Completion of a linear order that is a dense subspace of a compact space.

Question

Suppose $D$ is a linearly ordered space which is densely embedded in a compact Hausdorff space $K$.

What can we say about the relation between $K$ and $\overline D$, the completion of $D$. Is one a subspace of the other? Could we add other properties to $K$ to simplify the matter, or to even make $K $ homeomorphic to $\overline D$ ?

EDIT: I think I should clarify a few things.

• By "completion of $D$," $\overline D$, I mean the Dedekind completion of $D$ with the order topology.
• I want to know about the relation between the topological spaces $\overline D$ and $K$, assuming (i) $D$ is a dense subset of $K$, and (ii) The order topology on $D$ is the same as the topology $D$ inherits as a subspace of $K$.

Answer 1

Without additional hypotheses it need not be the case that $\overline D$ embeds in $K$.

$\Bbb N$ with its usual order and order topology is a LOTS that is densely embedded in the compact Hausdorff space $\beta\Bbb N$. The Dedekind completion $\overline{\Bbb N}$ of $\Bbb N$ is $\Bbb N\cup\{\infty\}$, where $n<\infty$ for each $n\in\Bbb N$, and the order topology makes this the one-point compactification of $\Bbb N$. However, $\beta\Bbb N$ contains no subspace homeomorphic to $\overline{\Bbb N}$: every infinite closed subset of $\beta\Bbb N$ contains a copy of $\beta\Bbb N$. (See, for instance, Theorem $3.6.14$ of Engelking, General Topology.)

Added: Let $D=\Bbb Q\cap(-1,1)$, so that $\overline{D}=[-1,1]$. Let $\tau$ be the usual topology on $[-1,1]$. Define

$$f:[-1,1]\to[-1,1]:x\mapsto\begin{cases} x,&\text{if }x\in D\\ -x,&\text{if }x\notin D\;. \end{cases}$$

Let $\tau'=\{f[U]:U\in\tau\}$; $f$ is a bijection, so it’s easy to check that $\tau'$ is a topology on $[-1,1]$ and indeed that $\big\langle[-1,1],\tau'\big\rangle$ is homeomorphic to $\big\langle[-1,1],\tau'\big\rangle$ and has $D$ as a dense subset. However, $\tau\nsubseteq\tau'\nsubseteq\tau$.

Added₂: Let $A=\{2^{-n}:n\in\Bbb Z^+\}$, and let $D=[0,1]\setminus A$ with the usual order; clearly $\overline{D}=[0,1]$. Let $X=\big([0,1]\times\{0\}\big)\cup\big(A\times\{1\}\big)$, ordered lexicographically; it’s not hard to check that $X$ is compact, and that $D\times\{0\}$ is dense in $X$ and homeomorphic to $D$. Note that $X$ is not connected; e.g., $\left\{\left\langle\frac12,1\right\rangle\right\}\cup\left(\left(\frac12,1\right]\times\{0\}\right)$ is a clopen subset of $X$, so $X$ is not homeomorphic to $\overline{D}$. (In fact $X$ is homeomorphic to the one-point compactification of $[0,1]\times\omega$.) Let

$$f:[0,1]\to X:x\mapsto\begin{cases} \langle x,0\rangle,&\text{if }x\in D\\ \langle 2^{-n},0\rangle,&\text{if }x=2^{-2n}\text{ for some }n\in\Bbb Z^+\\ \langle 2^{-n},1\rangle,&\text{if }x=2^{-(2n-1)}\text{ for some }n\in\Bbb Z^+\;. \end{cases}$$

clearly $f$ is a bijection. Let $\tau=\{f^{-1}[U]:U\text{ is open in }X\}$; then $\tau$ is a topology on $[0,1]$. Denote $\big\langle[0,1],\tau\big\rangle$ by $K$; $K$ is homeomorphic to $X$, and $D$ has the same relative topology in $K$ as in $[0,1]$ with its usual topology, but $K$ and $[0,1]$ are not homeomorphic. Moreover, if $\mathscr{E}$ is the Euclidean topology on $[0,1]$, then

$$\left\{\frac14\right\}\cup\left(\frac38,\frac12\right)\in\tau\setminus\mathscr{E}\;,$$

and

$$\left(\frac3{16},\frac38\right)\in\mathscr{E}\setminus\tau\;,$$

so $\mathscr{E}$ and $\tau$ are not comparable in the lattice of topologies on $[0,1]$.