Connected Subsets of the Real Line

From Connected Sets to Connected Components

The Characterization of Connected Subsets of $\mathbb{R}$, established earlier in this course, says that connected sets in the real line are exactly the intervals. We will use that result rather than prove it again. It gives a practical way to study an arbitrary set $E$: look for the largest intervals that lie entirely inside $E$. These intervals are the connected components of $E$.

A connected component is a maximal connected subset, where “maximal” means that it is not properly contained in a larger connected subset of $E$. Maximal does not mean that the component has the greatest size among all connected subsets. Instead, it means that no point can be added to that particular component while preserving connectedness and remaining inside $E$. Different components may have different sizes or shapes, but they will not overlap.

Definition: Let $E\subseteq\mathbb{R}$. A connected component of $E$ is a nonempty connected subset $C\subseteq E$ that is maximal with respect to inclusion: if $D$ is connected and $C\subseteq D\subseteq E$, then $D=C$. The connected component containing $x\in E$, when it exists, is denoted $C_x$.

Since connected subsets of $\mathbb{R}$ are intervals, a component cannot jump across a point missing from $E$. This observation also gives an exact description: the component containing $x$ consists of those points $y\in E$ for which the entire closed interval between $x$ and $y$ is contained in $E$. The closed interval between $x$ and $y$ is written $[\min\{x,y\},\max\{x,y\}]$.

The Component Containing a Point

Theorem (Explicit Description of Connected Components): Let $E\subseteq\mathbb{R}$ and $x\in E$. Define $$ C_x=\{y\in E:[\min\{x,y\},\max\{x,y\}]\subseteq E\}. $$ Then $C_x$ is the unique connected component of $E$ containing $x$. In particular, every point of $E$ belongs to a connected component.

Proof. First, $x\in C_x$, since $[x,x]=\{x\}\subseteq E$. By the definition of $C_x$, each $y\in C_x$ lies in $E$, so $C_x\subseteq E$.

We show that $C_x$ is an interval. Take $u,v\in C_x$, and let $z$ be any point between $u$ and $v$. The interval from $x$ to $z$ is contained in the union of the intervals from $x$ to $u$ and from $x$ to $v$: $$ [\min\{x,z\},\max\{x,z\}] \subseteq [\min\{x,u\},\max\{x,u\}] \cup [\min\{x,v\},\max\{x,v\}]. $$ Indeed, on the real line, if $z$ lies between $u$ and $v$, it cannot be farther from $x$ in either direction than both $u$ and $v$. Both intervals on the right are subsets of $E$, because $u,v\in C_x$. Thus the interval on the left is contained in $E$, and $z\in C_x$. This proves that $C_x$ is an interval. By the Characterization of Connected Subsets of $\mathbb{R}$, $C_x$ is connected.

Now let $D\subseteq E$ be any connected set containing $x$. If $y\in D$, then the Characterization of Connected Subsets of $\mathbb{R}$ implies that $D$ is an interval. Therefore the interval between $x$ and $y$ lies in $D$, and hence in $E$. It follows that $y\in C_x$. We have proved $D\subseteq C_x$. Since $C_x$ is itself connected and contains $x$, this inclusion shows that $C_x$ is maximal among connected subsets of $E$ containing $x$.

Finally, any connected component of $E$ containing $x$ must be contained in $C_x$, by the argument just given. Since $C_x$ is connected, maximality forces that component to equal $C_x$. This proves uniqueness as well as existence. $\square$

Worked Example: Two Unbounded Components

Let $E=(-\infty,-2]\cup[1,\infty)$. For $x=-5$, every $y\in(-\infty,-2]$ has the interval between $x$ and $y$ contained in $(-\infty,-2]$, so each such $y$ belongs to $C_x$. If $y\in[1,\infty)$, the interval between $x$ and $y$ contains $0$, which is not in $E$; therefore $y\notin C_x$. Thus the component containing $-5$ is $(-\infty,-2]$.

For $x=4$, every point $y\in[1,\infty)$ has the interval between $x$ and $y$ contained in $[1,\infty)$. Every $y\in(-\infty,-2]$ fails the test, since the interval between $x$ and $y$ contains $0\notin E$. So the other component is $[1,\infty)$. The missing points between the two rays prevent any connected subset from joining them.

Components Partition the Set

The explicit description immediately implies that components fit together without overlap. If $y\in C_x$, then the interval between $x$ and $y$ lies in $E$. The same condition, with the two points reversed, shows $x\in C_y$. More generally, any point of $C_x$ has the same component as $x$: the interval from $x$ to that point is already part of $C_x$, and the maximal connected set containing either point must be the same.

Theorem (Components Partition a Set): The connected components of $E\subseteq\mathbb{R}$ are pairwise disjoint, and their union is $E$. Each component is an interval, possibly unbounded or a singleton.

Proof. For every $x\in E$, the explicit description theorem gives a component $C_x$ containing $x$. Thus the components cover $E$. Each one is an interval by that theorem.

Suppose two components $C$ and $D$ have a common point $p$. Both are connected subsets of $E$ containing $p$. By uniqueness of the connected component containing $p$, both must equal $C_p$. Hence $C=D$. Therefore two distinct components are disjoint, and the components partition $E$. If $E$ is empty, it has no components and the empty family has union $E$, so the statement holds in that case as well. $\square$

Worked Example: A Singleton Component Between Intervals

Consider $E=[-4,-1)\cup\{1\}\cup(3,5]$. If $x=-2$, every point of $[-4,-1)$ can be joined to $x$ by an interval lying entirely within $[-4,-1)$. A point in either $\{1\}$ or $(3,5]$ cannot be joined to $x$ by an interval contained in $E$, since the interval would include points in the missing region. Hence the component containing $-2$ is $[-4,-1)$.

For $x=1$, the only $y\in E$ for which the interval between $x$ and $y$ stays in $E$ is $y=1$. For example, if $y\in(3,5]$, the interval from $1$ to $y$ contains points between $1$ and $3$, which are not in $E$. So $\{1\}$ is a component. Applying the same test to any $x\in(3,5]$ gives the component $(3,5]$. These three disjoint intervals cover $E$.

Components Are Closed Relative to the Set

A component need not be closed as a subset of the whole real line. For example, $(0,1)$ is not closed in $\mathbb{R}$. But it is closed relative to itself, and more generally every connected component is closed relative to the set whose components are being considered. Here “closed relative to $E$” means closed in the subspace $E$, or equivalently of the form $E\cap F$ for some closed set $F\subseteq\mathbb{R}$.

Theorem (Components Are Relatively Closed): Every connected component $C$ of $E\subseteq\mathbb{R}$ is closed relative to $E$.

Proof. Let $H$ be the closure of $C$ in $E$, that is, the set of points of $E$ whose every relative neighborhood in $E$ meets $C$. We first show that $H$ is connected. Suppose otherwise that $H=A\cup B$ is a separation, with $A$ and $B$ disjoint, nonempty, and relatively open in $H$. Since $C\subseteq H$ and $C$ is connected, $C$ cannot meet both pieces; the Restriction of a Separation theorem gives this conclusion. Thus $C$ lies wholly in one piece, say $A$.

Choose $b\in B$. Since $B$ is relatively open in $H$, there is an open set $G\subseteq\mathbb{R}$ such that $b\in G$ and $G\cap H\subseteq B$. Because $b\in H$, the relative neighborhood $G\cap E$ of $b$ must meet $C$. Choose $c\in G\cap E\cap C$. Since $C\subseteq H$, we have $c\in G\cap H\subseteq B$. But $C\subseteq A$, contradicting that $A$ and $B$ are disjoint. Therefore $H$ is connected.

The set $H$ contains $C$ and is contained in $E$. Since $C$ is a connected component, its maximality implies $H=C$. Thus $C$ equals its closure in $E$, which is exactly to say that $C$ is closed relative to $E$. $\square$

Worked Example: Relative Closedness Does Not Mean Ambient Closedness

Let $E=(0,1)\cup\{2\}$. Its components are $(0,1)$ and $\{2\}$: each is an interval contained in $E$, while any interval joining a point of $(0,1)$ to $2$ contains points of $(1,2)$, which are absent from $E$. The component $(0,1)$ is not closed in $\mathbb{R}$, since its closure in $\mathbb{R}$ is $[0,1]$. It is, however, closed relative to $E$, because its complement in $E$ is $\{2\}=E\cap(3/2,5/2)$, which is relatively open. This illustrates why the word “relative” matters.

How to Use the Component Test

To find a component containing a chosen point $x$, test how far you can move from $x$ while keeping every point between the old and new positions inside $E$. A gap blocks the move, even if the proposed endpoint itself belongs to $E$. If there is no such gap in a direction, the component may extend indefinitely in that direction. The resulting component is always an interval, but its endpoints may or may not belong to $E$.

The partition theorem is useful because it turns the potentially complicated collection of all connected subsets into a disjoint family of maximal pieces. It also prevents a common mistake: two intervals displayed in a description of $E$ are not necessarily different components. If they meet or fit together without omitting any point between them, their union may be one interval and therefore one component. Conversely, separated pieces of $E$ cannot belong to the same component if every interval joining them crosses a point outside $E$.

Check Your Understanding

Use the interval test and the definition of maximality to answer the following questions.

In the explicit description of $C_x$, why must the entire interval between $x$ and $y$ lie in $E$, rather than only the endpoints?
Find the connected components of $E=(-3,0]\cup[2,4)$.
Can two distinct connected components of the same set share a point? Explain using uniqueness.
What does it mean for a component to be closed relative to $E$, and how can that differ from being closed in $\mathbb{R}$?
If two intervals in a displayed union meet at an endpoint, what should you check before calling them different components?