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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 $L\times I$.

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    $\begingroup$ $X \times I$ 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). $\endgroup$ – Paul Frost Apr 12 at 14:05
  • $\begingroup$ @PaulFrost According to this, it looks like it's unsolved. $\endgroup$ – YuiTo Cheng Apr 12 at 14:24
  • $\begingroup$ @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. $\endgroup$ – 1830rbc03 Apr 12 at 14:53
  • $\begingroup$ @1830rbc03 You are right...Willard's General Topology is a bit outdated. $\endgroup$ – YuiTo Cheng Apr 12 at 14:56
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    $\begingroup$ 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. $\endgroup$ – Henno Brandsma Apr 12 at 16:13
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$X \times I$ 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).

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