# 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:

$\left({S, \preceq, \tau}\right)$ 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 $\struct {S, \preceq, \tau}$ 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

## 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:

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

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.

$\blacksquare$