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Does weakly sequentially closed imply weakly closed?
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As stated, I know of course weakly closed implies weakly sequentially closed, but is the converse also true? In other words, I want to know if in weak topology, a subset is weakly closed if and only if it is sequentially closed. aka they are equivalent definitions just analogous to the case in norm topology. I also know that it is not true to say sequentially closed imply closed in ALL topological spaces. But is this true specifically in weak topology? Answers appreciated.
general-topology functional-analysis
asked Jul 25 at 16:49
dantings
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up vote
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As stated, I know of course weakly closed implies weakly sequentially closed, but is the converse also true? In other words, I want to know if in weak topology, a subset is weakly closed if and only if it is sequentially closed. aka they are equivalent definitions just analogous to the case in norm topology. I also know that it is not true to say sequentially closed imply closed in ALL topological spaces. But is this true specifically in weak topology? Answers appreciated.
general-topology functional-analysis
asked Jul 25 at 16:49
dantings
185
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
As stated, I know of course weakly closed implies weakly sequentially closed, but is the converse also true? In other words, I want to know if in weak topology, a subset is weakly closed if and only if it is sequentially closed. aka they are equivalent definitions just analogous to the case in norm topology. I also know that it is not true to say sequentially closed imply closed in ALL topological spaces. But is this true specifically in weak topology? Answers appreciated.
general-topology functional-analysis
asked Jul 25 at 16:49
dantings
185
As stated, I know of course weakly closed implies weakly sequentially closed, but is the converse also true? In other words, I want to know if in weak topology, a subset is weakly closed if and only if it is sequentially closed. aka they are equivalent definitions just analogous to the case in norm topology. I also know that it is not true to say sequentially closed imply closed in ALL topological spaces. But is this true specifically in weak topology? Answers appreciated.
general-topology functional-analysis
asked Jul 25 at 16:49
dantings
185
asked Jul 25 at 16:49
dantings
185
asked Jul 25 at 16:49
dantings
185
asked Jul 25 at 16:49
dantings
185
185
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add a comment |Â
2 Answers
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The converse is false, even in Hilbert space. See Example 3.33 in Bauschke-Combettes's Convex Analysis and Monotone Operator Theory in Hilbert Spaces (second edition), which is a more general version of
@mechanodroid Example.
The converse is true in finite-dimensional Banach spaces or when the set considered is convex.
answered Jul 25 at 16:55
max_zorn
3,15151028
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It isn't true in general.
Let $H$ be an infinite-dimensional Hilbert space and let $(e_n)_n$ be an orthonormal sequence in $H$. Consider $S = sqrtne_n : n in mathbbN subseteq H$.
$S$ is weakly sequentially closed because every weakly convergent sequence in $S$ has to be bounded, so the sequence has to have at most finitely many distinct terms, and hence it stabilizes to an element of $S$.
On the other hand, $0$ is in the weak closure of $S$. Indeed, consider a basic weakly open neighbourhood $U(0, a_1,ldots, a_k, varepsilon) = < varepsilon, 1 le i le k$ with $a_1, ldots, a_k in H$ and $varepsilon > 0$.
Since $sum_n=1^infty |langle a_i, e_nrangle|^2 le |a_i| < +infty$, the series $sum_n=1^inftysum_i=1^k |langle a_i, e_nrangle|^2$ also converges so there exists $ninmathbbN$ such that
$$sum_i=1^k |langle a_i, e_nrangle|^2 < fracvarepsilon^2n$$
(otherwise we would have $sum_n=1^inftysum_i=1^k |langle a_i, e_nrangle|^2 ge sum_n=1^infty fracvarepsilon^2n = +infty$.)
For this $n$ we have $|langle a_i, e_nrangle| < fracvarepsilonsqrtn, forall 1 le i le k$ so $sqrtne_n in U(0, a_1,ldots, a_k, varepsilon)$.
Therefore every basic weakly open neighbourhood of $0$ intersects $S$ so $0 in overlineS^w$. We conclude that $S$ is not weakly closed.
answered Jul 25 at 17:13
mechanodroid
22.2k52041
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The converse is false, even in Hilbert space. See Example 3.33 in Bauschke-Combettes's Convex Analysis and Monotone Operator Theory in Hilbert Spaces (second edition), which is a more general version of
@mechanodroid Example.
The converse is true in finite-dimensional Banach spaces or when the set considered is convex.
answered Jul 25 at 16:55
max_zorn
3,15151028
add a comment |Â
up vote
2
down vote
accepted
The converse is false, even in Hilbert space. See Example 3.33 in Bauschke-Combettes's Convex Analysis and Monotone Operator Theory in Hilbert Spaces (second edition), which is a more general version of
@mechanodroid Example.
The converse is true in finite-dimensional Banach spaces or when the set considered is convex.
answered Jul 25 at 16:55
max_zorn
3,15151028
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The converse is false, even in Hilbert space. See Example 3.33 in Bauschke-Combettes's Convex Analysis and Monotone Operator Theory in Hilbert Spaces (second edition), which is a more general version of
@mechanodroid Example.
The converse is true in finite-dimensional Banach spaces or when the set considered is convex.
answered Jul 25 at 16:55
max_zorn
3,15151028
The converse is false, even in Hilbert space. See Example 3.33 in Bauschke-Combettes's Convex Analysis and Monotone Operator Theory in Hilbert Spaces (second edition), which is a more general version of
@mechanodroid Example.
The converse is true in finite-dimensional Banach spaces or when the set considered is convex.
answered Jul 25 at 16:55
max_zorn
3,15151028
edited Jul 26 at 5:52
answered Jul 25 at 16:55
max_zorn
3,15151028
answered Jul 25 at 16:55
max_zorn
3,15151028
answered Jul 25 at 16:55
max_zorn
3,15151028
3,15151028
add a comment |Â
add a comment |Â
up vote
1
down vote
It isn't true in general.
Let $H$ be an infinite-dimensional Hilbert space and let $(e_n)_n$ be an orthonormal sequence in $H$. Consider $S = sqrtne_n : n in mathbbN subseteq H$.
$S$ is weakly sequentially closed because every weakly convergent sequence in $S$ has to be bounded, so the sequence has to have at most finitely many distinct terms, and hence it stabilizes to an element of $S$.
On the other hand, $0$ is in the weak closure of $S$. Indeed, consider a basic weakly open neighbourhood $U(0, a_1,ldots, a_k, varepsilon) = < varepsilon, 1 le i le k$ with $a_1, ldots, a_k in H$ and $varepsilon > 0$.
Since $sum_n=1^infty |langle a_i, e_nrangle|^2 le |a_i| < +infty$, the series $sum_n=1^inftysum_i=1^k |langle a_i, e_nrangle|^2$ also converges so there exists $ninmathbbN$ such that
$$sum_i=1^k |langle a_i, e_nrangle|^2 < fracvarepsilon^2n$$
(otherwise we would have $sum_n=1^inftysum_i=1^k |langle a_i, e_nrangle|^2 ge sum_n=1^infty fracvarepsilon^2n = +infty$.)
For this $n$ we have $|langle a_i, e_nrangle| < fracvarepsilonsqrtn, forall 1 le i le k$ so $sqrtne_n in U(0, a_1,ldots, a_k, varepsilon)$.
Therefore every basic weakly open neighbourhood of $0$ intersects $S$ so $0 in overlineS^w$. We conclude that $S$ is not weakly closed.
answered Jul 25 at 17:13
mechanodroid
22.2k52041
add a comment |Â
up vote
1
down vote
It isn't true in general.
Let $H$ be an infinite-dimensional Hilbert space and let $(e_n)_n$ be an orthonormal sequence in $H$. Consider $S = sqrtne_n : n in mathbbN subseteq H$.
$S$ is weakly sequentially closed because every weakly convergent sequence in $S$ has to be bounded, so the sequence has to have at most finitely many distinct terms, and hence it stabilizes to an element of $S$.
On the other hand, $0$ is in the weak closure of $S$. Indeed, consider a basic weakly open neighbourhood $U(0, a_1,ldots, a_k, varepsilon) = < varepsilon, 1 le i le k$ with $a_1, ldots, a_k in H$ and $varepsilon > 0$.
Since $sum_n=1^infty |langle a_i, e_nrangle|^2 le |a_i| < +infty$, the series $sum_n=1^inftysum_i=1^k |langle a_i, e_nrangle|^2$ also converges so there exists $ninmathbbN$ such that
$$sum_i=1^k |langle a_i, e_nrangle|^2 < fracvarepsilon^2n$$
(otherwise we would have $sum_n=1^inftysum_i=1^k |langle a_i, e_nrangle|^2 ge sum_n=1^infty fracvarepsilon^2n = +infty$.)
For this $n$ we have $|langle a_i, e_nrangle| < fracvarepsilonsqrtn, forall 1 le i le k$ so $sqrtne_n in U(0, a_1,ldots, a_k, varepsilon)$.
Therefore every basic weakly open neighbourhood of $0$ intersects $S$ so $0 in overlineS^w$. We conclude that $S$ is not weakly closed.
answered Jul 25 at 17:13
mechanodroid
22.2k52041
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It isn't true in general.
Let $H$ be an infinite-dimensional Hilbert space and let $(e_n)_n$ be an orthonormal sequence in $H$. Consider $S = sqrtne_n : n in mathbbN subseteq H$.
$S$ is weakly sequentially closed because every weakly convergent sequence in $S$ has to be bounded, so the sequence has to have at most finitely many distinct terms, and hence it stabilizes to an element of $S$.
On the other hand, $0$ is in the weak closure of $S$. Indeed, consider a basic weakly open neighbourhood $U(0, a_1,ldots, a_k, varepsilon) = < varepsilon, 1 le i le k$ with $a_1, ldots, a_k in H$ and $varepsilon > 0$.
Since $sum_n=1^infty |langle a_i, e_nrangle|^2 le |a_i| < +infty$, the series $sum_n=1^inftysum_i=1^k |langle a_i, e_nrangle|^2$ also converges so there exists $ninmathbbN$ such that
$$sum_i=1^k |langle a_i, e_nrangle|^2 < fracvarepsilon^2n$$
(otherwise we would have $sum_n=1^inftysum_i=1^k |langle a_i, e_nrangle|^2 ge sum_n=1^infty fracvarepsilon^2n = +infty$.)
For this $n$ we have $|langle a_i, e_nrangle| < fracvarepsilonsqrtn, forall 1 le i le k$ so $sqrtne_n in U(0, a_1,ldots, a_k, varepsilon)$.
Therefore every basic weakly open neighbourhood of $0$ intersects $S$ so $0 in overlineS^w$. We conclude that $S$ is not weakly closed.
answered Jul 25 at 17:13
mechanodroid
22.2k52041
It isn't true in general.
Let $H$ be an infinite-dimensional Hilbert space and let $(e_n)_n$ be an orthonormal sequence in $H$. Consider $S = sqrtne_n : n in mathbbN subseteq H$.
$S$ is weakly sequentially closed because every weakly convergent sequence in $S$ has to be bounded, so the sequence has to have at most finitely many distinct terms, and hence it stabilizes to an element of $S$.
On the other hand, $0$ is in the weak closure of $S$. Indeed, consider a basic weakly open neighbourhood $U(0, a_1,ldots, a_k, varepsilon) = < varepsilon, 1 le i le k$ with $a_1, ldots, a_k in H$ and $varepsilon > 0$.
Since $sum_n=1^infty |langle a_i, e_nrangle|^2 le |a_i| < +infty$, the series $sum_n=1^inftysum_i=1^k |langle a_i, e_nrangle|^2$ also converges so there exists $ninmathbbN$ such that
$$sum_i=1^k |langle a_i, e_nrangle|^2 < fracvarepsilon^2n$$
(otherwise we would have $sum_n=1^inftysum_i=1^k |langle a_i, e_nrangle|^2 ge sum_n=1^infty fracvarepsilon^2n = +infty$.)
For this $n$ we have $|langle a_i, e_nrangle| < fracvarepsilonsqrtn, forall 1 le i le k$ so $sqrtne_n in U(0, a_1,ldots, a_k, varepsilon)$.
Therefore every basic weakly open neighbourhood of $0$ intersects $S$ so $0 in overlineS^w$. We conclude that $S$ is not weakly closed.
answered Jul 25 at 17:13
mechanodroid
22.2k52041
answered Jul 25 at 17:13
mechanodroid
22.2k52041
answered Jul 25 at 17:13
mechanodroid
22.2k52041
answered Jul 25 at 17:13
mechanodroid
22.2k52041
22.2k52041
add a comment |Â
add a comment |Â
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