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

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

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

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

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

By Lower Set with no Maximal Element:
 * $\displaystyle L = \bigcup \left\{{l^\prec: l \in L }\right\}$

where $l^\prec$ is the strict lower closure 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