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

## Theorem

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

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

## Proof

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.

$\blacksquare$