Closed unit ball in infinite dimensional normed linear space
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I have to prove that in any infinite dimension normed linear space we have that the closed unit ball is not compact.
I know that I have to construct a sequence such that $||x_n||=1$ and $|x_m-x_n|geq frac{1}{2}$. If I can do this then we have a bounded sequence with no convergent subsequence so that this space is not compact but I have no idea how to actually find such a sequence?
Thanks for any help
analysis hilbert-spaces
asked Apr 10 '13 at 14:55
hmmmmhmmmm
2,65922760
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add a comment |
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I have to prove that in any infinite dimension normed linear space we have that the closed unit ball is not compact.
I know that I have to construct a sequence such that $||x_n||=1$ and $|x_m-x_n|geq frac{1}{2}$. If I can do this then we have a bounded sequence with no convergent subsequence so that this space is not compact but I have no idea how to actually find such a sequence?
Thanks for any help
analysis hilbert-spaces
asked Apr 10 '13 at 14:55
hmmmmhmmmm
2,65922760
$endgroup$
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One approach is to use Riesz's Lemma. See this post. A proof of the lemma can be found here.
$endgroup$
– David Mitra
Apr 10 '13 at 15:08
add a comment |
$begingroup$
I have to prove that in any infinite dimension normed linear space we have that the closed unit ball is not compact.
I know that I have to construct a sequence such that $||x_n||=1$ and $|x_m-x_n|geq frac{1}{2}$. If I can do this then we have a bounded sequence with no convergent subsequence so that this space is not compact but I have no idea how to actually find such a sequence?
Thanks for any help
analysis hilbert-spaces
asked Apr 10 '13 at 14:55
hmmmmhmmmm
2,65922760
$endgroup$
I have to prove that in any infinite dimension normed linear space we have that the closed unit ball is not compact.
I know that I have to construct a sequence such that $||x_n||=1$ and $|x_m-x_n|geq frac{1}{2}$. If I can do this then we have a bounded sequence with no convergent subsequence so that this space is not compact but I have no idea how to actually find such a sequence?
Thanks for any help
analysis hilbert-spaces
analysis hilbert-spaces
asked Apr 10 '13 at 14:55
hmmmmhmmmm
2,65922760
asked Apr 10 '13 at 14:55
hmmmmhmmmm
2,65922760
asked Apr 10 '13 at 14:55
hmmmmhmmmm
2,65922760
asked Apr 10 '13 at 14:55
hmmmmhmmmm
2,65922760
asked Apr 10 '13 at 14:55
hmmmmhmmmm
2,65922760
2,65922760
$begingroup$
One approach is to use Riesz's Lemma. See this post. A proof of the lemma can be found here.
$endgroup$
– David Mitra
Apr 10 '13 at 15:08
add a comment |
$begingroup$
One approach is to use Riesz's Lemma. See this post. A proof of the lemma can be found here.
$endgroup$
– David Mitra
Apr 10 '13 at 15:08
$begingroup$
One approach is to use Riesz's Lemma. See this post. A proof of the lemma can be found here.
$endgroup$
– David Mitra
Apr 10 '13 at 15:08
$begingroup$
One approach is to use Riesz's Lemma. See this post. A proof of the lemma can be found here.
$endgroup$
– David Mitra
Apr 10 '13 at 15:08
add a comment |
1 Answer
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Idea for an indirect proof: Suposse unit ball is compact. Cover it by $cup_{xin B_1 (0)} B_{frac{1}{2}} (x) $. By compactness, there exist points finitely many points $a_i, i=1,dots,n$ such that balls of radius $frac{1}2$ cover unit ball. Then it should be that your space is infact closure of linear span of $a_i$.
answered Apr 29 '16 at 12:50
user133929user133929
164112
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
Idea for an indirect proof: Suposse unit ball is compact. Cover it by $cup_{xin B_1 (0)} B_{frac{1}{2}} (x) $. By compactness, there exist points finitely many points $a_i, i=1,dots,n$ such that balls of radius $frac{1}2$ cover unit ball. Then it should be that your space is infact closure of linear span of $a_i$.
answered Apr 29 '16 at 12:50
user133929user133929
164112
$endgroup$
add a comment |
$begingroup$
Idea for an indirect proof: Suposse unit ball is compact. Cover it by $cup_{xin B_1 (0)} B_{frac{1}{2}} (x) $. By compactness, there exist points finitely many points $a_i, i=1,dots,n$ such that balls of radius $frac{1}2$ cover unit ball. Then it should be that your space is infact closure of linear span of $a_i$.
answered Apr 29 '16 at 12:50
user133929user133929
164112
$endgroup$
add a comment |
$begingroup$
Idea for an indirect proof: Suposse unit ball is compact. Cover it by $cup_{xin B_1 (0)} B_{frac{1}{2}} (x) $. By compactness, there exist points finitely many points $a_i, i=1,dots,n$ such that balls of radius $frac{1}2$ cover unit ball. Then it should be that your space is infact closure of linear span of $a_i$.
answered Apr 29 '16 at 12:50
user133929user133929
164112
$endgroup$
Idea for an indirect proof: Suposse unit ball is compact. Cover it by $cup_{xin B_1 (0)} B_{frac{1}{2}} (x) $. By compactness, there exist points finitely many points $a_i, i=1,dots,n$ such that balls of radius $frac{1}2$ cover unit ball. Then it should be that your space is infact closure of linear span of $a_i$.
answered Apr 29 '16 at 12:50
user133929user133929
164112
edited Apr 29 '16 at 13:31
answered Apr 29 '16 at 12:50
user133929user133929
164112
answered Apr 29 '16 at 12:50
user133929user133929
164112
answered Apr 29 '16 at 12:50
user133929user133929
164112
164112
add a comment |
add a comment |
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$begingroup$
One approach is to use Riesz's Lemma. See this post. A proof of the lemma can be found here.
$endgroup$
– David Mitra
Apr 10 '13 at 15:08