GO-Space Embeds as Closed Subspace of Linearly Ordered Space

Theorem
Let $\struct {X, \preceq_X, \tau_X}$ be a generalized ordered space.

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

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

, assume $X$ is a subspace of $W$.

Let $Y = \set { \tuple {x, 0}: x \in X } \cup \paren {W \setminus X} \times \Z$.

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 $\map \phi x = \tuple {x, 0}$.

$\phi$ is clearly an order embedding.

Next, we show that it is a topological embedding:

If $$

Finally, we show that $\phi \sqbrk {X}$ 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.