Connected Components

Maximal Connected Pieces

Connectedness in a topological space is defined by the absence of a separation into two nonempty disjoint open sets. The previous tutorial established the clopen test for connectedness, along with preservation results for continuous images and products. We now use connectedness to divide a space into its largest connected pieces.

A connected component is not just any connected subset. It is a connected subset that cannot be enlarged to a strictly larger connected subset of the space. We will show that each point belongs to exactly one such piece, that these pieces partition the space, and that every component is closed. These conclusions hold in arbitrary topological spaces; no metric or distance is required.

Definition: Let $X$ be a topological space. A connected component of $X$ is a nonempty connected subset $C\subseteq X$ that is maximal under inclusion among connected subsets of $X$. For $x\in X$, the component containing $x$ will be denoted by $C_X(x)$.

Maximal means that no strictly larger connected subset of $X$ contains $C$; it does not mean that $C$ is the only connected subset near its points. A space can have many connected subsets inside a single component. The key construction is to collect all the connected subsets containing a particular point and show that their union remains connected.

Unions Sharing a Point

The union of connected sets is not always connected. For example, the disjoint closed intervals [0,1] and [2,3] on the real line have a disconnected union. A common point, however, prevents the sets in a family from lying on opposite sides of a separation.

Theorem (Union with a Common Point): Let $\{A_\lambda:\lambda\in\Lambda\}$ be a nonempty family of connected subsets of a topological space $X$. If there is a point $p\in\bigcap_{\lambda\in\Lambda}A_\lambda$, then $\bigcup_{\lambda\in\Lambda}A_\lambda$ is connected.

Proof. Set $A=\bigcup_{\lambda\in\Lambda}A_\lambda$. Suppose, for a contradiction, that $A$ is disconnected. Then it has a separation $U,V$, where $U$ and $V$ are disjoint, nonempty, relatively open subsets of $A$ and $A=U\cup V$. The point $p$ belongs to exactly one of these sets; suppose $p\in U$.

For each $\lambda$, the sets $A_\lambda\cap U$ and $A_\lambda\cap V$ are disjoint and relatively open in $A_\lambda$, and their union is $A_\lambda$. Since $p\in A_\lambda\cap U$, the first set is nonempty. If $A_\lambda\cap V$ were also nonempty, these two sets would form a separation of the connected space $A_\lambda$. Therefore $A_\lambda\cap V=\varnothing$, and $A_\lambda\subseteq U$. This holds for every $\lambda$, so $A\subseteq U$, contradicting that $V$ is nonempty. Hence $A$ is connected. $\square$

Worked Example: Joining Two Intervals at One Point

In $\mathbb{R}$ with its usual topology, let $A_1=[-2,1]$ and $A_2=[1,4]$. Each is an interval and therefore connected, and both contain $1$. The common-point theorem shows that their union is connected. Since $[-2,1]\cup[1,4]=[-2,4]$, this recovers the connectedness of the larger interval from two connected pieces joined at an endpoint.

The shared point is essential to this theorem. If instead the intervals were $[-2,-1]$ and $[1,4]$, they would have no common point, and their union would have a separation into those two relatively open pieces.

Constructing and Partitioning Components

For each $x\in X$, consider every connected subset of $X$ that contains $x$. This is a nonempty family because the singleton $\{x\}$ is connected. Their union is connected by the common-point theorem, since all members contain $x$.

Theorem (Construction and Partition of Components): For every $x\in X$, the set $$ C_X(x)=\bigcup\{A\subseteq X:A\text{ is connected and }x\in A\} $$ is the component containing $x$. Every point of $X$ belongs to exactly one component, and any two components are either equal or disjoint.

Proof. The family in the union is nonempty because it includes $\{x\}$, and all its members contain $x$. By the common-point theorem, $C_X(x)$ is connected. It contains $x$ because $\{x\}$ is one of the sets in the union.

Now let $A$ be any connected subset containing $C_X(x)$. Since $x\in C_X(x)\subseteq A$, the set $A$ is one of the connected subsets included in the defining union. Consequently $A\subseteq C_X(x)$. Together with $C_X(x)\subseteq A$, this gives $A=C_X(x)$. Thus $C_X(x)$ is maximal connected and is a component.

To prove uniqueness and the partition claim, suppose $C_X(x)\cap C_X(y)\ne\varnothing$. Both sets are connected and have a common point, so their union is connected. That union contains $x$, and therefore it is included in $C_X(x)$, by the definition of $C_X(x)$. In particular, $C_X(y)\subseteq C_X(x)$. Reversing the roles of $x$ and $y$ gives $C_X(x)\subseteq C_X(y)$, so the two components are equal. Thus components that intersect are equal. Finally, every point $x$ lies in $C_X(x)$, so the components cover $X$. These facts show that they partition $X$. $\square$

Worked Example: Components of Two Separated Intervals

Let $X=[-3,-1]\cup[1,3]$, with the subspace topology inherited from $\mathbb{R}$. Write $A=[-3,-1]$ and $B=[1,3]$. Both are intervals, hence connected. Each is also clopen in $X$: each is closed in $X$, and its complement in $X$ is the other closed interval, so each is open in $X$ as well.

If a connected subset $E\subseteq X$ met both $A$ and $B$, then $E\cap A$ and $E\cap B$ would be disjoint, nonempty, relatively open subsets of $E$ covering $E$. That would be a separation of $E$. Hence every connected subset of $X$ lies entirely in $A$ or entirely in $B$. Since $A$ and $B$ are themselves connected, neither can be enlarged to a connected subset of $X$. They are exactly the two components.

Worked Example: Components of a Finite Topological Space

Let $X=\{a,b,c,d\}$, and give $X$ the topology whose open sets are all unions of the two blocks $A=\{a,b\}$ and $B=\{c,d\}$. Thus the topology is $\{\varnothing,A,B,X\}$. The blocks are disjoint, nonempty, and clopen in $X$.

The subspace topology on $A$ is $\{\varnothing,A\}$, so $A$ is connected: its only open subsets are $\varnothing$ and $A$, and it has no separation. The same reasoning shows that $B$ is connected. Any subset of $X$ meeting both blocks has a separation obtained by intersecting it with $A$ and $B$, which are clopen in $X$. Thus no connected subset can meet both blocks, and the components are exactly $\{a,b\}$ and $\{c,d\}$. In particular, components need not be singletons.

Components Are Closed

A component is maximal among connected subsets, so one way to test whether it is closed is to examine its closure. Closure does not destroy connectedness: if the closure could be separated, each side of that separation would have to meet the original dense set, producing a separation there as well. This argument works in any topological space.

Theorem (Closure of a Connected Set): If $A$ is a connected subset of a topological space $X$, then its closure $\overline{A}$ in $X$ is connected.

Proof. Suppose, to the contrary, that $\overline{A}$ has a separation $U,V$. These are nonempty, disjoint, relatively open subsets of $\overline{A}$ whose union is $\overline{A}$. Because $A$ is dense in its closure, every nonempty relatively open subset of $\overline{A}$ meets $A$. Therefore $U\cap A\ne\varnothing$ and $V\cap A\ne\varnothing$.

The sets $U\cap A$ and $V\cap A$ are relatively open in $A$, are disjoint, and cover $A$. Their nonemptiness makes them a separation of $A$, contradicting that $A$ is connected. Thus $\overline{A}$ is connected. $\square$

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

Proof. Let $C$ be a component. By the closure theorem, $\overline{C}$ is connected. It contains $C$, so maximality of $C$ gives $\overline{C}=C$. Hence $C$ is closed. $\square$

Worked Example: Components of the Rational Numbers

Give $\mathbb{Q}$ the subspace topology inherited from $\mathbb{R}$. We show that every connected subset $E\subseteq\mathbb{Q}$ contains at most one point. Suppose $q,r\in E$ with $q<r$. Choose an irrational number $\alpha$ with $q<\alpha<r$. The sets $E\cap(-\infty,\alpha)$ and $E\cap(\alpha,\infty)$, where the intervals are taken in $\mathbb{R}$, are relatively open in $E$. They are disjoint and cover $E$, since no rational point equals $\alpha$. Both are nonempty because they contain $q$ and $r$, respectively. This separates $E$, a contradiction.

Every singleton in $\mathbb{Q}$ is connected, so the components are exactly the singletons $\{q\}$, for $q\in\mathbb{Q}$. None is open in $\mathbb{Q}$: every open neighborhood of a rational number contains other rational numbers. Thus components are always closed, but need not be open.

What Component Results Do—and Do Not—Say

Components collect points that cannot be separated by putting them into different connected subsets. The partition theorem guarantees a unique component for each point, while the closure theorem guarantees that each such piece is closed. The components can be large, as in the finite example, or singletons, as in $\mathbb{Q}$. Their size depends on the topology, not merely on the underlying set.

Closedness does not imply openness. The rational-number example shows why it is important not to assume components are open without an additional hypothesis. In an arbitrary topological space, the proofs above establish closedness but give no reason for a component to be a neighborhood of any of its points.

The common-point theorem is the main construction tool: it lets us join all connected subsets containing one point into a single largest connected set. When identifying components in a particular space, it is often efficient to combine this principle with clopen subsets. A clopen subset cannot be crossed by a connected set that meets both it and its complement, because those intersections would give a separation.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

Why does the common-point condition ensure that every connected set in the family lies on the same side of any proposed separation?
What makes $C_X(x)$ maximal among connected subsets containing $x$?
Why must two components that intersect be equal?
In the closure theorem, why must each nonempty relatively open side of a separation of $\overline{A}$ meet $A$?
Give an example from this tutorial showing that a component need not be open.

What You'll Learn