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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!

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    $\begingroup$ Crossposted to math.se: math.stackexchange.com/questions/3534793/… $\endgroup$
    – bof
    Feb 5 '20 at 6:57
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    $\begingroup$ See my answer on math stackexchange here. $\endgroup$ Feb 5 '20 at 22:16
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    $\begingroup$ E.g. the so-called double arrow is a classic separable non-metrisable compact LOTS. $\endgroup$ Feb 5 '20 at 22:46
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    $\begingroup$ @WholeFood, I rolled your edit back to the previous version because the post was too cryptic in the modified version, whereas the original one is more easily readable. For instance, I had no idea what the acronym "LOTS" was. $\endgroup$
    – Alex M.
    Jul 13 at 17:26
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Yes, because it is regular and has a countable base, namely the family of open intervals with the ends at the dense countable set. The connectedness guarantees that each open interval $(a,b)$ is not empty and hence contains a point from the countable dense set.

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    $\begingroup$ Metrizable? Isn't the space described in the question homeomorphic to an interval of the real line? $\endgroup$
    – bof
    Feb 5 '20 at 6:35
  • $\begingroup$ Yes, it is. But anyway some argument for proving this is necessary. Without connectedness you have a non-metrizable example of the double arrow space. $\endgroup$ Feb 5 '20 at 6:58
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    $\begingroup$ The general condition for LOTS's is having a $G_\delta$ diagonal or a $\sigma$-locally countable base. Theory from the 1970's when LOTS's where studied more. $\endgroup$ Feb 5 '20 at 22:18

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