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

## Theorem

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

$\blacksquare$