Lower Section with no Greatest Element is Open in GO-Space

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

Let $L$ be a lower set in $S$ with no greatest element.

Then $L$ is open in $(S, \preceq, \tau)$.

Proof
By Maximal Element in Toset is Unique and Greatest, $L$ has no maximal element.

By Lower Set with no Maximal Element:


 * $L = \bigcup \left\{{ {\dot\downarrow}l: l \in L }\right\}$

where ${\dot\downarrow}u$ is the strict down-set of $l$.

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

Also see

 * Upper Set with no Smallest Element is Open in GO-Space