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.