TITLE: Is the lexicographic ordering on the unit square perfectly normal?
QUESTION [6 upvotes]: It's known that order topology is completely normal, so the lexicographic ordering on the unit square is also completely normal. It's also known that the lexicographic ordering on the unit square is not metrizable. I am interested in whether it is perfectly normal. (A space is perfectly normal if for any two disjoint nonempty closed subsets, there is a continuous function $f$ to $[0,1]$ that "precisely separates" the two sets, meaning that the two closed sets are $f^{-1}(0)$ and $f^{-1}(1)$.) And how do we prove it?

REPLY [3 votes]: For a compact Hausdorff $X$ it is equivalent that $X$ is hereditarily Lindelöf or that $X$ is perfectly normal. Sketch: $X$ is hereditarily Lindelöf implies that every open set is an $F_\sigma$ (as $X$ is regular) so dually every closed set is a $G_\delta$. OTOH if $X$ is perfectly normal, every open set is an $F_\sigma$ so $\sigma$-compact and Lindelöf, making all open sets Lindelöf and $X$ hereditarily Lindelöf again. 
$[0,1]^2$ in the lexicographic order topology has a discrete subset $[0,1]\times \{\frac12\}$ so is definitely not hereditarily Lindelöf, so not perfectly normal either.