Does a continuous function embed a separable space into a separable closed subspace?
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Suppose that $X$ is a topological space and $Y$ is a normed vector space, and $f:Xrightarrow Y$ is continuous.
In general we know that if $X$ is separable, then the image $f[X]$ will be separable in $Y$. Does it then follow that there exists a closed subspace $Y_0$ of $Y$ which is separable and contains the $f[X]$?
(Basically I am asking this question with an eye towards the Pettis measurability criterion from Bochner integration.)
(Edited to fix a typo)
general-topology functional-analysis bochner-spaces
asked yesterday
pseudocydonia
370111
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up vote
0
down vote
favorite
Suppose that $X$ is a topological space and $Y$ is a normed vector space, and $f:Xrightarrow Y$ is continuous.
In general we know that if $X$ is separable, then the image $f[X]$ will be separable in $Y$. Does it then follow that there exists a closed subspace $Y_0$ of $Y$ which is separable and contains the $f[X]$?
(Basically I am asking this question with an eye towards the Pettis measurability criterion from Bochner integration.)
(Edited to fix a typo)
general-topology functional-analysis bochner-spaces
asked yesterday
pseudocydonia
370111
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose that $X$ is a topological space and $Y$ is a normed vector space, and $f:Xrightarrow Y$ is continuous.
In general we know that if $X$ is separable, then the image $f[X]$ will be separable in $Y$. Does it then follow that there exists a closed subspace $Y_0$ of $Y$ which is separable and contains the $f[X]$?
(Basically I am asking this question with an eye towards the Pettis measurability criterion from Bochner integration.)
(Edited to fix a typo)
general-topology functional-analysis bochner-spaces
asked yesterday
pseudocydonia
370111
Suppose that $X$ is a topological space and $Y$ is a normed vector space, and $f:Xrightarrow Y$ is continuous.
In general we know that if $X$ is separable, then the image $f[X]$ will be separable in $Y$. Does it then follow that there exists a closed subspace $Y_0$ of $Y$ which is separable and contains the $f[X]$?
(Basically I am asking this question with an eye towards the Pettis measurability criterion from Bochner integration.)
(Edited to fix a typo)
general-topology functional-analysis bochner-spaces
general-topology functional-analysis bochner-spaces
asked yesterday
pseudocydonia
370111
asked yesterday
pseudocydonia
370111
edited yesterday
asked yesterday
pseudocydonia
370111
asked yesterday
pseudocydonia
370111
asked yesterday
pseudocydonia
370111
370111
add a comment |
add a comment |
1 Answer
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There is a closed (separable) subset of $Y$ that contains $f[X]$ namely $overline{f[X]}$.
The closure of a separable subspace is still separable.
answered yesterday
Henno Brandsma
100k344107
$63$ to go now, and you are on $100,000$. Gefeliciteerd alvast, Henno!
– drhab
yesterday
1
@drhab Ach, arbitraire grenzen gebaseerd op tientallige systemen.... Maar dank desalniettemin.
– Henno Brandsma
yesterday
Sorry, there was a typo that made the question silly in the way you pointed out.
– pseudocydonia
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
There is a closed (separable) subset of $Y$ that contains $f[X]$ namely $overline{f[X]}$.
The closure of a separable subspace is still separable.
answered yesterday
Henno Brandsma
100k344107
$63$ to go now, and you are on $100,000$. Gefeliciteerd alvast, Henno!
– drhab
yesterday
1
@drhab Ach, arbitraire grenzen gebaseerd op tientallige systemen.... Maar dank desalniettemin.
– Henno Brandsma
yesterday
Sorry, there was a typo that made the question silly in the way you pointed out.
– pseudocydonia
yesterday
add a comment |
up vote
1
down vote
There is a closed (separable) subset of $Y$ that contains $f[X]$ namely $overline{f[X]}$.
The closure of a separable subspace is still separable.
answered yesterday
Henno Brandsma
100k344107
$63$ to go now, and you are on $100,000$. Gefeliciteerd alvast, Henno!
– drhab
yesterday
1
@drhab Ach, arbitraire grenzen gebaseerd op tientallige systemen.... Maar dank desalniettemin.
– Henno Brandsma
yesterday
Sorry, there was a typo that made the question silly in the way you pointed out.
– pseudocydonia
yesterday
add a comment |
up vote
1
down vote
up vote
1
down vote
There is a closed (separable) subset of $Y$ that contains $f[X]$ namely $overline{f[X]}$.
The closure of a separable subspace is still separable.
answered yesterday
Henno Brandsma
100k344107
There is a closed (separable) subset of $Y$ that contains $f[X]$ namely $overline{f[X]}$.
The closure of a separable subspace is still separable.
answered yesterday
Henno Brandsma
100k344107
edited yesterday
answered yesterday
Henno Brandsma
100k344107
answered yesterday
Henno Brandsma
100k344107
answered yesterday
Henno Brandsma
100k344107
100k344107
$63$ to go now, and you are on $100,000$. Gefeliciteerd alvast, Henno!
– drhab
yesterday
1
@drhab Ach, arbitraire grenzen gebaseerd op tientallige systemen.... Maar dank desalniettemin.
– Henno Brandsma
yesterday
Sorry, there was a typo that made the question silly in the way you pointed out.
– pseudocydonia
yesterday
add a comment |
$63$ to go now, and you are on $100,000$. Gefeliciteerd alvast, Henno!
– drhab
yesterday
1
@drhab Ach, arbitraire grenzen gebaseerd op tientallige systemen.... Maar dank desalniettemin.
– Henno Brandsma
yesterday
Sorry, there was a typo that made the question silly in the way you pointed out.
– pseudocydonia
yesterday
$63$ to go now, and you are on $100,000$. Gefeliciteerd alvast, Henno!
– drhab
yesterday
$63$ to go now, and you are on $100,000$. Gefeliciteerd alvast, Henno!
– drhab
yesterday
1
1
@drhab Ach, arbitraire grenzen gebaseerd op tientallige systemen.... Maar dank desalniettemin.
– Henno Brandsma
yesterday
@drhab Ach, arbitraire grenzen gebaseerd op tientallige systemen.... Maar dank desalniettemin.
– Henno Brandsma
yesterday
Sorry, there was a typo that made the question silly in the way you pointed out.
– pseudocydonia
yesterday
Sorry, there was a typo that made the question silly in the way you pointed out.
– pseudocydonia
yesterday
add a comment |
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