$\gamma(t) = (1-t)a+tb$$

is continuous and joins them.

Answer: path-connected, hence connected. ∎

Example 2: Show the circle $S^1$ is path-connected.

Solution. Given two points at angles $\theta_1$ and $\theta_2$, the path

$$\gamma(t) = \big(\cos((1-t)\theta_1+t\theta_2),\ \sin((1-t)\theta_1+t\theta_2)\big)$$

travels along the circle from one to the other.

Answer: path-connected. ∎

Note the circle is path-connected but not simply connected; the loop around it cannot be contracted. Those are different questions.

Example 3: Show $\mathbb{R}^2\setminus\{0\}$ is path-connected.

Solution. For any two nonzero points, if the straight segment between them misses the origin, use it. If it passes through the origin, the two points are antipodal in direction; go via a third point off that line, using two segments.

Answer: path-connected. ∎

Contrast $\mathbb{R}\setminus\{0\}$, which is not connected. Removing a point from the plane leaves room to go around; removing one from the line does not.

Example 4: Count the components and path components of the topologist’s sine curve.

Solution. Connected components: 1, since $T$ is connected.

Path components: 2, namely the segment $\{0\}\times[-1,1]$ and the curve $S$. No path crosses between them, and each piece is separately path-connected - the segment because it is convex, the curve because it is a continuous injective image of an interval.

$$1 \text{ component}, \quad 2 \text{ path components}$$

∎

Example 5: Show path components need not be open.

Solution. In the topologist’s sine curve, the path component $\{0\}\times[-1,1]$ is not open in $T$: every neighbourhood of a point on the segment contains points of the oscillating curve.

Connected components are always closed; path components need not be either open or closed. In a locally path-connected space path components are open, which is how the agreement theorem is proved. ∎

Example 6: Show an open connected subset of $\mathbb{R}^n$ is path-connected.

Solution. Fix $a\in U$ and let $A$ be the set of points reachable from $a$ by a path in $U$.

$A$ is open: if $x\in A$, some ball $B(x,r)\subseteq U$, and every point of that ball is reachable by appending a straight segment.

$U\setminus A$ is open by the same argument: if $y$ were reachable from a point of the ball around it, then $y$ would be reachable.

Both are open, they are disjoint, and $A\ne\varnothing$. Connectedness forces $U\setminus A=\varnothing$.

$$U = A, \text{ so } U \text{ is path-connected}$$

∎

This is the standard open-and-closed argument, and it is worth recognizing because the same shape of proof appears throughout topology.

Example 7: Give a star-shaped set that is not convex.

Solution. Take a plus sign, the union of a horizontal and a vertical bar crossing at the origin. Every point sees the origin along a straight segment inside the set, so it is star-shaped about 0.

It is not convex: the segment joining the tip of the top arm to the tip of the right arm leaves the set.

Answer: star-shaped, not convex. ∎

Motion planning. Two robot configurations are mutually achievable exactly when they lie in the same path component of the free configuration space, and the path itself is the motion. Connectedness alone would not be enough, since it does not produce a route.

Continuous deformation in manufacturing. Whether a part can be moved into place without collision is a path-connectedness question in configuration space, and it is why some assemblies are impossible despite fitting geometrically.

Homotopy in animation. Interpolating between two poses requires a path in the space of poses, and it fails when the space is disconnected - which is why some character rigs pop instead of blending.

Phase spaces in physics. Whether a system can evolve continuously from one state to another is a path question, and disconnected phase space means a transition is forbidden without a discontinuous event.

Network routing. A route is a path in the graph, and graph connectivity is the discrete analogue. Reachability queries in a network are path-component computations.

Molecular conformation. Whether a molecule can pass between two shapes without breaking bonds is a path-connectedness question in conformation space, and energy barriers are what make some paths practically unavailable even when they exist.