Every simply ordered set imbeds into a linear continuum? I was thinking recently about a seemingly simple way to construct, for any order topology, a linear continuum into which it imbeds. I have read about a concept that seems similar to this idea called the Dedekind-MacNeille completion, but that deals with posets more generally and different techniques, so I am left a bit uncertain.
Take any simply ordered set, $(X,<)$. For each member $a\times b$ of the set $S=\{x\times y| (x,y)=\emptyset,x,y\in X\}$, we choose a copy of $\mathbb Q$, $\mathbb Q_a$, which is just all the rationals with a little $a$ subscript. We assume for simplicity's sake each $Q_a$ is naturally disjoint from $X$. Now, we define a simply ordered set $(X',<')$, where $X'=X\cup_{x\in S}\mathbb Q_x$, and $<'$ agrees with $<$ on $X$. For any $r_x$ and any $y\in X'$, we have $r_x<'y$ if $y\geq \textrm{succ}(x)$, and $r_x>'y$ otherwise. For any $r_x\neq s_x$, we have $r_x<'s_x$ if $r<s$ in $\mathbb Q$ and $r_x>'s_x$ otherwise. For any $r_x$ and $s_y$ where $x\neq y$, we have $r_x<'s_y$ if $x<y$ and $r_x>'s_y$ otherwise. The effect of all of this is that $(X',<')$ is essentially an extension of $(X,<)$ which has a dense order.
The idea from here is just to further 'fill in' the gaps using Dedekind cuts on $(X',<')$, and produce a space $(X'',<'')$, which contains $(X',<')$ as a subspace (and hence also $(X,<)$ as a subspace) and has the l.u.b. property. So $(X'',<'')$ should be the desired linear continuum. I see no obvious reason why this construction shouldn't work, but I can't shake the feeling some degree of hand-waviness is at play.
Is the idea here valid? If so, how can it be made more rigorous?
A final question (assuming that this construction works on some level), can we talk reasonably about a minimal linear continuum containing $(X,<)$? In special cases perhaps? And if so, do these minimal linear continua have any of the typical uniqueness properties (unique up to homeomorphism or order-isomorphism, for instance)?
 A: A classic fact is that a LOTS $X$ (a linearly ordered set in the order topology) is connected iff it has no gaps and no jumps.
A gap is a non-trivial (both sets non-empty) partition $A \cup B = X$, such that $A < B$ (all $a \in A$ and all $b \in B$ obey $a < b$) such that $A$ has no maximum and $B$ has no minimum. (This is closely related to Dedekind cuts).
A jump is a pair of point $x < y$ such that $\langle x,y\rangle =\emptyset$, so that no point $z$ obeys $x < z < y$.
So starting with any LOTS $X$ we can form its order completion (aka Dedekind-McNeill completion) $\hat{X}$. (Adding a minimum and a maximum now, if not already present, gives an order compactification already). We've gotten rid of all gaps now, but jumps need to be filled. I suppose we can get away with filling these with copies of $[0,1]$, which gives us an essentially unique connected space in which $X$ embeds, but $X$ need not be dense in it, like it is in $\hat{X}$.
My hunch is that we probably cannot always embed densely. It is known that the Sorgenfrey line has no connected space in which it embeds but that's "only" a GO space (generalised ordered) not a LOTS.
A: I’ll begin by presenting the first part of your construction in a somewhat more compact form and demonstrating a problem with it.
Let $\langle X,\le_X\rangle$ be a LOTS, and let $$S=\{x\in X:x\text{ has an immediate successor in }\langle X,\le_X\rangle\}\;;$$ if $x\in S$, I’ll write $x^+$ for the successor of $x$. Let $\hat X=\{\langle x,q\rangle\in X\times\Bbb Q:x\in S\text{ or }q=0\}$, and let $\preceq$ be the lexicographic order on $\hat X$, so that $\langle x,p\rangle\preceq\langle y,q\rangle$ iff either $x<_Xy$, or $x=y$ and $p\le q$. Clearly $\langle\hat X,\preceq\rangle$ is a dense linear order, and the map
$$\iota:X\to\{\langle x,q\rangle\in\hat X:q=0\}:x\mapsto\langle x,0\rangle$$
is an order-isomorphism: $x\le_Xy$ iff $\langle x,0\rangle\preceq\langle y,0\rangle$ for $x,y\in X$. This is essentially your construction up to the point at which you want to take Dedekind cuts.
We’d like $\iota$ to be a topological embedding. It’s clear that $\iota$ is an open bijection from $X$ to $\hat X_0=\{\langle x,0\rangle\in\hat X:q=0\}$. To show that $\iota$ is continuous, let $\langle x,p\rangle,\langle y,q\rangle\in\hat X$ with $\langle x,p\rangle\prec\langle y,q\rangle$, and let $I=\big(\langle x,p\rangle,\langle y,q\rangle\big)$, so that $I\cap X_0$ is a basic open set in $\hat X_0$; we must show that $\iota^{-1}[I\cap\hat X_0]$ is open in $X$. If $p\ge 0$ and $q\le 0$ this is no problem: in that case $\iota^{-1}[I\cap\hat X_0]=(x,y)$. It’s also no problem if $p\ge 0$ and $q<0$: in that case $y\in S$, and $\iota^{-1}[I\cap\hat X_0]=(x,y]=(x,y^+)$. If $p<0$, however, we have a problem, because in that case $\langle x,0\rangle\in I$, and $\iota^{-1}[I\cap\hat X_0]$ is either $[x,y)$ or $[x,y^+)$, depending on $q$, and this is open in $X$ iff $x=\min X$, or $x$ has an immediate predecessor in $X$, neither of which can be guaranteed.
The basic idea will work, but we have to be a bit more careful in constructing $\hat X$. In particular, we also have to pay attention to whether an $x\in X$ has an immediate predecessor in $X$ (which I’ll denote by $x^-$ if it exists). Let
$$P=\{x\in X:x\text{ has an immediate predecessor in }\langle X,\le_X\rangle\}\;,$$
and let
$$\hat X=\{\langle x,q\rangle\in X\times\Bbb Q:q=0,\text{ or }x\in S\text{ and }q>0,\text{ or }x\in P\text{ and }q<0\}\;.$$
As before let $\preceq$ be the lexicographic order on $\hat X$; it’s not hard to check that $\langle\hat X,\preceq\rangle$ is dense, and now you’ll find that it’s possible to carry out the argument that I attempted above to show that $\iota$ is not just an order embedding, but in fact a homeomorphism from $X$ onto $\hat X_0$.
If you now imitate the construction of $\Bbb R$ from $\Bbb Q$ by Dedekind cuts, some of the new points will fill in the gaps in the intervals $\big(\langle x,0\rangle,\langle x^+,0\rangle\big)$ for $x\in S$ and the intervals $\big(\langle x^-,0\rangle,\langle x,0\rangle\big)$ for $x\in P$, as if we had simply defined $\hat X$ to be
$$\hat X=\{\langle x,r\rangle\in X\times\Bbb R:r=0,\text{ or }x\in S\text{ and }0<r\le1,\\\text{ or }x\in P\text{ and }r<0\}\;.$$
And if $\langle X,\le_X\rangle$ has any gaps, there will also be new points filling the corresponding gaps in $\langle\hat X,\preceq\rangle$. At this point you will indeed have a complete dense linear order containing a copy of the LOTS $X$.
