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In “Analyse fonctionnelle†of Brezis, in chapter III why do we need Banach spaces ? (especially for Kakutani's theorem)
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In the book of Brezis : "Analyse fonctionnelle : Théorie et application", chapter III (i.e. construction of weak topology, weak-* topology reflexives spaces...), why do we need "Banach spaces" ? Isn't normed spaces enough ? The particular example I have in mind if theorem III.16 (named as Kakutani) that says : Let $E$ a Banach spaces. Then $$B_E=x$$
is compact for the weak topology $sigma (E,E')$ $iff$ $E$ is reflexive.
I read the proof with attention, and I don't see where we use the fact that $E$ is complete for it's norm. So why do we need the assumption to be Banach ? The only reason for me would be that we use Banach-Steinhaus's theorem (BST) (and thus, we need completeness). But in the proof of Kakutani's theorem I don't see anywhere the used of (BST). So maybe the completeness is used somewhere I don't see ?
functional-analysis weak-topology
edited Aug 29 at 20:12
Bernard
111k635102
asked Aug 29 at 20:05
Peter
549113
add a comment |Â
up vote
5
down vote
favorite
In the book of Brezis : "Analyse fonctionnelle : Théorie et application", chapter III (i.e. construction of weak topology, weak-* topology reflexives spaces...), why do we need "Banach spaces" ? Isn't normed spaces enough ? The particular example I have in mind if theorem III.16 (named as Kakutani) that says : Let $E$ a Banach spaces. Then $$B_E=x$$
is compact for the weak topology $sigma (E,E')$ $iff$ $E$ is reflexive.
I read the proof with attention, and I don't see where we use the fact that $E$ is complete for it's norm. So why do we need the assumption to be Banach ? The only reason for me would be that we use Banach-Steinhaus's theorem (BST) (and thus, we need completeness). But in the proof of Kakutani's theorem I don't see anywhere the used of (BST). So maybe the completeness is used somewhere I don't see ?
functional-analysis weak-topology
edited Aug 29 at 20:12
Bernard
111k635102
asked Aug 29 at 20:05
Peter
549113
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
In the book of Brezis : "Analyse fonctionnelle : Théorie et application", chapter III (i.e. construction of weak topology, weak-* topology reflexives spaces...), why do we need "Banach spaces" ? Isn't normed spaces enough ? The particular example I have in mind if theorem III.16 (named as Kakutani) that says : Let $E$ a Banach spaces. Then $$B_E=x$$
is compact for the weak topology $sigma (E,E')$ $iff$ $E$ is reflexive.
I read the proof with attention, and I don't see where we use the fact that $E$ is complete for it's norm. So why do we need the assumption to be Banach ? The only reason for me would be that we use Banach-Steinhaus's theorem (BST) (and thus, we need completeness). But in the proof of Kakutani's theorem I don't see anywhere the used of (BST). So maybe the completeness is used somewhere I don't see ?
functional-analysis weak-topology
edited Aug 29 at 20:12
Bernard
111k635102
asked Aug 29 at 20:05
Peter
549113
In the book of Brezis : "Analyse fonctionnelle : Théorie et application", chapter III (i.e. construction of weak topology, weak-* topology reflexives spaces...), why do we need "Banach spaces" ? Isn't normed spaces enough ? The particular example I have in mind if theorem III.16 (named as Kakutani) that says : Let $E$ a Banach spaces. Then $$B_E=x$$
is compact for the weak topology $sigma (E,E')$ $iff$ $E$ is reflexive.
I read the proof with attention, and I don't see where we use the fact that $E$ is complete for it's norm. So why do we need the assumption to be Banach ? The only reason for me would be that we use Banach-Steinhaus's theorem (BST) (and thus, we need completeness). But in the proof of Kakutani's theorem I don't see anywhere the used of (BST). So maybe the completeness is used somewhere I don't see ?
functional-analysis weak-topology
edited Aug 29 at 20:12
Bernard
111k635102
asked Aug 29 at 20:05
Peter
549113
edited Aug 29 at 20:12
Bernard
111k635102
edited Aug 29 at 20:12
Bernard
111k635102
edited Aug 29 at 20:12
Bernard
111k635102
111k635102
asked Aug 29 at 20:05
Peter
549113
asked Aug 29 at 20:05
Peter
549113
asked Aug 29 at 20:05
Peter
549113
549113
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
7
down vote
accepted
Indeed the equivalence still holds if $E$ is an incomplete normed space (over $mathbbR$ or $mathbbC$): both sides are false. This is pretty easy to see directly and really misses the point of the theorem. So the authors probably just decided not to bother to include this relatively uninteresting case.
It's pretty common in functional analysis to write theorems that only cover Banach spaces, even when normed spaces could also be included. This can be for any of several reasons:
In most applications, you are working with Banach spaces
The theorem may become trivial for incomplete spaces
For an incomplete space $X$, the theorem gives you the "right" conclusion if you apply it to the completion of $X$.
answered Aug 29 at 20:15
Nate Eldredge
59.9k577162
Thank you very much for your answer. Could please point me where the author used the fact that $E$ is a Banach space ? (because if both side are false in non Banach space, to justify that at least one side is true in Banach side we must use completeness, don't we ?)
– Peter
Aug 29 at 20:29
As Eduardo point, on $mathbb R$ as a $mathbb Q-$vector space, the unit ball is compact. So it hold in incomplete vector space (as I understood, you said that both side are false in incomplete vector space). So could you please give me more explanation ? :)
– Peter
Aug 29 at 20:36
Leaving $mathbbQ$-vector spaces aside, I am not convinced that on a non-complete normed space the unit ball cannot be weakly compact. To me this does not seem 'easy to see' at least :(
– Lorenzo Quarisa
Aug 29 at 21:32
2
@Peter: No, you misunderstand me. The theorem is true for incomplete spaces as well as complete spaces. (If $E$ is incomplete, then it's true that $E$ has a weakly compact ball iff it is reflexive - because it can never have a weakly compact ball and it is never reflexive). So it's quite possible the author never used the assumption of completeness. (I don't have a copy of the book to see what's actually written.)
– Nate Eldredge
Aug 29 at 21:40
2
@LorenzoQuarisa Let $E$ be an incomplete normed space (real or complex), $tildeE$ its completion. Then $tildeE' = E'$, and $sigma(E,E')$ is the subspace topology induced from $sigma(tildeE,E')$. The closed unit ball of $E$ is norm-dense in the closed unit ball of $tildeE$, hence it is a fortiori $sigma(tildeE,E')$-dense in it. In particular, it is not $sigma(tildeE,E')$-closed, which it would be if it were $sigma(E,E')$-compact.
– Daniel Fischer♦
Aug 29 at 21:47
 |Â
show 2 more comments
up vote
2
down vote
Consider $mathbbR$ as an vector space over $mathbbQ$ the rationals space endowed with the norm of absolute value, then $mathbbR$ over $mathbbQ$ is an infinite dimensional normed vector space where the ball is compact.
answered Aug 29 at 20:14
Eduardo
396112
1
Interesting, but I think that in the definition of normed space we require either a real or complex vector space, so this is not a normed space.
– Lorenzo Quarisa
Aug 29 at 20:21
I think OP means "normed linear space," which at least Wikipedia takes to mean a vector space over $mathbb R$ or $mathbb C.$ en.wikipedia.org/wiki/Normed_vector_space
– Thomas Andrews
Aug 29 at 20:22
Can you provide me a reference, I meant if I'm looking for this definition in the Bourbaki or other classical book I will find this warning? I have this doubt, but I have never found a reference that said this with all words...
– Eduardo
Aug 29 at 20:22
1
@RobArthan I just trying to learn with you guys, If is told to me that I'm wrong It is my pleasure to kown why... I'm not pretend know everything, I'm free to be failure...
– Eduardo
Aug 29 at 20:35
1
How? The definition of the unit ball and its compactness only depend on the topology of $mathbbR$, and not on which field it is considered as a vector space.
– Lorenzo Quarisa
Aug 29 at 20:55
 |Â
show 9 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
Indeed the equivalence still holds if $E$ is an incomplete normed space (over $mathbbR$ or $mathbbC$): both sides are false. This is pretty easy to see directly and really misses the point of the theorem. So the authors probably just decided not to bother to include this relatively uninteresting case.
It's pretty common in functional analysis to write theorems that only cover Banach spaces, even when normed spaces could also be included. This can be for any of several reasons:
In most applications, you are working with Banach spaces
The theorem may become trivial for incomplete spaces
For an incomplete space $X$, the theorem gives you the "right" conclusion if you apply it to the completion of $X$.
answered Aug 29 at 20:15
Nate Eldredge
59.9k577162
Thank you very much for your answer. Could please point me where the author used the fact that $E$ is a Banach space ? (because if both side are false in non Banach space, to justify that at least one side is true in Banach side we must use completeness, don't we ?)
– Peter
Aug 29 at 20:29
As Eduardo point, on $mathbb R$ as a $mathbb Q-$vector space, the unit ball is compact. So it hold in incomplete vector space (as I understood, you said that both side are false in incomplete vector space). So could you please give me more explanation ? :)
– Peter
Aug 29 at 20:36
Leaving $mathbbQ$-vector spaces aside, I am not convinced that on a non-complete normed space the unit ball cannot be weakly compact. To me this does not seem 'easy to see' at least :(
– Lorenzo Quarisa
Aug 29 at 21:32
2
@Peter: No, you misunderstand me. The theorem is true for incomplete spaces as well as complete spaces. (If $E$ is incomplete, then it's true that $E$ has a weakly compact ball iff it is reflexive - because it can never have a weakly compact ball and it is never reflexive). So it's quite possible the author never used the assumption of completeness. (I don't have a copy of the book to see what's actually written.)
– Nate Eldredge
Aug 29 at 21:40
2
@LorenzoQuarisa Let $E$ be an incomplete normed space (real or complex), $tildeE$ its completion. Then $tildeE' = E'$, and $sigma(E,E')$ is the subspace topology induced from $sigma(tildeE,E')$. The closed unit ball of $E$ is norm-dense in the closed unit ball of $tildeE$, hence it is a fortiori $sigma(tildeE,E')$-dense in it. In particular, it is not $sigma(tildeE,E')$-closed, which it would be if it were $sigma(E,E')$-compact.
– Daniel Fischer♦
Aug 29 at 21:47
 |Â
show 2 more comments
up vote
7
down vote
accepted
Indeed the equivalence still holds if $E$ is an incomplete normed space (over $mathbbR$ or $mathbbC$): both sides are false. This is pretty easy to see directly and really misses the point of the theorem. So the authors probably just decided not to bother to include this relatively uninteresting case.
It's pretty common in functional analysis to write theorems that only cover Banach spaces, even when normed spaces could also be included. This can be for any of several reasons:
In most applications, you are working with Banach spaces
The theorem may become trivial for incomplete spaces
For an incomplete space $X$, the theorem gives you the "right" conclusion if you apply it to the completion of $X$.
answered Aug 29 at 20:15
Nate Eldredge
59.9k577162
Thank you very much for your answer. Could please point me where the author used the fact that $E$ is a Banach space ? (because if both side are false in non Banach space, to justify that at least one side is true in Banach side we must use completeness, don't we ?)
– Peter
Aug 29 at 20:29
As Eduardo point, on $mathbb R$ as a $mathbb Q-$vector space, the unit ball is compact. So it hold in incomplete vector space (as I understood, you said that both side are false in incomplete vector space). So could you please give me more explanation ? :)
– Peter
Aug 29 at 20:36
Leaving $mathbbQ$-vector spaces aside, I am not convinced that on a non-complete normed space the unit ball cannot be weakly compact. To me this does not seem 'easy to see' at least :(
– Lorenzo Quarisa
Aug 29 at 21:32
2
@Peter: No, you misunderstand me. The theorem is true for incomplete spaces as well as complete spaces. (If $E$ is incomplete, then it's true that $E$ has a weakly compact ball iff it is reflexive - because it can never have a weakly compact ball and it is never reflexive). So it's quite possible the author never used the assumption of completeness. (I don't have a copy of the book to see what's actually written.)
– Nate Eldredge
Aug 29 at 21:40
2
@LorenzoQuarisa Let $E$ be an incomplete normed space (real or complex), $tildeE$ its completion. Then $tildeE' = E'$, and $sigma(E,E')$ is the subspace topology induced from $sigma(tildeE,E')$. The closed unit ball of $E$ is norm-dense in the closed unit ball of $tildeE$, hence it is a fortiori $sigma(tildeE,E')$-dense in it. In particular, it is not $sigma(tildeE,E')$-closed, which it would be if it were $sigma(E,E')$-compact.
– Daniel Fischer♦
Aug 29 at 21:47
 |Â
show 2 more comments
up vote
7
down vote
accepted
up vote
7
down vote
accepted
Indeed the equivalence still holds if $E$ is an incomplete normed space (over $mathbbR$ or $mathbbC$): both sides are false. This is pretty easy to see directly and really misses the point of the theorem. So the authors probably just decided not to bother to include this relatively uninteresting case.
It's pretty common in functional analysis to write theorems that only cover Banach spaces, even when normed spaces could also be included. This can be for any of several reasons:
In most applications, you are working with Banach spaces
The theorem may become trivial for incomplete spaces
For an incomplete space $X$, the theorem gives you the "right" conclusion if you apply it to the completion of $X$.
answered Aug 29 at 20:15
Nate Eldredge
59.9k577162
Indeed the equivalence still holds if $E$ is an incomplete normed space (over $mathbbR$ or $mathbbC$): both sides are false. This is pretty easy to see directly and really misses the point of the theorem. So the authors probably just decided not to bother to include this relatively uninteresting case.
It's pretty common in functional analysis to write theorems that only cover Banach spaces, even when normed spaces could also be included. This can be for any of several reasons:
In most applications, you are working with Banach spaces
The theorem may become trivial for incomplete spaces
For an incomplete space $X$, the theorem gives you the "right" conclusion if you apply it to the completion of $X$.
answered Aug 29 at 20:15
Nate Eldredge
59.9k577162
answered Aug 29 at 20:15
Nate Eldredge
59.9k577162
answered Aug 29 at 20:15
Nate Eldredge
59.9k577162
answered Aug 29 at 20:15
Nate Eldredge
59.9k577162
59.9k577162
Thank you very much for your answer. Could please point me where the author used the fact that $E$ is a Banach space ? (because if both side are false in non Banach space, to justify that at least one side is true in Banach side we must use completeness, don't we ?)
– Peter
Aug 29 at 20:29
As Eduardo point, on $mathbb R$ as a $mathbb Q-$vector space, the unit ball is compact. So it hold in incomplete vector space (as I understood, you said that both side are false in incomplete vector space). So could you please give me more explanation ? :)
– Peter
Aug 29 at 20:36
Leaving $mathbbQ$-vector spaces aside, I am not convinced that on a non-complete normed space the unit ball cannot be weakly compact. To me this does not seem 'easy to see' at least :(
– Lorenzo Quarisa
Aug 29 at 21:32
2
@Peter: No, you misunderstand me. The theorem is true for incomplete spaces as well as complete spaces. (If $E$ is incomplete, then it's true that $E$ has a weakly compact ball iff it is reflexive - because it can never have a weakly compact ball and it is never reflexive). So it's quite possible the author never used the assumption of completeness. (I don't have a copy of the book to see what's actually written.)
– Nate Eldredge
Aug 29 at 21:40
2
@LorenzoQuarisa Let $E$ be an incomplete normed space (real or complex), $tildeE$ its completion. Then $tildeE' = E'$, and $sigma(E,E')$ is the subspace topology induced from $sigma(tildeE,E')$. The closed unit ball of $E$ is norm-dense in the closed unit ball of $tildeE$, hence it is a fortiori $sigma(tildeE,E')$-dense in it. In particular, it is not $sigma(tildeE,E')$-closed, which it would be if it were $sigma(E,E')$-compact.
– Daniel Fischer♦
Aug 29 at 21:47
 |Â
show 2 more comments
Thank you very much for your answer. Could please point me where the author used the fact that $E$ is a Banach space ? (because if both side are false in non Banach space, to justify that at least one side is true in Banach side we must use completeness, don't we ?)
– Peter
Aug 29 at 20:29
As Eduardo point, on $mathbb R$ as a $mathbb Q-$vector space, the unit ball is compact. So it hold in incomplete vector space (as I understood, you said that both side are false in incomplete vector space). So could you please give me more explanation ? :)
– Peter
Aug 29 at 20:36
Leaving $mathbbQ$-vector spaces aside, I am not convinced that on a non-complete normed space the unit ball cannot be weakly compact. To me this does not seem 'easy to see' at least :(
– Lorenzo Quarisa
Aug 29 at 21:32
2
@Peter: No, you misunderstand me. The theorem is true for incomplete spaces as well as complete spaces. (If $E$ is incomplete, then it's true that $E$ has a weakly compact ball iff it is reflexive - because it can never have a weakly compact ball and it is never reflexive). So it's quite possible the author never used the assumption of completeness. (I don't have a copy of the book to see what's actually written.)
– Nate Eldredge
Aug 29 at 21:40
2
@LorenzoQuarisa Let $E$ be an incomplete normed space (real or complex), $tildeE$ its completion. Then $tildeE' = E'$, and $sigma(E,E')$ is the subspace topology induced from $sigma(tildeE,E')$. The closed unit ball of $E$ is norm-dense in the closed unit ball of $tildeE$, hence it is a fortiori $sigma(tildeE,E')$-dense in it. In particular, it is not $sigma(tildeE,E')$-closed, which it would be if it were $sigma(E,E')$-compact.
– Daniel Fischer♦
Aug 29 at 21:47
Thank you very much for your answer. Could please point me where the author used the fact that $E$ is a Banach space ? (because if both side are false in non Banach space, to justify that at least one side is true in Banach side we must use completeness, don't we ?)
– Peter
Aug 29 at 20:29
Thank you very much for your answer. Could please point me where the author used the fact that $E$ is a Banach space ? (because if both side are false in non Banach space, to justify that at least one side is true in Banach side we must use completeness, don't we ?)
– Peter
Aug 29 at 20:29
As Eduardo point, on $mathbb R$ as a $mathbb Q-$vector space, the unit ball is compact. So it hold in incomplete vector space (as I understood, you said that both side are false in incomplete vector space). So could you please give me more explanation ? :)
– Peter
Aug 29 at 20:36
As Eduardo point, on $mathbb R$ as a $mathbb Q-$vector space, the unit ball is compact. So it hold in incomplete vector space (as I understood, you said that both side are false in incomplete vector space). So could you please give me more explanation ? :)
– Peter
Aug 29 at 20:36
Leaving $mathbbQ$-vector spaces aside, I am not convinced that on a non-complete normed space the unit ball cannot be weakly compact. To me this does not seem 'easy to see' at least :(
– Lorenzo Quarisa
Aug 29 at 21:32
Leaving $mathbbQ$-vector spaces aside, I am not convinced that on a non-complete normed space the unit ball cannot be weakly compact. To me this does not seem 'easy to see' at least :(
– Lorenzo Quarisa
Aug 29 at 21:32
2
2
@Peter: No, you misunderstand me. The theorem is true for incomplete spaces as well as complete spaces. (If $E$ is incomplete, then it's true that $E$ has a weakly compact ball iff it is reflexive - because it can never have a weakly compact ball and it is never reflexive). So it's quite possible the author never used the assumption of completeness. (I don't have a copy of the book to see what's actually written.)
– Nate Eldredge
Aug 29 at 21:40
@Peter: No, you misunderstand me. The theorem is true for incomplete spaces as well as complete spaces. (If $E$ is incomplete, then it's true that $E$ has a weakly compact ball iff it is reflexive - because it can never have a weakly compact ball and it is never reflexive). So it's quite possible the author never used the assumption of completeness. (I don't have a copy of the book to see what's actually written.)
– Nate Eldredge
Aug 29 at 21:40
2
2
@LorenzoQuarisa Let $E$ be an incomplete normed space (real or complex), $tildeE$ its completion. Then $tildeE' = E'$, and $sigma(E,E')$ is the subspace topology induced from $sigma(tildeE,E')$. The closed unit ball of $E$ is norm-dense in the closed unit ball of $tildeE$, hence it is a fortiori $sigma(tildeE,E')$-dense in it. In particular, it is not $sigma(tildeE,E')$-closed, which it would be if it were $sigma(E,E')$-compact.
– Daniel Fischer♦
Aug 29 at 21:47
@LorenzoQuarisa Let $E$ be an incomplete normed space (real or complex), $tildeE$ its completion. Then $tildeE' = E'$, and $sigma(E,E')$ is the subspace topology induced from $sigma(tildeE,E')$. The closed unit ball of $E$ is norm-dense in the closed unit ball of $tildeE$, hence it is a fortiori $sigma(tildeE,E')$-dense in it. In particular, it is not $sigma(tildeE,E')$-closed, which it would be if it were $sigma(E,E')$-compact.
– Daniel Fischer♦
Aug 29 at 21:47
 |Â
show 2 more comments
up vote
2
down vote
Consider $mathbbR$ as an vector space over $mathbbQ$ the rationals space endowed with the norm of absolute value, then $mathbbR$ over $mathbbQ$ is an infinite dimensional normed vector space where the ball is compact.
answered Aug 29 at 20:14
Eduardo
396112
1
Interesting, but I think that in the definition of normed space we require either a real or complex vector space, so this is not a normed space.
– Lorenzo Quarisa
Aug 29 at 20:21
I think OP means "normed linear space," which at least Wikipedia takes to mean a vector space over $mathbb R$ or $mathbb C.$ en.wikipedia.org/wiki/Normed_vector_space
– Thomas Andrews
Aug 29 at 20:22
Can you provide me a reference, I meant if I'm looking for this definition in the Bourbaki or other classical book I will find this warning? I have this doubt, but I have never found a reference that said this with all words...
– Eduardo
Aug 29 at 20:22
1
@RobArthan I just trying to learn with you guys, If is told to me that I'm wrong It is my pleasure to kown why... I'm not pretend know everything, I'm free to be failure...
– Eduardo
Aug 29 at 20:35
1
How? The definition of the unit ball and its compactness only depend on the topology of $mathbbR$, and not on which field it is considered as a vector space.
– Lorenzo Quarisa
Aug 29 at 20:55
 |Â
show 9 more comments
up vote
2
down vote
Consider $mathbbR$ as an vector space over $mathbbQ$ the rationals space endowed with the norm of absolute value, then $mathbbR$ over $mathbbQ$ is an infinite dimensional normed vector space where the ball is compact.
answered Aug 29 at 20:14
Eduardo
396112
1
Interesting, but I think that in the definition of normed space we require either a real or complex vector space, so this is not a normed space.
– Lorenzo Quarisa
Aug 29 at 20:21
I think OP means "normed linear space," which at least Wikipedia takes to mean a vector space over $mathbb R$ or $mathbb C.$ en.wikipedia.org/wiki/Normed_vector_space
– Thomas Andrews
Aug 29 at 20:22
Can you provide me a reference, I meant if I'm looking for this definition in the Bourbaki or other classical book I will find this warning? I have this doubt, but I have never found a reference that said this with all words...
– Eduardo
Aug 29 at 20:22
1
@RobArthan I just trying to learn with you guys, If is told to me that I'm wrong It is my pleasure to kown why... I'm not pretend know everything, I'm free to be failure...
– Eduardo
Aug 29 at 20:35
1
How? The definition of the unit ball and its compactness only depend on the topology of $mathbbR$, and not on which field it is considered as a vector space.
– Lorenzo Quarisa
Aug 29 at 20:55
 |Â
show 9 more comments
up vote
2
down vote
up vote
2
down vote
Consider $mathbbR$ as an vector space over $mathbbQ$ the rationals space endowed with the norm of absolute value, then $mathbbR$ over $mathbbQ$ is an infinite dimensional normed vector space where the ball is compact.
answered Aug 29 at 20:14
Eduardo
396112
Consider $mathbbR$ as an vector space over $mathbbQ$ the rationals space endowed with the norm of absolute value, then $mathbbR$ over $mathbbQ$ is an infinite dimensional normed vector space where the ball is compact.
answered Aug 29 at 20:14
Eduardo
396112
answered Aug 29 at 20:14
Eduardo
396112
answered Aug 29 at 20:14
Eduardo
396112
answered Aug 29 at 20:14
Eduardo
396112
396112
1
Interesting, but I think that in the definition of normed space we require either a real or complex vector space, so this is not a normed space.
– Lorenzo Quarisa
Aug 29 at 20:21
I think OP means "normed linear space," which at least Wikipedia takes to mean a vector space over $mathbb R$ or $mathbb C.$ en.wikipedia.org/wiki/Normed_vector_space
– Thomas Andrews
Aug 29 at 20:22
Can you provide me a reference, I meant if I'm looking for this definition in the Bourbaki or other classical book I will find this warning? I have this doubt, but I have never found a reference that said this with all words...
– Eduardo
Aug 29 at 20:22
1
@RobArthan I just trying to learn with you guys, If is told to me that I'm wrong It is my pleasure to kown why... I'm not pretend know everything, I'm free to be failure...
– Eduardo
Aug 29 at 20:35
1
How? The definition of the unit ball and its compactness only depend on the topology of $mathbbR$, and not on which field it is considered as a vector space.
– Lorenzo Quarisa
Aug 29 at 20:55
 |Â
show 9 more comments
1
Interesting, but I think that in the definition of normed space we require either a real or complex vector space, so this is not a normed space.
– Lorenzo Quarisa
Aug 29 at 20:21
I think OP means "normed linear space," which at least Wikipedia takes to mean a vector space over $mathbb R$ or $mathbb C.$ en.wikipedia.org/wiki/Normed_vector_space
– Thomas Andrews
Aug 29 at 20:22
Can you provide me a reference, I meant if I'm looking for this definition in the Bourbaki or other classical book I will find this warning? I have this doubt, but I have never found a reference that said this with all words...
– Eduardo
Aug 29 at 20:22
1
@RobArthan I just trying to learn with you guys, If is told to me that I'm wrong It is my pleasure to kown why... I'm not pretend know everything, I'm free to be failure...
– Eduardo
Aug 29 at 20:35
1
How? The definition of the unit ball and its compactness only depend on the topology of $mathbbR$, and not on which field it is considered as a vector space.
– Lorenzo Quarisa
Aug 29 at 20:55
1
1
Interesting, but I think that in the definition of normed space we require either a real or complex vector space, so this is not a normed space.
– Lorenzo Quarisa
Aug 29 at 20:21
Interesting, but I think that in the definition of normed space we require either a real or complex vector space, so this is not a normed space.
– Lorenzo Quarisa
Aug 29 at 20:21
I think OP means "normed linear space," which at least Wikipedia takes to mean a vector space over $mathbb R$ or $mathbb C.$ en.wikipedia.org/wiki/Normed_vector_space
– Thomas Andrews
Aug 29 at 20:22
I think OP means "normed linear space," which at least Wikipedia takes to mean a vector space over $mathbb R$ or $mathbb C.$ en.wikipedia.org/wiki/Normed_vector_space
– Thomas Andrews
Aug 29 at 20:22
Can you provide me a reference, I meant if I'm looking for this definition in the Bourbaki or other classical book I will find this warning? I have this doubt, but I have never found a reference that said this with all words...
– Eduardo
Aug 29 at 20:22
Can you provide me a reference, I meant if I'm looking for this definition in the Bourbaki or other classical book I will find this warning? I have this doubt, but I have never found a reference that said this with all words...
– Eduardo
Aug 29 at 20:22
1
1
@RobArthan I just trying to learn with you guys, If is told to me that I'm wrong It is my pleasure to kown why... I'm not pretend know everything, I'm free to be failure...
– Eduardo
Aug 29 at 20:35
@RobArthan I just trying to learn with you guys, If is told to me that I'm wrong It is my pleasure to kown why... I'm not pretend know everything, I'm free to be failure...
– Eduardo
Aug 29 at 20:35
1
1
How? The definition of the unit ball and its compactness only depend on the topology of $mathbbR$, and not on which field it is considered as a vector space.
– Lorenzo Quarisa
Aug 29 at 20:55
How? The definition of the unit ball and its compactness only depend on the topology of $mathbbR$, and not on which field it is considered as a vector space.
– Lorenzo Quarisa
Aug 29 at 20:55
 |Â
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