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

“Proof” that $mathbbR^J$ is not normal when $J$ is uncountableHow to prove that $mathbb R^omega$ with the box topology is completely regularcontinuity box topologyConvergence in the Box Topology.Normal spaces in box and uniform topologyWhat are the components and path components of $mathbbR^omega$ in the product, uniform, and box topologies?$mathbb R^omega$ in the box topology is not first countable.Uncountability of $mathbbR^I$ if $I$ is uncountableconnected components in box topologyMetrizability of infinite product space under box topologyIs $mathbbR^omega$ endowed with the box topology completely normal (or hereditarily normal)?

4

$begingroup$

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 $mathbbR^J$ 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.

general-topology separation-axioms box-topology

share|cite|improve this question

edited Mar 13 at 11:20

YuiTo Cheng

asked Mar 13 at 9:54

YuiTo ChengYuiTo Cheng

2,0592637

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


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  1
  

  
  
  
  
  
  

  
  $begingroup$
  @JeanMarie Maybe box product sounds better?
  $endgroup$
  – YuiTo Cheng
  Mar 13 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 at 14:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DanielWainfleet So you are suggesting this problem is likely to be open, right?
  $endgroup$
  – YuiTo Cheng
  Mar 13 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 at 16:49
  

  
  
  
  

 | 
show 2 more comments

4

$begingroup$

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 $mathbbR^J$ 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.

general-topology separation-axioms box-topology

share|cite|improve this question

edited Mar 13 at 11:20

YuiTo Cheng

asked Mar 13 at 9:54

YuiTo ChengYuiTo Cheng

2,0592637

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @JeanMarie Maybe box product sounds better?
  $endgroup$
  – YuiTo Cheng
  Mar 13 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 at 14:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DanielWainfleet So you are suggesting this problem is likely to be open, right?
  $endgroup$
  – YuiTo Cheng
  Mar 13 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 at 16:49
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

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 $mathbbR^J$ 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.

general-topology separation-axioms box-topology

share|cite|improve this question

edited Mar 13 at 11:20

YuiTo Cheng

asked Mar 13 at 9:54

YuiTo ChengYuiTo Cheng

2,0592637

$endgroup$

share|cite|improve this question

edited Mar 13 at 11:20

YuiTo Cheng

asked Mar 13 at 9:54

YuiTo ChengYuiTo Cheng

2,0592637

share|cite|improve this question

edited Mar 13 at 11:20

YuiTo Cheng

asked Mar 13 at 9:54

YuiTo ChengYuiTo Cheng

2,0592637

asked Mar 13 at 9:54

YuiTo ChengYuiTo Cheng

2,0592637

asked Mar 13 at 9:54

YuiTo ChengYuiTo Cheng

2,0592637