Unions of connected subspaces lead naturally to maximal connected pieces. Components of metric spaces give a precise way to decompose a space into connected parts, even when the whole space is disconnected. The connected component of a point is built by taking the union of all connected subsets containing that point. Because all these sets share that point, the union is connected. The focus keyword components of metric spaces is important for understanding how disconnected spaces still have internal connected structure. :::definition[Connected Component] Let $(X,d)$ be a metric space and let $x\in X$. The $connected component$ of $x$ in $X$, denoted by $C(x)$, is the union of all connected subsets of $X$ that contain $x$: $$\begin{equation*} C(x)=\bigcup\{A\subseteq X:A\ \text{is connected and}\ x\in A\}. \end{equation*}$$ ::: :::remark The set $C(x)$ is connected because it is the union of connected subsets having the common point $x$. It is maximal among connected subsets of $X$ that contain $x$. Thus, if $A$ is connected and $x\in A$, then $A\subseteq C(x)$. ::: :::example[Components of the Rational Numbers] Let $\mathbb{Q}$ have the usual metric. For each $x\in\mathbb{Q}$, the connected component of $x$ is $\{x\}$. Given two distinct rational numbers $x<y$, choose an irrational number $\alpha$ such that $x<\alpha<y$. Any subset of $\mathbb{Q}$ containing both $x$ and $y$ is separated by the cut at $\alpha$. Therefore no connected subset of $\mathbb{Q}$ can contain two distinct points. Hence the connected components of $\mathbb{Q}$ are singleton sets. $\square$ ::: :::example[A Connected Comb-Type Subspace] Let $Y\subseteq\mathbb{R}^2$ consist of the line segments joining $(0,0)$ to the points $(1,1/n)$ for $n\in\mathbb{N}$, together with the segment $\{(x,0):1/2<x\leq 1\}$. Let $Z$ be the union of the line segments joining $(0,0)$ to $(1,1/n)$. Each line segment is connected, and all the line segments contain $(0,0)$. Therefore $Z$ is connected. Also $Z\subseteq Y\subseteq\overline{Z}$. Hence $Y$ is connected. However, $Y\setminus\{(0,0)\}$ is disconnected. $\square$ ::: :::theorem[Properties of Connected Components] Let $(X,d)$ be a metric space. Then each connected subset of $X$ is contained in exactly one connected component; each nonempty connected subset of $X$ that is both open and closed in $X$ is a connected component of $X$; and each connected component of $X$ is closed. ::: :::proof Given that $(X,d)$ is a metric space. To prove the stated properties. Let $C(x)$ and $C(y)$ be two connected components of $X$. Suppose $C(x)\cap C(y)\neq\varnothing$. Then $C(x)\cup C(y)$ is connected because it is a union of connected sets with a common point. By maximality of connected components, $C(x)=C(y)$. Thus two connected components are either disjoint or identical. Therefore each connected subset of $X$ is contained in exactly one component. Now let $A$ be a nonempty connected subset of $X$ that is both open and closed in $X$. Choose $x\in A$. Since $A$ is connected and contains $x$, we have $A\subseteq C(x)$. Since $A$ is both open and closed in $X$, it is both open and closed in the subspace $C(x)$. Since $C(x)$ is connected and $A$ is a nonempty clopen subset of $C(x)$, we get $A=C(x)$. Hence $A$ is a connected component. Finally, let $C(x)$ be a connected component. Since the closure of a connected set is connected, $\overline{C(x)}$ is connected. By maximality of $C(x)$, $\overline{C(x)}\subseteq C(x)$. Since $C(x)\subseteq\overline{C(x)}$, we get $\overline{C(x)}=C(x)$. Therefore $C(x)$ is closed. $\square$ ::: :::example[A Component Need Not Be Open] Let $X=\{0\}\cup\{1/n:n\in\mathbb{N}\}$ with the metric induced from $\mathbb{R}$. The connected component of $0$ is $\{0\}$. However, $\{0\}$ is not open in $X$ because every neighbourhood of $0$ in $X$ contains points of the form $1/n$. Therefore connected components need not be open. $\square$ ::: :::solved-problem Show that the connected components of any metric space $X$ form a partition of $X$. ::: :::solution Given that $X$ is a metric space. To prove that the connected components of $X$ form a partition of $X$. For each $x\in X$, the singleton $\{x\}$ is connected. Therefore $x\in C(x)$. Hence $X=\bigcup_{x\in X}C(x)$. Now let $C(x)$ and $C(y)$ be two connected components. If $C(x)\cap C(y)\neq\varnothing$, then $C(x)\cup C(y)$ is connected. By maximality of connected components, $C(x)=C(y)$. Thus any two connected components are either identical or disjoint. Therefore the connected components form a partition of $X$. $\square$ ::: :::exercise [1] Define the connected component of a point. [2] Prove that every component is connected. [3] Show that components are closed. [4] Give an example of a component that is not open. ::: :::answer [1] It is the union of all connected subsets containing the point. [2] It is a union of connected sets with a common point. [3] The closure of a component is connected and hence lies in the same maximal component. [4] In $\{0\}\cup\{1/n:n\in\mathbb{N}\}$, the component $\{0\}$ is not open. ::: :::faq[Frequently Asked Questions] Q: Is a component always closed? A: Yes. In a metric space, every connected component is closed. Q: Is a component always open? A: No. Components need not be open unless additional local connectedness conditions hold. Q: Why do components form a partition? A: Every point belongs to its own component, and two components are either disjoint or identical. ::: :::call-to-action[Continue to Totally Disconnected Spaces] subtitle: After components of metric spaces, examine spaces whose components are only singletons. button: Next Lesson | /functional-analysis/metric-spaces/connectedness-and-path-structure-in-metric-spaces/totally-separated-metric-spaces button-ghost: Previous Lesson | /functional-analysis/metric-spaces/connectedness-and-path-structure-in-metric-spaces/unions-of-connected-subspaces :::