Must a linearly ordered, separable space be metrizable?

Question

Let $X$ be a linearly ordered topological space with a countable dense subset. Does it necessarily follow that $X$ is metrizable?

EDIT: Apollo's comment int he answers implies the answer is negative. Let $X$ be the open unit interval $(0,1)$ and adjoin to every real number $x$ a "ghost number" $x'$ such that $x'$ is the immediate successor of $x$. The "real rationals" are dense in this space. Simply note that sets of the form $(y, x]$ with $x$ and $y$ real and $[x',y)$ with $x'$ ghost and $y$ real form a basis, and these sets all contain a real rational. This space cannot be metrizable, because the subspace topology on the set of all ghost reals is exactly that of the Sorgenfrey line.

Answer 1

You already found a (classical) counterexample: the double arrow ($[0,1] \times \{0,1\}$, ordered lexicographically), which is even compact and separable. There is however a nice metrization theorem for linearly ordered spaces (due to Lutzer): a linearly ordered space $X$ is metrizable (in the order topology) iff the diagonal $D = \{(x,x) : x \in X\}$ is a countable intersection of open subsets of $X \times X$ (a $G_\delta$). This condition is also necessary and sufficient for countably compact regular spaces as well, not just the ordered ones.

Answer 2

No. Take $[0,1]\times\{0,1\}$ with the lexicographic order. This gives a counterexample --- it is separable (for example $\mathbb{Q}\times\{1\}$ is a countable dense set), yet it is not metrizable. One way to see this is to notice that the subspace $[0,1]\times\{1\}$ (homeomorphic to the Sorgenfrey line) is not second-countable, hence not metrizable. The counter-example can also be viewed as an example of an Alexandroff "double-point" construction, which is an example of the general construction of "(special) resolution" (which is a nice technique for generating counterexamples).

(Edited to incorporate comments --- original answer was incorrect, citing Sorgenfrey line as a counterexample.)

Answer 3

I believe this follows from the Urysohn metrization theorem. Order topologies are regular (even completely normal, according to Wikipedia), and your hypothesis of a dense countable subset should imply second countability, unless I'm missing something obvious. Namely open intervals with endpoints in your dense set will form a basis for the topology. So the two hypotheses of Urysohn's metrization theorem are satisfied.

Edit: This works if you have a countable dense subset $D$ in the sense that for every $x$ and $y$ there exists a $d\in D$ with $x < d < y$, but that's not the sense of density being used!

Answer 4

If you use the definition of dense used by Jim Conant, there is a fairly explicit way (apparently going back to Cantor) to metrize the space by embedding it in the unit interval. So let $(L,\leq)$ be a linearly ordered set and $C=(c_ 1,c_2,\ldots)$ a countable subset such that for all $x_1< x_2$ there is a $c\in C$ with $x_1< c < x_2$. Then it is easily seen that $u:L\to [0,1]$ given by $u(x)=\sum_{n\in L_x} 1/2^n$ with $L_x=(n\in\mathbb{N}:c_n< x)$ is an order-preserving function. So one can identify $L$ with a subset of $[0,1]$ and metrize it that way.

A similar construction can be used if $C$ just satisfies that there is a $c\in C$ with $x_1\leq c\leq x_2$ whenever $x_1 < x_2$ with a little more effort.

These results are used in mathematical economics to represent preference relations by utility functions. The book "Representations of Preference Orderings" by Bridges and Mehta contains a lot of related results.