# Equivalence of Definitions of Generalized Ordered Space

## Theorem

Let $\struct {S, \preceq}$ be a totally ordered set.

Let $\tau$ be a topology for $S$.

The following definitions of the concept of Generalized Ordered Space are equivalent:

### 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 $S$.

### 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.

### Definition 3

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

$(1): \quad \struct {S, \tau}$ is a Hausdorff space
$(2): \quad$ there exists a sub-basis for $\struct {S, \tau}$ each of whose elements is an upper section or lower section in $S$.

## Proof

### Definition $(1)$ implies Definition $(3)$

Let $\BB$ be a basis for $\tau$ consisting of convex sets.

Let:

$\SS = \set {U^\succeq: U \in \BB} \cup \set {U^\preceq: U \in \BB}$

where $U^\succeq$ and $U^\preceq$ denote the upper closure and lower closure respectively of $U$.

By Upper Closure is Upper Section and Lower Closure is Lower Section, the elements of $\SS$ are upper and lower sections.

It is to be shown that $\SS$ is a sub-basis for $\tau$.

$\SS \subseteq \tau$

By Convex Set Characterization (Order Theory), each element of $\BB$ is the intersection of its upper closure with its lower closure.

Thus each element of $\BB$ is generated by $\SS$.

Thus $\SS$ is a sub-basis for $\tau$.

$\blacksquare$

### Definition $(3)$ implies Definition $(1)$

Let $\SS$ be a sub-basis for $\tau$ consisting of upper sections and lower sections.

Let $\BB$ be the set of intersections of finite subsets of $\SS$.

the elements of $\BB$ are convex.

But $\BB$ is a basis for $\tau$.

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

$\blacksquare$

### Definition $(2)$ implies Definition $(1)$

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.

$\blacksquare$

### Definition $(3)$ implies Definition $(2)$

This follows from GO-Space Embeds Densely into Linearly Ordered Space.

$\blacksquare$