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. [definition: Path Connected Space] A topological space $(X,\tau)$ is path connected if for every $x,y\in X$ there exists a path $\gamma:[0,1]\to X$ from $x$ to $y$. [/definition] Path connectedness should prevent a separation of the space into two disjoint open regions. A route from a point in one region to a point in the other would give a continuous image of an interval crossing the proposed separation, which motivates the theorem below. [quotetheorem:9243] The converse fails, and the topologist's sine curve above is the standard warning. For analysis, however, many spaces have enough local room that connectedness and path connectedness coincide. ### Domains Analysis usually takes place on open connected subsets of Euclidean space. The word domain packages exactly those two hypotheses: openness gives small balls for local movement, and connectedness prevents independent pieces. [definition: Domain] A domain in $\mathbb{R}^n$ is a nonempty open connected subset of $\mathbb{R}^n$. [/definition] Open connected Euclidean sets should be navigable because each point has a small convex ball around it. Starting from one point, the set of points reachable by local straight-line moves is both open and closed inside the domain, so connectedness should force it to be the whole domain; this is the theorem below. [quotetheorem:9244] This result explains why analysis texts often move between connected open sets and path connected open sets. The theorem is special to open subsets of Euclidean space; it is not a theorem about arbitrary connected spaces. The proof idea also suggests a stronger and more constructive kind of path. Instead of allowing any continuous motion, we can ask for a route made from finitely many straight segments. [definition: Polygonal Path] Let $U\subset\mathbb{R}^n$. A polygonal path in $U$ is a path $\gamma:[0,1]\to U$ for which there exist points $x_0,\ldots,x_m\in U$ and a partition $0=t_0<t_1<\cdots<t_m=1$ such that for each $i\in\{1,\ldots,m\}$, the restriction of $\gamma$ to $[t_{i-1},t_i]$ parametrizes the line segment from $x_{i-1}$ to $x_i$. [/definition] For analysis, an arbitrary continuous route is often less useful than a finite route made from line segments. Euclidean domains have enough local convexity for every path-connected pair to be joined by such a finite polygonal route, and the following theorem states that constructive version. [quotetheorem:9245] The theorem lets arguments reduce to line segments, which is why it appears in proofs about primitives, [analytic continuation](/page/Analytic%20Continuation), and local-to-global properties of differentiable functions. [example: A Punctured Plane Path] In $\mathbb{R}^2\setminus\{0\}$, the straight line segment from $(-1,0)$ to $(1,0)$ is parametrized by \begin{align*} \sigma(t)=(1-t)(-1,0)+t(1,0)=(-1+2t,0). \end{align*} At $t=1/2$ this gives \begin{align*} \sigma(1/2)=(-1+2\cdot 1/2,0)=(0,0), \end{align*} so this segment is not a path in $\mathbb{R}^2\setminus\{0\}$. Instead define $\gamma:[0,1]\to\mathbb{R}^2$ by $\gamma(t)=(-1,3t)$ for $0\le t\le 1/3$, by $\gamma(t)=(6t-3,1)$ for $1/3\le t\le 2/3$, and by $\gamma(t)=(1,3-3t)$ for $2/3\le t\le 1$. At the joining times, \begin{align*} (-1,3\cdot 1/3)=(-1,1)=(6\cdot 1/3-3,1) \end{align*} and \begin{align*} (6\cdot 2/3-3,1)=(1,1)=(1,3-3\cdot 2/3). \end{align*} Thus the affine pieces match at their shared endpoints, so $\gamma$ is continuous. On the first piece the first coordinate is $-1$, on the second piece the second coordinate is $1$, and on the third piece the first coordinate is $1$; hence $\gamma(t)\ne(0,0)$ for every $t\in[0,1]$. Finally, \begin{align*} \gamma(0)=(-1,0) \end{align*} and \begin{align*} \gamma(1)=(1,0). \end{align*} Therefore $\gamma$ is a polygonal path in $\mathbb{R}^2\setminus\{0\}$ from $(-1,0)$ to $(1,0)$, with vertices $(-1,0)$, $(-1,1)$, $(1,1)$, and $(1,0)$. The route avoids the missing origin by moving around it, showing that path construction must respect holes even when the endpoints are close in Euclidean distance. [/example] ## Operations on Paths ### Concatenation A single trip is rarely enough. We want to travel from $x$ to $y$ and then from $y$ to $z$. This operation makes paths behave like composable arrows, with endpoints controlling which compositions are allowed. [definition: Concatenation of Paths] Let $(X,\tau)$ be a topological space. Let $\gamma:[0,1]\to X$ be a path from $x$ to $y$, and let $\eta:[0,1]\to X$ be a path from $y$ to $z$. The concatenation of $\gamma$ and $\eta$ is the function $\eta*\gamma:[0,1]\to X$ defined by $(\eta*\gamma)(t)=\gamma(2t)$ for $0\le t\le 1/2$, and $(\eta*\gamma)(t)=\eta(2t-1)$ for $1/2\le t\le 1$. [/definition] The only possible failure of the piecewise formula is a jump at the joining time $t=1/2$. Each half is continuous on its own subinterval, but the two halves define a genuine path only if they agree at the join. The shared endpoint condition $\gamma(1)=\eta(0)$ is exactly the condition that removes this obstruction and makes the piecewise route continuous from $x$ all the way to $z$. [remark: Order Convention for Concatenation] Throughout this page, $\eta*\gamma$ means: first traverse $\gamma$, then traverse $\eta$. Thus the right-hand factor is the earlier leg and the left-hand factor is the later leg. This is the convention used below in expressions such as $\overline{\gamma_-}*\gamma_+$, where $\gamma_+$ is traversed first and then $\overline{\gamma_-}$. [/remark] With this convention, concatenating several paths is composable only when adjacent endpoints match. It is not strictly associative as a parametrized function, because different bracketings spend different amounts of time on each leg. It becomes associative after passing to paths up to endpoint-preserving reparametrization or homotopy. ### Reversal and Reparametrization To compare a route with the same route traversed backwards, we need an operation that keeps the trace but inverts time. This is the basic orientation change for paths, and it is the prototype for sign changes in line integrals. [definition: Reverse Path] Let $(X,\tau)$ be a topological space and let $\gamma:[0,1]\to X$ be a path. The reverse path of $\gamma$ is the path $\overline{\gamma}:[0,1]\to X$ defined by $\overline{\gamma}(t)=\gamma(1-t)$. [/definition] The definition has introduced the formula for reversal, but it still leaves a basic well-definedness question: does this formula always produce a valid path with the intended endpoints? The next result verifies that reversing time preserves continuity and turns a path from $x$ to $y$ into a path from $y$ back to $x$. [quotetheorem:9158] Reversal is useful because it gives a canonical inverse direction for any path without requiring extra structure on the space. If $\gamma$ is a route from $x$ to $y$, then $\overline{\gamma}$ is the route available for returning from $y$ to $x$, and concatenations such as $\gamma*\overline{\gamma}$ become the basic loops used in homotopy arguments. The operation changes orientation but not the underlying image of the path; it also depends on the chosen parametrization, so it should not be confused with a new geometric curve. The same geometric route may also be traversed with many different clocks. To separate the route from its speed, we need a name for replacing the original time variable by a new continuous time variable. [definition: Reparametrization of a Path] Let $(X,\tau)$ be a topological space, let $\gamma:[0,1]\to X$ be a path, and let $\phi:[0,1]\to[0,1]$ be a continuous function. The path $\gamma\circ\phi:[0,1]\to X$ is called a reparametrization of $\gamma$ by $\phi$. [/definition] Endpoint-sensitive arguments need a narrower class of clock changes. The beginning, the end, and the direction of travel must be preserved, otherwise the reparametrized path may represent a different oriented trip. [definition: Orientation-Preserving Reparametrization] Let $(X,\tau)$ be a topological space and let $\gamma:[0,1]\to X$ be a path. An orientation-preserving reparametrization of $\gamma$ is a path $\gamma\circ\phi$ where $\phi:[0,1]\to[0,1]$ is continuous, nondecreasing, surjective, and satisfies $\phi(0)=0$ and $\phi(1)=1$. [/definition] This definition allows a path to slow down, speed up, or stop for a while, but it does not allow it to reverse direction. [example: The Same Trace with Different Speeds] Let $\gamma,\eta:[0,1]\to\mathbb{R}^2$ be defined by \begin{align*} \gamma(t)=(\cos(2\pi t),\sin(2\pi t)) \end{align*} and \begin{align*} \eta(t)=(\cos(2\pi t^2),\sin(2\pi t^2)). \end{align*} The functions $t\mapsto 2\pi t$ and $t\mapsto 2\pi t^2$ are continuous, and sine and cosine are continuous, so each coordinate function of $\gamma$ and $\eta$ is continuous. Hence both maps are paths in $\mathbb{R}^2$. For every $t\in[0,1]$, \begin{align*} |\gamma(t)|^2=\cos^2(2\pi t)+\sin^2(2\pi t)=1, \end{align*} so $\gamma([0,1])$ is contained in the unit circle. Conversely, if $(a,b)$ lies on the unit circle, then there is an angle $\theta\in[0,2\pi]$ with $(a,b)=(\cos\theta,\sin\theta)$, and with $t=\theta/(2\pi)$ we have $t\in[0,1]$ and \begin{align*} \gamma(t)=\gamma\left(\frac{\theta}{2\pi}\right)=(\cos\theta,\sin\theta)=(a,b). \end{align*} Thus $\gamma([0,1])$ is exactly the unit circle. Define $\phi:[0,1]\to[0,1]$ by $\phi(t)=t^2$. If $s\in[0,1]$, then $\sqrt{s}\in[0,1]$ and $\phi(\sqrt{s})=s$, so $\phi([0,1])=[0,1]$. Therefore \begin{align*} \eta([0,1])=\{\gamma(\phi(t)):t\in[0,1]\}=\{\gamma(s):s\in[0,1]\}=\gamma([0,1]). \end{align*} Also, for every $t\in[0,1]$, \begin{align*} (\gamma\circ\phi)(t)=\gamma(t^2)=(\cos(2\pi t^2),\sin(2\pi t^2))=\eta(t). \end{align*} So $\eta$ is a reparametrization of $\gamma$ by the clock $\phi(t)=t^2$. The different speeds are visible from the derivatives: \begin{align*} \gamma'(t)=(-2\pi\sin(2\pi t),2\pi\cos(2\pi t)) \end{align*} and \begin{align*} |\gamma'(t)|=\sqrt{4\pi^2\sin^2(2\pi t)+4\pi^2\cos^2(2\pi t)}=2\pi. \end{align*} Similarly, \begin{align*} \eta'(t)=(-4\pi t\sin(2\pi t^2),4\pi t\cos(2\pi t^2)) \end{align*} and, since $t\ge 0$, \begin{align*} |\eta'(t)|=\sqrt{16\pi^2t^2\sin^2(2\pi t^2)+16\pi^2t^2\cos^2(2\pi t^2)}=4\pi t. \end{align*} Thus $\eta$ has speed $0$ at $t=0$ and speed $4\pi$ at $t=1$, while $\gamma$ has constant speed $2\pi$. The two paths trace the same unit circle, but they traverse that trace with different clocks. [/example] ## Path Components ### Maximal Navigable Pieces When a space is not path connected, it may still decompose into maximal regions inside which navigation is possible. These pieces are more useful than connected components in many constructive arguments, because membership is witnessed by an actual path. The relation "can be joined by a path" behaves like equality at the level of components. It is reflexive by constant paths, symmetric by reversal, and transitive by concatenation. [definition: Path Component] Let $(X,\tau)$ be a topological space and let $x\in X$. The path component of $x$ in $X$ is the set \begin{align*} P_X(x)=\{y\in X:\text{there exists a path in }X\text{ from }x\text{ to }y\}. \end{align*} [/definition] Path components are intended to be the maximal navigable pieces of a space. The path operations from the previous section are precisely what is needed to prove that these pieces do not overlap except by being equal. [quotetheorem:9246] Path components need not be open in arbitrary spaces. Openness requires a local path connectedness hypothesis, which captures the idea that small neighbourhoods should themselves be navigable. ### Local Path Connectedness Global path connectedness says that any two points in the whole space can be joined. To control path components topologically, we also need small neighbourhoods around each point to contain smaller path connected neighbourhoods. [definition: Locally Path Connected Space] A topological space $(X,\tau)$ is locally path connected if for every $x\in X$ and every [open set](/page/Open%20Set) $U\subset X$ with $x\in U$, there exists an open path connected set $V\subset X$ such that $x\in V\subset U$. [/definition] If every point has arbitrarily small path connected open neighbourhoods, then membership in a path component should be stable under small movement. The possible obstruction is that a component might contain a point but no open neighbourhood around it; local path connectedness rules this out by letting nearby points connect back to the chosen point inside a small path connected neighbourhood. [quotetheorem:1058] This theorem is often the hidden reason why domains behave well. Open subsets of $\mathbb{R}^n$ are locally path connected because small Euclidean balls are convex. [example: Rational Points Have Tiny Path Components] In $\mathbb{Q}$ with the subspace topology inherited from $\mathbb{R}$, we show that every path $\gamma:[0,1]\to\mathbb{Q}$ is constant. Let $i:\mathbb{Q}\to\mathbb{R}$ be the inclusion map and set $h=i\circ\gamma$. The inclusion is continuous because, for every open set $O\subset\mathbb{R}$, one has $i^{-1}(O)=O\cap\mathbb{Q}$, which is open in the subspace topology on $\mathbb{Q}$. Hence $h:[0,1]\to\mathbb{R}$ is continuous. Suppose $h$ were not constant. Then there exist $s,t\in[0,1]$ with $h(s)\ne h(t)$. After interchanging $s$ and $t$ if necessary, assume $h(s)<h(t)$. By density of the irrational numbers in $\mathbb{R}$, choose $r\in\mathbb{R}\setminus\mathbb{Q}$ such that \begin{align*} h(s)<r<h(t). \end{align*} The *[Intermediate Value Theorem](/theorems/629)* applied to the continuous function $h$ on the interval between $s$ and $t$ gives some $c$ between $s$ and $t$ with \begin{align*} h(c)=r. \end{align*} But $h(c)=i(\gamma(c))=\gamma(c)$ as a real number, and $\gamma(c)\in\mathbb{Q}$, while $r\notin\mathbb{Q}$. This contradiction shows that $h$, and therefore $\gamma$, is constant. Now fix $q\in\mathbb{Q}$. The constant path at $q$ shows $q\in P_{\mathbb{Q}}(q)$. Conversely, if $y\in P_{\mathbb{Q}}(q)$, then there is a path in $\mathbb{Q}$ from $q$ to $y$; every such path is constant, so $y=q$. Therefore \begin{align*} P_{\mathbb{Q}}(q)=\{q\}. \end{align*} Thus every path component of $\mathbb{Q}$ is a single point, even though every nonempty open interval in $\mathbb{Q}$ contains infinitely many rational points. [/example] ## Length and Rectifiable Paths ### Metric Length Topology only asks whether a path is continuous. Analysis often asks how far the path travels. A wildly oscillating continuous path can have infinite length, so length is an extra property rather than a consequence of continuity. The definition of length should not depend on a differentiable parametrization. We measure by sampling finitely many times and summing the distances between successive positions, then taking the supremum over all partitions. [definition: Length of a Path] Let $(X,d)$ be a [metric space](/page/Metric%20Space). The length functional is \begin{align*} L:C([0,1],X)\to[0,\infty]. \end{align*} For $\gamma\in C([0,1],X)$, its value is \begin{align*} L(\gamma)=\sup\left\{\sum_{i=1}^m d(\gamma(t_i),\gamma(t_{i-1})):0=t_0<t_1<\cdots<t_m=1\right\}. \end{align*} For a path $\gamma:[0,1]\to X$, the length of $\gamma$ is $L(\gamma)$. [/definition] The supremum is taken over all finite partitions. This definition records the greatest polygonal distance that can be forced by observing the path at finitely many times. Finite arclength is the threshold at which a continuous path becomes suitable for length-based analysis. The length functional can take the value $\infty$, so we need a separate term for the paths whose accumulated partition lengths stay bounded. Differentiability is one source of this finiteness, but many nonsmooth paths also have it. [definition: Rectifiable Path] Let $(X,d)$ be a metric space. A path $\gamma:[0,1]\to X$ is rectifiable if $L(\gamma)<\infty$. [/definition] For differentiable Euclidean paths, the partition definition of length should recover the familiar integral of speed. The obstruction is that the metric definition is a supremum over finite samples, while calculus measures instantaneous velocity. Compatibility between these two measurements is what lets arclength be computed by integrating speed. [quotetheorem:9247] This theorem is the bridge between metric geometry and calculus. It lets us compute length using speed when a differentiable parametrization is available. [example: Length of a Circle Traversed Once] Let $\gamma:[0,1]\to\mathbb{R}^2$ be given by $\gamma(t)=(\cos(2\pi t),\sin(2\pi t))$. The coordinate functions are continuously differentiable, and by the chain rule, \begin{align*} \gamma'(t)=(-2\pi\sin(2\pi t),2\pi\cos(2\pi t)). \end{align*} Therefore \begin{align*} |\gamma'(t)|^2=(-2\pi\sin(2\pi t))^2+(2\pi\cos(2\pi t))^2. \end{align*} Expanding the squares gives \begin{align*} |\gamma'(t)|^2=4\pi^2\sin^2(2\pi t)+4\pi^2\cos^2(2\pi t). \end{align*} Factoring out $4\pi^2$ and using $\sin^2\theta+\cos^2\theta=1$, \begin{align*} |\gamma'(t)|^2=4\pi^2(\sin^2(2\pi t)+\cos^2(2\pi t))=4\pi^2. \end{align*} Since $|\gamma'(t)|\ge 0$, it follows that \begin{align*} |\gamma'(t)|=2\pi. \end{align*} By the *Length Formula for $C^1$ Paths*, \begin{align*} L(\gamma)=\int_0^1 |\gamma'(t)|\,d\mathcal{L}^1(t). \end{align*} Substituting the constant speed $|\gamma'(t)|=2\pi$ gives \begin{align*} L(\gamma)=\int_0^1 2\pi\,d\mathcal{L}^1(t). \end{align*} Because $\mathcal{L}^1([0,1])=1$, \begin{align*} \int_0^1 2\pi\,d\mathcal{L}^1(t)=2\pi\mathcal{L}^1([0,1])=2\pi. \end{align*} Thus \begin{align*} L(\gamma)=2\pi. \end{align*} This recovers the circumference of the unit circle from the speed of the parametrized traversal. [/example] ### Length Under Operations The definition of length was built from partitions, so changing the clock should not change the measured distance when the direction is preserved. This is what makes length a property of the oriented route rather than a property of the chosen speed. [quotetheorem:9248] After length is known to ignore orientation-preserving changes of clock, the next structural question is how length behaves when paths are assembled from consecutive pieces. The partition definition makes additivity nontrivial because partitions of the concatenated interval must be compared with partitions of the two halves; this is the result needed to analyse long paths stage by stage. [quotetheorem:9249] Finite length excludes many wild continuous behaviours but does not require smoothness. Polygonal paths, Lipschitz paths, and $C^1$ paths are all rectifiable. [example: A Continuous Path with Infinite Length] There exist continuous paths $\gamma:[0,1]\to\mathbb{R}^2$ with infinite length. Define target points by \begin{align*} q_{2k-1}=\left(\frac{1}{k},0\right),\qquad q_{2k}=(0,0)\quad\text{for }k\ge 1. \end{align*} Set $\gamma(0)=(0,0)$ and $\gamma(2^{-m})=q_m$ for every $m\ge 1$. On each interval $[2^{-(m+1)},2^{-m}]$, define $\gamma$ to be the affine line segment from $q_{m+1}$ to $q_m$, and on $[1/2,1]$ set $\gamma(t)=q_1$. The path is continuous away from $0$ because on each interval it is affine, and adjacent affine pieces agree at the shared dyadic endpoint: \begin{align*} \gamma(2^{-m})=q_m. \end{align*} To check continuity at $0$, let $\varepsilon>0$. Choose $K\ge 1$ with $1/K<\varepsilon$. If $m\ge 2K-1$, then $q_m$ is either $(0,0)$ or $q_m=(1/k,0)$ for some $k\ge K$, so \begin{align*} |q_m|\le \frac{1}{K}<\varepsilon. \end{align*} If $0<t\le 2^{-(2K-1)}$, then $t\in[2^{-(m+1)},2^{-m}]$ for some $m\ge 2K-1$, and $\gamma(t)$ lies on the line segment between $q_{m+1}$ and $q_m$. Since both endpoints have norm at most $1/K$, convexity of the Euclidean norm gives \begin{align*} |\gamma(t)|\le \frac{1}{K}<\varepsilon. \end{align*} Thus $\gamma(t)\to(0,0)=\gamma(0)$ as $t\downarrow 0$, so $\gamma$ is continuous on all of $[0,1]$. Now fix $N\ge 1$ and use the partition \begin{align*} 0<2^{-2N}<2^{-(2N-1)}<\cdots<2^{-2}<2^{-1}<1. \end{align*} The length of $\gamma$ is the supremum of all partition sums, so it is at least the sum over these dyadic sample points. In particular, \begin{align*} L(\gamma)\ge \sum_{m=1}^{2N-1}|q_m-q_{m+1}|. \end{align*} Keeping only the odd-indexed terms gives \begin{align*} L(\gamma)\ge \sum_{k=1}^{N}|q_{2k-1}-q_{2k}|. \end{align*} For each $k$, \begin{align*} |q_{2k-1}-q_{2k}|=\left|\left(\frac{1}{k},0\right)-(0,0)\right|=\left|\left(\frac{1}{k},0\right)\right|=\frac{1}{k}. \end{align*} Therefore \begin{align*} L(\gamma)\ge \sum_{k=1}^{N}\frac{1}{k}. \end{align*} Since this holds for every $N$ and the harmonic sums $\sum_{k=1}^{N}1/k$ are unbounded, $L(\gamma)=\infty$. Thus a continuous path can have infinite length even when its oscillations accumulate at a single point. [/example] ## Homotopy of Paths Sometimes two paths should count as the same route if one can be continuously deformed into the other while the endpoints stay fixed. This idea is not needed merely to define path connectedness, but it becomes central in complex analysis and topology, where holes obstruct deformations. The deformation parameter supplies a second variable. One variable moves along the path; the other moves through the family of paths. [definition: Path Homotopy] Let $(X,\tau)$ be a topological space. Let $\gamma_0:[0,1]\to X$ and $\gamma_1:[0,1]\to X$ be paths from $x$ to $y$. A path homotopy from $\gamma_0$ to $\gamma_1$ is a continuous function $H:[0,1]\times[0,1]\to X$ such that $H(t,0)=\gamma_0(t)$, $H(t,1)=\gamma_1(t)$, $H(0,s)=x$, and $H(1,s)=y$ for every $s,t\in[0,1]$. [/definition] The endpoint conditions prevent the deformation from sliding the start or finish through the space. The parameter $s$ indexes the deformation, while $t$ remains the path parameter. [definition: Homotopic Paths Relative to Endpoints] Let $(X,\tau)$ be a topological space. Two paths $\gamma_0,\gamma_1:[0,1]\to X$ from $x$ to $y$ are homotopic relative to endpoints if there exists a path homotopy from $\gamma_0$ to $\gamma_1$. [/definition] Homotopy sees holes that path connectedness ignores. The punctured plane is path connected, but loops winding around the origin cannot be deformed to a constant loop without crossing the missing point. [example: Two Paths Around a Hole] In $\mathbb{R}^2\setminus\{0\}$, define \begin{align*} \gamma_+(t)=(\cos(\pi t),\sin(\pi t)) \end{align*} and \begin{align*} \gamma_-(t)=(\cos(\pi t),-\sin(\pi t)). \end{align*} The coordinate functions are continuous, so $\gamma_+$ and $\gamma_-$ are continuous as maps into $\mathbb{R}^2$. For every $t\in[0,1]$, \begin{align*} |\gamma_+(t)|^2=\cos^2(\pi t)+\sin^2(\pi t)=1 \end{align*} and \begin{align*} |\gamma_-(t)|^2=\cos^2(\pi t)+(-\sin(\pi t))^2=\cos^2(\pi t)+\sin^2(\pi t)=1. \end{align*} Thus neither path ever hits $(0,0)$, so both are paths in $\mathbb{R}^2\setminus\{0\}$. Their endpoints agree: \begin{align*} \gamma_+(0)=(\cos 0,\sin 0)=(1,0) \end{align*} and \begin{align*} \gamma_+(1)=(\cos \pi,\sin \pi)=(-1,0), \end{align*} while \begin{align*} \gamma_-(0)=(\cos 0,-\sin 0)=(1,0) \end{align*} and \begin{align*} \gamma_-(1)=(\cos \pi,-\sin \pi)=(-1,0). \end{align*} So $\gamma_+$ passes above the missing origin and $\gamma_-$ passes below it, but both run from $(1,0)$ to $(-1,0)$. The reverse of $\gamma_-$ is \begin{align*} \overline{\gamma_-}(t)=\gamma_-(1-t)=(\cos(\pi-\pi t),-\sin(\pi-\pi t)). \end{align*} It starts at $(-1,0)$ and ends at $(1,0)$. Hence the concatenation \begin{align*} \ell=\overline{\gamma_-}*\gamma_+ \end{align*} is a loop based at $(1,0)$. By the definition of concatenation, \begin{align*} \ell(t)=\gamma_+(2t)=(\cos(2\pi t),\sin(2\pi t))\quad\text{for }0\le t\le 1/2. \end{align*} For $1/2\le t\le 1$, \begin{align*} \ell(t)=\overline{\gamma_-}(2t-1)=\gamma_-(2-2t). \end{align*} Substituting into $\gamma_-$ gives \begin{align*} \ell(t)=(\cos(2\pi-2\pi t),-\sin(2\pi-2\pi t)). \end{align*} Using $\cos(2\pi-\theta)=\cos\theta$ and $\sin(2\pi-\theta)=-\sin\theta$, this becomes \begin{align*} \ell(t)=(\cos(2\pi t),\sin(2\pi t)). \end{align*} Thus the concatenated loop traces the unit circle once around the missing origin, starting and ending at $(1,0)$. This loop is the visible obstruction: deforming the upper path into the lower path while keeping endpoints fixed would have to remove this one full winding without crossing the origin. [/example] Quotienting paths into deformation classes only makes sense if endpoint-preserving homotopy behaves like equality. Constant deformations, reversed deformations, and consecutive deformations should supply reflexivity, symmetry, and transitivity, which motivates the theorem below. For paths, the relevant special case is obtained by taking $X=[0,1]$ and $A=\{0,1\}$. Thus two paths $\gamma,\eta:[0,1]\to Y$ with the same endpoints are equivalent when there is a homotopy from $\gamma$ to $\eta$ that keeps both endpoint times fixed throughout the deformation. This raises a structural question independent of the circle example: does relative homotopy always form an [equivalence relation](/page/Equivalence%20Relation), so that deformation classes can be used as honest mathematical objects? The following theorem supplies exactly that bookkeeping principle for maps relative to a fixed subset, and hence for endpoint-preserving path homotopy. [quotetheorem:1872] Although the theorem is stated for maps $X\to Y$ relative to a subset $A\subseteq X$, its role here is exactly the endpoint-relative path case $A=\{0,1\}$. It turns deformation of paths into a stable notion of sameness: reflexivity uses the constant deformation, symmetry reverses the deformation parameter, and transitivity concatenates deformations in the deformation variable. ## Compactness and Uniform Control Because every path is defined on the compact interval $[0,1]$, paths inherit strong analytic control. This compactness is why paths are more manageable than maps defined on open intervals. The first consequence is that a metric-space path cannot oscillate too fast without warning. Continuity at each time becomes [uniform continuity](/page/Uniform%20Continuity) across the whole interval. [quotetheorem:9250] Uniform continuity means sufficiently close times have nearby positions along the whole path. In Euclidean space, this raises a useful approximation problem: can a continuous route be replaced by a polygonal one without moving any point of the route very far? Fine partitions control the motion between adjacent sample times, and chords between the samples turn that control into a uniformly close polygonal path. [quotetheorem:9251] If the original path lies in an open set $U\subset\mathbb{R}^n$, then compactness of $\gamma([0,1])$ and openness of $U$ give a positive distance from the trace to the boundary whenever the trace is contained in a compact subset of $U$. After taking a sufficiently fine partition, the polygonal approximation can then be chosen to remain in $U$. This is often how continuous paths are replaced by polygonal ones in analysis. [example: Approximating a Curved Path by Chords] Let $\gamma(t)=(t,t^2)$ for $t\in[0,1]$. Fix $m\ge 1$, and for each $i\in\{0,\ldots,m-1\}$ join $\gamma(i/m)$ to $\gamma((i+1)/m)$ by the affine segment. Thus, when $t\in[i/m,(i+1)/m]$, set $\lambda=mt-i$, so $0\le\lambda\le 1$, and define \begin{align*} p_m(t)=(1-\lambda)\gamma(i/m)+\lambda\gamma((i+1)/m). \end{align*} At a shared endpoint $t=i/m$, the formula from the interval on the left gives $\gamma(i/m)$, and the formula from the interval on the right also gives $\gamma(i/m)$, so the affine pieces match. Hence $p_m$ is a polygonal path through the points $\gamma(0),\gamma(1/m),\ldots,\gamma(1)$. Write $a=i/m$ and $h=1/m$. Since $t=a+\lambda h$, the first coordinate of $p_m(t)$ is \begin{align*} (1-\lambda)a+\lambda(a+h)=a+\lambda h=t. \end{align*} The second coordinate is \begin{align*} (1-\lambda)a^2+\lambda(a+h)^2=a^2+2\lambda ah+\lambda h^2. \end{align*} On the other hand, \begin{align*} t^2=(a+\lambda h)^2=a^2+2\lambda ah+\lambda^2h^2. \end{align*} Therefore \begin{align*} p_m(t)-\gamma(t)=\left(0,\lambda h^2-\lambda^2h^2\right)=\left(0,\lambda(1-\lambda)h^2\right). \end{align*} Since $0\le\lambda\le 1$, we have $\lambda(1-\lambda)\le 1/4$, and hence \begin{align*} |p_m(t)-\gamma(t)|=\lambda(1-\lambda)h^2\le\frac{1}{4m^2}. \end{align*} This bound is independent of $t$, so \begin{align*} \sup_{t\in[0,1]}|p_m(t)-\gamma(t)|\le\frac{1}{4m^2}. \end{align*} As $m\to\infty$, the quantity $1/(4m^2)$ tends to $0$, so the chordal polygonal paths $p_m$ converge uniformly to the curved path $\gamma$. [/example] The trace of a path should inherit compactness and connectedness from the interval that parametrizes it. These inherited properties explain why continuous functions along a path attain bounds and why open covers of the trace reduce to finite subcovers, so the following theorem records the compactness and connectedness of every path image. [quotetheorem:305] The compactness conclusion controls size and covering behaviour, but a path image also ought to remain in one piece. The next question is whether continuity from the connected interval $[0,1]$ prevents the trace from splitting into two separated open pieces; connectedness is the theorem that captures this navigational content. [quotetheorem:1056] This theorem summarizes the topological content of a path. A path traces a compact connected subset, but the opening example shows that not every compact connected subset should be mistaken for the image of a navigable route between arbitrary points inside it. ## Beyond and Connected Topics Paths are the entry point to several major themes in analysis and topology. In [Cambridge IA Analysis Notes](/page/Cambridge%20IA%20Analysis%20Notes), the relevant background is continuity on intervals, compactness, and the intermediate value theorem. In [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology), paths become a bridge between connectedness, compactness, and the topology of metric spaces. This is where path components and local path connectedness become part of the standard language. In [Cambridge IB Complex Analysis](/page/Cambridge%20IB%20Complex%20Analysis), paths become contours. Orientation, concatenation, reversal, and homotopy are the language behind contour integrals, [Cauchy's theorem](/theorems/797), residues, and [winding number](/page/Winding%20Number). In metric geometry, rectifiable paths lead to intrinsic distance. Given a metric space, one may define a new distance by taking the infimum of lengths of paths joining two points. This produces length spaces and geodesic spaces, where distance is realized or approximated by curves. In algebraic topology, endpoint-preserving homotopy classes of loops form the fundamental group. The fundamental group turns holes into algebraic data, and the punctured plane example is the first model for this transition. In analysis on manifolds, paths become curves in coordinate charts and later geodesics under a Riemannian metric. The same basic definition survives, but differentiability and length acquire coordinate-invariant formulations. ## References Androma, [Cambridge IA Analysis Notes](/page/Cambridge%20IA%20Analysis%20Notes). Androma, [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology). Androma, [Cambridge IB Complex Analysis](/page/Cambridge%20IB%20Complex%20Analysis). Androma, [Cambridge II Analysis of Functions](/page/Cambridge%20II%20Analysis%20of%20Functions). James R. Munkres, *Topology* (2000). Walter Rudin, *Principles of Mathematical Analysis* (1976). John B. Conway, *Functions of One Complex Variable I* (1978).