A Banach space is $sigma$-compact iff finite dimensional
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Reading this question I learned that a Banach space is $sigma$-compact iff finite dimensional. The $impliedby$ direction is obvious, but I cannot find a proof of the $implies$ direction, I checked Conway, Brezis and DiBenedetto, but they barely mention $sigma$-compactness.
How do you prove this fact or where can I find a proof of it?
general-topology functional-analysis reference-request compactness
asked Dec 28 '18 at 16:25
Alessandro CodenottiAlessandro Codenotti
3,83511639
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add a comment |
$begingroup$
Reading this question I learned that a Banach space is $sigma$-compact iff finite dimensional. The $impliedby$ direction is obvious, but I cannot find a proof of the $implies$ direction, I checked Conway, Brezis and DiBenedetto, but they barely mention $sigma$-compactness.
How do you prove this fact or where can I find a proof of it?
general-topology functional-analysis reference-request compactness
asked Dec 28 '18 at 16:25
Alessandro CodenottiAlessandro Codenotti
3,83511639
$endgroup$
$begingroup$
I believe (check this) that it is possible to find an uncountable set $S$ with no limit point in any infinite-dimensional Banach space $X$. If $X$ is $sigma$-compact, then it is the union of countably many compact sets. At least one of these must contain uncountably many elements of $S$. By compactness, $S$ has a limit point, a contradiction.
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– Ben W
Dec 28 '18 at 16:47
add a comment |
$begingroup$
Reading this question I learned that a Banach space is $sigma$-compact iff finite dimensional. The $impliedby$ direction is obvious, but I cannot find a proof of the $implies$ direction, I checked Conway, Brezis and DiBenedetto, but they barely mention $sigma$-compactness.
How do you prove this fact or where can I find a proof of it?
general-topology functional-analysis reference-request compactness
asked Dec 28 '18 at 16:25
Alessandro CodenottiAlessandro Codenotti
3,83511639
$endgroup$
Reading this question I learned that a Banach space is $sigma$-compact iff finite dimensional. The $impliedby$ direction is obvious, but I cannot find a proof of the $implies$ direction, I checked Conway, Brezis and DiBenedetto, but they barely mention $sigma$-compactness.
How do you prove this fact or where can I find a proof of it?
general-topology functional-analysis reference-request compactness
general-topology functional-analysis reference-request compactness
asked Dec 28 '18 at 16:25
Alessandro CodenottiAlessandro Codenotti
3,83511639
asked Dec 28 '18 at 16:25
Alessandro CodenottiAlessandro Codenotti
3,83511639
asked Dec 28 '18 at 16:25
Alessandro CodenottiAlessandro Codenotti
3,83511639
asked Dec 28 '18 at 16:25
Alessandro CodenottiAlessandro Codenotti
3,83511639
asked Dec 28 '18 at 16:25
Alessandro CodenottiAlessandro Codenotti
3,83511639
3,83511639
$begingroup$
I believe (check this) that it is possible to find an uncountable set $S$ with no limit point in any infinite-dimensional Banach space $X$. If $X$ is $sigma$-compact, then it is the union of countably many compact sets. At least one of these must contain uncountably many elements of $S$. By compactness, $S$ has a limit point, a contradiction.
$endgroup$
– Ben W
Dec 28 '18 at 16:47
add a comment |
$begingroup$
I believe (check this) that it is possible to find an uncountable set $S$ with no limit point in any infinite-dimensional Banach space $X$. If $X$ is $sigma$-compact, then it is the union of countably many compact sets. At least one of these must contain uncountably many elements of $S$. By compactness, $S$ has a limit point, a contradiction.
$endgroup$
– Ben W
Dec 28 '18 at 16:47
$begingroup$
I believe (check this) that it is possible to find an uncountable set $S$ with no limit point in any infinite-dimensional Banach space $X$. If $X$ is $sigma$-compact, then it is the union of countably many compact sets. At least one of these must contain uncountably many elements of $S$. By compactness, $S$ has a limit point, a contradiction.
$endgroup$
– Ben W
Dec 28 '18 at 16:47
$begingroup$
I believe (check this) that it is possible to find an uncountable set $S$ with no limit point in any infinite-dimensional Banach space $X$. If $X$ is $sigma$-compact, then it is the union of countably many compact sets. At least one of these must contain uncountably many elements of $S$. By compactness, $S$ has a limit point, a contradiction.
$endgroup$
– Ben W
Dec 28 '18 at 16:47
add a comment |
3 Answers
3
active
oldest
votes
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Hint: First show that in infinite dimension compact sets have no interior points (recall that the unit ball is not compact). Now assume that $X$ is $sigma$-compact and pick a family $(A_n)_{ngeq 1}$ of compact sets such that
$$ bigcup_{ngeq 1} A_n = X.$$
Now $X$ has an interior point. Use Baire's Theorem to show that this leads to a contradiction.
answered Dec 28 '18 at 17:01
Severin SchravenSeverin Schraven
6,4501935
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Suppose $X$ is a $sigma$-compact infinite-dimensional Banach space, with $X=cup_n K_n$. By the Baire category theorem, there exists some $ninmathbb N$ such that $K_n$ has nonempty interior. But compact sets in infinite-dimensional normed spaces always have empty interior.
To prove the claim made in this last sentence, assume $X$ is an infinite-dimensional normed space, and $Ksubset X$ is compact and has nonempty interior $K^circ$. Since scaling and translation are homeomorphisms in normed spaces, we may assume without loss of generality that $K$ contains the open ball of radius $>1$ centered at $0in X$. Then applying induction and the Riesz lemma, we obtain a sequence $(x_n)$ in $K$ with no convergent subsequence, contradicting the fact that $K$ is compact.
answered Dec 28 '18 at 17:02
AweyganAweygan
14.6k21442
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add a comment |
$begingroup$
From Chapter 25 of Willard's General Topology, we have the following nice theorem:
A $sigma$-compact space $X$ is Baire iff the set of points at which $X$ is locally compact is dense in $X$.
You can combine this with the fact that a topological vector space is locally compact iff it is finite-dimensional to prove that any $sigma$-compact Baire topological vector space (including any $sigma$-compact Banach space) must be finite-dimensional (since a topological vector space is locally compact iff it is locally compact at one point).
answered Dec 28 '18 at 17:02
Michael LeeMichael Lee
4,8381930
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: First show that in infinite dimension compact sets have no interior points (recall that the unit ball is not compact). Now assume that $X$ is $sigma$-compact and pick a family $(A_n)_{ngeq 1}$ of compact sets such that
$$ bigcup_{ngeq 1} A_n = X.$$
Now $X$ has an interior point. Use Baire's Theorem to show that this leads to a contradiction.
answered Dec 28 '18 at 17:01
Severin SchravenSeverin Schraven
6,4501935
$endgroup$
add a comment |
$begingroup$
Hint: First show that in infinite dimension compact sets have no interior points (recall that the unit ball is not compact). Now assume that $X$ is $sigma$-compact and pick a family $(A_n)_{ngeq 1}$ of compact sets such that
$$ bigcup_{ngeq 1} A_n = X.$$
Now $X$ has an interior point. Use Baire's Theorem to show that this leads to a contradiction.
answered Dec 28 '18 at 17:01
Severin SchravenSeverin Schraven
6,4501935
$endgroup$
add a comment |
$begingroup$
Hint: First show that in infinite dimension compact sets have no interior points (recall that the unit ball is not compact). Now assume that $X$ is $sigma$-compact and pick a family $(A_n)_{ngeq 1}$ of compact sets such that
$$ bigcup_{ngeq 1} A_n = X.$$
Now $X$ has an interior point. Use Baire's Theorem to show that this leads to a contradiction.
answered Dec 28 '18 at 17:01
Severin SchravenSeverin Schraven
6,4501935
$endgroup$
Hint: First show that in infinite dimension compact sets have no interior points (recall that the unit ball is not compact). Now assume that $X$ is $sigma$-compact and pick a family $(A_n)_{ngeq 1}$ of compact sets such that
$$ bigcup_{ngeq 1} A_n = X.$$
Now $X$ has an interior point. Use Baire's Theorem to show that this leads to a contradiction.
answered Dec 28 '18 at 17:01
Severin SchravenSeverin Schraven
6,4501935
edited Dec 29 '18 at 12:09
answered Dec 28 '18 at 17:01
Severin SchravenSeverin Schraven
6,4501935
answered Dec 28 '18 at 17:01
Severin SchravenSeverin Schraven
6,4501935
answered Dec 28 '18 at 17:01
Severin SchravenSeverin Schraven
6,4501935
6,4501935
add a comment |
add a comment |
$begingroup$
Suppose $X$ is a $sigma$-compact infinite-dimensional Banach space, with $X=cup_n K_n$. By the Baire category theorem, there exists some $ninmathbb N$ such that $K_n$ has nonempty interior. But compact sets in infinite-dimensional normed spaces always have empty interior.
To prove the claim made in this last sentence, assume $X$ is an infinite-dimensional normed space, and $Ksubset X$ is compact and has nonempty interior $K^circ$. Since scaling and translation are homeomorphisms in normed spaces, we may assume without loss of generality that $K$ contains the open ball of radius $>1$ centered at $0in X$. Then applying induction and the Riesz lemma, we obtain a sequence $(x_n)$ in $K$ with no convergent subsequence, contradicting the fact that $K$ is compact.
answered Dec 28 '18 at 17:02
AweyganAweygan
14.6k21442
$endgroup$
add a comment |
$begingroup$
Suppose $X$ is a $sigma$-compact infinite-dimensional Banach space, with $X=cup_n K_n$. By the Baire category theorem, there exists some $ninmathbb N$ such that $K_n$ has nonempty interior. But compact sets in infinite-dimensional normed spaces always have empty interior.
To prove the claim made in this last sentence, assume $X$ is an infinite-dimensional normed space, and $Ksubset X$ is compact and has nonempty interior $K^circ$. Since scaling and translation are homeomorphisms in normed spaces, we may assume without loss of generality that $K$ contains the open ball of radius $>1$ centered at $0in X$. Then applying induction and the Riesz lemma, we obtain a sequence $(x_n)$ in $K$ with no convergent subsequence, contradicting the fact that $K$ is compact.
answered Dec 28 '18 at 17:02
AweyganAweygan
14.6k21442
$endgroup$
add a comment |
$begingroup$
Suppose $X$ is a $sigma$-compact infinite-dimensional Banach space, with $X=cup_n K_n$. By the Baire category theorem, there exists some $ninmathbb N$ such that $K_n$ has nonempty interior. But compact sets in infinite-dimensional normed spaces always have empty interior.
To prove the claim made in this last sentence, assume $X$ is an infinite-dimensional normed space, and $Ksubset X$ is compact and has nonempty interior $K^circ$. Since scaling and translation are homeomorphisms in normed spaces, we may assume without loss of generality that $K$ contains the open ball of radius $>1$ centered at $0in X$. Then applying induction and the Riesz lemma, we obtain a sequence $(x_n)$ in $K$ with no convergent subsequence, contradicting the fact that $K$ is compact.
answered Dec 28 '18 at 17:02
AweyganAweygan
14.6k21442
$endgroup$
Suppose $X$ is a $sigma$-compact infinite-dimensional Banach space, with $X=cup_n K_n$. By the Baire category theorem, there exists some $ninmathbb N$ such that $K_n$ has nonempty interior. But compact sets in infinite-dimensional normed spaces always have empty interior.
To prove the claim made in this last sentence, assume $X$ is an infinite-dimensional normed space, and $Ksubset X$ is compact and has nonempty interior $K^circ$. Since scaling and translation are homeomorphisms in normed spaces, we may assume without loss of generality that $K$ contains the open ball of radius $>1$ centered at $0in X$. Then applying induction and the Riesz lemma, we obtain a sequence $(x_n)$ in $K$ with no convergent subsequence, contradicting the fact that $K$ is compact.
answered Dec 28 '18 at 17:02
AweyganAweygan
14.6k21442
answered Dec 28 '18 at 17:02
AweyganAweygan
14.6k21442
answered Dec 28 '18 at 17:02
AweyganAweygan
14.6k21442
answered Dec 28 '18 at 17:02
AweyganAweygan
14.6k21442
14.6k21442
add a comment |
add a comment |
$begingroup$
From Chapter 25 of Willard's General Topology, we have the following nice theorem:
A $sigma$-compact space $X$ is Baire iff the set of points at which $X$ is locally compact is dense in $X$.
You can combine this with the fact that a topological vector space is locally compact iff it is finite-dimensional to prove that any $sigma$-compact Baire topological vector space (including any $sigma$-compact Banach space) must be finite-dimensional (since a topological vector space is locally compact iff it is locally compact at one point).
answered Dec 28 '18 at 17:02
Michael LeeMichael Lee
4,8381930
$endgroup$
add a comment |
$begingroup$
From Chapter 25 of Willard's General Topology, we have the following nice theorem:
A $sigma$-compact space $X$ is Baire iff the set of points at which $X$ is locally compact is dense in $X$.
You can combine this with the fact that a topological vector space is locally compact iff it is finite-dimensional to prove that any $sigma$-compact Baire topological vector space (including any $sigma$-compact Banach space) must be finite-dimensional (since a topological vector space is locally compact iff it is locally compact at one point).
answered Dec 28 '18 at 17:02
Michael LeeMichael Lee
4,8381930
$endgroup$
add a comment |
$begingroup$
From Chapter 25 of Willard's General Topology, we have the following nice theorem:
A $sigma$-compact space $X$ is Baire iff the set of points at which $X$ is locally compact is dense in $X$.
You can combine this with the fact that a topological vector space is locally compact iff it is finite-dimensional to prove that any $sigma$-compact Baire topological vector space (including any $sigma$-compact Banach space) must be finite-dimensional (since a topological vector space is locally compact iff it is locally compact at one point).
answered Dec 28 '18 at 17:02
Michael LeeMichael Lee
4,8381930
$endgroup$
From Chapter 25 of Willard's General Topology, we have the following nice theorem:
A $sigma$-compact space $X$ is Baire iff the set of points at which $X$ is locally compact is dense in $X$.
You can combine this with the fact that a topological vector space is locally compact iff it is finite-dimensional to prove that any $sigma$-compact Baire topological vector space (including any $sigma$-compact Banach space) must be finite-dimensional (since a topological vector space is locally compact iff it is locally compact at one point).
answered Dec 28 '18 at 17:02
Michael LeeMichael Lee
4,8381930
answered Dec 28 '18 at 17:02
Michael LeeMichael Lee
4,8381930
answered Dec 28 '18 at 17:02
Michael LeeMichael Lee
4,8381930
answered Dec 28 '18 at 17:02
Michael LeeMichael Lee
4,8381930
4,8381930
add a comment |
add a comment |
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I believe (check this) that it is possible to find an uncountable set $S$ with no limit point in any infinite-dimensional Banach space $X$. If $X$ is $sigma$-compact, then it is the union of countably many compact sets. At least one of these must contain uncountably many elements of $S$. By compactness, $S$ has a limit point, a contradiction.
$endgroup$
– Ben W
Dec 28 '18 at 16:47