Verifying the Order-Topology Basis

Intervals and Rays as Basic Open Sets

An order topology is built from the order relation: its neighborhoods come from intervals and rays. To verify that these sets really provide a basis, we must check two separate requirements. They must cover every point, and whenever two of them overlap, their overlap must contain another basic set around each point in it. The coverage check deserves care at a first or last element, where an interval with points on both sides may not be available.

Let \(L\) be a linearly ordered set, with order relation \(<\). The order topology on \(L\) is the topology generated by the open rays \((-\infty,b)_L=\{x\in L:x<b\}\) and \((a,\infty)_L=\{x\in L:a<x\}\), for \(a,b\in L\). Here \(-\infty\) and \(+\infty\) are labels for unbounded sides, not elements of \(L\). For \(a,b\in L\), write \((a,b)_L=\{x\in L:a<x<b\}\). When the ordered set is understood, we omit the subscript \(L\).

Consider the family consisting of all open intervals \((a,b)_L\), all open rays \((-\infty,b)_L\) and \((a,\infty)_L\), and \(L\) itself. Empty intervals and rays are allowed in this family; they do not help cover points, but their presence causes no difficulty. Including \(L\) handles the empty and one-element ordered sets as well as the nontrivial cases.

The Basis Verification

Proof. First consider coverage. If \(L=\varnothing\), there are no points to cover. If \(L\) has one element, \(L\) itself covers that element. Now suppose \(L\) has at least two elements, and let \(x\in L\). If there are elements \(a<x\) and \(b>x\), then \(x\in(a,b)_L\). If there is no element below \(x\), then \(x\) is a first element. Since \(L\) has at least two elements, some \(b>x\), and \(x\in(-\infty,b)_L\). If there is no element above \(x\), then \(x\) is a last element. There is some \(a<x\), and \(x\in(a,\infty)_L\). These cases cover every point: a point that is neither first nor last has elements on both sides, while in a set with at least two elements no point can be both first and last.

For the intersection condition, each family member is described by a lower bound and an upper bound, either of which may be absent. For example, \((a,b)_L\) has lower bound \(a\) and upper bound \(b\), while \((-\infty,b)_L\) has no lower bound. Suppose \(x\in B_1\cap B_2\). Among the finite lower bounds of \(B_1\) and \(B_2\), take the greater one if there are any; call it \(a\). If there are none, leave the lower bound absent. Among their finite upper bounds, take the lesser one if there are any; call it \(b\). If there are none, leave the upper bound absent. The resulting interval or ray, or \(L\) if neither bound exists, is a member \(B_3\) of the family. Because \(x\in B_1\cap B_2\), every finite lower bound is less than \(x\), and every finite upper bound is greater than \(x\). Thus \(x\in B_3\). The chosen bounds are at least as restrictive as the bounds from each original set, so \(B_3\subseteq B_1\cap B_2\). This proves the basis-intersection condition.

It remains to check that this is a basis for the order topology, not merely a basis for some topology. Each ray is one of the sets used to generate the order topology. Each interval satisfies \((a,b)_L=(-\infty,b)_L\cap(a,\infty)_L\), so it is open in that topology; \(L\) is open as well. Therefore every union of family members is open in the order topology. Conversely, every generating ray belongs to the family, so every open set in the order topology, which is generated by those rays using finite intersections and arbitrary unions, is a union of family members. The two topologies agree. \(\square\)

The two basis conditions have different jobs. Coverage guarantees that every point has at least one basic neighborhood. The intersection condition guarantees that if a neighborhood is required to lie inside two basic open sets at once, a basic neighborhood can still be found there.

Worked Checks at Endpoints and in Finite Orders

Why the Intersection Check Matters

A common mistake is to combine interval endpoints in the wrong direction, or to assume that the intersection of two basic sets is itself a nonempty basic set. For example, in \(\mathbb{R}\), the incorrect calculation \((0,2)\cap(1,3)=(0,3)\) uses the smaller lower endpoint and the larger upper endpoint. That gives a set larger than either original interval, not their intersection. The correct calculation takes the greater lower endpoint and the lesser upper endpoint:

More generally, an intersection can be empty. For instance, \((0,1)\cap(2,3)=\varnothing\) in \(\mathbb{R}\). The basis condition does not require a basic set inside an empty intersection: it applies only when a point \(x\) belongs to the intersection. When there is such a point, the maximum-lower-bound and minimum-upper-bound construction gives the required neighborhood around that point.

Endpoint rays are not optional decorations. If one tried to use only intervals with two endpoints from \(L\), a first element would not belong to any such interval, because no lower endpoint could be chosen below it. The ray \((-\infty,b)_L\) handles a first element, and \((a,\infty)_L\) handles a last element. If \(L\) has no first or last element, each point has elements on both sides and an ordinary interval suffices. In a one-element order, no such interval or ray covers the point, which is why the basis family also includes \(L\) itself.