A path is the answer to a deceptively simple question: how can a point move through a space without jumping? In $\mathbb{R}^n$ we draw a curve and trust the picture, but analysis needs a definition that survives in metric spaces, function spaces, and complex domains where no preferred coordinates may be available. The interval $[0,1]$ supplies time, continuity forbids jumps, and the image records the trace of the motion.

The distinction between a set being in one piece and a set being navigable matters. A connected set cannot be separated into two open pieces, but connectedness alone does not always give a usable route between two points. Paths add motion to connectedness: they let us transport values, integrate along curves, compare branches of functions, and build homotopies.

A first warning comes from trying to replace paths by arbitrary connected sets. The graph below has no break, yet it becomes hard to navigate at the origin.

[example: The Topologist's Sine Curve] Let \begin{align*} S=\{(x,\sin(1/x)):0<x\le 1\}\cup(\{0\}\times[-1,1])\subsetneq\mathbb{R}^2. \end{align*} Let $G=\{(x,\sin(1/x)):0<x\le 1\}$. The map $f:(0,1]\to\mathbb{R}^2$ given by $f(x)=(x,\sin(1/x))$ is continuous, and $(0,1]$ is connected, so $G=f((0,1])$ is connected by *Continuous Images of Connected Sets are Connected*. Also $\overline{G}=S$: if $y\in[-1,1]$, choose $\theta\in[-\pi/2,\pi/2]$ with $\sin\theta=y$ and set $x_n=(\theta+2\pi n)^{-1}$ for all large $n$ for which the denominator is positive. Then $x_n\to 0$ and \begin{align*} (x_n,\sin(1/x_n))=\left((\theta+2\pi n)^{-1},\sin(\theta+2\pi n)\right)=\left((\theta+2\pi n)^{-1},y\right)\to(0,y). \end{align*} Conversely, every [limit point](/page/Limit%20Point) of $G$ either has first coordinate $x>0$, in which case continuity of $x\mapsto\sin(1/x)$ puts it back on the graph, or has first coordinate $0$ and second coordinate in $[-1,1]$. Hence $S=\overline{G}$, so $S$ is connected by *Closures of Connected Sets are Connected*. We show that $S$ is not path connected. Suppose, for contradiction, that $\gamma:[0,1]\to S$ is a path starting on the vertical segment and later reaching the oscillating graph. Write \begin{align*} \gamma(t)=(u(t),v(t)). \end{align*} Then $u$ and $v$ are continuous, $u(0)=0$, and $u(b)>0$ for some $b\in[0,1]$. Let $C$ be the [connected component](/page/Connected%20Component) of $\{t\in[0,1]:u(t)>0\}$ containing $b$, and let $a$ be the left endpoint of $C$. Since $\{u>0\}$ is open in $[0,1]$, we have $u(t)>0$ for $t\in(a,c)$ for some $c>a$, and continuity gives $u(a)=0$. For every $t\in(a,c)$, the point $\gamma(t)$ lies on the graph part of $S$, so \begin{align*} v(t)=\sin(1/u(t)). \end{align*} Choose $t_n\in(a,c)$ with $t_n\downarrow a$. Since $u(a)=0$ and $u(t_n)>0$, the [intermediate value theorem](/theorems/180) applied to $u$ on $[a,t_n]$ shows that $u$ takes every value between $0$ and $u(t_n)$. Pick integers $k_n$ so large that \begin{align*} 0<\frac{1}{\pi/2+2\pi k_n}<u(t_n)\quad\text{and}\quad 0<\frac{1}{3\pi/2+2\pi k_n}<u(t_n). \end{align*} Thus there exist $\alpha_n,\beta_n\in(a,t_n)$ such that \begin{align*} u(\alpha_n)=\frac{1}{\pi/2+2\pi k_n} \end{align*} and \begin{align*} u(\beta_n)=\frac{1}{3\pi/2+2\pi k_n}. \end{align*} Because $a<\alpha_n,\beta_n<t_n$ and $t_n\downarrow a$, both $\alpha_n\to a$ and $\beta_n\to a$. On the graph, \begin{align*} v(\alpha_n)=\sin(\pi/2+2\pi k_n)=1 \end{align*} and \begin{align*} v(\beta_n)=\sin(3\pi/2+2\pi k_n)=-1. \end{align*} Continuity of $v$ at $a$ would force $v(\alpha_n)\to v(a)$ and $v(\beta_n)\to v(a)$, so $v(a)=1$ and $v(a)=-1$, a contradiction. Therefore no path can join a point of the vertical segment to a point of the oscillating graph, and $S$ is connected but not path connected. [/example]

This example is the basic reason paths deserve their own theory. Connectedness detects whether a space splits, while [path connectedness](/page/Path%20Connectedness) asks whether points can actually be joined by continuous motion.

## Definition

The central object should allow motion in any [topological space](/page/Topological%20Space), not only in Euclidean space. We use the unit interval as a standard time parameter because every closed finite interval can be rescaled to it.

[definition: Path] Let $(X,\tau)$ be a topological space. A path in $X$ is a [continuous function](/page/Continuous%20Function) $\gamma:[0,1]\to X$. [/definition]

The image $\gamma([0,1])$ is the geometric trace of the path, but the path itself also remembers the speed and order in which the trace is traversed. This distinction becomes important when we concatenate paths, reverse them, or compute length.

A bare continuous map $[0,1]\to X$ does not by itself name the two points it is meant to connect. To compose routes, reverse them, or ask whether one route joins prescribed points, we must isolate the boundary data of the motion from its intermediate trace.

[definition: Endpoints of a Path] Let $(X,\tau)$ be a topological space and let $\gamma:[0,1]\to X$ be a path. The initial point of $\gamma$ is $\gamma(0)$, and the terminal point of $\gamma$ is $\gamma(1)$. [/definition]

Endpoints let us state the navigational problem with precision: given two points, can time be filled in continuously between them? That question is the local building block for every global connectivity statement involving paths.

[definition: Path from $x$ to $y$] Let $(X,\tau)$ be a topological space and let $x,y\in X$. A path from $x$ to $y$ is a path $\gamma:[0,1]\to X$ such that $\gamma(0)=x$ and $\gamma(1)=y$. [/definition]

The constant path is the smallest example and the identity element for later path operations. It says that staying still is allowed motion.

[example: Constant Paths] Let $(X,\tau)$ be a topological space and let $x\in X$. Define $c_x:[0,1]\to X$ by $c_x(t)=x$ for every $t\in[0,1]$. If $U\subset X$ is open, then \begin{align*} c_x^{-1}(U)=\{t\in[0,1]:c_x(t)\in U\}. \end{align*} Since $c_x(t)=x$ for every $t$, this preimage is $[0,1]$ when $x\in U$, and it is $\varnothing$ when $x\notin U$. Both $[0,1]$ and $\varnothing$ are open in the [subspace topology](/page/Subspace%20Topology) on $[0,1]$, so $c_x$ is continuous. Also \begin{align*} c_x(0)=x \end{align*} and \begin{align*} c_x(1)=x. \end{align*} Thus $c_x$ is a path from $x$ to $x$. Its image is \begin{align*} c_x([0,1])=\{c_x(t):t\in[0,1]\}=\{x\}, \end{align*} so the trace is the single point $x$, while the parametrized path records staying at that point for the whole time interval. [/example]

The Euclidean model is the line segment. It shows why convexity is a strong geometric condition: if the straight route between two points remains inside the set, then paths require no further construction.

[example: Straight-Line Paths in Convex Sets] Let $C\subset\mathbb{R}^n$ be convex, and let $x,y\in C$. Define $\gamma:[0,1]\to\mathbb{R}^n$ by \begin{align*} \gamma(t)=(1-t)x+ty. \end{align*} For each $t\in[0,1]$, the coefficients satisfy $1-t\ge 0$, $t\ge 0$, and \begin{align*} (1-t)+t=1. \end{align*} Since $C$ is convex and $x,y\in C$, this gives \begin{align*} \gamma(t)=(1-t)x+ty\in C. \end{align*} Thus $\gamma$ may be regarded as a map $\gamma:[0,1]\to C$. Writing $x=(x_1,\ldots,x_n)$ and $y=(y_1,\ldots,y_n)$, the $j$th coordinate of $\gamma$ is \begin{align*} \gamma_j(t)=(1-t)x_j+ty_j=x_j+t(y_j-x_j). \end{align*} This coordinate function is affine in $t$, hence continuous on $[0,1]$; therefore $\gamma$ is continuous as an $\mathbb{R}^n$-valued map, and hence continuous as a map into the subspace $C$. Finally, \begin{align*} \gamma(0)=(1-0)x+0y=x \end{align*} and \begin{align*} \gamma(1)=(1-1)x+1y=y. \end{align*} So $\gamma$ is a path in $C$ from $x$ to $y$, showing that any two points of a convex subset of $\mathbb{R}^n$ are joined by the straight line segment between them. [/example]

## Path Connectedness

### Global Navigation

A path is local data: it joins one ordered pair of points. The global question is whether this can be done for every pair. This turns navigation into a property of the whole space.