# Open Ray is Open in GO-Space

## Theorem

Let $\left({S, \preceq, \tau}\right)$ be a generalized ordered space.

Let $p \in S$.

Then:

$p^\prec$ and $p^\succ$ are $\tau$-open

where:

$p^\prec$ is the strict lower closure of $p$
$p^\succ$ is the strict upper closure of $p$.

## Proof for Definition 1 of GO-Space

We will prove that $U = p^\succ$ is $\tau$-open.

That $p^\prec$ is $\tau$-open will follow by duality.

Let $u \in U$.

Since $p \notin U$, $p \ne u$.

By the definition of GO-space, $\tau$ is Hausdorff, and therefore $T_1$.

Thus by the definition of GO-space, there is an open, convex set $M$ such that $u \in M$ and $p \notin M$.

Next we will show that $M \subseteq U$:

Let $x \in S \setminus U$.

Then $x \preceq p \preceq u$.

Suppose for the sake of contradiction that $x \in M$.

Since $x, u \in M$, $p \in M$ because $M$ is convex, contradicting the choice of $M$.

Thus $x \notin M$.

Since this hold for all $x \in S \setminus U$, $M \subseteq U$.

Thus $U$ contains a neighborhood of each of its points, so it is open.

$\blacksquare$

## Proof for Definition 2 of GO-Space

We will prove that $p^\succ$ is open.

That $p^\prec$ is open will follow by duality.

$\map {\phi^{-1} } {\map \phi p^\succ} = p^\succ$
$\map \phi p^\succ$ is an open ray in $S'$

Therefore $\tau'$-open by the definition of the order topology.

Since $\phi$ is a topological embedding, it is continuous.

Thus $p^\succ$ is $\tau$-open.

$\blacksquare$