Equivalence of Definitions of Generalized Ordered Space/Definition 1 implies Definition 3

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Theorem

Let $\struct {S, \preceq, \tau}$ be 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 $S$.


Then $\struct {S, \preceq, \tau}$ is a generalized ordered space by 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

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 Set and Lower Closure is Lower Set, the elements of $\SS$ are upper and lower sets.


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 $\SS$.

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

$\blacksquare$