The Order Topology and Its Basic Open Sets

From an Order to a Topology

An order tells us which points lie between others. The order topology turns that information into a notion of closeness: points near \(x\) are described by inequalities that place them between lower and upper bounds. In this tutorial, all orders are linear orders, so any two distinct points can be compared. The examples will show how the same interval-and-ray rule behaves on the real line, a finite ordered set, and the rationals.

Let \(X\) be a linearly ordered set, with its order relation written \(<\). For \(a,b\in X\) with \(a<b\), define the open order interval \((a,b)=\{x\in X:a<x<b\}\). Define the endpoint rays \((-\infty,b)=\{x\in X:x<b\}\) and \((a,\infty)=\{x\in X:a<x\}\). The symbols \(-\infty\) and \(\infty\) here are notation for the two directions of the order; they are not assumed to be points of \(X\).

This description gives a way to calculate open sets without first listing every open set. A finite intersection imposes finitely many strict lower bounds and finitely many strict upper bounds. Because the order is linear, the strongest lower bound is the largest of the finitely many lower bounds, and the strongest upper bound is the smallest of the finitely many upper bounds. Thus a nonempty finite intersection is an interval, a ray, or \(X\); an inconsistent collection of bounds gives the empty set.

Proof. Consider a finite intersection of intervals and rays. Collect every lower-bound condition in the form \(x>a_i\) and every upper-bound condition in the form \(x<b_j\). If there are lower bounds, their finite set has a largest member \(a\). All the lower-bound conditions together are then equivalent to \(x>a\). If there are upper bounds, their finite set has a smallest member \(b\), and all the upper-bound conditions are equivalent to \(x<b\). When both kinds of bounds occur, the intersection is \((a,b)\) if \(a<b\), and it is empty if \(a\ge b\). When only upper bounds occur, the intersection is \((-\infty,b)\); when only lower bounds occur, it is \((a,\infty)\). When neither occurs, the intersection is \(X\), including the case of an intersection of no sets. Conversely, each listed interval or ray is itself one of the generating sets, and \(X\) is the empty intersection. The empty set can also arise as an inconsistent intersection; it is open in any topology. \(\square\)

These sets are the basic open sets used to describe neighborhoods. More precisely, if an open set \(U\) contains \(x\), then \(U\) is a union of finite intersections, so at least one of those intersections contains \(x\) and is contained in \(U\). Thus, to find a small open neighborhood of \(x\), it is enough to find an interval or ray containing \(x\) and lying inside the open set.

Basic Neighborhoods on the Real Line

Give \(\mathbb{R}\) its usual order. Its order topology is the standard topology on the real line. The description above says that the basic sets are open intervals and the two kinds of open rays. The rays do not add new open sets: for example, \((-\infty,b)=\bigcup_{n\in\mathbb{N}}(-n,b)\), with the union understood to include sufficiently large \(n\) so that \(-n<b\). Similarly, \((a,\infty)\) is a union of open intervals. Conversely, every standard open neighborhood of a real number contains an open interval around that number.

Proof. Each generating interval \((a,b)\) is standard open. Each ray is standard open as well: if \(x<b\), then the positive number \((b-x)/2\) gives an interval around \(x\) contained in \((-\infty,b)\), and the corresponding argument applies to \((a,\infty)\). Therefore every finite intersection of generators, and every union of such intersections, is standard open. In the other direction, let \(U\) be standard open and let \(x\in U\). By the definition of the standard topology, there is an \(\varepsilon>0\) such that \((x-\varepsilon,x+\varepsilon)\subseteq U\). This interval is one of the order-topology generators, so every point of \(U\) belongs to an order-open set contained in \(U\). Taking the union of these intervals over all \(x\in U\) shows that \(U\) is order-open. The two topologies therefore have the same open sets. \(\square\)

At a point \(x\in\mathbb{R}\), intervals \((a,b)\) with \(a<x<b\) describe neighborhoods on both sides. Rays are also open, but every ray neighborhood of \(x\) contains an interval neighborhood of \(x\). For example, if \(x<b\), choose \(a=x-1\); then \(x\in(a,b)\subseteq(-\infty,b)\) when \(x-1<b\). This explains why intervals alone suffice for local calculations at real points, even though rays are part of the definition.

A Finite Ordered Set

Now let \(X=\{p_1,p_2,p_3,p_4\}\), ordered by \(p_1<p_2<p_3<p_4\), and give \(X\) its order topology. Unlike on the real line, consecutive points have no points between them. That makes it possible for an open interval to consist of a single point.

The same argument applies to any finite linearly ordered set with at least two points: list its elements in increasing order. At an interior point, the interval between its immediate predecessor and successor is its singleton. At the least and greatest points, use the ray bounded by the adjacent element. If the ordered set has just one point, its only subsets are \(\varnothing\) and \(X\), so its order topology is also discrete. This conclusion depends on finiteness: in an order with points between any two distinct points, such intervals need not be singletons.

Basic Neighborhoods in the Rationals

Give \(\mathbb{Q}\) its usual order as a subset of \(\mathbb{R}\), and give it the order topology from that order. The endpoints of its intervals must be rational, since they are elements of the ordered set. For \(a,b\in\mathbb{Q}\) with \(a<b\), the interval is \((a,b)_{\mathbb{Q}}=\{q\in\mathbb{Q}:a<q<b\}\); the subscript emphasizes that only rational points are included.

Every rational point \(q\) has intervals with rational endpoints around it, such as \((q-1,q+1)_{\mathbb{Q}}\). These intervals give a neighborhood base: every open neighborhood of \(q\) contains one of them with possibly smaller rational bounds. To see why the bounds can be chosen rational, first take a finite-intersection basic neighborhood containing \(q\). If it has both a lower and an upper bound, it is already an interval with rational endpoints. If it has just an upper bound \(b>q\), the rational number \((q+b)/2\) lies strictly between \(q\) and \(b\); a rational lower bound such as \(q-1\) then gives a smaller interval. The case of just a lower bound is handled by \((a+q)/2\) and an upper bound \(q+1\). If there are no bounds, use \((q-1,q+1)_{\mathbb{Q}}\).

There are no isolated points in \(\mathbb{Q}\) with this topology. In fact, any basic interval \((a,b)_{\mathbb{Q}}\) containing \(q\) has \(a<q<b\), and \(r=(q+b)/2\) is rational with \(q<r<b\). Hence that interval contains a rational point different from \(q\). A basic upper ray containing \(q\) also contains a point between \(q\) and its upper bound, and a basic lower ray contains a point between its lower bound and \(q\). The whole space contains points other than \(q\) as well. Thus no basic neighborhood containing \(q\) is a singleton.

A Common Mistake: Forgetting Strict Inequalities

Order intervals use strict inequalities. On the four-point chain, the incorrect calculation \((p_1,p_3)=\{p_1,p_2,p_3\}\) treats the interval as if its endpoints were included. The correction is \((p_1,p_3)=\{p_2\}\), because membership requires both \(p_1<x\) and \(x<p_3\). Neither endpoint satisfies both conditions.

Endpoint rays also use strict inequalities. In the same chain, \((-\infty,p_2)=\{p_1\}\), not \(\{p_1,p_2\}\), because \(p_2<p_2\) is false. When calculating any basic open set, write down the strict lower and upper bounds first; then retain exactly the points satisfying all of them.

The interval-and-ray description matters because the order itself determines the local open sets. A finite chain has singleton neighborhoods, while the rational order has intervals containing other rational points. On the real line, the order topology recovers the familiar standard topology. Keeping the strict inequalities and the ordered set in view prevents these different cases from being confused.