Order completion of a linear order I was considering the following construction in an attempt to "complete" a given linear order:
Given a linearly ordered set  $(X,\leq )$ I denote the set of lower sets in $X$ by $\mathcal{L}(X)$ and order it with respect to inclusion. Since $X$ is linearly ordered, so is $\mathcal{L}(X)$. I think that $X$ embeds in an order preserving sense into $\mathcal{L}(X)$ via
$$ x \overset{i}{\mapsto} \{ y\in x: y<x \}. $$
The notion of completeness I am using here is that every set bounded from above has a supremum and every set bounded from below has an infimum. I think that in the case of linear orders, we can say the following:

"Proposition"
$(i)$ If every lower set $\emptyset \neq A\ \subsetneq X $ is either an open intial segment $\{ y\in X: y<x_0 \}$ or a closed initial segment $\{ y\in X: y\leq x_0 \}$, then every bounded set from above in $X$ has a supremum.
$(ii)$ If every upper set $\emptyset \neq A\ \subsetneq X $ is either an open final segment $\{ y\in X: y>x_0 \}$ or a closed initial segment $\{ y\in X: y\geq x_0 \}$, then every bounded set from below in $X$ has an infimum.

$\mathcal{L}(X)$ satisfies $(i)$, and I think that in a linear order $A$ is a lower set if and only if $A^c$ is an upper set. I therefore think that $\mathcal{L}(X)$ is order complete.
I have a few questions about this:

*

*Does this seem generally correct?


*Is this construction essentially the same as Dedekind–MacNeille completion?


*It seems like $\mathcal{L}(X)$ is bounded and can therefore not be minimal necessarily. Will defining $\mathcal{L}^+(X)$ the collection of non empty lower sets yield a completion when $X$ is not bounded from below? Likewise, does looking at $\mathcal{L}_-(X)$, the collection of proper lower sets , yields a completion when $X$ is not bounded from above?
 A: Let $\langle X,\le\rangle$ be a linear order, and let $\mathscr{L}(X)$ be the family of lower sets of $X$; then $\langle\mathscr{L}(X),\subseteq\rangle$ is a complete linear order. If $\varnothing\ne\mathscr{A}\subseteq\mathscr{L}$, then $\bigcap\mathscr{A},\bigcup\mathscr{A}\in\mathscr{L}$, and it’s easy to check that $\bigcap\mathscr{A}=\inf\mathscr{A}$ and $\bigcup\mathscr{A}=\sup\mathscr{A}$.
For each $x\in X$ let $x^-=(\leftarrow,x)$ and $x^+=(\leftarrow,x]$; you are embedding $X$ in $\mathscr{L}$ by the map $x\mapsto x^-$, which is indeed order-preserving. You could just as well embed it by sending $x\in X$ to $x^+$, the immediate successor of $x^-$ in $\mathscr{L}(X)$. The fact that you have this choice and are doubling every point of $X$ is an indication that you are not getting the Dedekind-MacNeille completion of $X$. Here is a concrete example.

Let $X=[0,1]\cup[2,3]$, and let $\le$ be the usual linear order on $X$; the order $\langle X,\le\rangle$ is complete, but your construction produces an order isomorphic to $X\times\{0,1\}$ ordered lexicographically. In particular, if we use your embedding, it adds a wholly unnecessary point $1^+$ between the embedded copies, $1^-$ and $2^-$, of $1$ and $2$. The other embedding is no better: $1$ and $2$ embed as $1^+$ and $2^+$, respectively, with $2^-$ an isolated point between them.

The Dedekind-MacNeille completion avoids this by keeping only those $L\in\mathscr{L}(X)$ such that $(L^{\mathrm{u}})^{\mathrm{l}}=L$, where $A^{\mathrm{u}}$ is the set of upper bounds of $A$, and $A^{\mathrm{l}}$ is the set of lower bounds of $A$. It then embeds $x\in X$ as $x^+$. (It’s easily verified that $\left((x^+)^{\mathrm{u}}\right)^{\mathrm{l}}=x^+$, and that $\left((x^-)^{\mathrm{u}}\right)^{\mathrm{l}}=x^+$, so that $x^-$ isn’t in the Dedekind-MacNeille completion of $X$.)
If $L\in\mathscr{L}(X)$, it isn’t hard to verify that $\left(L^{\mathrm{u}}\right)^{\mathrm{l}}=L$ if and only if there is no $x\in X$ such that $L=x^-$: either $L$ is clopen in the order topology, so that it corresponds to a gap in $X$, or $L=x^+$ for some $x\in X$. Thus, the Dedekind-MacNeille completion of $X$ is $\langle\mathscr{L}(X)\setminus X^-,\subseteq\rangle$, where $X^-=\{x^-:x\in X\}$.
