Connectedness begins with the idea that a metric space is not made of two separated pieces. After open sets, closed sets, closures, and subspaces have been studied, the next question is whether a space remains in one piece under the metric topology. Separation criteria for connected spaces give the practical tests: instead of relying on a picture, we use open sets, closed sets, and clopen subsets. Students often confuse disjointness with separation; disjoint sets may still have one set touching the closure of the other. The focus keyword separation criteria for connected spaces will be used here to build the first rigorous test for connectedness in metric spaces. :::definition[Disconnected Metric Space] Let $(X,d)$ be a metric space. Then $(X,d)$ is said to be $disconnected$ if there exist two nonempty subsets $A$ and $B$ of $X$ such that $$\begin{equation*} X=A\cup B, \end{equation*}$$ and $$\begin{equation*} \overline{A}\cap B=\varnothing,\qquad A\cap\overline{B}=\varnothing. \end{equation*}$$ Here the closures are taken in the metric space $X$. The subsets $A$ and $B$ are then called separated subsets of $X$. ::: :::definition[Connected Metric Space] Let $(X,d)$ be a metric space. Then $(X,d)$ is said to be $connected$ if it is not disconnected. Equivalently, $X$ is connected if it cannot be represented as a union $X=A\cup B$ of two nonempty separated subsets. ::: :::definition[Connected Subset] Let $(X,d)$ be a metric space and let $Y\subseteq X$. Then $Y$ is said to be $connected$ if $Y$, with the metric induced from $X$, is connected as a metric space. ::: :::example[Separated Real Subset] Let $$\begin{equation*} X=\{x\in\mathbb{R}:-1\leq x<0\ \text{or}\ 0<x\leq 1\}. \end{equation*}$$ Put $A=[-1,0)$ and $B=(0,1]$. Then $X=A\cup B$, $A\cap B=\varnothing$, and the missing point $0$ does not belong to $X$. Therefore neither part meets the closure of the other in the subspace $X$. Hence $X$ is disconnected. $\square$ ::: :::example[Separated Discs in the Plane] Let $$\begin{equation*} X=\{z\in\mathbb{C}:|z|\leq 1\}\cup\{z\in\mathbb{C}:|z-3|\leq 1\}. \end{equation*}$$ The two closed discs are disjoint and have positive distance from each other. Thus each disc is separated from the other in the subspace $X$. Hence $X$ is disconnected. $\square$ ::: :::example[Rational Numbers Are Disconnected] Let $\mathbb{Q}$ have the usual metric. Define $$\begin{equation*} A=\{q\in\mathbb{Q}:q<\sqrt{2}\},\qquad B=\{q\in\mathbb{Q}:q>\sqrt{2}\}. \end{equation*}$$ Since $\sqrt{2}\notin\mathbb{Q}$, we have $\mathbb{Q}=A\cup B$, and the irrational cut separates the two subsets inside $\mathbb{Q}$. Hence $\mathbb{Q}$ is disconnected. $\square$ ::: :::example[Singleton Sets] Let $(X,d)$ be a metric space and let $x\in X$. The singleton set $\{x\}$ cannot be written as a union of two nonempty subsets. Hence $\{x\}$ is connected. $\square$ ::: :::theorem[Equivalent Conditions for Disconnectedness] Let $(X,d)$ be a metric space. Then the following statements are equivalent: [1] $X$ is disconnected. [2] There exist two nonempty disjoint subsets $A$ and $B$, both open in $X$, such that $X=A\cup B$. [3] There exist two nonempty disjoint subsets $A$ and $B$, both closed in $X$, such that $X=A\cup B$. [4] There exists a nonempty proper subset of $X$ that is both open and closed in $X$. ::: :::proof Given that $(X,d)$ is a metric space. To prove that the four statements are equivalent. [1] Assume that $X$ is disconnected. Then there exist nonempty subsets $A$ and $B$ of $X$ such that $X=A\cup B$, $\overline{A}\cap B=\varnothing$, and $A\cap\overline{B}=\varnothing$. Since $A\cap\overline{B}=\varnothing$, we get $A=X\setminus\overline{B}$. Therefore $A$ is open. Similarly, $B=X\setminus\overline{A}$, and therefore $B$ is open. Hence [1] implies [2]. [2] Assume that $X=A\cup B$, where $A$ and $B$ are nonempty disjoint open subsets of $X$. Since $A=X\setminus B$ and $B=X\setminus A$, both $A$ and $B$ are closed. Hence [2] implies [3]. [3] Assume that $X=A\cup B$, where $A$ and $B$ are nonempty disjoint closed subsets of $X$. Since $A=X\setminus B$, the set $A$ is open. It is also closed, nonempty, and proper. Hence [3] implies [4]. [4] Assume that $A$ is a nonempty proper subset of $X$ that is both open and closed. Let $B=X\setminus A$. Then $B$ is nonempty, $X=A\cup B$, and $A\cap B=\varnothing$. Since $A$ and $B$ are both closed, $\overline{A}=A$ and $\overline{B}=B$. Therefore $\overline{A}\cap B=\varnothing$ and $A\cap\overline{B}=\varnothing$. Hence $X$ is disconnected. Thus [4] implies [1]. Hence all four statements are equivalent. $\square$ ::: :::solved-problem Show that $$\begin{equation*} A=\{(x,y)\in\mathbb{R}^2:x^2-y^2\geq 4\} \end{equation*}$$ is disconnected. ::: :::solution Given that $A=\{(x,y)\in\mathbb{R}^2:x^2-y^2\geq 4\}$. To prove that $A$ is disconnected. From $x^2-y^2\geq 4$ and $y^2\geq 0$, we get $x^2\geq 4$. Therefore $|x|\geq 2$. Hence every point of $A$ satisfies either $x\leq -2$ or $x\geq 2$. Define $$\begin{equation*} A_1=A\cap\{(x,y):x<0\},\qquad A_2=A\cap\{(x,y):x>0\}. \end{equation*}$$ Then $A_1$ and $A_2$ are nonempty, disjoint, open in the subspace $A$, and $A=A_1\cup A_2$. Therefore $A$ is disconnected. $\square$ ::: :::lemma Let $X=A\cup B$, where $A$ and $B$ are separated subsets of a metric space $X$. If $C$ is a connected subset of $X$, then either $C\subseteq A$ or $C\subseteq B$. ::: :::proof Given that $C$ is connected and $X=A\cup B$, where $A$ and $B$ are separated. To prove that $C$ is contained in one of the two sets. We have $$\begin{equation*} C=(C\cap A)\cup(C\cap B). \end{equation*}$$ If both $C\cap A$ and $C\cap B$ are nonempty, then these two subsets form a separation of $C$. This contradicts the connectedness of $C$. Therefore one of $C\cap A$ and $C\cap B$ is empty. Hence either $C\subseteq A$ or $C\subseteq B$. $\square$ ::: :::exercise [1] Define a disconnected metric space using closures. [2] Prove that a metric space with a nonempty proper clopen subset is disconnected. [3] Show that $\{x\}$ is connected in every metric space. [4] Decide whether $[-1,0)\cup(0,1]$ is connected as a subspace of $\mathbb{R}$. ::: :::answer [1] A metric space is disconnected when it is the union of two nonempty separated subsets. [2] If $A$ is nonempty, proper, open, and closed, then $X=A\cup(X\setminus A)$ is a separation. [3] A singleton cannot be split into two nonempty subsets. [4] It is disconnected because the two parts are separated in the subspace. ::: :::faq[Frequently Asked Questions] Q: Is every disjoint union a separation? A: No. Separation also requires that neither set meet the closure of the other. Q: Why are clopen sets important? A: A nonempty proper clopen set gives a direct open-and-closed decomposition of the space. Q: Are singletons always connected? A: Yes. A singleton has no two nonempty pieces into which it can be separated. ::: :::call-to-action[Continue to Connected Subsets of the Real Line] subtitle: Use the separation criteria for connected spaces to recognise why intervals are the connected subsets of the real line. button: Next Lesson | /functional-analysis/metric-spaces/connectedness-and-path-structure-in-metric-spaces/intervals-and-connected-subsets-of-the-real-line :::