GO-Space Embeds Densely into Linearly Ordered Space

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

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

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

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

$\preceq'$ is an ordering:

Note that $\subseteq$ is total on the set of lower sets.

Reflexive: obvious

Transitive:

There are eight possibilities to consider.

Two follow immediately from transitivity of $\preceq$ and $\subseteq$.

If $y_1 \preceq' y_2$ and $y_2 \preceq' L$, then $y_1 \preceq L$ because $L$ is a lower set.

If $y_1 \preceq' L$ and $L \preceq' y_2$ then $y_1 \preceq' y_2$ because $L$ is a lower set.

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 $L_1 \preceq' y$.