Is a totally ordered, separable and connected topological space metrizable (in the order topology)?
If we relax the assumption of connectedness, I know the counterexamples, but if we have a linear continuum that is also separable, can we say it is metrizable?
Thanks!
Yes, if $D$ is countable and dense in $X$ then $\{(a,b): a,b \in D, a < b\}$ together with (if applicable) $\{[m,a), (a,M], a \in D\}$ (if $m = \min(X)$ and/or $M = \max(X)$ exists) forms a countable base for $X$. Urysohn then finishes the job (as all ordered spaces are normal $T_1$).
This happens by connectedness: this ensures there are no (gaps and) jumps in $X$. So if e.g. $x \in (a,b)$, the open interval $(a,x)$ cannot be empty by connectedness (it would be a jump otherwise) and so contains a point $a' \in D$, and the same can be said for a $b' \in D \cap (x,b)$ and then $x \in (a',b') \subseteq (a,b)$, and we've almost shown the collection to be a base (we also need to consider the two other base set types when $m,M$ exist, in general).
The connectedness is not necessary and neither is the separability: an ordered topological space is metrisable iff it has a $G_\delta$ diagonal (i.e. $\Delta_X = \{(x,x): x \in X\}$ is a countable intersection of open sets in $X \times X$ (in the product topology). This is equivalent to having a $\sigma$-locally countable base, (which is more in spirit with traditional metrisation theorems like Bing-Nagata-Smirnov).