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

Question:

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.

• What do you mean by "Box topology" ? It's not a standard naming... – Jean Marie Mar 13 '19 at 12:12
• @JeanMarie Maybe box product sounds better? – YuiTo Cheng Mar 13 '19 at 12:13
• 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. – DanielWainfleet Mar 13 '19 at 14:49
• @DanielWainfleet So you are suggesting this problem is likely to be open, right? – YuiTo Cheng Mar 13 '19 at 14:53
• If the countable product isn't normal (which is open) then so would the higher powers be. – 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.