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

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 $$ℝ^𝜔$$ 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?

• About a year ago on this site I asked whether it was normal and the answer was that it was still unknown. – DanielWainfleet Dec 15 '18 at 6:04
• Yeah, but completely normality is a strictly stronger notion. – YuiTo Cheng Dec 15 '18 at 6:27
• @DanielWainfleet not completely normal has been known for quite a long time now. Normal is stil open AFAIK (in ZFC). – Henno Brandsma Dec 15 '18 at 9:48
• @HennoBrandsma. I haven't checked but it may have been you who answered this to me on my posted Q about this. – DanielWainfleet Dec 15 '18 at 18:08
• @DanielWainfleet I’m not sure but I don’t think so. – Henno Brandsma Dec 15 '18 at 18:30

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.