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.



Must a linearly ordered, separable space be metrizable?
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](http://www.ams.org/journals/proc/1969-022-02/S0002-9939-1969-0248761-1/S0002-9939-1969-0248761-1.pdf)): 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.


