Open Rays as a Subbasis for the Real Line

From Open Rays to the Usual Topology

Intervals describe the usual topology on \(\mathbb{R}\), but there is another way to build the same open sets: start with the two kinds of open ray and combine them. A finite intersection of rays can impose both a lower and an upper bound, producing a bounded open interval. Arbitrary unions of the resulting sets then assemble every usual open subset of \(\mathbb{R}\), including sets with many separate components.

Throughout, \(\mathbb{R}\) has its usual topology, whose open sets are arbitrary unions of open intervals. For \(a\in\mathbb{R}\), the two open rays with endpoint \(a\) are \((-\infty,a)=\{x\in\mathbb{R}:x<a\}\) and \((a,\infty)=\{x\in\mathbb{R}:x>a\}\).

For the real line, consider the collection of all open rays \(\mathcal{S}=\{(-\infty,a),(a,\infty):a\in\mathbb{R}\}\). Each ray is open in the usual topology. For example, if \(x<a\), then \(\varepsilon=(a-x)/2>0\), and \((x-\varepsilon,x+\varepsilon)\subseteq(-\infty,a)\). If \(x>a\), then \(\varepsilon=(x-a)/2>0\), and \((x-\varepsilon,x+\varepsilon)\subseteq(a,\infty)\). Thus the rays are legitimate usual open sets. The central question is whether they generate all of them.

What Finite Intersections of Rays Look Like

Proof. Consider a finite collection of upper rays \((-\infty,a_i)\) and lower rays \((b_j,\infty)\). If there is at least one upper ray, let \(u\) be the smallest of its finitely many endpoints. A real number belongs to every upper ray exactly when it is less than \(u\). If there is at least one lower ray, let \(\ell\) be the largest of its finitely many endpoints. A real number belongs to every lower ray exactly when it is greater than \(\ell\). If there are only upper rays, the intersection is \((-\infty,u)\); if there are only lower rays, it is \((\ell,\infty)\). If there are neither, the intersection is the empty intersection \(\mathbb{R}\). If both kinds occur, the intersection is \((\ell,u)\) when \(\ell<u\), and is empty when \(\ell\ge u\), since no real number can be strictly greater than \(\ell\) and strictly less than \(u\) in that case. This proves the classification. Taking one upper ray with endpoint \(b\) and one lower ray with endpoint \(a\), where \(a<b\), gives exactly the points \(x\) satisfying \(a<x<b\). \(\square\)

The strict inequalities matter: the endpoints are not included. If the two bounds coincide, then \((-\infty,a)\cap(a,\infty)=\varnothing\), not \(\{a\}\). Intersections of rays therefore give intervals and rays with open endpoints, as required for the usual topology.

Why Arbitrary Unions Are Needed

Finite intersections create the local pieces, but they do not by themselves produce every open set. The subbasis construction also allows arbitrary unions, including unions with infinitely many terms. That second operation is what lets us combine separated intervals or cover an open set with as many interval pieces as it needs.

Proof. First, let \(\mathcal{B}\) be the collection of finite intersections of members of \(\mathcal{S}\), including the empty intersection \(\mathbb{R}\). The arbitrary unions of members of \(\mathcal{B}\) form a topology: the empty union gives \(\varnothing\), and \(\mathbb{R}\in\mathcal{B}\) gives the whole space. An arbitrary union of arbitrary unions is again a union of members of \(\mathcal{B}\). For two such unions, their intersection is the union of all pairwise intersections of their terms; each pairwise intersection is again a finite intersection of rays and therefore belongs to \(\mathcal{B}\). Thus these unions form a topology containing \(\mathcal{S}\). Any topology containing \(\mathcal{S}\) must contain its finite intersections and arbitrary unions, so it must contain this topology. It is therefore the topology generated by \(\mathcal{S}\).

Every member of \(\mathcal{B}\) is usual open: the proposition shows that it is \(\mathbb{R}\), empty, an open ray, or an open interval, all of which are usual open. Arbitrary unions of usual open sets are usual open, so every set in the generated topology is usual open. For the reverse inclusion, let \(U\) be any usual open subset of \(\mathbb{R}\). By the definition of the usual topology, \(U\) is a union of open intervals. Each bounded open interval \((a,b)\) is the finite intersection \((a,\infty)\cap(-\infty,b)\) of two rays. An unbounded open interval is itself a ray, and \(\mathbb{R}\) is the empty intersection. Thus every interval in the union belongs to the generated topology, and the union \(U\) belongs to it as well. The two topologies contain exactly the same sets. \(\square\)

A Common Mistake: Stopping at Finite Intersections

A tempting but incorrect conclusion is that every open set must be one finite intersection of open rays. The proposition shows why this cannot work for a set with separated components: a finite intersection of rays is always an interval, a ray, \(\mathbb{R}\), or \(\varnothing\). For a concrete example, \((-1,0)\cup(1,2)\) contains both \(-1/2\) and \(3/2\), but does not contain \(0\). Any interval containing \(-1/2\) and \(3/2\) must contain every real number between them, including \(0\). So this union is not a single finite intersection of rays.

The correction is to use the two operations in their proper roles. Finite intersections produce interval pieces, and arbitrary unions combine those pieces: \((-1,0)\cup(1,2)=\bigl((-\infty,0)\cap(-1,\infty)\bigr)\cup\bigl((-\infty,2)\cap(1,\infty)\bigr)\). The first intersection imposes \(-1<x<0\); the second imposes \(1<x<2\). Their union excludes \(0\) and includes both stated components.

This distinction is useful whenever a topology is specified by a subbasis. The subbasic sets are not usually expected to be the whole list of open sets. Instead, finite intersections combine their conditions, and arbitrary unions assemble the resulting pieces. For open rays on \(\mathbb{R}\), the finite step makes intervals and the union step recovers every usual open set.