Path-Connectedness

Path-connectedness is the load-bearing hypothesis under the entire machinery of algebraic topology: the Fundamental Group $\pi_1(X,x_0)$ is only independent of the choice of basepoint $x_0$ (up to isomorphism) when $X$ is path-connected, and the theory of Covering Spaces — used to classify subgroups of $\pi_1$ and to build universal covers — is stated for path-connected, locally path-connected base spaces. In complex analysis, path-connectedness of domains underlies the existence of analytic continuations and single-valued branches of functions like $\log z$ along a path avoiding a branch point. In differential geometry, a connected Riemannian manifold is automatically path-connected (even geodesically connected under completeness, by the Hopf–Rinow theorem), which is why "connected manifold" and "path-connected manifold" are used interchangeably by geometers. In dynamical systems and control theory, path-connectedness of a reachable set answers whether a system can be steered continuously between two states without discrete jumps. In computer graphics and mesh processing, testing path-connectedness of a discretized surface (via graph connectivity of a triangulation) is a basic sanity check before texture-mapping or physical simulation. In data analysis, "connected components" of a point cloud built from a neighborhood graph are really discrete analogues of path components, feeding into clustering algorithms and topological data analysis (persistent homology pipelines begin by identifying such components).

Ask the tutor: Can you show me step by step why removing just one point from the topologist's sine curve's limit segment still leaves it path-disconnected from the oscillating part?

Ordinary connectedness is a negative condition: a space is connected if it cannot be broken into two disjoint nonempty open sets. It says nothing constructive about how points relate to each other. Path-connectedness replaces this with a positive, hands-on condition: to certify that $x$ and $y$ "belong together," you must produce an actual continuous map from $[0,1]$ landing on $x$ at time $0$ and $y$ at time $1$. This is a much more demanding — but also much more usable — requirement, because paths can be concatenated, reversed, and reparametrized, giving path-connectedness an algebraic flavor that pure connectedness lacks. Picture the plane. Any two points can obviously be joined by a straight line segment, so $\mathbb{R}^2$ is path-connected, and hence (as we will prove) connected. Now picture a graph like $y = \sin(1/x)$ for $x \in (0,1]$, together with the limiting segment $\{0\}\times[-1,1]$ that the curve wiggles toward but never touches as $x\to 0^+$. The closure of the graph is connected — you cannot separate it into two disjoint open pieces, because the oscillating curve has points arbitrarily close to every point of the segment. But no path can travel from a point on the curve to a point on the segment: any such path would have to cross infinitely many oscillations in a finite amount of "time," which forces a discontinuity at the endpoint. This is the topologist's sine curve, and it is the standard proof that path-connectedness is strictly stronger.

Example 1: The comb space is path-connected but shows how fragile the property can be under removing a point. Let

the union of the vertical segment at $x=0$, vertical "teeth" at $x = 1, \tfrac12, \tfrac13,\dots$, and the horizontal base segment, with the subspace topology from $\mathbb{R}^2$. Claim: $C$ is path-connected. Take any two points $p=(a,b), q=(c,d)\in C$. Each point of $C$ lies either on the base $[0,1]\times\{0\}$ or on a tooth $\{1/n\}\times[0,1]$ or on the limit segment $\{0\}\times[0,1]$. Construct a path from $p$ to $q$ in three (or fewer) straight-line stages, each an explicit continuous map: 1. From $p=(a,b)$ move straight down to $(a,0)$ along $\{a\}\times[0,b]$: the map $\gamma_1(t) = (a, b(1-t))$ for $t\in[0,1]$ is continuous (each coordinate is a polynomial in $t$, hence continuous, and continuity into a product/subspace is checked coordinatewise), and its image lies in $C$ since $a\in\{0,1,\tfrac12,\tfrac13,\dots\}\cup[0,1]$ is one of the teeth-or-base $x$-values, wait — more carefully: since $(a,b)\in C$, the vertical fiber $\{a\}\times[0,1]$ is entirely contained in $C$ whenever $a=0$ or $a=1/n$; if instead $p$ lies on the base itself ($b=0$) this step is trivial (constant path). 2. Travel along the base from $(a,0)$ to $(c,0)$: the map $\gamma_2(t) = (a+t(c-a),0)$, $t\in[0,1]$, is continuous and its image lies in the base $[0,1]\times\{0\}\subset C$. 3. From $(c,0)$ move straight up to $(c,d)=q$ along $\{c\}\times[0,d]$, by the same reasoning as step 1. Concatenating $\gamma_1,\gamma_2,\gamma_3$ (reparametrized to $[0,\tfrac13],[\tfrac13,\tfrac23],[\tfrac23,1]$ exactly as in the transitivity proof above, extended to three pieces by the same pasting-lemma argument) gives a single continuous path from $p$ to $q$ in $C$. Hence $C$ is path-connected. Note the subtlety: this construction crucially uses that every tooth, including the limiting one at $x=0$, is a full vertical segment attached to the base. If one deletes even a single point of the base — say removes $(0,0)$ but keeps the rest of $\{0\}\times[0,1]$ — the resulting "deleted comb space" is connected (a limit-point argument, since every neighborhood of a point on the limit segment meets infinitely many teeth) but fails to be path-connected: no path can reach the segment $\{0\}\times(0,1]$ from a tooth, by an oscillation argument entirely analogous to the topologist's sine curve below. This shows how a single point's removal can silently destroy path-connectedness while connectedness survives — exactly the kind of asymmetric, easy-to-miss failure the theorem's converse is guarding against.

Example 2: Products of path-connected spaces are path-connected — verified explicitly for a genuinely asymmetric case. Let $X = S^1$ (the unit circle, with its subspace topology from $\mathbb{R}^2$, known to be path-connected via $\gamma(t) = (\cos 2\pi t, \sin 2\pi t)$) and let $Y = [0,1)$ with the subspace topology from $\mathbb{R}$ (path-connected via straight-line paths $\gamma(t) = a + t(b-a)$ for $a,b\in[0,1)$, since $[0,1)$ is convex). We verify directly, without quoting a general product theorem as a black box, that $X\times Y$ is path-connected, to display the mechanism. Take two points $p = (e^{i\theta_0}, s_0)$ and $q=(e^{i\theta_1}, s_1)$ in $X\times Y$ (writing points of $S^1$ as $e^{i\theta}$). Since $X$ is path-connected, fix a path $\alpha: I \to X$ with $\alpha(0) = e^{i\theta_0}$, $\alpha(1) = e^{i\theta_1}$ — explicitly $\alpha(t) = e^{i(\theta_0 + t(\theta_1-\theta_0))}$ works (continuous since $t\mapsto \theta_0+t(\theta_1-\theta_0)$ is continuous and $\theta \mapsto e^{i\theta}$ is continuous). Since $Y$ is path-connected, fix $\beta(t) = s_0 + t(s_1-s_0)$, a straight-line path in $[0,1)$ from $s_0$ to $s_1$ (valid because $[0,1)$ is convex, so the segment stays inside $Y$). Define $\Gamma: I \to X\times Y$ by

By the universal property of the product topology (a map into a product is continuous iff each coordinate map is continuous — a standard fact about The Product Topology), and since $\alpha,\beta$ are each continuous, $\Gamma$ is continuous. Also $\Gamma(0) = (\alpha(0),\beta(0)) = (e^{i\theta_0}, s_0) = p$ and $\Gamma(1) = q$. So $\Gamma$ is a path from $p$ to $q$ in $X\times Y$. Since $p,q$ were arbitrary, $X\times Y = S^1 \times [0,1)$ is path-connected. This exact coordinatewise construction generalizes verbatim to any product $\prod_\alpha X_\alpha$ of path-connected spaces (finite or infinite, using the same universal-property argument coordinate by coordinate), giving the general fact that arbitrary products of path-connected spaces are path-connected.