A space can be connected in the sense that it cannot be torn into two separated open pieces, while still refusing to let a point move continuously from one region to another. Path-connectedness rules out this failure. It asks for more than absence of separation: it asks for an actual continuous journey between any two points. The distinction matters because many arguments in analysis and topology do not merely need a space to be in one piece. They need to transport values along curves, compare local data along a route, integrate along paths, deform maps, or speak about homotopy. In those settings, connectedness is often too weak, while path-connectedness is exactly the workable replacement. The standard warning sign is a connected subset of the plane with an oscillating end. It is visually one object, but no continuous path can pass from its limiting vertical segment into its oscillating graph without being trapped by infinitely many swings. This example gives the right problem before the definition appears: connectedness can see one piece, while paths may still be unable to travel through it. [example: The Topologist's Sine Curve] Let \begin{align*} S=\{(x,\sin(1/x)):0<x\le 1\}\subset \mathbb R^2. \end{align*} 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 $S=f((0,1])$ is connected by *Continuous Image of a [Connected Space](/page/Connected%20Space)*. Now define \begin{align*} T=\overline{S}=S\cup(\{0\}\times[-1,1]). \end{align*} The equality holds because if $y\in[-1,1]$, choose $\theta\in \mathbb R$ with $\sin \theta=y$, and choose integers $k$ so large that $\theta+2\pi k>1$; then $x_k=1/(\theta+2\pi k)$ satisfies $x_k\to 0$ and \begin{align*} (x_k,\sin(1/x_k))=(x_k,\sin(\theta+2\pi k))=(x_k,y)\to(0,y). \end{align*} Conversely, every [limit point](/page/Limit%20Point) of $S$ with first coordinate $0$ has second coordinate in $[-1,1]$, because $\sin(1/x)\in[-1,1]$ for every $x>0$. Hence $T$ is the closure of the connected set $S$, so $T$ is connected by *[Closure of a Connected Set Is Connected](/theorems/297)*. We show that $T$ is not path-connected by proving that no path can join the vertical segment $\{0\}\times[-1,1]$ to the oscillating graph $S$. Suppose, toward a contradiction, that $\gamma:[0,1]\to T$ is such a path, oriented so that $\gamma(0)\in\{0\}\times[-1,1]$ and $\gamma(1)\in S$. Write $\gamma(t)=(u(t),v(t))$. The coordinate functions $u$ and $v$ are continuous because they are compositions of $\gamma$ with the two coordinate projections. Let \begin{align*} A=\{t\in[0,1]:u(t)=0\}. \end{align*} Then $A=u^{-1}(\{0\})$ is closed, $0\in A$, and $1\notin A$ because $\gamma(1)\in S$ has positive first coordinate. Set $a=\sup A$. Since $A$ is closed in $[0,1]$, we have $a\in A$, so $u(a)=0$; also $a<1$. If $t>a$, then $t\notin A$ by the definition of supremum, and since every point of $T$ has nonnegative first coordinate, $u(t)>0$. Therefore, for every $t\in(a,1]$, the point $\gamma(t)$ lies in $S$, and so \begin{align*} v(t)=\sin(1/u(t)). \end{align*} Fix any $b$ with $a<b\le 1$. Since $u(b)>0$ and $u$ is continuous on $[a,b]$, the intermediate value property gives, for every $r\in(0,u(b))$, some $t_r\in(a,b)$ such that $u(t_r)=r$. Choose $k$ large enough that \begin{align*} 0<\frac{1}{\pi/2+2\pi k}<u(b). \end{align*} For the corresponding $t_k^+\in(a,b)$ with $u(t_k^+)=1/(\pi/2+2\pi k)$, we get \begin{align*} v(t_k^+)=\sin(1/u(t_k^+))=\sin(\pi/2+2\pi k)=1. \end{align*} Choose $k$ large enough that \begin{align*} 0<\frac{1}{3\pi/2+2\pi k}<u(b). \end{align*} For the corresponding $t_k^-\in(a,b)$ with $u(t_k^-)=1/(3\pi/2+2\pi k)$, we get \begin{align*} v(t_k^-)=\sin(1/u(t_k^-))=\sin(3\pi/2+2\pi k)=-1. \end{align*} Thus every interval $(a,b)$ contains points where $v=1$ and points where $v=-1$. This contradicts continuity of $v$ at $a$. Indeed, $v(a)\in[-1,1]$, so at least one of $|1-v(a)|$ and $|-1-v(a)|$ is at least $1$. But continuity at $a$ with $\varepsilon=1/2$ would give a right-neighbourhood of $a$ on which all values of $v$ lie within $1/2$ of $v(a)$, while every such neighbourhood contains a point with $v=1$ and a point with $v=-1$. Hence no such path exists. The space $T$ is therefore connected, but not path-connected. [/example] This example sets the agenda. Connectedness detects whether open sets separate the space. Path-connectedness detects whether the space supports continuous motion. The two notions agree in many familiar geometric settings, but they diverge in spaces with bad local behaviour. ## Definition ### The Global Condition The topologist's sine curve shows that the central question is global: can any point of the space be reached from any other by continuous motion? We begin with the page's main definition, phrased directly in terms of continuous maps from the unit interval. The supporting language of paths is then isolated immediately afterward. [definition: Path-Connected Space] A [topological space](/page/Topological%20Space) $(X,\tau)$ is path-connected if for every pair of points $x,y \in X$, there exists a continuous map $\gamma : [0,1] \to X$ such that $\gamma(0)=x$ and $\gamma(1)=y$. [/definition] This definition includes the one-point and empty cases by quantification. A one-point space is path-connected. The empty space is also path-connected under the usual vacuous convention, since there are no pairs of points to test. ### Paths The definition names a global property, but its basic object deserves its own name. The parameter interval $[0,1]$ is not important because of its length. It is important because it is connected, ordered, compact, and has two distinguished endpoints. [definition: Path] Let $(X, \tau)$ be a topological space. A path in $X$ from $x$ to $y$ is a continuous map $\gamma : [0,1] \to X$ such that $\gamma(0)=x$ and $\gamma(1)=y$. [/definition] The endpoints matter, but the speed along the path usually does not. A path records that a continuous route exists; it does not record a preferred parametrisation unless the problem has extra metric, differentiable, or variational structure. The most familiar paths are straight-line paths inside intervals and convex sets. [example: Paths in an Interval] Let $a,b \in \mathbb R$ with $a<b$, and let $[a,b]$ have the [subspace topology](/page/Subspace%20Topology) from $\mathbb R$. Given $x,y \in [a,b]$, define $\gamma:[0,1]\to \mathbb R$ by \begin{align*} \gamma(t)=(1-t)x+ty. \end{align*} For $t\in[0,1]$, we have $1-t\ge 0$ and $t\ge 0$. Since $a\le x,y\le b$, \begin{align*} \gamma(t)-a=(1-t)(x-a)+t(y-a)\ge 0. \end{align*} Also, \begin{align*} b-\gamma(t)=(1-t)(b-x)+t(b-y)\ge 0. \end{align*} Thus $a\le \gamma(t)\le b$, so $\gamma(t)\in[a,b]$ for every $t\in[0,1]$. The formula can be rewritten as \begin{align*} \gamma(t)=x+t(y-x). \end{align*} This is continuous as a real-valued function because it is obtained from the identity function $t\mapsto t$ by multiplying by the constant $y-x$ and adding the constant $x$; since its image lies in $[a,b]$, it is continuous as a map into the subspace $[a,b]$. Finally, \begin{align*} \gamma(0)=(1-0)x+0y=x. \end{align*} and \begin{align*} \gamma(1)=(1-1)x+1y=y. \end{align*} Hence every two points of the interval are joined by this straight-line path. [/example] [remark: Naming Convention] Some books write pathwise connected instead of path-connected. The two terms mean the same condition: any two points can be joined by a continuous map from $[0,1]$. [/remark] ### First Geometric Sources To build examples rather than merely name the property, we need a structural condition that manufactures paths. Convexity is the strongest and most concrete such condition in Euclidean space: it says that the most direct route between two points stays in the set. This makes convex sets the basic laboratory for path-connectedness. [definition: Convex Subset] A set $C$ in $\mathbb R^n$ is convex if for every $x,y \in C$ and every $t \in [0,1]$, \begin{align*} (1-t)x + ty \in C. \end{align*} [/definition] Convexity supplies a path formula, so it deserves to be converted into a theorem. The result is the first permanence principle: a strong geometric closure condition implies a topological navigability condition. It also explains why balls, intervals, half-spaces, affine subspaces, and simplices behave as single pieces for path arguments. [quotetheorem:9157] The theorem is useful, but it is not a characterisation. Path-connectedness allows curves to bend, detour, or wind around holes. The circle is the first example where straight segments fail but continuous travel remains possible. [example: A Circle Is Path-Connected but Not Convex] Let $S^1=\{z\in\mathbb C:|z|=1\}$ with the subspace topology from $\mathbb C$. Given $z,w\in S^1$, choose [real numbers](/page/Real%20Numbers) $\alpha,\beta\in\mathbb R$ such that $z=e^{i\alpha}$ and $w=e^{i\beta}$. Define $\gamma:[0,1]\to\mathbb C$ by \begin{align*} \gamma(t)=e^{i((1-t)\alpha+t\beta)}. \end{align*} For each $t\in[0,1]$, the number $(1-t)\alpha+t\beta$ is real, so \begin{align*} |\gamma(t)|=|e^{i((1-t)\alpha+t\beta)}|=1. \end{align*} Hence $\gamma(t)\in S^1$ for every $t\in[0,1]$, so $\gamma$ may be regarded as a map $[0,1]\to S^1$. The function \begin{align*} t\mapsto (1-t)\alpha+t\beta=\alpha+t(\beta-\alpha) \end{align*} is continuous as a real-valued function, and $r\mapsto e^{ir}$ is continuous from $\mathbb R$ to $\mathbb C$; therefore their composition $\gamma$ is continuous as a map into $\mathbb C$, and hence continuous as a map into the subspace $S^1$ because its image lies in $S^1$. At the endpoints, \begin{align*} \gamma(0)=e^{i((1-0)\alpha+0\beta)}=e^{i\alpha}=z. \end{align*} Also, \begin{align*} \gamma(1)=e^{i((1-1)\alpha+1\beta)}=e^{i\beta}=w. \end{align*} Thus every two points of $S^1$ are joined by a path in $S^1$, so the circle is path-connected. It is not convex as a subset of $\mathbb R^2\cong\mathbb C$: the points $1$ and $-1$ both lie in $S^1$, but the midpoint of the line segment between them is \begin{align*} (1-\tfrac12)1+\tfrac12(-1)=\tfrac12-\tfrac12=0. \end{align*} Since $|0|=0\ne1$, this midpoint is not in $S^1$. Thus straight-line convexity fails even though continuous travel around the circle is always possible. [/example] ## Algebra of Paths ### Reversal and Concatenation Path-connectedness treats paths as evidence that points lie in the same navigable piece. For this purpose, the exact parametrisation of a path is secondary. What matters is that paths can be reversed, delayed, and glued without leaving the category of continuous maps. A path should be usable in either direction. If movement from $x$ to $y$ is available, then the same route run backwards gives movement from $y$ to $x$. This symmetry is the first reason path-joinability behaves like an [equivalence relation](/page/Equivalence%20Relation). [quotetheorem:9158] Reversal is only half the algebra of paths. To move from $x$ to $z$ through an intermediate point $y$, we need to glue a path from $x$ to $y$ to a path from $y$ to $z$. The matching endpoint is the datum that makes the resulting piecewise formula continuous. [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 $\gamma * \eta : [0,1] \to X$ is defined by \begin{align*} (\gamma * \eta)(t)=\gamma(2t) \quad \text{for } 0 \le t \le 1/2, \end{align*} and \begin{align*} (\gamma * \eta)(t)=\eta(2t-1) \quad \text{for } 1/2 \le t \le 1. \end{align*} [/definition] The definition gives a candidate route, but topology still has to certify that the route is continuous at the joining time. That certification is needed before concatenation can be used in later arguments about equivalence relations, path components, and basepoint tests. The next result supplies exactly this guarantee. [quotetheorem:9159] ### Basepoints Reversal and concatenation make it unnecessary to construct paths separately for every ordered pair. If every point can be reached from one chosen point, then any two points can be joined by going back to the chosen point and then out again. This reduces many proofs of path-connectedness to a single-hub construction. [quotetheorem:9160] The basepoint test is often the most efficient way to prove examples. Some sets have a distinguished centre from which every point is visible by a straight segment. These sets need not be convex, but the basepoint still supplies paths to all points. [example: Star-Shaped Sets] Let $A \subset \mathbb R^n$ and suppose there is a point $a_0 \in A$ such that for every $x \in A$ and every $s \in [0,1]$, \begin{align*} (1-s)a_0+sx \in A. \end{align*} We show directly that any two points $x,y \in A$ can be joined by a path in $A$. Define $\gamma:[0,1]\to \mathbb R^n$ by \begin{align*} \gamma(t)=(1-2t)x+2t a_0 \quad \text{for } 0\le t\le \tfrac12, \end{align*} and \begin{align*} \gamma(t)=(2-2t)a_0+(2t-1)y \quad \text{for } \tfrac12\le t\le 1. \end{align*} For $0\le t\le \tfrac12$, set $s=1-2t$. Then $s\in[0,1]$, and \begin{align*} \gamma(t)=(1-2t)x+2t a_0=(1-s)x+s a_0=s a_0+(1-s)x. \end{align*} Since $1-s\in[0,1]$, the star-shaped hypothesis applied to the point $x$ shows that $\gamma(t)\in A$. For $\tfrac12\le t\le 1$, set $r=2t-1$. Then $r\in[0,1]$, and \begin{align*} \gamma(t)=(2-2t)a_0+(2t-1)y=(1-r)a_0+ry. \end{align*} The star-shaped hypothesis applied to the point $y$ shows that $\gamma(t)\in A$. Thus $\gamma([0,1])\subset A$. The two formulae agree at the joining time because \begin{align*} (1-2\cdot\tfrac12)x+2\cdot\tfrac12 a_0=0x+a_0=a_0, \end{align*} and \begin{align*} (2-2\cdot\tfrac12)a_0+(2\cdot\tfrac12-1)y=a_0+0y=a_0. \end{align*} Each piece is affine in $t$, hence continuous as a map into $\mathbb R^n$, and the pieces agree at $t=\tfrac12$, so $\gamma$ is continuous by the *[Pasting Lemma](/theorems/1037)*. Since its image lies in $A$, it is continuous as a map $[0,1]\to A$ with the subspace topology. At the endpoints, \begin{align*} \gamma(0)=(1-0)x+0a_0=x, \end{align*} and \begin{align*} \gamma(1)=(2-2)a_0+(2-1)y=y. \end{align*} Hence every pair $x,y\in A$ can be joined by a path in $A$, so $A$ is path-connected. Star-shaped sets need not be convex. For example, in $\mathbb R^2$ the set \begin{align*} A=\{s(1,0):0\le s\le1\}\cup\{s(0,1):0\le s\le1\} \end{align*} is star-shaped with centre $(0,0)$, because every segment from $(0,0)$ to a point of $A$ stays in the same arm. But $(1,0),(0,1)\in A$, while their midpoint is \begin{align*} \tfrac12(1,0)+\tfrac12(0,1)=(\tfrac12,\tfrac12), \end{align*} which lies on neither coordinate arm, so it is not in $A$. Thus path-connectedness can come from a common centre even when convexity fails. [/example] ## Behaviour under Constructions ### Continuous Images A useful connectedness notion should survive the constructions that occur in ordinary mathematics. The most important construction is applying a continuous map. Since a path is itself a continuous map, composing it with another continuous map sends journeys in the domain to journeys in the image. [quotetheorem:1056] This theorem is the path-connected analogue of the continuous image theorem for connected spaces. It is often used in contrapositive form: if an image has two path components, then no path-connected domain can map onto it continuously. The smallest target where this obstruction appears is a discrete two-point space. [example: No Continuous Surjection from an Interval onto a Two-Point Discrete Space] Let $D=\{0,1\}$ with the [discrete topology](/page/Discrete%20Topology), and let $f:[0,1]\to D$ be continuous. We first show that $f([0,1])$ is path-connected. If $p,q\in f([0,1])$, choose $s,t\in[0,1]$ with $f(s)=p$ and $f(t)=q$. Define $\alpha:[0,1]\to[0,1]$ by \begin{align*} \alpha(r)=(1-r)s+rt. \end{align*} Since $0\le s,t\le 1$ and $0\le r\le 1$, the same convexity calculation used for intervals gives $0\le \alpha(r)\le 1$. The map $\alpha$ is continuous because it is affine in $r$, and therefore $f\circ\alpha:[0,1]\to f([0,1])$ is continuous. Its endpoints are \begin{align*} (f\circ\alpha)(0)=f((1-0)s+0t)=f(s)=p. \end{align*} and \begin{align*} (f\circ\alpha)(1)=f((1-1)s+1t)=f(t)=q. \end{align*} Thus any two points of $f([0,1])$ are joined by a path in $f([0,1])$. Now no path in $D$ can join $0$ to $1$. Indeed, if $\gamma:[0,1]\to D$ were continuous with $\gamma(0)=0$ and $\gamma(1)=1$, then \begin{align*} [0,1]=\gamma^{-1}(\{0\})\cup\gamma^{-1}(\{1\}) \end{align*} because $D=\{0,1\}$. The sets $\{0\}$ and $\{1\}$ are open in the discrete topology on $D$, so both preimages are open in $[0,1]$. They are disjoint, since no point of $[0,1]$ can be mapped to both $0$ and $1$. They are nonempty, because $0\in\gamma^{-1}(\{0\})$ and $1\in\gamma^{-1}(\{1\})$. This separates $[0,1]$, contradicting the connectedness of the interval. Therefore a path-connected subspace of $D$ cannot contain both $0$ and $1$. Since $[0,1]$ is nonempty, $f([0,1])$ is nonempty, so $f([0,1])$ is either $\{0\}$ or $\{1\}$. Hence every continuous map $f:[0,1]\to D$ is constant, and in particular there is no continuous surjection from $[0,1]$ onto $D$. [/example] ### Products and Unions Products allow independent movement in each coordinate. To move from $(x_1,y_1)$ to $(x_2,y_2)$, one should be able to move in $X$ and in $Y$ at the same time. The [product topology](/page/Product%20Topology) is designed so that coordinatewise continuous paths assemble into a continuous path in the product. [quotetheorem:1057] Products combine spaces coordinate by coordinate, while unions combine subspaces by overlap. The next permanence question is whether two path-connected regions remain path-connected after they are glued together inside a larger space. A common point gives the needed transfer point for any path travelling from one region to the other. [quotetheorem:9161] The overlap hypothesis cannot simply be removed. Without a common transfer point, there may be no way for a path to pass from one piece to the other. The failure is already visible in a finite discrete space. [example: Two Isolated Points] Let $X=\{0,1\}$ with the discrete topology. Each singleton subspace is path-connected: for $\{0\}$, the map $c_0:[0,1]\to\{0\}$ defined by $c_0(t)=0$ is continuous and joins $0$ to $0$; similarly, $c_1:[0,1]\to\{1\}$ defined by $c_1(t)=1$ is continuous and joins $1$ to $1$. We show that the union $X=\{0\}\cup\{1\}$ is not path-connected. Suppose, toward a contradiction, that there is a path $\gamma:[0,1]\to X$ with $\gamma(0)=0$ and $\gamma(1)=1$. Since $X$ has the discrete topology, the subsets $\{0\}$ and $\{1\}$ are open in $X$. Therefore \begin{align*} U=\gamma^{-1}(\{0\}) \end{align*} and \begin{align*} V=\gamma^{-1}(\{1\}) \end{align*} are open in $[0,1]$ by continuity of $\gamma$. The sets $U$ and $V$ are nonempty because $0\in U$ and $1\in V$. They are disjoint because no point $t\in[0,1]$ can satisfy both $\gamma(t)=0$ and $\gamma(t)=1$. They cover $[0,1]$ because every value of $\gamma$ lies in $X=\{0,1\}$, so for each $t\in[0,1]$, either $\gamma(t)=0$ and $t\in U$, or $\gamma(t)=1$ and $t\in V$. Hence $U$ and $V$ form a separation of $[0,1]$, contradicting *Connectedness of Interval*. Thus no path in $X$ joins $0$ to $1$, so $X$ is not path-connected. [/example] ## Path Components ### Path Equivalence When a space is not path-connected, the next useful question is not only that it fails, but how it fails. We need a systematic way to group points that can reach each other by paths. The algebra of reversal and concatenation is exactly what turns this grouping into equivalence classes. [definition: Path Equivalence] Let $(X,\tau)$ be a topological space. For $x,y \in X$, write $x \sim_p y$ if there exists a path in $X$ from $x$ to $y$. [/definition] For this relation to define meaningful pieces of the space, it must satisfy the three axioms of an equivalence relation. Each axiom corresponds to a basic path operation: constant paths for staying put, reversal for changing direction, and concatenation for passing through an intermediate point. The next theorem verifies that the relation really partitions the space. [quotetheorem:9162] ### Components Equivalence relations divide a set into classes, and here the classes are the largest regions detected by paths. This definition records the exact pieces inside which continuous travel is possible. It also gives a replacement for connected components when path behaviour is the primary concern. [definition: Path Component] Let $(X,\tau)$ be a topological space. A path component of $X$ is an equivalence class of the relation $\sim_p$. [/definition] After naming the pieces, there is a possible ambiguity: an equivalence class is defined set-theoretically, while a path-connected piece is defined by the existence of paths between all of its points. These two descriptions need to agree if path components are to serve as the natural pieces of the space. The structural question is whether the classes cover the space without overlap and whether every path-connected subset is forced to lie inside one of them. [quotetheorem:9163] Path components need not be open or closed in an arbitrary topological space. This is one of the places where local behaviour becomes decisive. The topologist's sine curve shows that path components can be much smaller than connected components. [example: Path Components of the Topologist's Sine Curve] For the topologist's sine curve closure \begin{align*} T=\{(x,\sin(1/x)):0<x\le 1\}\cup(\{0\}\times[-1,1]), \end{align*} write \begin{align*} S=\{(x,\sin(1/x)):0<x\le1\} \end{align*} and \begin{align*} V=\{0\}\times[-1,1]. \end{align*} The set $S$ is path-connected: if $p=(x_0,\sin(1/x_0))$ and $q=(x_1,\sin(1/x_1))$ lie in $S$, define \begin{align*} \lambda(t)=(1-t)x_0+tx_1. \end{align*} Since $0<x_0,x_1\le1$ and $0\le t\le1$, we have $0<\lambda(t)\le1$. Hence \begin{align*} \gamma(t)=(\lambda(t),\sin(1/\lambda(t))) \end{align*} lies in $S$ for every $t\in[0,1]$. The map $\lambda$ is affine, so it is continuous, and the maps $r\mapsto 1/r$ on $(0,\infty)$ and $r\mapsto\sin r$ are continuous; therefore $\gamma$ is continuous. Its endpoints are \begin{align*} \gamma(0)=(x_0,\sin(1/x_0))=p. \end{align*} and \begin{align*} \gamma(1)=(x_1,\sin(1/x_1))=q. \end{align*} The vertical segment $V$ is also path-connected. If $(0,y_0),(0,y_1)\in V$, define \begin{align*} \delta(t)=(0,(1-t)y_0+ty_1). \end{align*} Because $[-1,1]$ is an interval, $(1-t)y_0+ty_1\in[-1,1]$ for every $t\in[0,1]$, so $\delta([0,1])\subset V$. The coordinate formula is affine in $t$, hence continuous, and \begin{align*} \delta(0)=(0,y_0). \end{align*} and \begin{align*} \delta(1)=(0,y_1). \end{align*} It remains to show that no path in $T$ joins $V$ to $S$. Suppose, for contradiction, that $\gamma:[0,1]\to T$ is such a path, oriented so that $\gamma(0)\in V$ and $\gamma(1)\in S$. Write $\gamma(t)=(u(t),v(t))$. The coordinate functions $u$ and $v$ are continuous because they are compositions of $\gamma$ with the coordinate projections. Let \begin{align*} A=\{t\in[0,1]:u(t)=0\}. \end{align*} Then $A$ is closed, $0\in A$, and $1\notin A$. Set $a=\sup A$. Since $A$ is closed in $[0,1]$, we have $a\in A$, so $u(a)=0$, and since $1\notin A$, we have $a<1$. If $t>a$, then $t\notin A$, so $u(t)\ne0$. Every point of $T$ has nonnegative first coordinate, hence $u(t)>0$. Therefore $\gamma(t)\in S$ for every $t\in(a,1]$, and so \begin{align*} v(t)=\sin(1/u(t)). \end{align*} Fix $b$ with $a<b\le1$. Since $u(a)=0$ and $u(b)>0$, the intermediate value property gives, for every $r\in(0,u(b))$, some $t_r\in(a,b)$ such that $u(t_r)=r$. Choose $k$ so large that \begin{align*} 0<\frac{1}{\pi/2+2\pi k}<u(b). \end{align*} Then for some $t_k^+\in(a,b)$, \begin{align*} u(t_k^+)=\frac{1}{\pi/2+2\pi k}. \end{align*} Thus \begin{align*} v(t_k^+)=\sin(1/u(t_k^+))=\sin(\pi/2+2\pi k)=1. \end{align*} Similarly, choose $k$ so large that \begin{align*} 0<\frac{1}{3\pi/2+2\pi k}<u(b). \end{align*} Then for some $t_k^-\in(a,b)$, \begin{align*} u(t_k^-)=\frac{1}{3\pi/2+2\pi k}. \end{align*} Hence \begin{align*} v(t_k^-)=\sin(1/u(t_k^-))=\sin(3\pi/2+2\pi k)=-1. \end{align*} So every interval $(a,b)$ contains points where $v=1$ and points where $v=-1$. This contradicts continuity of $v$ at $a$. Since $v(a)\in[-1,1]$, at least one of $|1-v(a)|$ and $|-1-v(a)|$ is at least $1$. But continuity at $a$ with $\varepsilon=1/2$ gives a number $\delta>0$ such that $|v(t)-v(a)|<1/2$ whenever $t\in[0,1]$ and $|t-a|<\delta$. Taking $b\in(a,\min\{1,a+\delta\})$ gives points in $(a,b)$ where $v=1$ and $v=-1$, one of which has distance at least $1$ from $v(a)$, a contradiction. Thus no path joins $S$ to $V$. Consequently the path components of $T$ are exactly the oscillating graph $S$ and the vertical segment $V$, while the connectedness of $T$ shows that connected components do not detect this finer path splitting. [/example] ## Local Path-Connectedness ### Local Control The bad behaviour in the topologist's sine curve comes from neighbourhoods that are connected in a limiting sense but do not contain small path-connected neighbourhoods. To rule out this pathology, the space must have enough path-connected open sets near each point. Local path-connectedness packages exactly that local control. [definition: Locally Path-Connected Space] A topological space $(X,\tau)$ is locally path-connected if for every point $x \in X$ and every [open set](/page/Open%20Set) $U \in \tau$ with $x \in U$, there exists an open set $V \in \tau$ such that $x \in V \subset U$ and $V$ is path-connected. [/definition] The definition is local because it only asks for arbitrarily small path-connected neighbourhoods. It does not say that the whole space is path-connected. A space can be locally path-connected while having several separate global pieces. [example: Disjoint Union of Two Intervals] Let $X=[0,1]\cup[2,3]$ with the subspace topology from $\mathbb R$. We first verify local path-connectedness. Let $x\in X$, and let $U$ be open in $X$ with $x\in U$. Since $X$ has the subspace topology, there is an open set $O\subset\mathbb R$ such that $U=O\cap X$. Because $x\in O$ and $O$ is open in $\mathbb R$, choose $\varepsilon>0$ with $(x-\varepsilon,x+\varepsilon)\subset O$. If $x\in[0,1]$, choose \begin{align*} r=\min\{\varepsilon,\tfrac12\}. \end{align*} Then \begin{align*} V=(x-r,x+r)\cap X=(x-r,x+r)\cap[0,1], \end{align*} because $(x-r,x+r)$ cannot meet $[2,3]$. The set $V$ is open in $X$, contains $x$, and satisfies $V\subset U$. If $y,z\in V$ and $t\in[0,1]$, then \begin{align*} (1-t)y+tz-y=t(z-y). \end{align*} If $y\le z$, this gives $y\le(1-t)y+tz\le z$ because $0\le t\le1$; if $z\le y$, the same calculation gives $z\le(1-t)y+tz\le y$. Since $V$ is an interval in $\mathbb R$, the straight-line path \begin{align*} \gamma(t)=(1-t)y+tz \end{align*} lies in $V$ for every $t\in[0,1]$, is continuous as an affine real-valued function, and satisfies \begin{align*} \gamma(0)=y \end{align*} and \begin{align*} \gamma(1)=z. \end{align*} Thus $V$ is path-connected. The same argument applies when $x\in[2,3]$, using $V=(x-r,x+r)\cap[2,3]$. Hence $X$ is locally path-connected. The space $X$ is not path-connected. Suppose, for contradiction, that there is a path $\gamma:[0,1]\to X$ with $\gamma(0)=0$ and $\gamma(1)=2$. Regard $\gamma$ as a real-valued map by composing with the inclusion $X\hookrightarrow\mathbb R$; this map is continuous because inclusions of subspaces are continuous. Since \begin{align*} \gamma(0)=0<\tfrac32<2=\gamma(1), \end{align*} the *[Intermediate Value Theorem](/theorems/180)* gives $t_0\in[0,1]$ with \begin{align*} \gamma(t_0)=\tfrac32. \end{align*} This is impossible because $\tfrac32\notin[0,1]$ and $\tfrac32\notin[2,3]$. Therefore no path joins $0$ to $2$, so $X$ is locally path-connected but not path-connected. [/example] ### Open Path Components Local path-connectedness is powerful because it aligns path components with the open-set structure. If every point has small path-connected neighbourhoods, then a point inside a path component carries an open neighbourhood that stays inside the same component. This makes path components visible to the topology rather than hidden as thin subsets. [quotetheorem:1058] The theorem has two linked consequences. First, local path-connectedness makes path components open, so they behave like visible pieces of the topology. Second, if the whole space is connected, those open pieces cannot form more than one nonempty component without producing a separation. This is why the local hypothesis is exactly what lets connectedness force path-connectedness. The theorem explains why path-connectedness and connectedness are often indistinguishable in elementary examples. The missing hypothesis is local path-connectedness, and the topologist's sine curve shows that this hypothesis carries real content. It is a local condition with global consequences. [example: A Connected Space That Is Not Locally Path-Connected] Let \begin{align*} T=\{(x,\sin(1/x)):0<x\le 1\}\cup(\{0\}\times[-1,1]). \end{align*} The earlier topologist's sine curve example established two facts: $T$ is connected, and no path in $T$ joins a point of the vertical segment $\{0\}\times[-1,1]$ to a point of the oscillating graph $\{(x,\sin(1/x)):0<x\le1\}$. Hence $T$ is not path-connected. Suppose, for contradiction, that $T$ were locally path-connected. Since $T$ is connected and locally path-connected, *Connected Locally Path-Connected Spaces Are Path-Connected* would imply that $T$ is path-connected. This contradicts the fact that no path in $T$ can join the vertical segment to the oscillating graph. Therefore $T$ is connected but not locally path-connected. The obstruction is local: every neighbourhood in $T$ of a point on the limiting vertical segment contains infinitely many nearby arcs of the oscillating graph, but those arcs cannot be connected by paths to the vertical segment inside $T$. [/example] ## Euclidean Domains ### Open Sets in $\mathbb R^n$ In subsets of $\mathbb R^n$, path-connectedness often appears through polygonal paths. Open sets are locally flexible: once a point lies in an open set, a small ball around it also lies in the set. Since balls are convex, open Euclidean sets have path-connected neighbourhoods at every point. [quotetheorem:9164] This local result matters because analysts often start with a connected open domain and need actual paths inside it. Connectedness alone rules out a decomposition into separated open pieces, but it does not by itself produce curves. In an open subset of Euclidean space, the small balls around each point supply the local path-connectedness needed to turn the absence of separation into the existence of paths. [quotetheorem:301] ### Polygonal Paths For analysis, it is often not enough to know that some continuous path exists. Estimates along line segments are easier to control than arbitrary curves. In open connected subsets of Euclidean space, the available paths can be chosen as finite chains of straight segments. [definition: Polygonal Path] Let $U \subset \mathbb R^n$. A polygonal path in $U$ from $x$ to $y$ is a path $\gamma : [0,1] \to U$ for which there exist an integer $m \in \mathbb N$, points $x=p_0,p_1,\ldots,p_m=y$ in $U$, and a partition \begin{align*} 0=t_0<t_1<\cdots<t_m=1 \end{align*} such that $[p_{j-1},p_j] \subset U$ for each $j \in \{1,\ldots,m\}$ and, for every $j \in \{1,\ldots,m\}$, the restricted map $\gamma|_{[t_{j-1},t_j]} : [t_{j-1},t_j] \to U$ is given by \begin{align*} \gamma(t)=\left(1-\frac{t-t_{j-1}}{t_j-t_{j-1}}\right)p_{j-1}+\frac{t-t_{j-1}}{t_j-t_{j-1}}p_j \quad \text{for } t \in [t_{j-1},t_j]. \end{align*} [/definition] Polygonal paths are central in multivariable analysis because they allow local estimates to be added along finitely many line segments. This is why connected open sets support arguments that first prove a statement in a ball and then propagate it across the domain. The next theorem records the exact equivalence. [quotetheorem:9165] The openness assumption is essential. A subset of $\mathbb R^n$ may be connected without allowing polygonal paths, or even without allowing any nonconstant paths between certain points. The original motivating example is precisely such a failure. [example: Why Openness Matters] Let \begin{align*} T=\{(x,\sin(1/x)):0<x\le 1\}\cup(\{0\}\times[-1,1])\subset\mathbb R^2. \end{align*} The earlier topologist's sine curve example shows that $T$ is connected and that no path in $T$ joins a point of the vertical segment $\{0\}\times[-1,1]$ to a point of the oscillating graph $\{(x,\sin(1/x)):0<x\le1\}$. A polygonal path is, in particular, a path, so if every two points of $T$ could be joined by a polygonal path in $T$, then every two points of $T$ could be joined by a path in $T$. This would make $T$ path-connected, contradicting the established separation between the vertical segment and the graph. The openness hypothesis fails visibly at every point of the vertical segment. Fix $y\in[-1,1]$ and put $p=(0,y)\in T$. For any $\varepsilon>0$, define \begin{align*} q=(-\varepsilon/2,y). \end{align*} Then \begin{align*} \|q-p\|=\|(-\varepsilon/2,0)\|=\varepsilon/2<\varepsilon, \end{align*} so $q$ lies in the Euclidean ball $B_\varepsilon(p)$. But $q\notin T$, because every point of $T$ has first coordinate either $0$ or a positive number, while $q$ has first coordinate $-\varepsilon/2<0$. Hence no Euclidean ball around $p$ is contained in $T$, so $T$ is not open in $\mathbb R^2$. Thus $T$ is a connected subset of the plane, but it lies outside the scope of the polygonal path characterisation because it is not an open subset of $\mathbb R^2$. [/example] ### Holes and Dimension Punctured Euclidean spaces show a different phenomenon: holes may prevent convexity without preventing path-connectedness. Whether a point removal disconnects the space depends on dimension. This example marks the boundary between elementary connectedness and the richer questions studied by algebraic topology. [example: Punctured Euclidean Space] For $n\ge 2$, we show that $\mathbb R^n\setminus\{0\}$ is path-connected. Let $x,y\in\mathbb R^n\setminus\{0\}$. If \begin{align*} (1-t)x+ty\ne 0 \quad \text{for every } t\in[0,1], \end{align*} then the straight-line map $\gamma(t)=(1-t)x+ty$ has image in $\mathbb R^n\setminus\{0\}$, is continuous because each coordinate is affine in $t$, and satisfies \begin{align*} \gamma(0)=(1-0)x+0y=x. \end{align*} Also, \begin{align*} \gamma(1)=(1-1)x+1y=y. \end{align*} Thus $x$ and $y$ are joined by a path. It remains to handle the case where the segment from $x$ to $y$ contains $0$. Then there is some $t_0\in[0,1]$ such that \begin{align*} (1-t_0)x+t_0y=0. \end{align*} Since $x\ne0$ and $y\ne0$, this cannot happen at $t_0=0$ or $t_0=1$, so $0<t_0<1$. Rearranging gives \begin{align*} t_0y=-(1-t_0)x. \end{align*} Dividing by $t_0>0$ gives \begin{align*} y=-\frac{1-t_0}{t_0}x. \end{align*} Hence $x$ and $y$ lie on the one-dimensional line $L=\{\lambda x:\lambda\in\mathbb R\}$. Because $n\ge2$, this line is a proper subset of $\mathbb R^n$, so choose $z\in\mathbb R^n\setminus L$. In particular $z\ne0$. The segment from $x$ to $z$ avoids $0$. Indeed, if for some $s\in[0,1]$ we had \begin{align*} (1-s)x+sz=0, \end{align*} then $s\ne0$ because $x\ne0$, and $s\ne1$ because $z\ne0$. Thus $0<s<1$, and rearranging gives \begin{align*} sz=-(1-s)x. \end{align*} Dividing by $s>0$ gives \begin{align*} z=-\frac{1-s}{s}x. \end{align*} This puts $z$ in $L$, contradicting the choice of $z$. The same argument shows that the segment from $z$ to $y$ avoids $0$: if \begin{align*} (1-s)z+sy=0 \end{align*} for some $s\in[0,1]$, then $0<s<1$, so \begin{align*} z=-\frac{s}{1-s}y. \end{align*} Since $y\in L$, this again implies $z\in L$, a contradiction. Define $\gamma:[0,1]\to\mathbb R^n\setminus\{0\}$ by \begin{align*} \gamma(t)=(1-2t)x+2tz \quad \text{for } 0\le t\le \tfrac12, \end{align*} and \begin{align*} \gamma(t)=(2-2t)z+(2t-1)y \quad \text{for } \tfrac12\le t\le 1. \end{align*} The first formula parametrises the segment from $x$ to $z$, and the second parametrises the segment from $z$ to $y$, so the image avoids $0$. At the joining time, \begin{align*} (1-2\cdot\tfrac12)x+2\cdot\tfrac12 z=0x+z=z. \end{align*} Also, \begin{align*} (2-2\cdot\tfrac12)z+(2\cdot\tfrac12-1)y=z+0y=z. \end{align*} Each piece is affine in $t$, and the two pieces agree at $t=\tfrac12$, so $\gamma$ is continuous by the *Pasting Lemma*. Its endpoints are \begin{align*} \gamma(0)=(1-0)x+0z=x. \end{align*} and \begin{align*} \gamma(1)=(2-2)z+(2-1)y=y. \end{align*} Thus every two points of $\mathbb R^n\setminus\{0\}$ can be joined by a path when $n\ge2$. For $n=1$, the punctured space is \begin{align*} \mathbb R\setminus\{0\}=(-\infty,0)\cup(0,\infty). \end{align*} Each interval is path-connected by straight-line paths. No path can join a negative point to a positive point: if $\gamma:[0,1]\to\mathbb R\setminus\{0\}$ were continuous with $\gamma(0)<0$ and $\gamma(1)>0$, then the *[Intermediate Value Theorem](/theorems/629)* would give some $t_0\in[0,1]$ with \begin{align*} \gamma(t_0)=0. \end{align*} This is impossible because $0\notin\mathbb R\setminus\{0\}$. Hence the path components of $\mathbb R\setminus\{0\}$ are exactly $(-\infty,0)$ and $(0,\infty)$. [/example] This dimension dependence is a first glimpse of algebraic topology. Path-connectedness detects whether points lie in one piece, but it does not detect the loop around the missing origin in $\mathbb R^2 \setminus \{0\}$. For that, one needs homotopy of paths and loops. ## Relation with Connectedness Path-connectedness implies connectedness in every topological space. The reason is conceptual: the image of each path is connected, and a path-connected space can be viewed as being swept out by connected images sharing a common point. This result explains why path-connectedness is the stronger condition. [quotetheorem:300] The converse fails, and the failure is exactly why this page exists. Connectedness cannot always be upgraded to path-connectedness without additional local assumptions. The topologist's sine curve is the standard example because it lives inside the plane while still displaying pathological local behaviour. [example: Connected Does Not Imply Path-Connected] Let \begin{align*} T=\{(x,\sin(1/x)):0<x\le 1\}\cup(\{0\}\times[-1,1])\subset\mathbb R^2. \end{align*} The earlier topologist's sine curve example established that $T$ is connected. It also established that no path in $T$ can join a point of the vertical segment $\{0\}\times[-1,1]$ to a point of the oscillating graph $\{(x,\sin(1/x)):0<x\le1\}$. These two subsets of $T$ are nonempty: for instance, \begin{align*} (0,0)\in \{0\}\times[-1,1], \end{align*} and \begin{align*} (1,\sin 1)\in \{(x,\sin(1/x)):0<x\le1\}. \end{align*} If $T$ were path-connected, then there would be a path in $T$ from $(0,0)$ to $(1,\sin 1)$ by the definition of path-connectedness. This contradicts the established fact that no path in $T$ joins the vertical segment to the oscillating graph. Thus $T$ is connected but not path-connected. Consequently connectedness does not imply path-connectedness, even for subspaces of $\mathbb R^2$. [/example] Still, connectedness remains useful. In locally path-connected settings, proving connectedness is enough to prove path-connectedness. For open subsets of $\mathbb R^n$, manifolds, graphs, and CW complexes, local path-connectedness is built into the objects, so connectedness and path-connectedness usually coincide. [remark: Practical Test] In a locally path-connected space, connectedness and path-connectedness are equivalent. This is the working form used in many analysis and topology arguments. [/remark] ## Beyond and Connected Topics Path-connectedness is the entry point to homotopy theory. Once paths exist, one can ask when two paths with the same endpoints can be continuously deformed into each other. That question leads to the fundamental group, covering spaces, and the algebraic invariants developed in [Cambridge II Algebraic Topology](/page/Cambridge%20II%20Algebraic%20Topology) and [Cambridge III Algebraic Topology](/page/Cambridge%20III%20Algebraic%20Topology). In analysis, path-connectedness is often used through connected open subsets of Euclidean space. It justifies propagating local identities across a domain, for example when a derivative vanishes on a connected interval or when [analytic continuation](/page/Analytic%20Continuation) follows chains of overlapping neighbourhoods. The background on connectedness, compactness, and continuity fits naturally with [Cambridge IA Analysis Notes](/page/Cambridge%20IA%20Analysis%20Notes) and [Cambridge IB Analysis and Topology](/page/Cambridge%20IB%20Analysis%20and%20Topology). Path components also prepare the reader for quotient constructions and classification problems. Instead of asking whether a space is path-connected, one studies the set of all path components and how maps act on them. In algebraic topology, this set of path components is often denoted $\pi_0(X)$; the notation fits into the broader pattern of homotopy groups, where higher indices record higher-dimensional deformation data. Local path-connectedness is a useful warning that global theorems often require local hypotheses. Many familiar spaces satisfy it automatically, but pathological subspaces of Euclidean space may not. Studying those examples clarifies why topology distinguishes connected components from path components. ## 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 II Algebraic Topology](/page/Cambridge%20II%20Algebraic%20Topology). Androma, [Cambridge III Algebraic Topology](/page/Cambridge%20III%20Algebraic%20Topology). James R. Munkres, *Topology* (2000). Allen Hatcher, *Algebraic Topology* (2002). John L. Kelley, *General Topology* (1955).