GO-Space Embeds Densely into Linearly Ordered Space

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Theorem

Let $\struct {Y, \preceq, \tau}$ be a generalized ordered space (GO-space) by Definition 3.

That is:

let $\struct {Y, \tau}$ be a Hausdorff space

and:

let $\tau$ have a sub-basis consisting of upper sections and lower sections relative to $\preceq$.

Then $\struct {Y, \preceq, \tau}$ is a GO-space by Definition 2.

That is, there is a linearly ordered space $\struct {X, \preceq', \tau'}$ and a mapping from $Y$ to $X$ which is a order embedding and a topological embedding.

Proof

Let $X$ be the disjoint union of $Y$ with the set of all lower sections $L$ in $Y$ such that $L$ and $Y \setminus L$ are open and nonempty and either:

$L$ has a maximum, and $Y\setminus L$ does not have a minimum, or

$Y \setminus L$ has a minimum, and $L$ does not have a maximum.

Let $\phi: Y \to X$ be the inclusion mapping.

In the following, let $y$, $y_1$, $y_2$, and $y_3$ represent elements of $Y$, and let $L$, $L_1$, $L_2$, and $L_3$ represent lower sections of $Y$ that are members of $X$.

Define a relation $\preceq'$ extending $\preceq$ by letting:

$y_1 \preceq' y_2 \iff y_1 \preceq y_2$

$y \preceq' L \iff y \in L$

$L_1 \preceq' L_2 \iff L_1 \subseteq L_2$

$L \preceq' y \iff y \in Y \setminus L$

By Union of Total Ordering with Lower Sections is Total Ordering and Restriction of Total Ordering is Total Ordering, $\preceq'$ is a total ordering.

Because $y_1 \preceq' y_2$ iff $y_1 \preceq y_2$, $\phi$ is an order embedding.

Let $\tau'$ be the $\preceq'$ order topology on $X$.

$\phi$ is a topological embedding of $\struct {Y, \tau}$ into $\struct {X, \tau'}$

Let $L^\succeq$ and $L^\succ$ represent the upper closure and strict upper closure of $L$ with respect to $\preceq$.

Let $L^{\succeq'}$ and $L^{\succ'}$ represent the upper closure and strict upper closure with respect to $\preceq'$.

Let $L^\preceq$ and $L^\prec$ represent the lower closure and strict lower closure of $L$ with respect to $\preceq$.

Let $L^{\preceq'}$ and $L^{\prec'}$ represent the lower closure and strict lower closure with respect to $\preceq'$.

Open rays from elements of $Y$ are $\tau$-open by Open Ray is Open in GO-Space.

$L^{\preceq'} \cap Y = L$, which is open.

$L^{\succeq'} \cap Y = Y \setminus L$, which is open.

Thus the subspace topology is coarser than $\tau$.

Let $U$ be an open upper section in $Y$ with $\O \subsetneqq U \subsetneqq Y$.

Let $U$ have no minimum.

Then by Upper Section with no Smallest Element is Open in GO-Space, $U$ is open in the subspace topology.

Let $Y \setminus U$ has a maximum $p$.

Then $U = p^\succ = Y \cap p^{\succ'}$, which is open in the subspace topology.

Otherwise, by Lower Section with no Greatest Element is Open in GO-Space, $Y \setminus U$ is open.

Therefore $Y \setminus U \in X$.

Then:

$U = Y \cap \paren {Y \setminus U}^{\succ'}$

A similar argument works for open lower sections.

Thus it follows that the subspace topology is finer than $\tau$.

Thus they are equal by definition of set equality.

$\Box$

$Y$ is dense in $X$

Let $L \in X \setminus Y$.

By the definition of $X$, $L$ and $Y \setminus L$ are non-empty.

So:

$L$ is $\preceq'$-preceded by at least one element of $Y$

$L$ is $\preceq'$-succeeded by at least one element of $Y$.

Thus every open ray in $X$ containing $L$ contains an element of $Y$.

Let $x_1, x_2 \in X$ such that $x_1 \prec' L \prec' x_2$.

By the definition of $X$, either:

$L$ has a $\preceq$-greatest element and $Y \setminus L$ has no $\preceq$-smallest element

or:

$L$ has no $\preceq$-greatest element and $Y \setminus L$ has a $\preceq$-smallest element.

Suppose that $L$ has a $\preceq$-greatest element and $Y \setminus L$ has no $\preceq$-smallest element.

Let $x_2 \in Y$.

Then $x_2 \in Y \setminus L$.

Since $Y \setminus L$ has no $\preceq$-smallest element, there is an element $q\in Y \setminus L$ such that $q \prec x_2$.

Then $x_1 \prec L \prec' q \prec' x_2$.

Therefore:

$q \in Y \cap \openint {x_1} {x_2}$

On the other hand, let $x_2 \notin Y$.

Then $L \subsetneqq x_2$.

Then there is some $q \in x_2 \setminus L$.

Therefore:

$x_1 \prec' L \prec' q \prec' x_2$

In particular:

$q \in Y \cap \openint {x_1} {x_2}$

If on the other hand we had supposed that $L$ has no $\preceq$-greatest element and $Y \setminus L$ has a $\preceq$-smallest element, we could obtain similar results, with $x_1$ taking on the role we have given $x_2$.

Thus in any case, $L$ is an adherent point of $Y$.

Since every element of $X \setminus Y$ is an adherent point of $Y$, $Y$ is dense in $X$.

Thus the inclusion map from $Y$ to $X$ is a topological embedding and an order embedding of $\struct {Y, \preceq, \tau}$ as a dense subspace of $\struct {X, \preceq', \tau'}$.

$\blacksquare$