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

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

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.

Definition 3

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

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

Proof

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

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

Let:

$\mathcal S = \left\{ {U^\succeq: U \in \mathcal B}\right\} \cup \left\{ {U^\preceq: U \in \mathcal B}\right\}$

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

By Upper Closure is Upper Set and Lower Closure is Lower Set, the elements of $\mathcal S$ are upper and lower sets.

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

$\mathcal S \subseteq \tau$

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

Thus each element of $\mathcal B$ is generated by $\mathcal S$.

Thus $\mathcal S$ is a sub-basis for $\tau$.

$\blacksquare$

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

Let $\mathcal S$ be a sub-basis for $\tau$ consisting of upper sets and lower sets.

Let $\mathcal B$ be the set of intersections of finite subsets of $\mathcal S$.

the elements of $\mathcal B$ are convex.

But $\mathcal B$ 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:

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

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.

$\blacksquare$