# Is $L^{n}$ normal, where $L$ denotes the closed long ray?

1.I am trying to prove that $$L^{n}$$, the $$n$$-$$th$$ product of closed long ray is normal, so that I can apply Tietze extension theorem to its closed subset and prove something else. I think I am able to prove this (if it is true) for $$n\leq3$$ in a very elementary and ugly method, but I cannot convince myself about generalizing this in a staightforward way to higher dimensions. Could anyone offer me hint?

2.Possibly related to 1: I found this 1963 paper. At the beginning the author says whether the pruduct of a normal space $$X$$ and unit interval $$I$$ must be normal was then an open question. Is it still open now? Is there some condition which ensures the normality of $$X\times I$$?

Briefly explain my idea for $$n=2$$: I like to use the following equivalent criterion for normality: for any closed $$E$$ contained in open $$U$$, there exists open $$V$$ such that $$E\subset V\subset\overline{V}\subset U$$. It can be proved that if $$F\subset U\subset L$$ where $$U$$ is open and $$F$$ is closed unbounded (closed in topology sense and unbounded in order sense) then $$U$$ is "big", namely $$[x,+\infty)\subset U$$ for some $$x\in L$$. The proof is via Fodor's lemma and the property of $$L$$ near limit ordinals. Similarly if $$F\subset U\subset L^{2}$$ where $$U$$ is open and $$F$$ is closed unbounded in both directions, then $$[x,+\infty)^{2}\subset U$$ for some $$x$$. Deleting $$(x,+\infty)^{2}$$ from $$L^{2}$$ we obtain something homeomorphic to gluing two $$L\times I$$, and we would be almost done if $$L\times I$$ is normal, which I believe I am able to prove by similar augument. The case when $$F$$ is unbounded only in one direction can be reduced to $$L\times I$$.

• $X \times I$ is in general not normal (unfortunately I couldn't find a reference). In fact, the class of spaces such that $X \times I$ is normal even has an own name (binormal spaces). – Paul Frost Apr 12 at 14:05
• @PaulFrost According to this, it looks like it's unsolved. – YuiTo Cheng Apr 12 at 14:24
• @YuiToCheng But according to wikipedia there are counterexamples under ZFC......Although they were constructed quite late and weird even compared to L, so I guess at least I am trying to prove something true. – 1830rbc03 Apr 12 at 14:53
• @1830rbc03 You are right...Willard's General Topology is a bit outdated. – YuiTo Cheng Apr 12 at 14:56
• A normal space $X$ such that $X \times I$ is not normal is called a Dowker space. Such spaces exist in ZFC (ME Rudin). $X \times I$ is normal iff $X$ is normal and countably paracompact (Dowker theorem). $L$ is countably paracompact ( being ordered) and so $L \times I$ is normal. – Henno Brandsma Apr 12 at 16:13

$$X \times I$$ is normal iff $$X$$ is normal and countably paracompact (every countable open cover has a locally finite refinement). This is due to Dowker (On countably paracompact spaces, Canad. J. Math. 3 (1951) 219-224 ), see Engelking General Topology, Theorem 5.2.8 for an accessible modern reference.
A normal non-countably paracompact space $$X$$ (so that $$X \times I$$ is not normal) exists in ZFC, as shown by M.E. Rudin in 1971 (so that problem was open for about 20 years).
But ordered spaces are countably paracompact so $$L \times I$$ is certainly normal.
It is also well-known that $$\omega_1^n$$ is normal (and pseudocompact) so I think (but I have a PhD thesis by van Dalen that has a lot of results on this problem that I could look into to) that $$L^n$$ will probably also be normal.
If we add the compactifying point to $$L$$, so we get $$L^+$$ (as its sometimes called) though, we can see that $$L \times L^+$$ is not normal, so it's subtle. This has the same reason that $$\omega_1 \times (\omega_1 +1)$$ is not normal: $$\omega_1$$ (and $$L$$) is not paracompact, and $$\omega_1 +1 = \beta \omega_1$$ and $$L^+=\beta L$$ (the Cech-Stone compactification).