Proof That Intervals Are Connected

A Proof Built from Chains and Compactness

The previous tutorial showed that any two points of an interval can be joined by a finite chain whose steps are smaller than a prescribed positive number. Here we use that fact to prove that intervals are connected. The idea is to compare two properties: a chain can cross from one side of a separation to the other only in a step that crosses the gap between the pieces, while compactness will force that gap to have positive size on a suitable closed subinterval.

This proof gives a useful reason to focus on a closed segment even when the interval under consideration is open or unbounded. If a separation existed, choose one point from each side. The segment between those two points is a compact interval contained in the original interval. On that segment, the two sides become disjoint nonempty compact sets, and disjoint compact sets in the real line have positive distance. A chain with sufficiently small steps cannot cross between them.

We use the standard definition: a separation of a set \(E\subseteq\mathbb{R}\) is a representation \(E=A\cup B\), where \(A\) and \(B\) are disjoint, nonempty, and relatively open in \(E\). To prove that \(E\) is connected, it is enough to show that no such separation exists.

Separated Pieces Cannot Be Crossed by a Fine Chain

Proof. Suppose \(x_0,x_1,\ldots,x_n\) were such a chain, with \(x_0\in A\) and \(x_n\in B\). Since the chain starts in \(A\) and ends in \(B\), there is a first index \(k\) for which \(x_k\in B\). Then \(x_{k-1}\in A\): it cannot be in \(B\), by the choice of \(k\), and every point of \(E=A\cup B\) belongs to one of the two sets. By the definition of distance between sets, every point of \(A\) is at distance at least \(d\) from every point of \(B\). Hence \(|x_k-x_{k-1}|\geq d\geq\varepsilon\). But an \(\varepsilon\)-chain requires \(|x_k-x_{k-1}|<\varepsilon\). This contradiction proves the lemma. \(\square\)

The strict inequality in the chain definition matters here. Even if \(\varepsilon=d\), a step from \(A\) to \(B\) has length at least \(d=\varepsilon\), so it still cannot be an allowed step. In the proof below, we can choose \(\varepsilon\) strictly smaller than \(d\), which also makes the contradiction immediate.

Restricting a Proposed Separation to a Compact Segment

Suppose an interval \(J\) had a separation \(J=A\cup B\). Choose \(p\in A\) and \(q\in B\). If \(p<q\), put \(a=p\) and \(b=q\), retaining the names \(A\) and \(B\). If \(q<p\), interchange the names \(A\) and \(B\), then put \(a=q\) and \(b=p\). In either case, \(a<b\), \(a\) belongs to the set currently named \(A\), and \(b\) belongs to the set currently named \(B\). This ordering ensures that the endpoints of the segment we use lie on opposite sides.

Because \(J\) is an interval and \(a,b\in J\), the whole closed segment \(K=[a,b]\) lies in \(J\). Define \(A_0=A\cap K\) and \(B_0=B\cap K\). Both sets are nonempty, since \(a\in A_0\) and \(b\in B_0\). Also, \(A_0\) and \(B_0\) are disjoint and their union is \(K\).

Each of \(A\) and \(B\) is relatively closed in \(J\), because it is the complement in \(J\) of the other relatively open set. Consequently, \(A_0\) and \(B_0\) are closed in \(K\). The closed interval \(K\) is compact, so its closed subsets \(A_0\) and \(B_0\) are compact. The Positive Separation of a Compact Set and a Closed Set theorem from earlier in the course now applies: \(A_0\) is nonempty and compact, and \(B_0\) is nonempty and closed in \(\mathbb{R}\). It follows that \(\operatorname{dist}(A_0,B_0)=d>0\).

The use of compactness is local. The interval \(J\) itself need not be bounded or closed. The selected points \(a\) and \(b\) always determine a bounded closed segment \(K\), and that is enough to obtain the positive distance needed for the argument.

Theorem: Intervals Are Connected

Proof. The empty set has no separation, and a singleton cannot be the union of two disjoint nonempty sets. It remains to consider an interval with at least two points. Suppose, for a contradiction, that \(J=A\cup B\) is a separation. Choose \(a\in A\) and \(b\in B\), relabeling \(A\) and \(B\) if necessary so that \(a<b\). As described above, \(K=[a,b]\) is contained in \(J\), and the restrictions \(A_0=A\cap K\) and \(B_0=B\cap K\) are disjoint nonempty compact sets whose union is \(K\).

By the Positive Separation of a Compact Set and a Closed Set theorem, \(d=\operatorname{dist}(A_0,B_0)>0\). Set \(\varepsilon=d/2\), so \(0<\varepsilon<d\). The interval \(K\) is chain connected by the theorem “Intervals Are Chain Connected” from the previous tutorial. Therefore there is an \(\varepsilon\)-chain in \(K\) from \(a\) to \(b\). It starts in \(A_0\) and ends in \(B_0\), while every pair of points from opposite pieces is at distance at least \(d\). This contradicts the lemma: no \(\varepsilon\)-chain can cross between \(A_0\) and \(B_0\) when \(\varepsilon<d\). The assumed separation cannot exist, so \(J\) is connected. \(\square\)

What the Proof Uses—and What It Does Not

The proof combines two earlier ideas in a way that is useful beyond this particular theorem. Chain connectedness provides a finite route with controllably small steps. Compactness, after restricting to a segment, prevents the two sides of a separation from approaching each other arbitrarily closely. Together they rule out the possibility that the chain changes sides.

A common pitfall is to use chain connectedness alone to infer connectedness. That inference is not valid: as the previous tutorial explained, the rational numbers admit arbitrarily fine finite chains between any two of their points, yet they are disconnected as a subset of \(\mathbb{R}\). The additional feature used here is not merely that chains exist. A hypothetical separation of an interval, when restricted to the segment between two points on opposite sides, would yield two disjoint compact pieces at positive distance. That positive gap is what makes a sufficiently fine chain impossible.

The argument also shows why the endpoint type of an interval is irrelevant. Whether \(J\) is open, closed, half-open, bounded, or unbounded, any two selected points determine a closed segment contained in \(J\). The proof uses compactness on that segment, not on the whole interval.