5
$\begingroup$

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.

$\endgroup$
7
  • 1
    $\begingroup$ What do you mean by "Box topology" ? It's not a standard naming... $\endgroup$ – Jean Marie Mar 13 '19 at 12:12
  • 1
    $\begingroup$ @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
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.