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Is there a norm $N$ such that the unit ball is not compact ? even not closed?
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Is there a normed space $(E,|cdot |)$ such that the unit ball $$xin Emid $$ is not compact ? not closed ? I would say no, but it's a question of an exercise of mine (please not go in weak topology).
functional-analysis
asked Jul 15 at 14:17
Peter
358112
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Is there a normed space $(E,|cdot |)$ such that the unit ball $$xin Emid $$ is not compact ? not closed ? I would say no, but it's a question of an exercise of mine (please not go in weak topology).
functional-analysis
asked Jul 15 at 14:17
Peter
358112
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Is there a normed space $(E,|cdot |)$ such that the unit ball $$xin Emid $$ is not compact ? not closed ? I would say no, but it's a question of an exercise of mine (please not go in weak topology).
functional-analysis
asked Jul 15 at 14:17
Peter
358112
Is there a normed space $(E,|cdot |)$ such that the unit ball $$xin Emid $$ is not compact ? not closed ? I would say no, but it's a question of an exercise of mine (please not go in weak topology).
functional-analysis
asked Jul 15 at 14:17
Peter
358112
asked Jul 15 at 14:17
Peter
358112
asked Jul 15 at 14:17
Peter
358112
asked Jul 15 at 14:17
Peter
358112
358112
add a comment |Â
add a comment |Â
2 Answers
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The unit ball will always be closed: If $x$ is not in the unit ball because $|x|>1$, then the open ball of radius $|x|-1$ around $x$ is disjoint to the uint ball.
However, it is often not compact: Let $E$ be the space of bounded real sequences and $|x|=supx_n$. Then the open balls of radius $frac12$ around arbitrary centers cover the unit ball, but there is no finite subcover.
answered Jul 15 at 14:24
Hagen von Eitzen
265k20258477
Thank you for your answer. I'm unfortunately not allowed to prove the second point. But What could be a sequence of the unit ball that is not bounded in $ell^infty $ ?
– Peter
Jul 15 at 14:27
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As Hagen von Eitzen points out, the unit ball $leq 1$ is always closed, because the map $xmapsto|x|$ is continuous, and $[0,1]$ is a closed subset of $[0,infty)$.
But compactness is different. The unit ball is compact if and only if $E$ is finite-dimensional. This is a consequence of the Riesz Lemma, which can be used to inductively define a sequence in the unit sphere $x$, a closed subset of the unit ball, which has to convergent subsequence.
answered Jul 15 at 15:02
Aweygan
11.9k21437
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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
The unit ball will always be closed: If $x$ is not in the unit ball because $|x|>1$, then the open ball of radius $|x|-1$ around $x$ is disjoint to the uint ball.
However, it is often not compact: Let $E$ be the space of bounded real sequences and $|x|=supx_n$. Then the open balls of radius $frac12$ around arbitrary centers cover the unit ball, but there is no finite subcover.
answered Jul 15 at 14:24
Hagen von Eitzen
265k20258477
Thank you for your answer. I'm unfortunately not allowed to prove the second point. But What could be a sequence of the unit ball that is not bounded in $ell^infty $ ?
– Peter
Jul 15 at 14:27
add a comment |Â
up vote
2
down vote
The unit ball will always be closed: If $x$ is not in the unit ball because $|x|>1$, then the open ball of radius $|x|-1$ around $x$ is disjoint to the uint ball.
However, it is often not compact: Let $E$ be the space of bounded real sequences and $|x|=supx_n$. Then the open balls of radius $frac12$ around arbitrary centers cover the unit ball, but there is no finite subcover.
answered Jul 15 at 14:24
Hagen von Eitzen
265k20258477
Thank you for your answer. I'm unfortunately not allowed to prove the second point. But What could be a sequence of the unit ball that is not bounded in $ell^infty $ ?
– Peter
Jul 15 at 14:27
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The unit ball will always be closed: If $x$ is not in the unit ball because $|x|>1$, then the open ball of radius $|x|-1$ around $x$ is disjoint to the uint ball.
However, it is often not compact: Let $E$ be the space of bounded real sequences and $|x|=supx_n$. Then the open balls of radius $frac12$ around arbitrary centers cover the unit ball, but there is no finite subcover.
answered Jul 15 at 14:24
Hagen von Eitzen
265k20258477
The unit ball will always be closed: If $x$ is not in the unit ball because $|x|>1$, then the open ball of radius $|x|-1$ around $x$ is disjoint to the uint ball.
However, it is often not compact: Let $E$ be the space of bounded real sequences and $|x|=supx_n$. Then the open balls of radius $frac12$ around arbitrary centers cover the unit ball, but there is no finite subcover.
answered Jul 15 at 14:24
Hagen von Eitzen
265k20258477
answered Jul 15 at 14:24
Hagen von Eitzen
265k20258477
answered Jul 15 at 14:24
Hagen von Eitzen
265k20258477
answered Jul 15 at 14:24
Hagen von Eitzen
265k20258477
265k20258477
Thank you for your answer. I'm unfortunately not allowed to prove the second point. But What could be a sequence of the unit ball that is not bounded in $ell^infty $ ?
– Peter
Jul 15 at 14:27
add a comment |Â
Thank you for your answer. I'm unfortunately not allowed to prove the second point. But What could be a sequence of the unit ball that is not bounded in $ell^infty $ ?
– Peter
Jul 15 at 14:27
Thank you for your answer. I'm unfortunately not allowed to prove the second point. But What could be a sequence of the unit ball that is not bounded in $ell^infty $ ?
– Peter
Jul 15 at 14:27
Thank you for your answer. I'm unfortunately not allowed to prove the second point. But What could be a sequence of the unit ball that is not bounded in $ell^infty $ ?
– Peter
Jul 15 at 14:27
add a comment |Â
up vote
2
down vote
As Hagen von Eitzen points out, the unit ball $leq 1$ is always closed, because the map $xmapsto|x|$ is continuous, and $[0,1]$ is a closed subset of $[0,infty)$.
But compactness is different. The unit ball is compact if and only if $E$ is finite-dimensional. This is a consequence of the Riesz Lemma, which can be used to inductively define a sequence in the unit sphere $x$, a closed subset of the unit ball, which has to convergent subsequence.
answered Jul 15 at 15:02
Aweygan
11.9k21437
add a comment |Â
up vote
2
down vote
As Hagen von Eitzen points out, the unit ball $leq 1$ is always closed, because the map $xmapsto|x|$ is continuous, and $[0,1]$ is a closed subset of $[0,infty)$.
But compactness is different. The unit ball is compact if and only if $E$ is finite-dimensional. This is a consequence of the Riesz Lemma, which can be used to inductively define a sequence in the unit sphere $x$, a closed subset of the unit ball, which has to convergent subsequence.
answered Jul 15 at 15:02
Aweygan
11.9k21437
add a comment |Â
up vote
2
down vote
up vote
2
down vote
As Hagen von Eitzen points out, the unit ball $leq 1$ is always closed, because the map $xmapsto|x|$ is continuous, and $[0,1]$ is a closed subset of $[0,infty)$.
But compactness is different. The unit ball is compact if and only if $E$ is finite-dimensional. This is a consequence of the Riesz Lemma, which can be used to inductively define a sequence in the unit sphere $x$, a closed subset of the unit ball, which has to convergent subsequence.
answered Jul 15 at 15:02
Aweygan
11.9k21437
As Hagen von Eitzen points out, the unit ball $leq 1$ is always closed, because the map $xmapsto|x|$ is continuous, and $[0,1]$ is a closed subset of $[0,infty)$.
But compactness is different. The unit ball is compact if and only if $E$ is finite-dimensional. This is a consequence of the Riesz Lemma, which can be used to inductively define a sequence in the unit sphere $x$, a closed subset of the unit ball, which has to convergent subsequence.
answered Jul 15 at 15:02
Aweygan
11.9k21437
answered Jul 15 at 15:02
Aweygan
11.9k21437
answered Jul 15 at 15:02
Aweygan
11.9k21437
answered Jul 15 at 15:02
Aweygan
11.9k21437
11.9k21437
add a comment |Â
add a comment |Â
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