2.3Order Topology

Updated 24 May 2026Chapter (PDF)

If a set is totally ordered, then the ordering actually induces a canonical topology on $X$ called the order topology. We investigate its construction and behavior in this section.

First, considering open/closed/half-open intervals in $\bR$ , we may define them equivalently in an arbitrary totally ordered set, using its order in place of $\bR$ 's standard ordering. We name them equivalently, suggesting that open intervals in a totally ordered set should be open with respect to the order topology. This is true.

[2.3.1]Definition(Order Topology)#

Suppose $(X, <)$ is a totally ordered set. Let $\mathcal{B}$ be the collection of all sets of the following forms.

All open intervals $(a, b) \subseteq X$ , for every $a, b \in X$ .
The half-open interval $[a_0, b) \subseteq X$ , for some minimum element (if any) $a_0 \in X$ and any $b \in X$ .
The half-open interval $(a, b_0] \subseteq X$ , for any $a \in X$ and some maximum element (if any) $b_0 \in X$ .

Then $\mathcal{B}$ forms a basis for a topology on $X$ called the ordered topology.

Proof.

We show $\mathcal{B}$ is, indeed, a basis. First, note that every element of $X$ lies in one of the three prescribed sets: the smallest lies in the upper-open set, the largest lies in the lower-open set, and all other elements lie in an open interval—thus $\mathcal{B}$ covers $X$ as promised. With a long argument by cases, it follows that the intersection of two basis elements can be locally refined.

Note the standard topology on $\bR$ is simply the topology induced by the standard order. Note the order topology on $\bZ_+$ is simply the discrete topology, since every subset of $\bZ_+$ falls into an interval (the singleton $\{n\} = (n - 1, n + 1)$ ).

[2.3.2]Definition(Rays)#

If $X$ is a totally ordered set, an element $a \in X$ determines four rays. They are $(a, \infty), [a, \infty), (-\infty, a), (-\infty, a]$ .

[2.3.3]Remark(Open Rays are Open)#

The rays of form $(a, \infty)$ and $(-\infty, a)$ are called open rays. Its name is accurate—this interval forms an open set with respect to the order topology. For a set $X$ whose maximum $b_0$ , $(a, \infty) = (a, b_0]$ ; if no maximum exists, then $(a, \infty)$ is simply the union of all basis elements $(a, x)$ for $x: x > a$ . Similar argumentation follows for $(-\infty, a)$ .

Please note that, for a totally ordered set $X$ with maximum (resp. minimum) $b_0$ (resp. $a_0$ ), then $(a, \infty) = (a, b_0]$ (resp. $(-\infty, a) = [a_0, b)$ ).

[2.3.4]Theorem(Open Rays Form Subbasis for Order Topology)#

Suppose $X$ is a totally ordered set. Then the set

S = \{(-\infty, a): a \in X\} \cup \{(a, \infty): a \in X\}

forms a subbasis for the order topology on $X$ .

Proof.

It should follow immediately that every basis element can be generated by a finite intersection of one or two open rays.