<section  notitle="True" name="proof"> Let $X$ be the disjoint union of $Y$ with the set of all lower sets $L$ 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 sets 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 Sets 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 $\left({Y, \tau}\right)$ into $\left({X, \tau'}\right)$ === 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 set in $Y$ with $\varnothing \subsetneqq U \subsetneqq Y$. Let $U$ have no minimum. Then by Upper Set 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 Set with no Greatest Element is Open in GO-Space, $Y \setminus U$ is open. Therefore $Y \setminus U \in X$. Then: :$U = Y \cap \left({Y \setminus U}\right)^{\succ'}$ A similar argument works for open lower sets. Thus it follows that the subspace topology is finer than $\tau$. Thus they are equal by definition of set equality. {{qed|lemma}} === $Y$ 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: :$q \in Y \cap \left({x_1 \,.\,.\, x_2}\right)$ 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: :$q \in Y \cap \left({x_1 \,.\,.\, x_2}\right)$ 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 $\left({Y, \preceq, \tau}\right)$ as a dense subspace of $\left({X, \preceq', \tau'}\right)$. {{qed}} </section> Category:Generalized Ordered Spaces Category:Linearly Ordered Spaces hn4l61n82lai0amm0ngzstkocscdrn5