From Completeness to Connectedness
The connectedness of intervals rests on a basic feature of the real numbers: every nonempty set that is bounded above has a least upper bound. A proposed separation of an interval would have to divide points on opposite sides of a transition. The least-upper-bound property produces a point at that transition, and relative openness prevents either side of the separation from containing it.
All subsets of \(\mathbb{R}\) in this tutorial have the subspace topology inherited from the standard topology on \(\mathbb{R}\). In particular, an open subset of an interval \(I\) means a set open in that subspace topology; it need not be open in \(\mathbb{R}\).
This definition of interval covers open, closed, and half-open intervals, as well as intervals unbounded in either direction. It also includes \(\mathbb{R}\) itself. The empty set and every singleton satisfy the defining condition too. We will check those two cases separately; for any interval containing at least two points, the order-convexity condition supplies the segments needed in the proof.
The Key Step: A Closed Segment Is Connected
Proof. Suppose, for contradiction, that \([a,b]\) has a separation \(U,V\). Since \(U\cup V=[a,b]\) and both sets are nonempty, choose a point from each. These two points are distinct because \(U\cap V=\varnothing\). Relabel the two sets if necessary, and call the point in the first set \(a_0\) and the point in the second set \(b_0\), where \(a_0<b_0\). The sets \(U\cap[a_0,b_0]\) and \(V\cap[a_0,b_0]\) are a separation of \([a_0,b_0]\): they are disjoint, cover that segment, are nonempty because they contain \(a_0\) and \(b_0\) respectively, and are relatively open as intersections of relatively open sets with the segment. Thus it is enough to rule out a separation whose left endpoint belongs to \(U\) and whose right endpoint belongs to \(V\). Rename these endpoints \(a\) and \(b\), and the resulting separating sets \(U\) and \(V\).
Consider the set \(S=U\subseteq[a,b]\). Since \(U\) is relatively open and contains \(a\), there is a \(\delta>0\) such that \((a-\delta,a+\delta)\cap[a,b]\subseteq U\). The point \(a+\min(\delta,b-a)/2\) is strictly greater than \(a\), is at most \(b\), and belongs to \(U\). Therefore \(S\) is nonempty and contains a point greater than \(a\). It is bounded above by \(b\), so completeness of \(\mathbb{R}\) gives \(c=\sup S\), with \(a<c\le b\).
The separation covers \([a,b]\), so \(c\in U\) or \(c\in V\). If \(c\in U\), then \(c\ne b\), because \(b\in V\) and the two sets are disjoint. Relative openness of \(U\) gives a \(\delta>0\) such that \((c-\delta,c+\delta)\cap[a,b]\subseteq U\). The point \(t=c+\min(\delta,b-c)/2\) satisfies \(c<t\le b\) and belongs to \(U=S\), contradicting that \(c\) is an upper bound for \(S\).
If \(c\in V\), relative openness gives a \(\delta>0\) such that \((c-\delta,c+\delta)\cap[a,b]\subseteq V\). Since \(c=\sup S\), there exists \(s\in S\) with \(c-\delta<s\le c\); otherwise \(c-\delta\) would be an upper bound for \(S\) smaller than \(c\). But \(s\in S=U\), while the displayed neighborhood of \(c\) lies in \(V\). This contradicts \(U\cap V=\varnothing\). Both cases are impossible, so \([a,b]\) has no separation and is connected. \(\square\)
The proof uses the supremum to locate a point \(c\) in the interval, then uses the separation to put \(c\) on one side or the other. It is relative openness at that point—not any assumption that \(c\) is an endpoint or belongs to \(U\)—that yields the contradiction.
Every Interval Is Connected
Proof. If \(I=\varnothing\), there are no two nonempty subsets that could form a separation. If \(I\) is a singleton, its only subsets are itself and the empty set, so again it has no separation.
Now suppose \(I\) contains at least two points and, for contradiction, has a separation \(U,V\). Choose \(x\in U\) and \(y\in V\). They are distinct. Relabel the two sets and the chosen points if necessary so that \(x<y\), with \(x\in U\) and \(y\in V\). Because \(I\) is an interval, every point between \(x\) and \(y\) belongs to \(I\), so \([x,y]\subseteq I\). The sets \(U\cap[x,y]\) and \(V\cap[x,y]\) are disjoint, nonempty, and cover \([x,y]\). They are open in \([x,y]\): for example, relative openness of \(U\) in \(I\) means \(U=O\cap I\) for some open \(O\subseteq\mathbb{R}\), and then \(U\cap[x,y]=O\cap[x,y]\). The same argument applies to \(V\). Thus these sets would be a separation of \([x,y]\), contradicting Theorem 1. Therefore \(I\) is connected. \(\square\)
Only the segment between two points in opposite sides of a proposed separation is needed. This is why the same proof handles every endpoint convention: the selected points are already in \(I\), and interval convexity guarantees that the entire segment between them is in \(I\).
Worked Example: A Bounded Closed Interval
Consider \(I=[-2,3]\). To apply Theorem 2, take any \(x,y\in I\) with \(x<y\). Then \(-2\le x<y\le3\). For any \(z\) with \(x<z<y\), these inequalities give \(-2\le z\le3\), so \(z\in[-2,3]\). Thus \(I\) is an interval and Theorem 2 gives its connectedness.
For a concrete segment check, choose \(x=-1\) and \(y=2\). They lie in \(I\), since \(-2\le-1\le3\) and \(-2\le2\le3\). Every \(z\) with \(-1\le z\le2\) also satisfies \(-2\le z\le3\), so \([-1,2]\subseteq I\). If \(I\) had a separation, its restrictions to this segment could be used whenever the two chosen points of the separation are \(-1\) and \(2\); Theorem 1 rules out that separation of the segment. Theorem 2 makes the same argument for any pair of points on opposite sides. As a check on the endpoint convention, the argument includes both endpoints \(-2\) and \(3\); it does not need them to be interior points.
Worked Example: A Half-Open Interval
Consider \(I=(0,1]\). If \(x,y\in I\) and \(x<y\), then \(0<x<z<y\le1\) for every \(z\) strictly between them. Hence \(0<z\le1\), so \(z\in I\). Theorem 2 therefore proves that \((0,1]\) is connected, despite excluding its left endpoint and including its right endpoint.
For a specific segment, take \(x=\tfrac14\) and \(y=1\). The calculations \(0<\tfrac14\le1\) and \(0<1\le1\) verify that both points belong to \(I\). If \(\tfrac14\le z\le1\), then \(0<\tfrac14\le z\le1\), so \(z\in(0,1]\), and \([\tfrac14,1]\subseteq I\). Any proposed separation of \(I\) with these points on opposite sides would restrict to a separation of \([\tfrac14,1]\), which Theorem 1 forbids. More generally, the interval proof uses the segment between whichever two points a proposed separation supplies; it never needs the excluded point \(0\).
Worked Example: An Unbounded Interval
Let \(I=(-\infty,4)\). For \(x,y\in I\) with \(x<y\), we have \(y<4\). Every \(z\) with \(x<z<y\) then satisfies \(z<4\), so \(z\in I\). Thus \(I\) is an interval, and Theorem 2 proves it is connected. No argument requires a left endpoint: any two points selected from a proposed separation are real numbers and determine a bounded segment between them.
For example, take \(x=-3\) and \(y=2\). The checks \(-3<4\) and \(2<4\) show that both points lie in \(I\), and every \(z\) with \(-3\le z\le2\) satisfies \(z\le2<4\). Consequently, \([-3,2]\subseteq(-\infty,4)\). If these points belonged to opposite sides of a separation, restriction to this segment would contradict Theorem 1. The same reasoning applies to any two points chosen from opposite sides, which is the contradiction required by Theorem 2.
Endpoint Conventions and a Common Pitfall
The proof does not depend on whether an interval is open or closed at either finite endpoint, or whether it has a finite endpoint at all. The only property used in extending Theorem 1 to Theorem 2 is that a segment between two points of \(I\) stays inside \(I\). For instance, \((0,1)\), \([0,1)\), \((0,1]\), and \([0,1]\) all have this property. The empty set and singletons are connected by the separate cases in the theorem’s proof.
Common pitfall. A supremum need not be a maximum. For a concrete calculation, take \(A=(0,\tfrac12)\subseteq\mathbb{R}\). The number \(\tfrac12\) is an upper bound for \(A\), and every number less than \(\tfrac12\) fails to be an upper bound: if \(r<\tfrac12\), then \(s=(\max(r,0)+\tfrac12)/2\) satisfies \(r<s<\tfrac12\), so \(s\in A\). Thus \(\sup A=\tfrac12\), but \(\tfrac12\notin A\). The incorrect step “\(c=\sup S\), so \(c\in S\)” cannot be used in the segment proof. Instead, the proof places \(c\) in one of the two sets because those sets cover the whole segment; whichever set contains it, relative openness contradicts the supremum property or disjointness.
Connectedness here is a statement about the topology inherited from the real line. The argument proves it directly from the definition of separation and the least-upper-bound property; it does not rely on a picture of an interval or on any assertion about the connected subsets of \(\mathbb{R}\).
Check Your Understanding
Use the separation definition and the least-upper-bound argument to justify your answers.
- Why must a separation of \([a,b]\) place \(a\) and \(b\) in different sets after restricting to a suitable segment?
- In Theorem 1, why is \(S\) nonempty and why is it bounded above?
- Which part of the proof for an arbitrary interval uses the interval property?
- Explain why \((-\infty,0]\) is an interval and state the connectedness conclusion.
- Why would the statement “every supremum belongs to the set” be an invalid step in the proof?