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)?
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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)?
$begingroup$
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
$endgroup$
|
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
$endgroup$
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
$endgroup$
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
general-topology separation-axioms box-topology
edited Mar 13 at 11:20
YuiTo Cheng
asked Mar 13 at 9:54
YuiTo ChengYuiTo Cheng
2,0592637
2,0592637
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
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
1
1
$begingroup$
What do you mean by "Box topology" ? It's not a standard naming...
$endgroup$
– Jean Marie
Mar 13 at 12:12
$begingroup$
What do you mean by "Box topology" ? It's not a standard naming...
$endgroup$
– Jean Marie
Mar 13 at 12:12
1
1
$begingroup$
@JeanMarie Maybe box product sounds better?
$endgroup$
– YuiTo Cheng
Mar 13 at 12:13
$begingroup$
@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
$begingroup$
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
$begingroup$
@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
$begingroup$
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
0
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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