Normal Subspaces of Tychonoff Cubes Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Useful sufficient conditions for a topological space to be the underlying space of a topological group?What are the requirements for separability inheritanceAre Hausdorff compactifications of a Tychonoff space $X$ in one-to-one correspondence with completely regular subalgebras of $BC(X)$?A locally metrizable, Lindelöf Hausdorff space that is not metrizableDo Hausdorff spaces that aren't completely regular appear in practice?Show that every locally compact Hausdorff space is embedded in a compact Hausdorff spaceWhen is uniform space normalCan you add a single point to any completely regular (not normal) space to get a normal space?Which of the following spaces are completely normal?Can every locally compact Hausdorff space be recognized as a subspace of a cube that has an open underlying set?
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Normal Subspaces of Tychonoff Cubes
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Useful sufficient conditions for a topological space to be the underlying space of a topological group?What are the requirements for separability inheritanceAre Hausdorff compactifications of a Tychonoff space $X$ in one-to-one correspondence with completely regular subalgebras of $BC(X)$?A locally metrizable, Lindelöf Hausdorff space that is not metrizableDo Hausdorff spaces that aren't completely regular appear in practice?Show that every locally compact Hausdorff space is embedded in a compact Hausdorff spaceWhen is uniform space normalCan you add a single point to any completely regular (not normal) space to get a normal space?Which of the following spaces are completely normal?Can every locally compact Hausdorff space be recognized as a subspace of a cube that has an open underlying set?
$begingroup$
I know that the completely regular spaces are exactly the subspaces of Tychonoff cubes (possibly uncountable powers of $[0,1]$). Out of these, the compact Hausdorff spaces are exactly the closed subspaces of Tychonoff cubes. Is there a way to describe the class of subspaces of Tychonoff cubes that are exactly the class of normal spaces?
Where I've gotten so far:
Sufficient conditions for normality: Second countability, Closedness, Countability of $lambda$.
Necessary conditions for normality: It isn't this counterexample.
general-topology compactification
asked Oct 29 '17 at 7:20
Joshua MeyersJoshua Meyers
801513
$endgroup$
add a comment |
$begingroup$
I know that the completely regular spaces are exactly the subspaces of Tychonoff cubes (possibly uncountable powers of $[0,1]$). Out of these, the compact Hausdorff spaces are exactly the closed subspaces of Tychonoff cubes. Is there a way to describe the class of subspaces of Tychonoff cubes that are exactly the class of normal spaces?
Where I've gotten so far:
Sufficient conditions for normality: Second countability, Closedness, Countability of $lambda$.
Necessary conditions for normality: It isn't this counterexample.
general-topology compactification
asked Oct 29 '17 at 7:20
Joshua MeyersJoshua Meyers
801513
$endgroup$
$begingroup$
The example only implies that $(0,1)^I subseteq [0,1]^I$ is not normal.
$endgroup$
– Henno Brandsma
Oct 29 '17 at 8:36
add a comment |
$begingroup$
I know that the completely regular spaces are exactly the subspaces of Tychonoff cubes (possibly uncountable powers of $[0,1]$). Out of these, the compact Hausdorff spaces are exactly the closed subspaces of Tychonoff cubes. Is there a way to describe the class of subspaces of Tychonoff cubes that are exactly the class of normal spaces?
Where I've gotten so far:
Sufficient conditions for normality: Second countability, Closedness, Countability of $lambda$.
Necessary conditions for normality: It isn't this counterexample.
general-topology compactification
asked Oct 29 '17 at 7:20
Joshua MeyersJoshua Meyers
801513
$endgroup$
I know that the completely regular spaces are exactly the subspaces of Tychonoff cubes (possibly uncountable powers of $[0,1]$). Out of these, the compact Hausdorff spaces are exactly the closed subspaces of Tychonoff cubes. Is there a way to describe the class of subspaces of Tychonoff cubes that are exactly the class of normal spaces?
Where I've gotten so far:
Sufficient conditions for normality: Second countability, Closedness, Countability of $lambda$.
Necessary conditions for normality: It isn't this counterexample.
general-topology compactification
general-topology compactification
asked Oct 29 '17 at 7:20
Joshua MeyersJoshua Meyers
801513
asked Oct 29 '17 at 7:20
Joshua MeyersJoshua Meyers
801513
asked Oct 29 '17 at 7:20
Joshua MeyersJoshua Meyers
801513
asked Oct 29 '17 at 7:20
Joshua MeyersJoshua Meyers
801513
asked Oct 29 '17 at 7:20
Joshua MeyersJoshua Meyers
801513
801513
$begingroup$
The example only implies that $(0,1)^I subseteq [0,1]^I$ is not normal.
$endgroup$
– Henno Brandsma
Oct 29 '17 at 8:36
add a comment |
$begingroup$
The example only implies that $(0,1)^I subseteq [0,1]^I$ is not normal.
$endgroup$
– Henno Brandsma
Oct 29 '17 at 8:36
$begingroup$
The example only implies that $(0,1)^I subseteq [0,1]^I$ is not normal.
$endgroup$
– Henno Brandsma
Oct 29 '17 at 8:36
$begingroup$
The example only implies that $(0,1)^I subseteq [0,1]^I$ is not normal.
$endgroup$
– Henno Brandsma
Oct 29 '17 at 8:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You're really just asking for conditions for a space $X$ to to be $T_4$ (i.e. normal and $T_1$). Any such space embeds as a (thus normal) subspace of $[0,1]^w(X)$. E.g. any $F_sigma$ subset of $[0,1]^I$ is normal (as it is $sigma$-compact hence Lindelöf and a Lindelöf and regular space is normal.
But this is not at all a necessary condition. Even $G_delta$ subsets need not be normal.
The description is just : the normal subspaces of $[0,1]^I$ are the subspaces that are normal (insert definition here). You cannot expect better. Or you'd just have a characterisation of normality in general.
answered Oct 29 '17 at 7:39
Henno BrandsmaHenno Brandsma
117k349127
$endgroup$
add a comment |
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$begingroup$
You're really just asking for conditions for a space $X$ to to be $T_4$ (i.e. normal and $T_1$). Any such space embeds as a (thus normal) subspace of $[0,1]^w(X)$. E.g. any $F_sigma$ subset of $[0,1]^I$ is normal (as it is $sigma$-compact hence Lindelöf and a Lindelöf and regular space is normal.
But this is not at all a necessary condition. Even $G_delta$ subsets need not be normal.
The description is just : the normal subspaces of $[0,1]^I$ are the subspaces that are normal (insert definition here). You cannot expect better. Or you'd just have a characterisation of normality in general.
answered Oct 29 '17 at 7:39
Henno BrandsmaHenno Brandsma
117k349127
$endgroup$
add a comment |
$begingroup$
You're really just asking for conditions for a space $X$ to to be $T_4$ (i.e. normal and $T_1$). Any such space embeds as a (thus normal) subspace of $[0,1]^w(X)$. E.g. any $F_sigma$ subset of $[0,1]^I$ is normal (as it is $sigma$-compact hence Lindelöf and a Lindelöf and regular space is normal.
But this is not at all a necessary condition. Even $G_delta$ subsets need not be normal.
The description is just : the normal subspaces of $[0,1]^I$ are the subspaces that are normal (insert definition here). You cannot expect better. Or you'd just have a characterisation of normality in general.
answered Oct 29 '17 at 7:39
Henno BrandsmaHenno Brandsma
117k349127
$endgroup$
add a comment |
$begingroup$
You're really just asking for conditions for a space $X$ to to be $T_4$ (i.e. normal and $T_1$). Any such space embeds as a (thus normal) subspace of $[0,1]^w(X)$. E.g. any $F_sigma$ subset of $[0,1]^I$ is normal (as it is $sigma$-compact hence Lindelöf and a Lindelöf and regular space is normal.
But this is not at all a necessary condition. Even $G_delta$ subsets need not be normal.
The description is just : the normal subspaces of $[0,1]^I$ are the subspaces that are normal (insert definition here). You cannot expect better. Or you'd just have a characterisation of normality in general.
answered Oct 29 '17 at 7:39
Henno BrandsmaHenno Brandsma
117k349127
$endgroup$
You're really just asking for conditions for a space $X$ to to be $T_4$ (i.e. normal and $T_1$). Any such space embeds as a (thus normal) subspace of $[0,1]^w(X)$. E.g. any $F_sigma$ subset of $[0,1]^I$ is normal (as it is $sigma$-compact hence Lindelöf and a Lindelöf and regular space is normal.
But this is not at all a necessary condition. Even $G_delta$ subsets need not be normal.
The description is just : the normal subspaces of $[0,1]^I$ are the subspaces that are normal (insert definition here). You cannot expect better. Or you'd just have a characterisation of normality in general.
answered Oct 29 '17 at 7:39
Henno BrandsmaHenno Brandsma
117k349127
edited Mar 26 at 15:45
answered Oct 29 '17 at 7:39
Henno BrandsmaHenno Brandsma
117k349127
answered Oct 29 '17 at 7:39
Henno BrandsmaHenno Brandsma
117k349127
answered Oct 29 '17 at 7:39
Henno BrandsmaHenno Brandsma
117k349127
117k349127
add a comment |
add a comment |
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$begingroup$
The example only implies that $(0,1)^I subseteq [0,1]^I$ is not normal.
$endgroup$
– Henno Brandsma
Oct 29 '17 at 8:36