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

Question

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
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
@HennoBrandsma What about assuming the continuum hypothesis? — YuiTo Cheng, Mar 13 '19 at 23:41
If we assume CH we know nothing more about the uncountable case. — Henno Brandsma, Mar 14 '19 at 04:53

score 5 · Answer 1 · answered Aug 29 '19 at 17:42

5

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.

answered Aug 29 '19 at 17:42

Hector Alonzo

51

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

1 Answers1