Connected Components

Components as Maximal Connected Pieces

A connected set cannot be split into two separated pieces, but an arbitrary set may contain many connected pieces. The connected components of a set identify those pieces that cannot be enlarged while remaining connected. Earlier in this course, the Explicit Description of Connected Components gave a formula for the component containing a point, and the Components Partition a Set theorem showed that the components are disjoint and cover the set. Here we use that description to establish a useful maximality property, then examine what more can be said when the set is open.

“Maximal” does not mean “largest” in the sense of containing every other component. Distinct components are disjoint. Rather, each component is as large as possible while remaining connected. The following characterization makes this precise for each point \(x\in E\).

Proof. By the Explicit Description of Connected Components, the component containing \(x\) is \(C_x=\{y\in E:[\min\{x,y\},\max\{x,y\}]\subseteq E\}\), and \(C_x\) is connected. Let \(D\subseteq E\) be any connected set containing \(x\). For each \(y\in D\), the Characterization of Connected Subsets of \(\mathbb{R}\) says that \(D\) is an interval. Therefore every point between \(x\) and \(y\) belongs to \(D\), and hence to \(E\). This means \(y\in C_x\). Since this holds for every \(y\in D\), we have \(D\subseteq C_x\).

In particular, any connected subset of \(E\) containing \(C_x\) also contains \(x\), so it must be contained in \(C_x\). Thus \(C_x\) is maximal. If another maximal connected subset \(M\) contains \(x\), the same argument gives \(M\subseteq C_x\). Since \(C_x\) is connected and contains \(x\), maximality of \(M\) gives \(C_x\subseteq M\). Hence \(M=C_x\), proving uniqueness. \(\square\)

This theorem provides a practical test: to show that a connected set \(I\subseteq E\) is a component, it is enough to show that no larger connected subset of \(E\) can contain it. On the real line, any connected set is an interval. Thus a missing point between \(I\) and any proposed enlargement prevents that enlargement from being connected.

Components of Open Sets

For a general set \(E\), its components need not be open, even relative to \(E\). When \(E\) itself is open, however, every component is open in \(\mathbb{R}\). The key observation is that every point of \(E\) has a small interval around it that remains inside \(E\). That interval is connected and meets the point’s component, so maximality forces the whole interval to lie in that component.

Proof. Suppose \(E\ne\varnothing\), and let \(C\) be a component of \(E\). By the Characterization of Connected Subsets of \(\mathbb{R}\), \(C\) is an interval. Take any \(x\in C\). Since \(E\) is open, there is an \(\varepsilon>0\) such that \((x-\varepsilon,x+\varepsilon)\subseteq E\). This interval is connected and intersects \(C\), because it contains \(x\). The Union of Pairwise-Intersecting Connected Sets theorem implies that \(C\cup(x-\varepsilon,x+\varepsilon)\) is connected. It is a subset of \(E\) containing \(C\), so maximality of \(C\) gives \(C\cup(x-\varepsilon,x+\varepsilon)=C\). In particular, \((x-\varepsilon,x+\varepsilon)\subseteq C\).

Every point of \(C\) therefore has an open interval around it contained in \(C\), so \(C\) is open in \(\mathbb{R}\). As \(C\) is both open and an interval, it is an open interval, with the possibilities of an unbounded interval such as \((a,\infty)\) or \(\mathbb{R}\) itself. \(\square\)

The theorem has a useful consequence: an open subset of the real line is a union of disjoint open intervals, namely its components. The intervals may be finite or unbounded, and there may be finitely many, countably many, or no components. In fact, there cannot be uncountably many.

Proof. The empty set has no components, so suppose \(E\ne\varnothing\). By the Components of Open Sets Are Open Intervals theorem, every component is a nonempty open interval. The density of the rationals in \(\mathbb{R}\) implies that each such interval contains a rational number. Fix an enumeration \(q_1,q_2,\ldots\) of the rational numbers. Assign to each component \(C\) the first \(q_k\) in this enumeration that belongs to \(C\).

This assignment sends each component to a rational number in that component. Distinct components are disjoint, by the Components Partition a Set theorem, so no rational number can be assigned to two different components. The assignment is therefore one-to-one from the collection of components into the rational numbers. Since the rationals are countable, the collection of components is at most countable. \(\square\)

Relative Closedness Does Not Mean Relative Openness

Every component is closed relative to its containing set, by the Components Are Relatively Closed theorem. This does not imply that a component is relatively open. In an open set, the theorem above does give openness in \(\mathbb{R}\); but for other sets, components may fail to be open even relative to the set itself. The distinction matters: relative closedness says limits within \(E\) stay in a component, while relative openness would require every point of a component to have a neighborhood in \(E\) containing no points from other components.

How to Identify Components

For subsets of \(\mathbb{R}\), the interval characterization supplies a direct method. Start with a point \(x\in E\), and extend an interval around \(x\) as far as possible without leaving \(E\). The resulting maximal interval is the component containing \(x\). A gap in \(E\) blocks any connected enlargement across that gap. For an open set, the resulting maximal intervals are open intervals, and the rational-number argument guarantees that there are at most countably many of them.