[Munkres Topology] 14. The Order Topology

 This article is one of the posts in the Textbook Commentary Project.

If $X$ is a simply ordered set, there is a standard topology for $X$, defined using the order relation. It is called the order topology; in this section, we consider it and study some of its properties.

Suppose that $X$ is a set having a simple order relation $<$. Given elements $a$ and $b$ of $X$ such that $a<b$, there are four subsets of $X$ that are called the intervals determined by $a$ and $b$. They are the following: $$\begin{align*}(a,b)&=& \{x| a<x<b\} ,\\ (a,b] &=& \{ x| a< x \le b\} ,\\ [a,b) &=& \{ x| a\le x < b \} ,\\ [a,b] &=& \{ x| a \le x \le b \}.\end{align*}$$

The notation used here is familiar to you already in the case where $X$ is the real line, but these are intervals in an arbitrary ordered set. A set of the first type is called an open interval in $X$, a set of the last type is called a closed interval in $X$, and sets of the second and third types are called half-open intervals. The use of the term 'open' in this connection suggests that open intervals in $X$ should turn out to be open sets when we put a topology on $X$. And so they will.

Definition. Let $X$ be a set with a simple order relation; assume $X$ has more than one element. Let $\mathcal{B}$ be the collection of all sets of the following types:

(1) All open intervals $(a,b)$ in $X$.

(2) All intervals of the form $[a_0,b)$, where $a_0$ is the smallest element (if any) of $X$.

(3) All intervals of the form $(a,b_0]$, where $b_0$ is the largest element (if any) of $X$.

The collection $\mathcal{B}$ is a basis for a topology on $X$, which is called the order topology.

(If $X$ has no smallest element, there are no sets of type (2), and if $X$ has no largest element, there are no sets of type (3).)

One has to check that $\mathcal{B}$ satisfies the requirements for a basis. First, note that every element $x$ of $X$ lies in at least one element of $\mathcal{B}$: The smallest element (if any) lies in all sets of type (2), the largest element (if any) lies in all sets of type (3), and every other element lies in a set of type (1). Second, note that the intersection of any two sets of the preceding types is again a set of these types, or is empty. Other several cases need to be checked.

Definition. If $X$ is an ordered set, and $a$ is an element of $X$, there are four subsets of $X$ that are called the rays determined by $a$. They are the following: $$\begin{align*}(a,+\infty)&=& \{ x| x >a\},\\ (-\infty, a) &=& \{ x| x < a \},\\ [a, +\infty ) &=& \{x | x \ge a \} ,\\ (-\infty, a] &=& \{x | x \le a \}.\end{align*}$$ Sets of the first two types are called open rays, and sets of the last two types are called closed rays.

The use of the term 'open' suggests that open rays in $X$ are open sets in the order topology. And so they are. Consider, for example, the ray $(a,+\infty)$. If $X$ has a largest element $b_0$, then $(a,+\infty)$ equals the basis element $(a,b_0]$. If $X$ has no largest element, then $(a,+\infty)$ euqals the union of all basis elements of the form $(a,x)$, for $x>a$. In either case, $(a,+\infty)$ is open. A similar argument applies to the ray $(-\infty,a)$.

The open rays, in factm form a subbasis for the order topology on $X$, as we now show. Because the open rays are open in the order topology, the topology they generate is contained in the order topology. On the other hand, every basis element for the order topology equals a finite intersection of open rays; the interval $(a,b)$ equals the intersection of $(-\infty,b)$ and $(a,+\infty)$, while $[a_0,b)$ and $(a,b_0]$m if they exist, are themselves open rays. Hence the topology generated by the open rays contains the order topology.