GO-Space Embeds as Closed Subspace of Linearly Ordered Space

Theorem
Let $(X, \preceq_X, \tau_X)$ be a generalized ordered space.

Then there is a linearly ordered space $(Y, \preceq_Y, \tau_Y)$ and a mapping $\phi: X \to Y$ such that $\phi$ is a topological embedding and an order embedding, and $\phi(X)$ is closed in $Y$.

Proof
By GO-Space Embeds Densely into Linearly Ordered Space, there is a linearly ordered space $(W, \preceq_W, \tau_W)$ and a mapping $\psi:X \to W$ which is an order embedding and a topological embedding.

Assume without loss of generality that $X$ is a subspace of $W$.

Let $Y = \left\{{ (x, 0): x \in X }\right\} \cup (W \setminus X) \times \Z$.

Let $\preceq_Y$ be the restriction to $Y$ of the lexicographic ordering on $W \times \Z$.

Let $\tau_Y$ be the $\preceq_Y$-order topology on $Y$.

Let $\phi:X \to Y$ be given by $\phi(x) = (x,0)$.

$\phi$ is clearly an order embedding.

Next, we show that it is a topological embedding:

If $$

Finally, we show that $\phi(X)$ is closed in $Y$:

Remark
The set of integers used in the proof above was chosen because it is the simplest non-empty, totally ordered set with no maximum and no minimum. However, any such set will do. Some may find the space easier to visualize if they substitute the real interval $(-1 ,\,.\,.\, 1)$ for $\Z$ and make the necessary (minor) adjustments to the proof.