Open Ray is Open in GO-Space/Definition 1

Theorem
Let $(X, \prec, \tau)$ be a Generalized Ordered Space/Definition 1.

Let $p \in X$.

Then ${\downarrow}p$ and ${\uparrow}p$ are $\tau$-open.

Proof
We will prove that $U = {\uparrow} p$ is open.

That ${\downarrow}p$ is open will follow by duality.

Let $u \in U$.

Since $p \notin U$, $p \ne u$.

By the definition of GO-space, $\tau$ is Hausdorff, and therefore $\mathrm T_1$.

Thus by the definition of GO-space, there is an open, convex set $M$ such that $u \in M$ and $p \notin M$.

$M \subseteq U$:

Let $x \in X \setminus U$.

Then $x \preceq p$.

If $x$ were in $M$, then $p$ would be in $U$ because Strict Upper Closure is Convex, a contradiction,

Thus $x \notin M$.

Since this hold for all $x \in X \setminus U$, $M \subseteq U$.

Thus $U$ contains a neighborhood of each of its points, so it is open.