Classifying Connected Subsets of the Real Line

From the Interval Test to Concrete Classifications

For subsets of the real line, connectedness can be decided by checking whether the set has a gap between any two of its points. This is especially useful when a set is presented as a union of intervals, or when one or more points have been removed. The details of endpoint inclusion matter: two intervals can approach the same endpoint while still leaving that point out of their union.

Throughout, \(\mathbb{R}\) has its standard topology, and every subset \(A\subseteq\mathbb{R}\) has the subspace topology. A separation of a space \(A\) is a pair of disjoint, nonempty, relatively open subsets whose union is \(A\). The space \(A\) is connected if it has no separation; otherwise it is disconnected. A subset \(A\) of \(\mathbb{R}\) is an interval if, whenever \(x,y\in A\) and \(x<z<y\), then \(z\in A\). This order-convex definition includes singletons and the empty set.

To use the characterization, choose two points of the set and inspect the points strictly between them. If even one such intermediate point is missing, the set is not an interval and therefore is disconnected. Conversely, if every intermediate point is present, the set is an interval and therefore connected. This test does not require the set to be open, closed, bounded, or to have included endpoints.

A Missing Intermediate Point Gives a Separation

Proof. Suppose \(a,b\in A\), \(a<c<b\), and \(c\notin A\). Set \(U=A\cap(-\infty,c)\) and \(V=A\cap(c,\infty)\). The sets \((-\infty,c)\) and \((c,\infty)\) are open in \(\mathbb{R}\), so \(U\) and \(V\) are open in the subspace \(A\). They are disjoint, and because \(c\notin A\), their union is \(A\). Also \(a\in U\) and \(b\in V\), so both are nonempty. Thus \(U,V\) form a separation of \(A\), which proves that \(A\) is disconnected. \(\square\)

This proposition gives a direct way to verify disconnection, not just a label from the interval test. The cut at \(c\) splits the set into the points below \(c\) and the points above it. Conversely, if no such missing intermediate point exists, then \(A\) is an interval and the interval characterization gives connectedness.

Endpoint Inclusion and Punctured Intervals

A missing point inside an interval creates a gap. Removing a point that was already outside the interval does not. For unions, a shared endpoint fills the gap only if at least one interval includes it; if both intervals exclude it, that point remains missing.

Unions with a Common Point

Proof. Let \(\{I_\alpha\}_{\alpha\in J}\) be a collection of intervals with a common point \(p\), and let \(x,y\in\bigcup_{\alpha\in J}I_\alpha\) satisfy \(x<y\). Choose indices \(\alpha,\beta\in J\) such that \(x\in I_\alpha\) and \(y\in I_\beta\). If \(x\leq p\leq y\), then \(x,p\in I_\alpha\), so every point of \([x,p]\) is in \(I_\alpha\); also \(p,y\in I_\beta\), so every point of \([p,y]\) is in \(I_\beta\). Together these intervals contain every point of \([x,y]\). If \(p<x<y\), then \(x,p\in I_\alpha\) and \(y,p\in I_\beta\); the interval property gives \([p,x]\subseteq I_\alpha\) and \([p,y]\subseteq I_\beta\), which cover \([x,y]\). If \(x<y<p\), the same direct check gives \([x,p]\subseteq I_\alpha\) and \([y,p]\subseteq I_\beta\), again covering \([x,y]\). Thus the union contains every point between any two of its points, so it is an interval. By the interval characterization, it is connected. \(\square\)

The requirement of a common point is sufficient, but it is not necessary for a union of two intervals to be connected. For instance, \([0,1)\cup[1,2]=[0,2]\): the point \(1\) belongs to the second interval, so there is no gap, even though the two intervals do not intersect. By contrast, \([0,1)\cup(1,2]\) omits \(1\) and is disconnected. The interval test handles both cases without relying on whether the displayed pieces overlap.

Singletons, Finite Sets, and a Common Mistake

A singleton \(\{x\}\) is an interval: there are no two distinct points in it with a point strictly between them. Therefore every singleton is connected. The empty set is also an interval under the definition above, and is connected because it cannot be written as a union of two nonempty sets.

More generally, every finite subset of \(\mathbb{R}\) with at least two points is disconnected. To see this, take two distinct points \(a<b\) in the set. The finite set has only finitely many points strictly between \(a\) and \(b\), while the open interval \((a,b)\) has infinitely many points. Thus some \(c\) satisfies \(a<c<b\) and is not in the finite set. The missing-intermediate-point proposition gives a separation. For example, \(\{-2,0,3\}\) is disconnected: \(-2,3\) belong to it and \(1\) does not, with \(-2<1<3\).

Common mistake. One might write \([0,1)\cup(1,2]=[0,2]\) on the grounds that the pieces have endpoints \(1\) in common. That equality is false: \(1\notin[0,1)\) and \(1\notin(1,2]\), so \(1\) is absent from the union. For a direct test, \(0,2\in[0,1)\cup(1,2]\), but \(0<1<2\) and \(1\) is not in the set. The union is therefore disconnected. In contrast, \([0,1)\cup[1,2]=[0,2]\), since the second interval includes \(1\). Always check membership at each candidate joining endpoint before combining interval notation.

In practice, classification can be organized into three checks:

• For a proposed connected set, verify that every point strictly between any two of its points is included.
• For a proposed disconnected set, exhibit two included points and a missing intermediate point; the cut at that point gives a separation.
• When joining interval pieces, inspect the endpoints themselves rather than relying on how close the pieces appear.