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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?

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    $\begingroup$ Wikipedia on this space: en.wikipedia.org/wiki/…. (Very interesting question!) $\endgroup$ – Joel David Hamkins Jan 28 at 14:55
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    $\begingroup$ Steen and Seebach, Counterexamples in Topology, 2e, say that it isn't, but the proof is left as an exercise (problem 71, page 209). $\endgroup$ – Nate Eldredge Jan 28 at 15:09
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    $\begingroup$ I'm not sure about your definition of "perfectly normal". Doesn't Urysohn's lemma say that, in any $T_4$ space, any disjoint closed sets are separated by a continuous function? The definition I know of "perfectly normal" is that every closed set is a $G_\delta$. This is equivalent to "every nonempty closed set can be separated from its complement by a continuous function onto $[0,1]$". Maybe this is the definition/characterization you're thinking of? $\endgroup$ – Will Brian Jan 28 at 15:17
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    $\begingroup$ Thanks @JoelDavidHamkins. The wikipedia entry uses the phrase "precisely separated" which I think makes a difference here -- I'll edit the question to make it more clear. $\endgroup$ – Will Brian Jan 28 at 15:41
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    $\begingroup$ @Gro-Tsen, the open unit square would be metrizable (it is just a disjoint union of open intervals). $\endgroup$ – Ramiro de la Vega Jan 28 at 16:53
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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.

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  • $\begingroup$ I think you missed one: "horizontal line" at the end should be "vertical line"? $\endgroup$ – Nate Eldredge Jan 28 at 15:51
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    $\begingroup$ Just to note a slight variation (since I had been thinking about a similar proof before I saw yours): the complement of $A_U$ is actually at most countable, since every point of it is isolated in the usual topology of $[0,1]$. So any $G_\delta$ containing $C$ must in fact contain co-countably many vertical lines, and you don't need Baire. $\endgroup$ – Nate Eldredge Jan 28 at 15:54
  • $\begingroup$ @NateEldredge: Good observation! It's not exactly using a nuke to kill a fly, but it is using more than just the flyswatter. I've edited to simplify things. $\endgroup$ – Will Brian Jan 29 at 12:54
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    $\begingroup$ In the the Moore plane/Niemytskii plane every closed set is $G_\delta$, but it is not normal. Call me old-fashioned, but I think "perfectly normal" should imply "normal". It is true, via a proof using Urysohn's lemma countably many times and carefully summing up the resulting functions, that a perfectly normal space is the same thing as a normal space in which every closed set is $G_\delta$. Of course, this does not affect the argument for the lexicographic ordering not being normal, because only the direction "perfectly normal $\Rightarrow$ all closed sets are $G_\delta$" is used. $\endgroup$ – Robert Furber Jan 30 at 3:03
  • $\begingroup$ @RobertFurber: You're perfectly right, of course. A perfectly normal space is a normal space in which every closed set is $G_\delta$. (I was thinking of this whole problem in the context of normal spaces, but apparently that's not clear. I'll edit to make it clear.) $\endgroup$ – Will Brian Jan 30 at 13:57
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It is a known fact that any perfectly normal (countably) compact space $X$ is ccc. The proof is what you would try: start with an uncountable cellular family $\mathcal{U}$, choose a point $p_U \in U$ for each $U \in \mathcal{U}$ and consider the closed set $C=X \setminus \bigcup \mathcal{U}$. Since the space is perfectly normal, $C$ is a $G_\delta$ (say $C= \bigcap_nV_n$) and since $\mathcal{U}$ is uncountable you can find $n$ such that infinitely many of the $p_U$'s are outside of $V_n$. Now this infinite set of $p_U$´s has no limit points, contradicting the compactness of $X$.

Since $I^2_{lex}$ is compact and not ccc (both standard facts), it cannot be perfectly normal.

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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.

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