GO-Space Embeds as Closed Subspace of Linearly Ordered Space
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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.
Without loss of generality, 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 $Y$:
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