Connectedness says that a space cannot be separated into two pieces, but it does not always provide an explicit curve joining two points. Paths and path-connected sets introduce a stronger and more constructive idea. A path is a continuous map from the interval $[0,1]$ into the space, and path-connectedness means that any two points can be joined by such a path. The focus keyword paths and path-connected sets is important because many spaces in analysis and geometry are proved connected by first proving path-connectedness. :::definition[Path] Let $(X,d)$ be a metric space and let $Y\subseteq X$. Let $I=[0,1]$. A $path$ or $arc$ in $Y$ is a continuous mapping $f:I\to Y$. The point $f(0)$ is called the initial point of the path, and $f(1)$ is called the final point of the path. If $f(0)=a$ and $f(1)=b$, then $f$ is said to join $a$ to $b$. ::: :::definition[Path-Connected Subset] Let $(X,d)$ be a metric space and let $Y\subseteq X$. Then $Y$ is said to be $path-connected$, or $arcwise connected$, if for every two points $a,b\in Y$, there exists a path $f:[0,1]\to Y$ such that $f(0)=a$ and $f(1)=b$. ::: :::remark If $f:[0,1]\to Y$ is a path from $a$ to $b$, then $g(t)=f(1-t)$ is a path from $b$ to $a$. This reverse path is useful because it lets us change the direction of a path without changing its image. ::: :::example[Connected but Not Path-Connected] Let $A=\{(0,y):-1\leq y\leq 1\}$ and $B=\{(x,y):y=\sin(1/x),\ 0<x\leq 1\}$. Let $Y=A\cup B$. The set $Y$ is connected, but it is not path-connected. There is no path in $Y$ joining a point of the vertical segment $A$ to a point of the oscillating curve $B$. Hence connectedness does not imply path-connectedness in general. ::: :::example[The Open Unit Disc Is Path-Connected] Let $S(0,1)=\{z\in\mathbb{C}:|z|<1\}$. Let $z_1,z_2\in S(0,1)$. Define $f(t)=tz_2+(1-t)z_1$ for $0\leq t\leq 1$. Then $f(0)=z_1$ and $f(1)=z_2$. Also, $|f(t)|\leq t|z_2|+(1-t)|z_1|<t+(1-t)=1$. Therefore $f(t)\in S(0,1)$ for every $t\in[0,1]$. Hence the open unit disc is path-connected. $\square$ ::: :::lemma Let $(X,d)$ be a metric space and let $x_0\in X$. Then $X$ is path-connected if and only if each point of $X$ can be joined to $x_0$ by a path. ::: :::proof Given that $(X,d)$ is a metric space and $x_0\in X$. To prove the equivalence. First assume that $X$ is path-connected. Then, for each $x\in X$, there exists a path joining $x$ to $x_0$. Conversely, assume that each point of $X$ can be joined to $x_0$ by a path. Let $x_1,x_2\in X$. There exists a path $f$ from $x_1$ to $x_0$, and there exists a path $g$ from $x_0$ to $x_2$. Define $$\begin{equation*} h(t)= \begin{cases} f(2t), & 0\leq t\leq \frac12,\\ g(2t-1), & \frac12\leq t\leq 1. \end{cases} \end{equation*}$$ At $t=1/2$, the two parts agree because $f(1)=x_0=g(0)$. Therefore $h$ is continuous. Also $h(0)=x_1$ and $h(1)=x_2$. Hence $h$ is a path from $x_1$ to $x_2$. Since $x_1$ and $x_2$ are arbitrary, $X$ is path-connected. $\square$ ::: :::solved-problem Show that every convex subset of a normed linear space is path-connected. ::: :::solution Given that $C$ is a convex subset of a normed linear space. To prove that $C$ is path-connected. Let $x,y\in C$. Define $f:[0,1]\to C$ by $f(t)=(1-t)x+ty$. Since $C$ is convex, $f(t)\in C$ for every $t\in[0,1]$. The map $f$ is continuous with respect to the norm metric. Also $f(0)=x$ and $f(1)=y$. Therefore any two points of $C$ can be joined by a path in $C$. Hence $C$ is path-connected. $\square$ ::: :::exercise [1] Define a path in a subset of a metric space. [2] Define path-connectedness. [3] Write the reverse path associated with $f(t)$. [4] Prove that the open unit disc is path-connected. ::: :::answer [1] A path is a continuous map $f:[0,1]\to Y$. [2] A set is path-connected when every two points can be joined by a path. [3] The reverse path is $g(t)=f(1-t)$. [4] Use the straight-line path $f(t)=tz_2+(1-t)z_1$. ::: :::faq[Frequently Asked Questions] Q: Is every connected space path-connected? A: No. The oscillating sine curve closure is connected but not path-connected. Q: Why is the interval $[0,1]$ used? A: It provides a standard connected parameter interval for paths. Q: Is path-connectedness stronger than connectedness? A: Yes. Every path-connected metric space is connected, but not conversely. ::: :::call-to-action[Continue to Path-Connectedness and Connectedness] subtitle: Use paths and path-connected sets to prove the general theorem that path-connected spaces are connected. button: Next Lesson | /functional-analysis/metric-spaces/connectedness-and-path-structure-in-metric-spaces/path-connectedness-and-ordinary-connectedness button-ghost: Previous Lesson | /functional-analysis/metric-spaces/connectedness-and-path-structure-in-metric-spaces/local-forms-of-connectedness :::