A set $C$ that is not closed in the weak* topology
$begingroup$
I have a problem, but first I would like to understand your statement. The statement is about a subset $Csubset l^{infty}$ that is convex and closed in $C(l^{infty}, |.|_{sup})$. I know that if $C$ is convex, then $C$ is closed in the weak topology. So far, so good. The problem now asks to show that C is not closed in the weak* topology.
So, how can I say that $Csubset l^{infty}$ is closed in weak* topology if this topology is defined in its dual $l^{infty '}$?
With this well-known, I would like a tip to show that
$$C = {x =(x_i)_{iin mathbb{N}} in l^{infty} : 0leq x_i leq 1, x_i to 1 }$$
is not closed in the topology weak*.
functional-analysis weak-topology
edited Dec 13 '18 at 17:34
user1101010
9011830
asked Dec 13 '18 at 15:22
FamFam
295
$endgroup$
|
show 2 more comments
$begingroup$
I have a problem, but first I would like to understand your statement. The statement is about a subset $Csubset l^{infty}$ that is convex and closed in $C(l^{infty}, |.|_{sup})$. I know that if $C$ is convex, then $C$ is closed in the weak topology. So far, so good. The problem now asks to show that C is not closed in the weak* topology.
So, how can I say that $Csubset l^{infty}$ is closed in weak* topology if this topology is defined in its dual $l^{infty '}$?
With this well-known, I would like a tip to show that
$$C = {x =(x_i)_{iin mathbb{N}} in l^{infty} : 0leq x_i leq 1, x_i to 1 }$$
is not closed in the topology weak*.
functional-analysis weak-topology
edited Dec 13 '18 at 17:34
user1101010
9011830
asked Dec 13 '18 at 15:22
FamFam
295
$endgroup$
$begingroup$
It is implicitly assumed that the weak* topology on $l^infty$ means that we view it as the dual space of $l^1$.
$endgroup$
– BigbearZzz
Dec 13 '18 at 15:28
$begingroup$
"So, how can I say that C⊂l∞ is closed in weak* topology if this topology is defined in its dual $(ell^infty)^*$?" Here they are considering the weak* topology of $ell^infty$ as the dual of $ell^1$.
$endgroup$
– Federico
Dec 13 '18 at 15:28
$begingroup$
Can you find an element not in $C$ and a sequence in $C$ converging to it?
$endgroup$
– Federico
Dec 13 '18 at 15:33
$begingroup$
Hint: try to find a sequence converging to $0$, which is not in $C$
$endgroup$
– Federico
Dec 13 '18 at 15:34
$begingroup$
It was not clear yet, the convergence sequence has to be in $C$?
$endgroup$
– Fam
Dec 13 '18 at 16:10
|
show 2 more comments
$begingroup$
I have a problem, but first I would like to understand your statement. The statement is about a subset $Csubset l^{infty}$ that is convex and closed in $C(l^{infty}, |.|_{sup})$. I know that if $C$ is convex, then $C$ is closed in the weak topology. So far, so good. The problem now asks to show that C is not closed in the weak* topology.
So, how can I say that $Csubset l^{infty}$ is closed in weak* topology if this topology is defined in its dual $l^{infty '}$?
With this well-known, I would like a tip to show that
$$C = {x =(x_i)_{iin mathbb{N}} in l^{infty} : 0leq x_i leq 1, x_i to 1 }$$
is not closed in the topology weak*.
functional-analysis weak-topology
edited Dec 13 '18 at 17:34
user1101010
9011830
asked Dec 13 '18 at 15:22
FamFam
295
$endgroup$
I have a problem, but first I would like to understand your statement. The statement is about a subset $Csubset l^{infty}$ that is convex and closed in $C(l^{infty}, |.|_{sup})$. I know that if $C$ is convex, then $C$ is closed in the weak topology. So far, so good. The problem now asks to show that C is not closed in the weak* topology.
So, how can I say that $Csubset l^{infty}$ is closed in weak* topology if this topology is defined in its dual $l^{infty '}$?
With this well-known, I would like a tip to show that
$$C = {x =(x_i)_{iin mathbb{N}} in l^{infty} : 0leq x_i leq 1, x_i to 1 }$$
is not closed in the topology weak*.
functional-analysis weak-topology
functional-analysis weak-topology
edited Dec 13 '18 at 17:34
user1101010
9011830
asked Dec 13 '18 at 15:22
FamFam
295
edited Dec 13 '18 at 17:34
user1101010
9011830
asked Dec 13 '18 at 15:22
FamFam
295
edited Dec 13 '18 at 17:34
user1101010
9011830
edited Dec 13 '18 at 17:34
user1101010
9011830
edited Dec 13 '18 at 17:34
user1101010
9011830
9011830
asked Dec 13 '18 at 15:22
FamFam
295
asked Dec 13 '18 at 15:22
FamFam
295
asked Dec 13 '18 at 15:22
FamFam
295
295
$begingroup$
It is implicitly assumed that the weak* topology on $l^infty$ means that we view it as the dual space of $l^1$.
$endgroup$
– BigbearZzz
Dec 13 '18 at 15:28
$begingroup$
"So, how can I say that C⊂l∞ is closed in weak* topology if this topology is defined in its dual $(ell^infty)^*$?" Here they are considering the weak* topology of $ell^infty$ as the dual of $ell^1$.
$endgroup$
– Federico
Dec 13 '18 at 15:28
$begingroup$
Can you find an element not in $C$ and a sequence in $C$ converging to it?
$endgroup$
– Federico
Dec 13 '18 at 15:33
$begingroup$
Hint: try to find a sequence converging to $0$, which is not in $C$
$endgroup$
– Federico
Dec 13 '18 at 15:34
$begingroup$
It was not clear yet, the convergence sequence has to be in $C$?
$endgroup$
– Fam
Dec 13 '18 at 16:10
|
show 2 more comments
$begingroup$
It is implicitly assumed that the weak* topology on $l^infty$ means that we view it as the dual space of $l^1$.
$endgroup$
– BigbearZzz
Dec 13 '18 at 15:28
$begingroup$
"So, how can I say that C⊂l∞ is closed in weak* topology if this topology is defined in its dual $(ell^infty)^*$?" Here they are considering the weak* topology of $ell^infty$ as the dual of $ell^1$.
$endgroup$
– Federico
Dec 13 '18 at 15:28
$begingroup$
Can you find an element not in $C$ and a sequence in $C$ converging to it?
$endgroup$
– Federico
Dec 13 '18 at 15:33
$begingroup$
Hint: try to find a sequence converging to $0$, which is not in $C$
$endgroup$
– Federico
Dec 13 '18 at 15:34
$begingroup$
It was not clear yet, the convergence sequence has to be in $C$?
$endgroup$
– Fam
Dec 13 '18 at 16:10
$begingroup$
It is implicitly assumed that the weak* topology on $l^infty$ means that we view it as the dual space of $l^1$.
$endgroup$
– BigbearZzz
Dec 13 '18 at 15:28
$begingroup$
It is implicitly assumed that the weak* topology on $l^infty$ means that we view it as the dual space of $l^1$.
$endgroup$
– BigbearZzz
Dec 13 '18 at 15:28
$begingroup$
"So, how can I say that C⊂l∞ is closed in weak* topology if this topology is defined in its dual $(ell^infty)^*$?" Here they are considering the weak* topology of $ell^infty$ as the dual of $ell^1$.
$endgroup$
– Federico
Dec 13 '18 at 15:28
$begingroup$
"So, how can I say that C⊂l∞ is closed in weak* topology if this topology is defined in its dual $(ell^infty)^*$?" Here they are considering the weak* topology of $ell^infty$ as the dual of $ell^1$.
$endgroup$
– Federico
Dec 13 '18 at 15:28
$begingroup$
Can you find an element not in $C$ and a sequence in $C$ converging to it?
$endgroup$
– Federico
Dec 13 '18 at 15:33
$begingroup$
Can you find an element not in $C$ and a sequence in $C$ converging to it?
$endgroup$
– Federico
Dec 13 '18 at 15:33
$begingroup$
Hint: try to find a sequence converging to $0$, which is not in $C$
$endgroup$
– Federico
Dec 13 '18 at 15:34
$begingroup$
Hint: try to find a sequence converging to $0$, which is not in $C$
$endgroup$
– Federico
Dec 13 '18 at 15:34
$begingroup$
It was not clear yet, the convergence sequence has to be in $C$?
$endgroup$
– Fam
Dec 13 '18 at 16:10
$begingroup$
It was not clear yet, the convergence sequence has to be in $C$?
$endgroup$
– Fam
Dec 13 '18 at 16:10
|
show 2 more comments
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$begingroup$
It is implicitly assumed that the weak* topology on $l^infty$ means that we view it as the dual space of $l^1$.
$endgroup$
– BigbearZzz
Dec 13 '18 at 15:28
$begingroup$
"So, how can I say that C⊂l∞ is closed in weak* topology if this topology is defined in its dual $(ell^infty)^*$?" Here they are considering the weak* topology of $ell^infty$ as the dual of $ell^1$.
$endgroup$
– Federico
Dec 13 '18 at 15:28
$begingroup$
Can you find an element not in $C$ and a sequence in $C$ converging to it?
$endgroup$
– Federico
Dec 13 '18 at 15:33
$begingroup$
Hint: try to find a sequence converging to $0$, which is not in $C$
$endgroup$
– Federico
Dec 13 '18 at 15:34
$begingroup$
It was not clear yet, the convergence sequence has to be in $C$?
$endgroup$
– Fam
Dec 13 '18 at 16:10