Is $\mathbb R^J$ normal in the box topology when $J$ is uncountable?

I know $\mathbb R^J$ is not normal in the product topology, see "Proof" that $\mathbb{R}^J$ is not normal when $J$ is uncountable ;

I also know $\mathbb R^{\omega}$ is normal in the box topology assuming the continuum hypothesis, see Is it still an open problem whether $\mathbb R^{\omega}$ is normal in the box topology?.

That's the motivation for this problem. Unfortunately, the above two theorems don't imply anything about the normality of $\mathbb R^J$. Any hint would be appreciated.

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    $\begingroup$ What do you mean by "Box topology" ? It's not a standard naming... $\endgroup$ – Jean Marie Mar 13 '19 at 12:12
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    $\begingroup$ @JeanMarie Maybe box product sounds better? $\endgroup$ – YuiTo Cheng Mar 13 '19 at 12:13
  • $\begingroup$ A closed subspace of a normal space is normal. $\Bbb R^{\omega}$ is homeomorphic to a closed subspace of $\Bbb R^k$ if $k$ is uncountable. $\endgroup$ – DanielWainfleet Mar 13 '19 at 14:49
  • $\begingroup$ @DanielWainfleet So you are suggesting this problem is likely to be open, right? $\endgroup$ – YuiTo Cheng Mar 13 '19 at 14:53
  • $\begingroup$ If the countable product isn't normal (which is open) then so would the higher powers be. $\endgroup$ – Henno Brandsma Mar 13 '19 at 16:49

It is known that the space $\square (\omega +1)^{\omega_1}$ is not normal. This is an amazing result of B. Lawrence, 1996: Failure of normality in the box product of uncountably many real lines.


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