GO-Space Embeds Densely into Linearly Ordered Space

Theorem
Let $(Y, \preceq, \tau)$ be a GO-space by definition 3.

That is, let $\tau$ have a sub-basis consisting of open rays.

Then $(Y, \preceq, \tau)$ is a GO-space by definition 2.

That is, there is a linearly ordered space $(X, \preceq', \tau')$ and a mapping from $Y$ to $X$ which is a order embedding and a topological embedding.

Proof
Let $X$ be the disjoint union of $Y$ with the set of all lower sets $L$ in $Y$ such that $L$ and $Y \setminus L$ are open and nonempty and either:
 * $L$ has a maximum, and $Y\setminus L$ does not have a minimum, or
 * $Y \setminus L$ has a minimum, and $L$ does not have a maximum.

In the following, let $y$, $y_1$, $y_2$, and $y_3$ represent elements of $Y$, and let $L$, $L_1$, $L_2$, and $L_3$ represent lower sets of $Y$ that are members of $X$.

Define a relation $\preceq'$ extending $\preceq$ by letting:
 * $y_1 \preceq' y_2 \iff y_1 \preceq y_2$
 * $y \preceq' L \iff y \in L$
 * $L_1 \preceq' L_2 \iff L_1 \subseteq L_2$
 * $L \preceq' y \iff y \in Y \setminus L$

Lemma
$\preceq'$ is an ordering.

Proof
First note that by Lower Sets in Totally Ordered Set form Nest:
 * $\subseteq$ is a total ordering on the set of lower sets.

$\preceq'$ is reflexive:

This follows immediately from the fact that $\preceq$ and $\subseteq$ are reflexive.

$\preceq'$ is transitive:

There are eight possibilities to consider.

If $y_1 \preceq' y_2$ and $y_2 \preceq' y_3$, then $y_1 \preceq' y_3$ because $\preceq$ is transitive.

If $L_1 \preceq' L_2$ and $L_2 \preceq' L_3$, then $L_1 \preceq' L_3$ because $\subseteq$ is transitive.

If $y_1 \preceq' y_2$ and $y_2 \preceq' L$, then: $y_1 \preceq y_2$ and $y_2 \in L$

Since $L$ is a lower set in $Y$: $y_1 \in L$, so $y_1 \preceq' L$.

If $y_1 \preceq' L$ and $L \preceq' y_2$ then
 * $y_1 \in L$ and $y_2 \in Y \setminus L$.

Since $L$ is a lower set, $y_2 \not\preceq y_1$.

Since $\preceq$ is a total ordering, $y_1 \preceq y_2$, so:
 * $y_1 \preceq' y_2$

If $L \preceq' y_1$ and $y_1 \preceq' y_2$ then again....

If $y \preceq' L_1$ and $L_1 \preceq' L_2$, ...

If $L_1 \preceq' y$ and $y \preceq' L_2$, then $y \in L_2$ but $y \notin L_1$, so $L_1 \preceq' L_2$.

If $L_1 \preceq' L_2$ and $L_2 \preceq' y$, then $y \notin L_2$ and $L_2 \supseteq L_1$, so $y \notin L_1$, so $L_1 \preceq' y$.

$\preceq'$ is antisymmetric:

There are three cases. Two follow immediately from antisymmetry of $\preceq$ and $\subseteq$.

But by the definition of $\preceq'$, it's impossible for $y \preceq' L$ and $L \preceq' y$.

$\preceq'$ is clearly total. Thus it is a total ordering of $X$. Furthermore, it obviously induces $\preceq$ on $Y$.

Next we show that the $\preceq'$ order topology induces $\tau$ as the subspace topology.

Open rays from elements of $Y$ are $\tau$-open by Open Ray is Open in GO-Space.

${\downarrow'}L \cap Y = L$, which is open.

${\uparrow'}L \cap Y = Y \setminus L$, which is open.

Thus the induced topology is coarser than $\tau$.

Let $U$ be an open upper set in $Y$ with $\varnothing \subsetneqq U \subsetneqq Y$.

If $U$ has no minimum, then it is a union of open rays, so it is open in the subspace topology.

If $Y \setminus U$ has a maximum $p$, then $U = {\uparrow}p = Y \cap {\uparrow'}p$.

Otherwise, $Y \setminus U$ is open (EXPLAIN) so $Y \setminus U \in X$.

Then $U = Y \cap {\uparrow'}(Y \setminus U)$.

A similar argument works for lower sets.

Density: