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

Theorem
Let $\struct {S, \preceq, \tau}$ be a generalized ordered space by Definition 2:

Then $\struct {S, \preceq, \tau}$ is a generalized ordered space by Definition 1:

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

Then by the definition of topological embedding:
 * $\map \phi U$ is an open neighborhood of $\map \phi x$ in $\map \phi S$ 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:
 * $\map \phi x \in I' \cap \map \phi S \subseteq \map \phi U$

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

Then:

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

So:
 * $\phi^{-1} \sqbrk {\phi \sqbrk U} = U$

Thus:
 * $x \in \phi^{-1} \sqbrk {I'} \subseteq U$

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

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