The last result in this section explains why connected open subsets of the complex plane behave better than arbitrary connected subsets. Open connected sets in the plane are path-connected. This theorem is especially important in complex analysis, where regions are usually defined as open connected sets, but proofs often use paths. The focus keyword open connected sets in the plane appears here because openness is the condition that turns ordinary connectedness into path-connectedness. Without openness, the oscillating sine curve example shows that the conclusion can fail. :::theorem[Open Connected Subsets of the Complex Plane Are Path-Connected] Let $V$ be a nonempty open connected subset of the complex plane $\mathbb{C}$. Then $V$ is path-connected. ::: :::proof Given that $V$ is a nonempty open connected subset of $\mathbb{C}$. To prove that $V$ is path-connected. Choose a point $z_0\in V$. Let $G=\{z\in V:z\ \text{can be joined to}\ z_0\ \text{by a path in}\ V\}$. It is enough to prove that $G=V$. First, prove that $G$ is open in $V$. Let $z\in G$. Since $V$ is open, there exists $r>0$ such that $S(z,r)\subseteq V$. The open disc $S(z,r)$ is path-connected. Therefore every point of $S(z,r)$ can be joined to $z$ by a path inside $S(z,r)$. Since $z\in G$, the point $z$ can be joined to $z_0$ by a path in $V$. By concatenating paths, every point of $S(z,r)$ can be joined to $z_0$ by a path in $V$. Hence $S(z,r)\subseteq G$. Thus $G$ is open in $V$. Now let $H=V\setminus G$. We prove that $H$ is open in $V$. Let $w\in H$. Since $V$ is open, there exists $s>0$ such that $S(w,s)\subseteq V$. Suppose that some point $u\in S(w,s)$ belongs to $G$. Then $u$ can be joined to $z_0$ by a path in $V$. Also $w$ can be joined to $u$ by a path inside $S(w,s)$. Therefore $w$ can be joined to $z_0$ by a path in $V$. This contradicts $w\in H$. Hence $S(w,s)\subseteq H$. Therefore $H$ is open in $V$. Now $V=G\cup H$ and $G\cap H=\varnothing$. Also $G$ is nonempty because $z_0\in G$. Since $V$ is connected, $H$ must be empty. Therefore $G=V$. Hence every point of $V$ can be joined to $z_0$ by a path in $V$. Thus $V$ is path-connected. $\square$ ::: :::remark The openness assumption cannot be omitted. The set $\{(0,y):-1\leq y\leq 1\}\cup\{(x,y):y=\sin(1/x),\ 0<x\leq 1\}$ is connected but not path-connected. ::: :::solved-problem Prove that $\mathbb{R}^2\setminus\{(0,0)\}$ is path-connected and hence connected. ::: :::solution Given that $X=\mathbb{R}^2\setminus\{(0,0)\}$. To prove that $X$ is path-connected. Let $a,b\in X$. Choose a point $p\in X$ such that the line segments from $a$ to $p$ and from $p$ to $b$ do not pass through the origin. Define $f(t)=(1-t)a+tp$ and $g(t)=(1-t)p+tb$ for $0\leq t\leq 1$. Both paths lie in $X$ by the choice of $p$. Define $$\begin{equation*} h(t)= \begin{cases} f(2t), & 0\leq t\leq \frac12,\\ g(2t-1), & \frac12\leq t\leq 1. \end{cases} \end{equation*}$$ Then $h$ is a path in $X$ from $a$ to $b$. Therefore $X$ is path-connected. Since every path-connected metric space is connected, $X$ is connected. $\square$ ::: :::solved-problem Prove that $S^1=\{z\in\mathbb{C}:|z|=1\}$ is connected. ::: :::solution Given that $S^1=\{z\in\mathbb{C}:|z|=1\}$. To prove that $S^1$ is connected. Define $F:\mathbb{R}^2\setminus\{(0,0)\}\to S^1$ by $F(z)=z/|z|$. The mapping $F$ is continuous and onto. Since $\mathbb{R}^2\setminus\{(0,0)\}$ is connected, its continuous image is connected. Therefore $F(\mathbb{R}^2\setminus\{(0,0)\})=S^1$ is connected. $\square$ ::: :::solved-problem Let $$Y=\left\{\begin{pmatrix}\cos u & \sin u\\-\sin u & \cos u\end{pmatrix}:u\in\mathbb{R}\right\}$$ with the metric induced from $\mathbb{R}^4$. Show that $Y$ is connected. ::: :::solution Given that $$Y=\left\{\begin{pmatrix}\cos u & \sin u\\-\sin u & \cos u\end{pmatrix}:u\in\mathbb{R}\right\}$$. To prove that $Y$ is connected. Define $F:\mathbb{R}\to Y$ by $$\begin{equation*} F(u)=\begin{pmatrix}\cos u & \sin u\\-\sin u & \cos u\end{pmatrix}. \end{equation*}$$ Since $\sin u$ and $\cos u$ are continuous functions, $F$ is continuous. By the definition of $Y$, the mapping $F$ is onto. Since $\mathbb{R}$ is connected, the continuous image $F(\mathbb{R})$ is connected. Therefore $Y$ is connected. $\square$ ::: :::exercise [1] Prove that every nonempty open connected subset of $\mathbb{C}$ is path-connected. [2] Explain why openness is needed in the theorem. [3] Prove that $\mathbb{R}^2\setminus\{(0,0)\}$ is connected. [4] Prove that the unit circle is connected. ::: :::answer [1] Fix $z_0$, let $G$ be the set of points path-joinable to $z_0$, and prove both $G$ and its complement in $V$ are open. [2] The oscillating sine curve closure is connected but not path-connected. [3] It is path-connected by joining two points through a third point avoiding the origin. [4] It is the continuous image of $\mathbb{R}^2\setminus\{(0,0)\}$ under $z\mapsto z/|z|$. ::: :::faq[Frequently Asked Questions] Q: Why does an open disc help in the proof? A: A small open disc around a point is path-connected and stays inside the open set. Q: Does connected imply path-connected in every subset of the plane? A: No. Openness is essential. Q: Why is this theorem important in complex analysis? A: Regions are open connected sets, and the theorem allows paths to be used inside them. ::: :::call-to-action[Complete the Connectedness Section] subtitle: Review open connected sets in the plane together with components, local connectedness, and path-connectedness before moving onward. button: Review First Lesson | /functional-analysis/metric-spaces/connectedness-and-path-structure-in-metric-spaces/separation-criteria-for-connected-spaces button-ghost: Previous Lesson | /functional-analysis/metric-spaces/connectedness-and-path-structure-in-metric-spaces/path-connected-images-and-unions :::