Let $X$ be a linearly ordered topological space with a countable dense subset. Does it necessarily follow that $X$ is metrizable?
A linearly ordered space
EDIT: Apollo's comment int he answers implies the answer is a set negative. Let $X$ with be the open unit interval $(0,1)$ and adjoin to every real number $x$ a total order "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 open intervals form $(y, x]$ with $x$ and $y$ real and $[x',y)$ with $x'$ ghost and $y$ real form a basisfor $X$. For example, every ordinal can be given the order topology and the usual Euclidean topology on the these sets all contain a real line is the order topologyrational. See This space cannot be metrizable, because the Order subspace topology on Wikipedia. I am not asking for an example the set of a separable space that all ghost reals is not metrizable - there are many examples for exactly that (there's no need for of the Sorgenfrey line, you can even pick the trivial topology...).