Equivalence of Definitions of Generalized Ordered Space/Definition 3 implies Definition 1

Theorem
Let $\left({S, \preceq, \tau}\right)$ be a generalized ordered space by Definition 3:

Then $\left({S, \preceq, \tau}\right)$ is a generalized ordered space by Definition 1:

Proof
 Let $\mathcal S$ be a sub-basis for $\tau$ consisting of upper sets and lower sets.

Let $\mathcal B$ be the set of intersections of finite subsets of $\mathcal S$.

By Upper Set is Convex, Lower Set is Convex and Intersection of Convex Sets is Convex Set (Order Theory) :
 * the elements of $\mathcal B$ are convex.

But $\mathcal B$ is a basis for $\tau$.

Therefore $\tau$ has a basis consisting of convex sets.