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 $L\times I$.

  • 2
    $\begingroup$ $X \times I$ 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  • $\begingroup$ @1830rbc03 You are right...Willard's General Topology is a bit outdated.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
  • 2
    $\begingroup$ 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

$X \times I$ 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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.