From Connected Sets to Their Components
The Characterization of Connected Subsets of \(\mathbb{R}\) gives a direct test: a subset of the real line is connected exactly when it is an interval. This turns the study of connectedness into a question about order. Whenever a point lies between two points of an interval, it must also belong to that interval.
For a set that is not itself an interval, its connected components describe the largest connected pieces it contains. The Components Partition a Set theorem says that these pieces are pairwise disjoint and cover the set. The Maximality Characterization of Components says that each is a maximal connected subset. We will use these earlier results to examine additional structure: what happens at the endpoints of a component, and what this implies when the original set is closed.
Since a connected subset of \(\mathbb{R}\) is an interval, every component is an interval. But components need not all have the same endpoint form. A component may include one endpoint and omit the other, include both, or omit both. The membership of its endpoints is controlled by the set \(E\) itself.
Finite Endpoints of Components
Components are relatively closed in their containing set, by the Components Are Relatively Closed theorem. This does not say that a component must be closed in all of \(\mathbb{R}\): the containing set might not be closed. It does, however, determine exactly what happens at a finite endpoint of a component.
Proof. First consider the lower endpoint \(\alpha=\inf C\). If \(\alpha\in C\), then \(C\subseteq E\) gives \(\alpha\in E\).
Conversely, suppose \(\alpha\in E\). For each positive integer \(n\), the definition of infimum gives a point \(c_n\in C\) such that \(\alpha\leq c_n<\alpha+1/n\). Hence \(c_n\to\alpha\). Because \(C\) is closed relative to \(E\), its relative closure satisfies \(\overline{C}\cap E=C\). The convergence \(c_n\to\alpha\) implies \(\alpha\in\overline{C}\), and the assumption \(\alpha\in E\) then gives \(\alpha\in C\).
For the upper endpoint \(\beta=\sup C\), if \(\beta\in C\), then \(\beta\in E\). If instead we assume \(\beta\in E\), the definition of supremum gives, for each positive integer \(n\), a point \(d_n\in C\) with \(\beta-1/n<d_n\leq\beta\). Thus \(d_n\to\beta\). Relative closedness gives \(\beta\in\overline{C}\cap E=C\). This proves both endpoint claims. \(\square\)
The theorem is useful because it rules out endpoint choices that might otherwise seem possible. If a finite endpoint of a component belongs to \(E\), it must belong to the component. If it is not in the component, then it is not in \(E\). In particular, a component cannot stop just short of a point of \(E\) at its endpoint.
Worked Example: Endpoint Membership Determines the Brackets
Let \(E=(-4,-1]\cup[2,5)\cup\{8\}\). Each displayed piece is an interval, so each is connected. The gaps between them show that no connected subset of \(E\) can contain points from two different pieces: if it did, the interval property would require it to contain points in one of the missing gaps. Thus these three pieces are exactly the components of \(E\).
The first component has finite endpoints \(-4\) and \(-1\). Since \(-4\notin E\), that endpoint is omitted; since \(-1\in E\), it is included. The component is therefore \((-4,-1]\). The same endpoint check gives \([2,5)\) for the second component. The third component is the singleton \(\{8\}\). Each endpoint decision agrees with the theorem: a finite endpoint belongs to its component exactly when it belongs to \(E\).
Components of Closed Sets
For an arbitrary set \(E\), its components are relatively closed in \(E\), but may fail to be closed in \(\mathbb{R}\). When \(E\) is closed in \(\mathbb{R}\), relative closedness becomes ordinary closedness. Combining that fact with the interval characterization gives a particularly simple description.
Proof. Let \(C\) be a connected component of the closed set \(E\). By the Components Are Relatively Closed theorem, \(C\) is closed relative to \(E\). To show it is closed in \(\mathbb{R}\), take any \(x\in\overline{C}\), where the closure is in \(\mathbb{R}\). Since \(C\subseteq E\) and \(E\) is closed, \(\overline{C}\subseteq E\), so \(x\in E\). Relative closedness now gives \(x\in C\). Thus \(\overline{C}\subseteq C\); the reverse inclusion always holds, so \(C=\overline{C}\) and \(C\) is closed in \(\mathbb{R}\).
A component is connected by definition. The Characterization of Connected Subsets of \(\mathbb{R}\) therefore says that \(C\) is an interval. Since \(C\) is also closed in \(\mathbb{R}\), it is a closed interval, allowing an unbounded interval or a singleton. \(\square\)
“Closed interval” here includes intervals such as \([a,b]\), \([a,\infty)\), \((-\infty,b]\), and \(\mathbb{R}\), as well as \([a,a]=\{a\}\). It does not mean that every component of a closed set is bounded. Nor does it say that the whole closed set is one interval: a closed set may have many components.
Worked Example: Components of a Closed Set
Consider \(F=[-3,-1]\cup\{1\}\cup[4,\infty)\). This set is closed: each of its three pieces is closed, and a finite union of closed sets is closed. Each piece is an interval, hence connected. The missing intervals between the pieces prevent a connected subset from meeting two pieces. For example, if such a subset contained \(-1\) and \(1\), the interval property would require it to contain \(0\), but \(0\notin F\). The same reasoning applies to pairs chosen from the other distinct pieces.
Therefore the components are exactly \([-3,-1]\), \(\{1\}\), and \([4,\infty)\). Each is a closed interval, in accordance with the theorem. The example also illustrates why a singleton can be a component: it is connected, and the gaps on both sides prevent it from being enlarged to a connected subset of \(F\).
Missing Points and the Shape of Components
A missing point between two members of a set is an obstruction to putting those members in the same connected subset. Indeed, any connected subset of \(\mathbb{R}\) containing \(x<y\) must contain the entire interval \([x,y]\). If \(c\) lies strictly between \(x\) and \(y\) and \(c\notin E\), no connected subset of \(E\) can contain both \(x\) and \(y\).
This is a convenient way to identify components: first find intervals contained in the set, then check whether missing points prevent those intervals from being enlarged. It also explains why a single omitted point can split a set even when the parts on either side approach that point as closely as desired. A positive-width gap is not required.
Worked Example: Removing the Integers
Let \(G=\mathbb{R}\setminus\mathbb{Z}\), the set of nonintegers. For each integer \(n\), the interval \((n,n+1)\) lies in \(G\) and is connected. These intervals cover \(G\): every noninteger real number lies strictly between two consecutive integers.
No connected subset of \(G\) can contain points from two different such intervals. If \(x\in(n,n+1)\) and \(y\in(m,m+1)\) with \(n<m\), then the integer \(n+1\) lies between \(x\) and \(y\), but \(n+1\notin G\). The interval property rules out a connected subset of \(G\) containing both points. Consequently the components of \(G\) are exactly the intervals \((n,n+1)\), for \(n\in\mathbb{Z}\).
Each component \((n,n+1)\) is open in \(\mathbb{R}\) because \(G\) is open, and is not closed because its endpoints \(n\) and \(n+1\) are limit points that it omits. Its two finite endpoints are integers, and neither belongs to \(G\). This matches the endpoint theorem: an endpoint omitted from a component must be absent from the containing set.
Reading the Structure Carefully
The component description depends on the set and on the order of the real line. A component is an interval because connected subsets of \(\mathbb{R}\) are intervals. It is maximal among intervals contained in the set because it is a maximal connected subset. Its finite endpoints are included precisely when they belong to the set. If the set is closed, relative closedness upgrades each component to a closed interval.
These facts should not be confused with the claim that distinct components must be separated by a gap of positive length. In the example \(\mathbb{R}\setminus\mathbb{Z}\), neighboring components are separated by an omitted integer, but their distance is zero: points in the two intervals can be chosen arbitrarily close to that integer. The relevant obstruction is the missing point and the interval property, not positive distance.
For an open subset of \(\mathbb{R}\), the earlier Components of Open Sets Are Open Intervals theorem gives the corresponding open-set description: its components are open intervals, possibly unbounded. For a closed subset, the theorem proved here gives closed intervals. For a set that is neither open nor closed, endpoint inclusion can vary from component to component, as the endpoint theorem and the first example show.
Check Your Understanding
Use the interval characterization, the endpoint theorem, and the component results to answer the following questions.
- If \(\alpha=\inf C\) is finite for a component \(C\) of \(E\), why does \(\alpha\in E\) imply \(\alpha\in C\)?
- What are the components of \((-4,-1]\cup[2,5)\cup\{8\}\), and which finite endpoints are omitted?
- Why must a component of a closed subset of \(\mathbb{R}\) be closed in \(\mathbb{R}\), rather than only relatively closed in the subset?
- Why can no connected subset of \(\mathbb{R}\setminus\mathbb{Z}\) contain both a point in \((2,3)\) and a point in \((3,4)\)?
- Does the theorem about components of closed sets require the components to be bounded? Explain.