Is $\mathbb{R}^\omega$ endowed with the box topology completely normal (or hereditarily normal)?

Question

Just out of curiosity, I'd like to know more properties of box topology. I found Is $\mathbb{R}^\omega$ a completely normal space, in the box topology? quite interesting, but unfortunately, it hasn't attracted too much attention. I also searched it in MathOverflow, the comment by Ramiro de la Vega in Is it still an open problem whether $ℝ^$ is normal in the box topology? asserted it's been known the answer is negative. However, I can't obtain any further information on the Internet. Could somebody provide a disproof, or at least offer some useful links?

Answer 1

It is not. Erik van Douwen showed ("The box product of countably many metrizable spaces need not be normal", Fund. Math., link) that if $X_0$ is the irrationals (as a subspace of the reals) and for $n \ge 1$, $X_n = \omega+1$ (a compact space: a convergent sequence in the reals is homeomorphic to it) then $\Box_{n \in \omega} X_n$ is not normal.

This space can be seen as a subspace of $\mathbb{R}^\omega$ in the box topology, so the latter space is not hereditarily normal (so not completely normal). Erik himself showed in the paper (as a "byproduct") that a box product of metrisable spaces cannot be hereditarily normal if infinitely many of them are non-discrete.

A proof of the first result can also be found in Mary Ellen Rudin's "Lectures on Set theoretic topology" (a very nice book that should be read by anyone interested in research in general topology, IMHO), in the chapter on box products; this is where I found it.