Upper Section with no Smallest Element is Open in GO-Space

Theorem
Let $\left({S, \preceq, \tau}\right)$ be a generalized ordered space.

Let $U$ be an upper set in $S$ with no smallest element.

Then $U$ is open in $\left({S, \preceq, \tau}\right)$.

Proof
By Minimal Element in Toset is Unique and Smallest, $U$ has no minimal element.

By Upper Set with no Minimal Element:
 * $U = \bigcup \left\{{ {\dot\uparrow} u: u \in U }\right\}$

where ${\dot\uparrow}u$ is the strict up-set of $u$.

By Open Ray is Open in GO-Space and the fact that a union of open sets is open, $U$ is open.

Also see

 * Lower Set with no Greatest Element is Open in GO-Space