# 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:

$\struct {S, \preceq, \tau}$ is a generalized ordered space if and only if:

$(1): \quad$ there exists a linearly ordered space $\struct {S', \preceq', \tau'}$
$(2): \quad$ there exists a mapping $\phi: S \to S'$ such that $\phi$ is both an order embedding and a topological embedding.

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

$\struct {S, \preceq, \tau}$ is a generalized ordered space if and only if:

$(1): \quad \struct {S, \tau}$ is a Hausdorff space
$(2): \quad$ there exists a basis for $\struct {S, \tau}$ whose elements are convex in $S$.

## 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:

 $\ds x \in \phi^{-1} \sqbrk {I'}$ $=$ $\ds \phi^{-1} \sqbrk {I' \cap \phi \sqbrk S}$ $\ds$ $\subseteq$ $\ds \phi^{-1} \sqbrk {\phi \sqbrk U}$

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

So:

$\phi^{-1} \sqbrk {\phi \sqbrk U} = U$

Thus:

$x \in \phi^{-1} \sqbrk {I'} \subseteq U$
$\phi^{-1} \sqbrk {I'}$ is convex.

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

$\blacksquare$