Equivalence of Definitions of Generalized Ordered Space/Definition 2 implies Definition 1
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Theorem
Let $\struct {S, \preceq, \tau}$ be a generalized ordered space by Definition 2:
$\struct {S, \preceq, \tau}$ is a generalized ordered space if and only if:
- $(1): \quad$ there exists a linearly ordered space $\struct {S', \preceq', \tau'}$
- $(2): \quad$ there exists a mapping $\phi: S \to S'$ such that $\phi$ is both an order embedding and a topological embedding.
Then $\struct {S, \preceq, \tau}$ is a generalized ordered space by Definition 1:
$\struct {S, \preceq, \tau}$ is a generalized ordered space if and only if:
- $(1): \quad \struct {S, \tau}$ is a Hausdorff space
Proof
Let $x \in U \in \tau$.
Then by the definition of topological embedding:
- $\map \phi U$ is an open neighborhood of $\map \phi x$ in $\map \phi S$ with the subspace topology.
Thus by Basis for Topological Subspace and the definition of the order topology, there is an open interval or open ray $I' \in \tau'$ such that:
- $\map \phi x \in I' \cap \map \phi S \subseteq \map \phi U$
Since $I'$ is an interval or ray, it is convex in $S'$ by Interval of Ordered Set is Convex or Ray is Convex, respectively.
Then:
| \(\ds x \in \phi^{-1} \sqbrk {I'}\) | \(=\) | \(\ds \phi^{-1} \sqbrk {I' \cap \phi \sqbrk S}\) | ||||||||||||
| \(\ds \) | \(\subseteq\) | \(\ds \phi^{-1} \sqbrk {\phi \sqbrk U}\) |
Because $\phi$ is a topological embedding, it is injective by definition.
So:
- $\phi^{-1} \sqbrk {\phi \sqbrk U} = U$
Thus:
- $x \in \phi^{-1} \sqbrk {I'} \subseteq U$
By Inverse Image of Convex Set under Monotone Mapping is Convex:
- $\phi^{-1} \sqbrk {I'}$ is convex.
Thus $\tau$ has a basis consisting of convex sets.
$\blacksquare$