Mixing
If $X times X$ is normal, then is $X times X times X$ normal?
I am looking at some topological dimension theory for product spaces, and in trying to construct a certain type of counterexample it's become relevant to consider the question in the title above. I am interested in finding a normal space $X$ whose products with itself is eventually non-normal, but not immediately.
It's not actually important for my application that it happens in three steps as opposed to more. An alternative question would be: Is there a normal space $X$ with $X times X = Y$ normal, but $Y times Y$ is not normal?
The original problem is here:
https://mathoverflow.net/questions/315657/if-textdimx-times-x-2-textdimx-does-textdimxn-n-textdim
Thanks for any help!
As mentioned in a comment below, if we assume that $X$ is a compact Hausdorff space and that $X times X times X$ is completely normal, then $X$ is metrizable. Thus it stands to reason that a compact counterexample may be harder (if not impossible) to construct. The author in the linked paper wonders aloud if the complete normality of $X times X$ is sufficient for the metrizability of $X$, so it may also be advisable to avoid cases where $X times X$ is completely normal.
general-topology examples-counterexamples dimension-theory separation-axioms product-space
asked Nov 21 '18 at 5:36
John Samples
1,086516
add a comment |
I am looking at some topological dimension theory for product spaces, and in trying to construct a certain type of counterexample it's become relevant to consider the question in the title above. I am interested in finding a normal space $X$ whose products with itself is eventually non-normal, but not immediately.
It's not actually important for my application that it happens in three steps as opposed to more. An alternative question would be: Is there a normal space $X$ with $X times X = Y$ normal, but $Y times Y$ is not normal?
The original problem is here:
https://mathoverflow.net/questions/315657/if-textdimx-times-x-2-textdimx-does-textdimxn-n-textdim
Thanks for any help!
As mentioned in a comment below, if we assume that $X$ is a compact Hausdorff space and that $X times X times X$ is completely normal, then $X$ is metrizable. Thus it stands to reason that a compact counterexample may be harder (if not impossible) to construct. The author in the linked paper wonders aloud if the complete normality of $X times X$ is sufficient for the metrizability of $X$, so it may also be advisable to avoid cases where $X times X$ is completely normal.
general-topology examples-counterexamples dimension-theory separation-axioms product-space
asked Nov 21 '18 at 5:36
John Samples
1,086516
1
This is not an answer, but somehow related: math.stackexchange.com/q/2872102
– Paul Frost
Nov 21 '18 at 8:57
Interesting result, I will mention it in an edit.
– John Samples
Nov 21 '18 at 10:00
add a comment |
I am looking at some topological dimension theory for product spaces, and in trying to construct a certain type of counterexample it's become relevant to consider the question in the title above. I am interested in finding a normal space $X$ whose products with itself is eventually non-normal, but not immediately.
It's not actually important for my application that it happens in three steps as opposed to more. An alternative question would be: Is there a normal space $X$ with $X times X = Y$ normal, but $Y times Y$ is not normal?
The original problem is here:
https://mathoverflow.net/questions/315657/if-textdimx-times-x-2-textdimx-does-textdimxn-n-textdim
Thanks for any help!
As mentioned in a comment below, if we assume that $X$ is a compact Hausdorff space and that $X times X times X$ is completely normal, then $X$ is metrizable. Thus it stands to reason that a compact counterexample may be harder (if not impossible) to construct. The author in the linked paper wonders aloud if the complete normality of $X times X$ is sufficient for the metrizability of $X$, so it may also be advisable to avoid cases where $X times X$ is completely normal.
general-topology examples-counterexamples dimension-theory separation-axioms product-space
asked Nov 21 '18 at 5:36
John Samples
1,086516
I am looking at some topological dimension theory for product spaces, and in trying to construct a certain type of counterexample it's become relevant to consider the question in the title above. I am interested in finding a normal space $X$ whose products with itself is eventually non-normal, but not immediately.
It's not actually important for my application that it happens in three steps as opposed to more. An alternative question would be: Is there a normal space $X$ with $X times X = Y$ normal, but $Y times Y$ is not normal?
The original problem is here:
https://mathoverflow.net/questions/315657/if-textdimx-times-x-2-textdimx-does-textdimxn-n-textdim
Thanks for any help!
As mentioned in a comment below, if we assume that $X$ is a compact Hausdorff space and that $X times X times X$ is completely normal, then $X$ is metrizable. Thus it stands to reason that a compact counterexample may be harder (if not impossible) to construct. The author in the linked paper wonders aloud if the complete normality of $X times X$ is sufficient for the metrizability of $X$, so it may also be advisable to avoid cases where $X times X$ is completely normal.
general-topology examples-counterexamples dimension-theory separation-axioms product-space
general-topology examples-counterexamples dimension-theory separation-axioms product-space
asked Nov 21 '18 at 5:36
John Samples
1,086516
asked Nov 21 '18 at 5:36
John Samples
1,086516
edited Nov 21 '18 at 10:04
asked Nov 21 '18 at 5:36
John Samples
1,086516
asked Nov 21 '18 at 5:36
John Samples
1,086516
asked Nov 21 '18 at 5:36
John Samples
1,086516
1,086516
1
This is not an answer, but somehow related: math.stackexchange.com/q/2872102
– Paul Frost
Nov 21 '18 at 8:57
Interesting result, I will mention it in an edit.
– John Samples
Nov 21 '18 at 10:00
add a comment |
1
This is not an answer, but somehow related: math.stackexchange.com/q/2872102
– Paul Frost
Nov 21 '18 at 8:57
Interesting result, I will mention it in an edit.
– John Samples
Nov 21 '18 at 10:00
1
1
This is not an answer, but somehow related: math.stackexchange.com/q/2872102
– Paul Frost
Nov 21 '18 at 8:57
This is not an answer, but somehow related: math.stackexchange.com/q/2872102
– Paul Frost
Nov 21 '18 at 8:57
Interesting result, I will mention it in an edit.
– John Samples
Nov 21 '18 at 10:00
Interesting result, I will mention it in an edit.
– John Samples
Nov 21 '18 at 10:00
add a comment |
1 Answer
1
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oldest
votes
A construction can be found in (or weaned from)
Przymusinski, Teodor C., Normality and paracompactness in finite and countable Cartesian products, Fundam. Math. 105, 87-104 (1980). ZBL0438.54021.
wherein the following remarkable result is proved:
Theorem 1. For every $k$ and $m$ such that $1 leq k leq m leq omega$ there exists a separable and first coutnable space $X = X(k,m)$ such that
$X^n$ is paracompact (Lindelöf, subparacompact) if and only if $n < k$,
$X^n$ is normal (collectionwise normal) if and only if $n < m$.
In particular, we can construct a (normal) space such that the failure of normality of its powers happens first at any prescribed finite power.
answered Nov 21 '18 at 9:53
1-3-7-Trimethylxanthine
4,662927
Wow, what a result!
– John Samples
Nov 21 '18 at 10:06
add a comment |
Your Answer
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1 Answer
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A construction can be found in (or weaned from)
Przymusinski, Teodor C., Normality and paracompactness in finite and countable Cartesian products, Fundam. Math. 105, 87-104 (1980). ZBL0438.54021.
wherein the following remarkable result is proved:
Theorem 1. For every $k$ and $m$ such that $1 leq k leq m leq omega$ there exists a separable and first coutnable space $X = X(k,m)$ such that
$X^n$ is paracompact (Lindelöf, subparacompact) if and only if $n < k$,
$X^n$ is normal (collectionwise normal) if and only if $n < m$.
In particular, we can construct a (normal) space such that the failure of normality of its powers happens first at any prescribed finite power.
answered Nov 21 '18 at 9:53
1-3-7-Trimethylxanthine
4,662927
Wow, what a result!
– John Samples
Nov 21 '18 at 10:06
add a comment |
A construction can be found in (or weaned from)
Przymusinski, Teodor C., Normality and paracompactness in finite and countable Cartesian products, Fundam. Math. 105, 87-104 (1980). ZBL0438.54021.
wherein the following remarkable result is proved:
Theorem 1. For every $k$ and $m$ such that $1 leq k leq m leq omega$ there exists a separable and first coutnable space $X = X(k,m)$ such that
$X^n$ is paracompact (Lindelöf, subparacompact) if and only if $n < k$,
$X^n$ is normal (collectionwise normal) if and only if $n < m$.
In particular, we can construct a (normal) space such that the failure of normality of its powers happens first at any prescribed finite power.
answered Nov 21 '18 at 9:53
1-3-7-Trimethylxanthine
4,662927
Wow, what a result!
– John Samples
Nov 21 '18 at 10:06
add a comment |
A construction can be found in (or weaned from)
Przymusinski, Teodor C., Normality and paracompactness in finite and countable Cartesian products, Fundam. Math. 105, 87-104 (1980). ZBL0438.54021.
wherein the following remarkable result is proved:
Theorem 1. For every $k$ and $m$ such that $1 leq k leq m leq omega$ there exists a separable and first coutnable space $X = X(k,m)$ such that
$X^n$ is paracompact (Lindelöf, subparacompact) if and only if $n < k$,
$X^n$ is normal (collectionwise normal) if and only if $n < m$.
In particular, we can construct a (normal) space such that the failure of normality of its powers happens first at any prescribed finite power.
answered Nov 21 '18 at 9:53
1-3-7-Trimethylxanthine
4,662927
A construction can be found in (or weaned from)
Przymusinski, Teodor C., Normality and paracompactness in finite and countable Cartesian products, Fundam. Math. 105, 87-104 (1980). ZBL0438.54021.
wherein the following remarkable result is proved:
Theorem 1. For every $k$ and $m$ such that $1 leq k leq m leq omega$ there exists a separable and first coutnable space $X = X(k,m)$ such that
$X^n$ is paracompact (Lindelöf, subparacompact) if and only if $n < k$,
$X^n$ is normal (collectionwise normal) if and only if $n < m$.
In particular, we can construct a (normal) space such that the failure of normality of its powers happens first at any prescribed finite power.
answered Nov 21 '18 at 9:53
1-3-7-Trimethylxanthine
4,662927
answered Nov 21 '18 at 9:53
1-3-7-Trimethylxanthine
4,662927
answered Nov 21 '18 at 9:53
1-3-7-Trimethylxanthine
4,662927
answered Nov 21 '18 at 9:53
1-3-7-Trimethylxanthine
4,662927
4,662927
Wow, what a result!
– John Samples
Nov 21 '18 at 10:06
add a comment |
Wow, what a result!
– John Samples
Nov 21 '18 at 10:06
Wow, what a result!
– John Samples
Nov 21 '18 at 10:06
Wow, what a result!
– John Samples
Nov 21 '18 at 10:06
add a comment |
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1
This is not an answer, but somehow related: math.stackexchange.com/q/2872102
– Paul Frost
Nov 21 '18 at 8:57
Interesting result, I will mention it in an edit.
– John Samples
Nov 21 '18 at 10:00