Components describe the largest connected pieces of a metric space. The opposite extreme occurs when no connected piece contains more than one point. Totally separated metric spaces, commonly called totally disconnected metric spaces in standard terminology, make this idea precise by requiring every pair of distinct points to be separated by a disconnection. The rational numbers provide the main example. The focus keyword totally separated metric spaces is used here to emphasize how strong this form of disconnectedness is. :::definition[Totally Disconnected Metric Space] Let $(X,d)$ be a metric space. Then $X$ is said to be $totally disconnected$ if for every pair of distinct points $x,y\in X$, there exist two nonempty subsets $A$ and $B$ of $X$ such that $X=A\cup B$, $\overline{A}\cap B=\varnothing$, $A\cap\overline{B}=\varnothing$, $x\in A$, and $y\in B$. ::: :::solved-problem Show that $\mathbb{Q}$ with the usual metric is totally disconnected. ::: :::solution Given that $\mathbb{Q}$ has the usual metric. To prove that $\mathbb{Q}$ is totally disconnected. Let $x,y\in\mathbb{Q}$ with $x\neq y$. Without loss of generality, assume $x<y$. Choose an irrational number $\alpha$ such that $x<\alpha<y$. Define $A=\{q\in\mathbb{Q}:q<\alpha\}$ and $B=\{q\in\mathbb{Q}:q>\alpha\}$. Since $\alpha\notin\mathbb{Q}$, we have $\mathbb{Q}=A\cup B$. Also $A\cap B=\varnothing$, and the sets $A$ and $B$ are separated in $\mathbb{Q}$. Moreover, $x\in A$ and $y\in B$. Therefore any two distinct points of $\mathbb{Q}$ can be separated by a disconnection. Hence $\mathbb{Q}$ is totally disconnected. $\square$ ::: :::theorem[Components of a Totally Disconnected Space] Let $X$ be a totally disconnected metric space. Then every connected component of $X$ is a singleton subset. ::: :::proof Given that $X$ is totally disconnected. To prove that every connected component of $X$ is a singleton subset. Let $C$ be a connected component of $X$. Suppose, if possible, that $C$ contains two distinct points $x$ and $y$. Since $X$ is totally disconnected, there exist nonempty separated subsets $A$ and $B$ of $X$ such that $X=A\cup B$, $x\in A$, and $y\in B$. Then $C=(C\cap A)\cup(C\cap B)$. Both $C\cap A$ and $C\cap B$ are nonempty because $x\in C\cap A$ and $y\in C\cap B$. Also, they are separated in $C$. Therefore $C$ is disconnected. This contradicts the connectedness of $C$. Hence $C$ cannot contain two distinct points. Therefore $C$ is a singleton subset. $\square$ ::: :::corollary The connected components of $\mathbb{Q}$ with the usual metric are singleton subsets. ::: :::proof Given that $\mathbb{Q}$ is totally disconnected. To prove that every connected component of $\mathbb{Q}$ is a singleton subset. By the theorem on components of a totally disconnected space, every connected component of a totally disconnected space is a singleton subset. Hence every connected component of $\mathbb{Q}$ is a singleton subset. $\square$ ::: :::example[Integers with the Usual Metric] The set $\mathbb{Z}$ with the usual metric is totally disconnected. If $m<n$ are integers, then the sets $\{k\in\mathbb{Z}:k\leq m\}$ and $\{k\in\mathbb{Z}:k\geq m+1\}$ separate $m$ from $n$. Hence every component of $\mathbb{Z}$ is a singleton. $\square$ ::: :::solved-problem Show that a totally disconnected metric space cannot contain a connected subset with two distinct points. ::: :::solution Given that $X$ is totally disconnected. To prove that no connected subset of $X$ contains two distinct points. Suppose, if possible, that $C\subseteq X$ is connected and contains distinct points $x$ and $y$. Since $X$ is totally disconnected, there exists a separation $X=A\cup B$ such that $x\in A$ and $y\in B$. Then $C=(C\cap A)\cup(C\cap B)$ is a separation of $C$, because both pieces are nonempty and separated in $C$. This contradicts the connectedness of $C$. Hence no connected subset of $X$ contains two distinct points. $\square$ ::: :::exercise [1] Define a totally disconnected metric space. [2] Prove that $\mathbb{Q}$ is totally disconnected. [3] Show that the components of a totally disconnected metric space are singletons. [4] Explain why totally disconnected implies disconnected when the space has at least two points. ::: :::answer [1] Any two distinct points can be separated by a separation of the space. [2] Use an irrational cut between two rational points. [3] A component with two distinct points would be disconnected by the defining separation. [4] Choose two points and separate them; this gives a nontrivial disconnection. ::: :::faq[Frequently Asked Questions] Q: Does totally disconnected mean empty of connected subsets? A: No. Singleton subsets are connected. Q: Is $\mathbb{Q}$ connected? A: No. It is totally disconnected with singleton components. Q: Is every disconnected space totally disconnected? A: No. A space can be disconnected while still having connected components with many points. ::: :::call-to-action[Continue to Local Connectedness] subtitle: Compare totally separated metric spaces with spaces that have connected neighbourhoods near each point. button: Next Lesson | /functional-analysis/metric-spaces/connectedness-and-path-structure-in-metric-spaces/local-forms-of-connectedness button-ghost: Previous Lesson | /functional-analysis/metric-spaces/connectedness-and-path-structure-in-metric-spaces/components-of-metric-spaces :::