Connected Components

From Connected Sets to Components

A space may fail to be connected while still having large connected parts. For example, two intervals separated by a gap form a disconnected space, even though each interval is connected. The natural question is whether every point belongs to a largest connected piece, and whether those pieces fit together in a useful way. Connected components answer both questions.

Throughout, closures are taken in the ambient metric space unless a subspace is explicitly specified. We will use the Connectedness and Separated Pairs Theorem and the theorem that a connected subset of a union of two separated sets lies on one side, both established in the previous tutorial. In particular, if a space is split into two separated sets, a connected subset of that space cannot meet both sets.

Definition: A connected component of a metric space $X$ is a connected subset of $X$ that is maximal with respect to inclusion among connected subsets of $X$. For $x\in X$, the connected component containing $x$ is called the component of $x$.

“Maximal” means that the connected subset cannot be enlarged to a strictly larger connected subset of $X$. It does not mean that it contains every other connected subset of $X$. Different components may be distinct, but we will prove that they do not overlap.

There is a direct way to construct the component of $x$: take the union of every connected subset containing $x$. To see why this union works, we first need the following result.

Unions of Connected Sets with a Common Point

Theorem (Union with a Common Point): Let $\{C_\lambda:\lambda\in\Lambda\}$ be a nonempty family of connected subsets of a metric space $X$. If there is a point $p$ such that $p\in C_\lambda$ for every $\lambda\in\Lambda$, then $\bigcup_{\lambda\in\Lambda}C_\lambda$ is connected.

Proof. Let $U=\bigcup_{\lambda\in\Lambda}C_\lambda$. Suppose, to obtain a contradiction, that $U$ is disconnected. By the Connectedness and Separated Pairs Theorem, there are nonempty separated sets $A,B\subseteq U$ such that $U=A\cup B$. The point $p$ belongs to $U$, so it belongs to exactly one of $A$ and $B$; suppose $p\in A$.

Each $C_\lambda$ is a connected subset of $A\cup B$. By the theorem that a connected subset lies on one side of two separated sets, $C_\lambda$ is contained in $A$ or contained in $B$. Because $p\in C_\lambda\cap A$, it cannot be contained in $B$. Thus $C_\lambda\subseteq A$ for every $\lambda$, and consequently $U\subseteq A$. This contradicts $B\ne\varnothing$ and $B\subseteq U$. Therefore $U$ is connected. $\square$

The common point is essential to this argument: it ensures that every connected set in the family must lie on the same side of any proposed separation. The result does not require the family to be finite, and it does not require the sets to be open or closed.

Worked Example: A Connected Planar Cross

In $\mathbb{R}^2$ with its usual metric, consider $H=[-1,1]\times\{0\}$ and $V=\{0\}\times[-2,2]$. The map $t\mapsto(t,0)$ sends the interval $[-1,1]$ continuously onto $H$, so $H$ is connected by the theorem that continuous images of connected spaces are connected. Similarly, $s\mapsto(0,s)$ sends $[-2,2]$ continuously onto $V$, so $V$ is connected.

The two sets have the common point $(0,0)$, since $(0,0)\in H$ and $(0,0)\in V$. The Union with a Common Point Theorem therefore shows that $H\cup V$ is connected. Since the space under consideration is $H\cup V$ itself, it is its only component: it is connected and cannot be enlarged within that space.

Constructing the Component of a Point

Theorem (Construction of Components): Let $X$ be a metric space and $x\in X$. The set $$ C(x)=\bigcup\{C\subseteq X:C\text{ is connected and }x\in C\} $$ is the connected component of $x$. In particular, it is connected and contains every connected subset of $X$ that contains $x$.

Proof. The family in this union is nonempty because the singleton $\{x\}$ is connected. Every set in the family contains $x$, so the Union with a Common Point Theorem shows that $C(x)$ is connected. By its definition, $x\in C(x)$, and every connected subset of $X$ that contains $x$ is included in $C(x)$.

It remains to show maximality. Suppose $D$ is connected and $C(x)\subseteq D$. Then $x\in D$, so $D$ is one of the connected subsets in the family defining $C(x)$. Hence $D\subseteq C(x)$. Together with $C(x)\subseteq D$, this gives $D=C(x)$. Thus $C(x)$ cannot be enlarged to a strictly larger connected subset, and is a component of $x$. $\square$

This construction also shows that every point of a nonempty metric space belongs to a component. Unlike a choice of one connected subset containing $x$, the union collects all such subsets at once. Their common point guarantees that this larger set remains connected.

Worked Example: Components of the Rationals

Consider $\mathbb{Q}$ with the usual metric. We show that every component is a singleton. Let $C\subseteq\mathbb{Q}$ be connected, and suppose it contains distinct points $a<b$. Choose a positive integer $n$ large enough that $\sqrt{2}/n<(b-a)/2$, and put $\alpha=(a+b)/2+\sqrt{2}/n$. Then \[ \frac{a+b}{2}<\alpha<b, \] so $a<\alpha<b$. Also $\alpha$ is irrational: $(a+b)/2$ is rational, and $\sqrt{2}/n$ is irrational, so their sum is irrational. In particular, $\alpha\notin\mathbb{Q}$.

The sets $C_-=C\cap(-\infty,\alpha)$ and $C_+=C\cap(\alpha,\infty)$ are disjoint and cover $C$, because no element of $C\subseteq\mathbb{Q}$ equals $\alpha$. They are both nonempty: $a\in C_-$ and $b\in C_+$. Each is open in $C$, as it is the intersection of $C$ with an open subset of $\mathbb{R}$. They therefore give a separation of $C$, contradicting connectedness. Thus no connected subset of $\mathbb{Q}$ contains two distinct points.

Every singleton is connected, so the component of each rational $q$ is exactly $\{q\}$. The example illustrates that the components of a space can be very small even when the space contains many points arbitrarily close to one another.

Components Form a Partition

Theorem (Components Partition the Space): Every point of a metric space belongs to exactly one connected component. Any two components are either equal or disjoint.

Proof. Let $x,y\in X$, and suppose their components $C(x)$ and $C(y)$ intersect. Choose $z\in C(x)\cap C(y)$. Both components are connected and contain $z$. By the Union with a Common Point Theorem, $C(x)\cup C(y)$ is connected. It contains $x$, so it is included in $C(x)$ by the construction of $C(x)$. Similarly, it contains $y$, so it is included in $C(y)$. Hence $C(x)=C(y)$.

Every point $x\in X$ belongs to $C(x)$, so the components cover $X$. The argument just given shows that two components with a common point must be equal. Therefore distinct components are disjoint, and each point belongs to exactly one component. $\square$

The partition property gives a convenient test for identifying components: find connected subsets that cover the space, then determine whether any connected subset could join two of them. Separation can rule out such a joining.

Worked Example: Two Components in a Subspace of the Real Line

Let $X=[-3,-1]\cup[2,5]$, with the metric inherited from $\mathbb{R}$. Write $A=[-3,-1]$ and $B=[2,5]$. Each is connected because it is an interval in $\mathbb{R}$. In the subspace $X$, they are disjoint and closed: each is the complement of the other, which is closed in $X$. Thus $A$ and $B$ are separated in $X$.

A connected subset $C\subseteq X=A\cup B$ cannot meet both $A$ and $B$, by the theorem that a connected subset of a union of two separated sets lies on one side. Therefore no connected subset of $X$ can strictly enlarge $A$ by adding points of $B$, or strictly enlarge $B$ by adding points of $A$. Since each of $A$ and $B$ is already connected, they are maximal connected subsets of $X$. They are exactly the two components.

The ambient space matters here. The intervals $A$ and $B$ are not components of $\mathbb{R}$, where they can be enlarged to larger connected intervals. They are components of the particular subspace $X$.

Components Are Closed

A component is defined by connectedness and maximality, not by closedness. Nevertheless, every component is closed in its ambient space. The key step is that taking the closure of a connected set does not destroy connectedness.

Theorem (Closure of a Connected Set): If $C$ is a connected subset of a metric space $X$, then $\overline C$ is connected.

Proof. Suppose instead that $\overline C$ is disconnected. By the Connectedness and Separated Pairs Theorem, there are nonempty separated subsets $A,B\subseteq\overline C$ such that $\overline C=A\cup B$. The characterization of separated sets from the previous tutorial implies that $A$ and $B$ are open in the subspace $\overline C$.

Both $A$ and $B$ meet $C$. To check this for $A$, choose $a\in A$. Since $A$ is open in $\overline C$, there is an open set $O\subseteq X$ with $a\in O$ and $O\cap\overline C\subseteq A$. Because $a\in\overline C$, every open neighborhood of $a$, including $O$, meets $C$. Thus $O\cap C\ne\varnothing$, and $O\cap\overline C\subseteq A$ gives $A\cap C\ne\varnothing$. The same reasoning gives $B\cap C\ne\varnothing$.

The sets $A\cap C$ and $B\cap C$ are disjoint, nonempty, and cover $C$. They are open in $C$, since $A$ and $B$ are open in $\overline C$ and $C\subseteq\overline C$. They would therefore disconnect $C$, contrary to its connectedness. This contradiction proves that $\overline C$ is connected. $\square$

Corollary (Components Are Closed): Every connected component of a metric space $X$ is closed in $X$.

Proof. Let $C$ be a component of $X$. The Closure of a Connected Set Theorem shows that $\overline C$ is connected, and $\overline C$ contains $C$. Since $C$ is maximal among connected subsets of $X$, it cannot be properly contained in the connected set $\overline C$. Hence $\overline C=C$, so $C$ is closed. $\square$

In particular, distinct components are separated as subsets of $X$: they are disjoint by the partition theorem, and each is closed, so each has empty intersection with the closure of the other. This does not assert that there is a positive distance between distinct components.

Why Components Matter

Components turn the question “Is this space connected?” into a more detailed description of its connected pieces. A space is connected exactly when it has one component: if it is connected, it is itself maximal connected; if it has one component, the partition theorem says that this component covers the space. When there is more than one component, the space is disconnected.

A common pitfall is to confuse a connected subset with a component. A connected subset need not be maximal. For instance, a one-point subset of an interval is connected, but it is not a component of that interval because it can be enlarged while remaining connected. To establish that a proposed connected set is a component, one must also show that no larger connected subset of the ambient space contains it.

Another useful distinction is between components and a positive-distance decomposition. Components are closed and pairwise disjoint, but their distances need not have a common positive lower bound. The definition concerns connected subsets and maximality, not a uniform metric gap. The closure and separation results provide tools for ruling out connected sets that cross between proposed pieces without imposing such a gap.

Check Your Understanding

Use the construction and theorems above to answer the following questions.

Why is the family of connected subsets used to define $C(x)$ nonempty?
Where does the common point enter the proof that a union of connected sets is connected?
If two components intersect, why must their union be connected, and why does maximality force them to be equal?
In the rationals example, why does an irrational number between two distinct rational points give a separation of any subset containing both?
Why does maximality imply that the closure of a component equals the component itself?

What You'll Learn