Equivalence of Definitions of Order Topology

Proof
By definition, an open ray of $S$ is defined as a subset of $S$ of one of the two forms:


 * An upward-pointing ray:
 * $\left\{{x \in S: s \prec x}\right\}$ (the strict upper closure of $s$), denoted $s^\succ$


 * A downward-pointing ray:
 * $\left\{{x \in S: x \prec s}\right\}$ (the strict lower closure of $s$), denoted $s^\prec$

from a given point $s \in S$.

Thus the set of open rays of $S$ can be expressed as:
 * the set of all subsets of $S$ of the form:
 * $\left\{ {s^\succ: s \in S}\right\}$

and:
 * the set of all subsets of $S$ of the form:
 * $\left\{ {s^\prec: s \in S}\right\}$

This is precisely the set ${\Uparrow} \left({S}\right) \cup {\Downarrow} \left({S}\right)$, where:


 * ${\Uparrow} \left({S}\right) = \left\{{s^\succ: s \in S}\right\}$
 * ${\Downarrow} \left({S}\right) = \left\{{s^\prec: s \in S}\right\}$

Hence the result, by definition of the topology on $S$ generated by a sub-basis of $S$.