Open and Closed Intervals

Endpoint Brackets and Interval Types

In the previous tutorial, intervals were defined by the between-points property, and bracket notation recorded whether specified endpoints belong to an interval. This tutorial focuses on two important forms: open intervals, which exclude both finite endpoints, and closed intervals, which include both. We will also connect these names to the broader ideas of open and closed sets on the real line.

The distinction is useful, but it requires care. An open interval is an open set, and a closed interval is a closed set. However, the words “open” and “closed” can describe properties of arbitrary sets, not just intervals. We will define those properties precisely and prove the connection rather than relying only on the appearance of the brackets.

A parenthesis marks an excluded endpoint, while a square bracket marks an included endpoint. For example, in \((a,b)\), neither \(a\) nor \(b\) belongs to the set. In \([a,b]\), both do. The points strictly between \(a\) and \(b\) belong to both intervals.

The notation \((a,a)\) describes the empty set, since no real number satisfies \(a<x<a\). By contrast, \([a,a]=\{a\}\). Although a singleton is an interval by the between-points definition, it is not an open interval in the usual notation \((a,b)\) with \(a<b\). It is a closed interval with equal endpoints.

Open and Closed Sets

To state what it means for a set to be open, we use the usual distance on the real line, \(d(x,y)=|x-y|\). A neighborhood of \(x\) with radius \(\varepsilon>0\) is the open interval \((x-\varepsilon,x+\varepsilon)\). A set is open if every one of its points has some neighborhood contained entirely in the set.

The radius may depend on the point \(x\). An endpoint of an interval matters here: a point just inside an excluded endpoint can be close to that endpoint, but it still has some positive distance from it and can have a sufficiently small neighborhood contained in the interval.

These definitions concern neighborhoods and complements, not the shape of the brackets in a formula. In particular, “closed” does not mean simply that a set is bounded, and “open” does not mean that a set has no endpoints in every possible sense. The following results explain why the familiar open and closed intervals have their names.

Open Intervals Are Open Sets

Proof. Let \(x\in(a,b)\). Then \(a<x<b\), so both \(x-a\) and \(b-x\) are positive. Set \(\varepsilon=\frac12\min\{x-a,b-x\}\), which is positive. If \(y\in(x-\varepsilon,x+\varepsilon)\), then \(|y-x|<\varepsilon\), and therefore

Since \(\varepsilon\leq\frac{x-a}{2}\), we have \(x-\varepsilon\geq\frac{x+a}{2}>a\). Since \(\varepsilon\leq\frac{b-x}{2}\), we have \(x+\varepsilon\leq\frac{x+b}{2}<b\). It follows that \(a<y<b\), so \(y\in(a,b)\). Thus every \(x\in(a,b)\) has a neighborhood contained in \((a,b)\), proving that the interval is open. \(\square\)

The choice of radius is the key technique: take a positive radius no larger than the distance from \(x\) to either endpoint. Using half the smaller distance makes both strict bounds immediate. The radius can vary with \(x\); there is no requirement that one radius work for every point in the interval.

Closed Intervals Are Closed Sets

For a closed interval, the useful test is its complement. If \(a\leq b\), the points outside \([a,b]\) lie either to the left of \(a\) or to the right of \(b\). We show that every such point has a neighborhood that stays outside the interval.

Proof. Let \(x\in\mathbb{R}\setminus[a,b]\). By the definition of \([a,b]\), either \(x<a\) or \(x>b\). If \(x<a\), choose \(\varepsilon=(a-x)/2>0\). For any \(y\) with \(|y-x|<\varepsilon\), we have

Thus \(y\notin[a,b]\). If instead \(x>b\), choose \(\varepsilon=(x-b)/2>0\). Then \(|y-x|<\varepsilon\) implies

so again \(y\notin[a,b]\). Every point in the complement therefore has a neighborhood contained in the complement. Hence \(\mathbb{R}\setminus[a,b]\) is open, and \([a,b]\) is closed by definition. This proof also covers \(a=b\): a point outside the singleton \(\{a\}\) lies either below or above \(a\). \(\square\)

This argument illustrates why it is useful to define closedness by the complement. To prove that a set is closed, it is enough to show that points outside it have neighborhoods that remain outside. No assumption that the set has positive length is needed.

Half-Open Intervals and a Common Pitfall

An interval with one included endpoint and one excluded endpoint is called half-open (or half-closed). The next tutorial will examine these intervals further. Here, one distinction is important: when \(a<b\), \([a,b)\) is neither an open set nor a closed set. Its endpoint notation records which boundary point belongs to it, but it does not make the set open or closed.

Proof. The point \(a\) belongs to \([a,b)\). For any \(\varepsilon>0\), the point \(y=a-\varepsilon/2\) satisfies \(|y-a|=\varepsilon/2<\varepsilon\), but \(y<a\), so \(y\notin[a,b)\). Thus no neighborhood of \(a\) is contained in \([a,b)\), and the set is not open.

To show that the set is not closed, consider its complement and the point \(b\), which belongs to that complement because the right endpoint is excluded. For any \(\varepsilon>0\), choose \(\delta=\frac12\min\{\varepsilon,b-a\}>0\) and let \(y=b-\delta\). Then \(|y-b|=\delta<\varepsilon\), while \(a<y<b\), so \(y\in[a,b)\). Thus every neighborhood of \(b\) meets \([a,b)\), and no neighborhood of \(b\) is contained in the complement. The complement is not open, so \([a,b)\) is not closed. \(\square\)

The proof checks both properties separately. To disprove openness, it finds a point of the set with no neighborhood contained in the set. To disprove closedness, it finds a point of the complement with no neighborhood contained in the complement. A failure of one property alone would not establish failure of the other.

Unbounded Intervals and Endpoint Conventions

The same endpoint distinction applies to intervals that extend indefinitely. The symbols \(\infty\) and \(-\infty\) are not real numbers that can belong to an interval, so they are always written with parentheses. For example, \((c,\infty)=\{x\in\mathbb{R}:x>c\}\) is open: for any \(x>c\), a radius smaller than \(x-c\) keeps a neighborhood to the right of \(c\). The ray \([c,\infty)=\{x\in\mathbb{R}:x\geq c\}\) is closed, since its complement \((-\infty,c)\) is open.

A common pitfall is to infer open or closed status from boundedness. The interval \((c,\infty)\) is unbounded but open, while \([c,\infty)\) is unbounded but closed. Conversely, a bounded half-open interval such as \([1,6)\) is neither. Boundedness and openness or closedness are different properties.