Equivalence of Definitions of Generalized Ordered Space/Definition 2 implies Definition 1

Theorem
Let $\left({S, \preceq, \tau}\right)$ be a generalized ordered space by Definition 2:

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

Proof
Let $x \in U \in \tau$.

Then by the definition of topological embedding:
 * $\phi \left({U}\right)$ is an open neighborhood of $\phi \left({x}\right)$ in $\phi \left({S}\right)$ with the subspace topology.

Thus by Basis for Topological Subspace and the definition of the order topology, there is an open interval or open ray $I' \in \tau'$ such that:
 * $\phi \left({x}\right) \in I' \cap \phi \left({S}\right) \subseteq \phi \left({U}\right)$

Since $I'$ is an interval or ray, it is convex in $S'$ by Interval is Convex or Ray is Convex, respectively.

Then:

Since $\phi$ is a topological embedding, it is injective by definition.

So:
 * $\phi^{-1} \left({\phi \left({U}\right)}\right) = U$

Thus:
 * $x \in \phi^{-1} \left({I'}\right) \subseteq U$

By Inverse Image of Convex Set under Monotone Mapping is Convex:
 * $\phi^{-1}\left({I'}\right)$ is convex.

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