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Compactness in normed vector spaces.

Are Arzelà–Ascoli theorems results of similar theorems on normed spaces, metric spaces or other spaces?Finiteness of the dimension of a normed space and compactnessProof of compactness for sets of norm equal to one in finite-dimensional normed vector spacesConvex hull, compactness, normed spacesAre all the finite dimensional vector spaces with a metric isometric to $mathbb R^n$Compactness of subspaces of finite-dimensional vector spacesCounterexample showing that weakly* compact sets might not be norm bounded in a normed spaces.Dual of a normed vector spaceCompactness of normed linear spacesProblem about strictly normed spaces.

3

$begingroup$

Let $(X,VertcdotVert)$ be a normed $mathbbK$-vector space and $A subset X$ be closed and bounded. My problem is how to determine whether $A$ is compact?

I know that a compact subset is always closed and bounded. And that in $mathbbR$ the converse holds due to Bolzano–Weierstraß.

But I think I am missing something. Do there exist subsets in normed vector spaces which are closed and bounded but not compact?

functional-analysis compactness normed-spaces

share|cite|improve this question

edited 1 hour ago

Grazel

asked 1 hour ago

GrazelGrazel

155115

$endgroup$

add a comment | 

3

$begingroup$

Let $(X,VertcdotVert)$ be a normed $mathbbK$-vector space and $A subset X$ be closed and bounded. My problem is how to determine whether $A$ is compact?

I know that a compact subset is always closed and bounded. And that in $mathbbR$ the converse holds due to Bolzano–Weierstraß.

But I think I am missing something. Do there exist subsets in normed vector spaces which are closed and bounded but not compact?

functional-analysis compactness normed-spaces

share|cite|improve this question

edited 1 hour ago

Grazel

asked 1 hour ago

GrazelGrazel

155115

$endgroup$

add a comment | 

$begingroup$

Let $(X,VertcdotVert)$ be a normed $mathbbK$-vector space and $A subset X$ be closed and bounded. My problem is how to determine whether $A$ is compact?

I know that a compact subset is always closed and bounded. And that in $mathbbR$ the converse holds due to Bolzano–Weierstraß.

But I think I am missing something. Do there exist subsets in normed vector spaces which are closed and bounded but not compact?

functional-analysis compactness normed-spaces

share|cite|improve this question

edited 1 hour ago

Grazel

asked 1 hour ago

GrazelGrazel

155115

$endgroup$

share|cite|improve this question

edited 1 hour ago

Grazel

asked 1 hour ago

GrazelGrazel

155115

share|cite|improve this question

edited 1 hour ago

Grazel

asked 1 hour ago

GrazelGrazel

155115

asked 1 hour ago

GrazelGrazel

155115

asked 1 hour ago

GrazelGrazel

155115

3 Answers
3

active

oldest

votes

2

$begingroup$

Let $c_0$ denote the infinite-dimensional normed vector space consisting of sequences that converge to $0$. The norm on $c_0$ is given by:

$$
|(a_1,a_1,a_3,ldots)|:=sup_kinmathbbN|a_k|.
$$

Let $A$ denote the set of all vectors in $c_0$ of norm at most one. Clearly $A$ is bounded and it is easy to show that $A$ is closed. To see that $A$ is not compact, let, for each $xin A$, $B_1/2(x):=yin c_0:$. Then, $B_1/2(x):xin A$ is an open cover of $A$. To see that it has no finite subcover, let, for each $kinmathbbN$, $e_k$ denote the sequence $(0,ldots,0,1,0,ldots)$, where the $1$ is in position $k$. Since $|e_k-e_ell|=1$ if $knot=ell$, it follows that no two such vectors can belong to the same $B_1/2(x)$ for any $x$.

Note that in general, a set in a metric space is compact if and only if it is complete and totally bounded.

share|cite|improve this answer

answered 1 hour ago

ervxervx

10.4k31438

$endgroup$

add a comment | 

4

$begingroup$

Yes.

The equivalence of "compact" and "closed and bounded" holds only in finite-dimensional spaces.

In a general, infinite-dimensional normed space (e.g., over the real scalars), for a set to be compact, it is not enough that it be closed and bounded. It must be closed and totally bounded.

share|cite|improve this answer

answered 1 hour ago

avsavs

4,6701515

$endgroup$

add a comment | 

1

$begingroup$

Example. The unit ball $B$ in Hilbert space $l^2$. It is closed and bounded, but not compact. Let the orthonormal unit vectors be $e_n, n=1,2,3,dots$. Then sequence $e_n$ is a sequence in $B$, but it has no convergent subsequence. This is because all the distances are $|e_n-e_m| = sqrt2$ so in fact no subsequence is even a Cauchy sequence.

share|cite|improve this answer

answered 1 hour ago

GEdgarGEdgar

63.9k369177

$endgroup$

add a comment | 

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2

$begingroup$

Let $c_0$ denote the infinite-dimensional normed vector space consisting of sequences that converge to $0$. The norm on $c_0$ is given by:

$$
|(a_1,a_1,a_3,ldots)|:=sup_kinmathbbN|a_k|.
$$

Let $A$ denote the set of all vectors in $c_0$ of norm at most one. Clearly $A$ is bounded and it is easy to show that $A$ is closed. To see that $A$ is not compact, let, for each $xin A$, $B_1/2(x):=yin c_0:$. Then, $B_1/2(x):xin A$ is an open cover of $A$. To see that it has no finite subcover, let, for each $kinmathbbN$, $e_k$ denote the sequence $(0,ldots,0,1,0,ldots)$, where the $1$ is in position $k$. Since $|e_k-e_ell|=1$ if $knot=ell$, it follows that no two such vectors can belong to the same $B_1/2(x)$ for any $x$.

Note that in general, a set in a metric space is compact if and only if it is complete and totally bounded.

share|cite|improve this answer

answered 1 hour ago

ervxervx

10.4k31438

$endgroup$

add a comment | 

2

$begingroup$

Let $c_0$ denote the infinite-dimensional normed vector space consisting of sequences that converge to $0$. The norm on $c_0$ is given by:

$$
|(a_1,a_1,a_3,ldots)|:=sup_kinmathbbN|a_k|.
$$

Let $A$ denote the set of all vectors in $c_0$ of norm at most one. Clearly $A$ is bounded and it is easy to show that $A$ is closed. To see that $A$ is not compact, let, for each $xin A$, $B_1/2(x):=yin c_0:$. Then, $B_1/2(x):xin A$ is an open cover of $A$. To see that it has no finite subcover, let, for each $kinmathbbN$, $e_k$ denote the sequence $(0,ldots,0,1,0,ldots)$, where the $1$ is in position $k$. Since $|e_k-e_ell|=1$ if $knot=ell$, it follows that no two such vectors can belong to the same $B_1/2(x)$ for any $x$.

Note that in general, a set in a metric space is compact if and only if it is complete and totally bounded.

share|cite|improve this answer

answered 1 hour ago

ervxervx

10.4k31438

$endgroup$

add a comment | 

$begingroup$

Let $c_0$ denote the infinite-dimensional normed vector space consisting of sequences that converge to $0$. The norm on $c_0$ is given by:

$$
|(a_1,a_1,a_3,ldots)|:=sup_kinmathbbN|a_k|.
$$

Let $A$ denote the set of all vectors in $c_0$ of norm at most one. Clearly $A$ is bounded and it is easy to show that $A$ is closed. To see that $A$ is not compact, let, for each $xin A$, $B_1/2(x):=yin c_0:$. Then, $B_1/2(x):xin A$ is an open cover of $A$. To see that it has no finite subcover, let, for each $kinmathbbN$, $e_k$ denote the sequence $(0,ldots,0,1,0,ldots)$, where the $1$ is in position $k$. Since $|e_k-e_ell|=1$ if $knot=ell$, it follows that no two such vectors can belong to the same $B_1/2(x)$ for any $x$.

Note that in general, a set in a metric space is compact if and only if it is complete and totally bounded.

share|cite|improve this answer

answered 1 hour ago

ervxervx

10.4k31438

$endgroup$

share|cite|improve this answer

answered 1 hour ago

ervxervx

10.4k31438

answered 1 hour ago

ervxervx

10.4k31438

answered 1 hour ago

ervxervx

10.4k31438

4

$begingroup$

Yes.

The equivalence of "compact" and "closed and bounded" holds only in finite-dimensional spaces.

In a general, infinite-dimensional normed space (e.g., over the real scalars), for a set to be compact, it is not enough that it be closed and bounded. It must be closed and totally bounded.

share|cite|improve this answer

answered 1 hour ago

avsavs

4,6701515

$endgroup$

add a comment | 

4

$begingroup$

Yes.

The equivalence of "compact" and "closed and bounded" holds only in finite-dimensional spaces.

In a general, infinite-dimensional normed space (e.g., over the real scalars), for a set to be compact, it is not enough that it be closed and bounded. It must be closed and totally bounded.

share|cite|improve this answer

answered 1 hour ago

avsavs

4,6701515

$endgroup$

add a comment | 

$begingroup$

Yes.

The equivalence of "compact" and "closed and bounded" holds only in finite-dimensional spaces.

In a general, infinite-dimensional normed space (e.g., over the real scalars), for a set to be compact, it is not enough that it be closed and bounded. It must be closed and totally bounded.

share|cite|improve this answer

answered 1 hour ago

avsavs

4,6701515

$endgroup$

share|cite|improve this answer

answered 1 hour ago

avsavs

4,6701515

answered 1 hour ago

avsavs

4,6701515

answered 1 hour ago

avsavs

4,6701515

1

$begingroup$

Example. The unit ball $B$ in Hilbert space $l^2$. It is closed and bounded, but not compact. Let the orthonormal unit vectors be $e_n, n=1,2,3,dots$. Then sequence $e_n$ is a sequence in $B$, but it has no convergent subsequence. This is because all the distances are $|e_n-e_m| = sqrt2$ so in fact no subsequence is even a Cauchy sequence.

share|cite|improve this answer

answered 1 hour ago

GEdgarGEdgar

63.9k369177

$endgroup$

add a comment | 

1

$begingroup$

Example. The unit ball $B$ in Hilbert space $l^2$. It is closed and bounded, but not compact. Let the orthonormal unit vectors be $e_n, n=1,2,3,dots$. Then sequence $e_n$ is a sequence in $B$, but it has no convergent subsequence. This is because all the distances are $|e_n-e_m| = sqrt2$ so in fact no subsequence is even a Cauchy sequence.

share|cite|improve this answer

answered 1 hour ago

GEdgarGEdgar

63.9k369177

$endgroup$

add a comment | 

$begingroup$

Example. The unit ball $B$ in Hilbert space $l^2$. It is closed and bounded, but not compact. Let the orthonormal unit vectors be $e_n, n=1,2,3,dots$. Then sequence $e_n$ is a sequence in $B$, but it has no convergent subsequence. This is because all the distances are $|e_n-e_m| = sqrt2$ so in fact no subsequence is even a Cauchy sequence.

share|cite|improve this answer

answered 1 hour ago

GEdgarGEdgar

63.9k369177

$endgroup$

share|cite|improve this answer

answered 1 hour ago

GEdgarGEdgar

63.9k369177

answered 1 hour ago

GEdgarGEdgar

63.9k369177

answered 1 hour ago

GEdgarGEdgar

63.9k369177