# 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

### 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 $\tau$.

By Upper and Lower Closures of Open Set in GO-Space are Open:

- $\SS \subseteq \tau$

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 $\SS$.

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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 $\SS$.

By Upper Section is Convex, Lower Section is Convex and Intersection of Convex Sets is Convex Set (Order Theory) :

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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:

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

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.

$\blacksquare$

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

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

$\blacksquare$