# 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:

$\left({S, \preceq, \tau}\right)$ is a generalized ordered space if and only if:

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

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

$\left({S, \preceq, \tau}\right)$ is a generalized ordered space if and only if:

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

## 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 of Ordered Set is Convex or Ray is Convex, respectively.

Then:

 $\displaystyle x \in \phi^{-1} \left({I'}\right)$ $=$ $\displaystyle \phi^{-1}\left({I' \cap \phi \left({S}\right)}\right)$ $\displaystyle$ $\subseteq$ $\displaystyle \phi^{-1}\left({\phi \left({U}\right)}\right)$

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

So:

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

Thus:

$x \in \phi^{-1} \left({I'}\right) \subseteq U$
$\phi^{-1}\left({I'}\right)$ is convex.

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

$\blacksquare$