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?



Is the lexicographic ordering on the unit square perfectly normal?
As Nate Eldredge points out in the comments, the book *Counterexamples in Topology* states that this space is not perfectly normal, but does not provide a proof. Here is the idea for a proof.


A *perfectly normal* space is a normal space in which every closed set is a $G\_\delta$. So to prove this space is not perfectly normal, we'd like to find a closed set that is not a $G\_\delta$. I claim that the subspace $C = [0,1] \times \{0,1\}$ works. ($C$ is the two horizontal lines on the top and bottom.) The reason is that if $U \subseteq [0,1] \times [0,1]$ is an open set containing $C$, then 
$$A\_U = \{x \in [0,1] \ : \ \{x\} \times [0,1] \subseteq U\}$$
(i.e., the set of all vertical lines contained in $U$) is a co-countable subset of $[0,1]$. (This is because for every $p \in (0,1]$, $U$ must contain a neighborhood of $(p,1)$, and therefore $A\_U$ must contain an interval of the form $(p-\varepsilon,p)$. Likewise, $U$ must contain a neighborhood of $(p,0)$, and therefore $A\_U$ must contain an interval of the form $(p,p+\varepsilon)$. So every point of $[0,1] \setminus A\_U$ is isolated.) Therefore any countable intersection of open neighborhoods of $C$ contains a horizontal line.


**EDIT:** Thanks to Nate Eldredge for helping me to simplify my original argument.


