compactness of closed unit ball in normed spaces

compactness of closed unit ball in normed spaces

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finite dimensional

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infinite-dimensional

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subspace

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convergent

Theorem

-

Let

X

be a

normed space

and

B

1

⁢

(

0

)

¯

⊂

X

the

closed ball

(

http://planetmath.org/Ball

). Then

B

1

⁢

(

0

)

¯

is

compact

if and only if

X

is finite dimensional (

http://planetmath.org/Dimension2

).

The above result is false, in general, if one is considering other

topologies

in

X

besides the norm topology (see, for example, the

Banach-Alaoglu theorem

). It follows that infinite dimensional (

http://planetmath.org/Dimension2

) normed spaces are not

locally compact

.

Proof:

•

(

⟸

)

This is the easy part. Since

X

is finite dimensional it is

isomorphic

to some

ℝ

n

(with the

standard topology

). The result then follows from the

Heine-Borel theorem

.

•

(

⟹

)

Suppose that

X

is not finite dimensional. Pick an element

x

1

∈

X

such that

∥

x

1

∥

=

1

and denote by

S

1

the subspace (

http://planetmath.org/VectorSubspace

) generated by

x

1

.

Recall that, according to the Riesz Lemma (

http://planetmath.org/RiezsLemma

), there exists an element

x

2

∈

X

such that

∥

x

2

∥

=

1

and

d

⁢

(

x

2

,

S

1

)

≥

1

2

, where

d

⁢

(

y

,

M

)

:=

inf

⁡

{

∥

y

-

z

∥

:

z

∈

M

}

denotes the distance between an element

y

∈

X

and a subspace

M

⊂

X

.

Now consider the subspace

S

2

generated by

x

1

,

x

2

. Since

X

is infinite dimensional

S

2

is a proper subspace and we can still apply the Riesz Lemma to find an element

x

3

such that

∥

x

3

∥

=

1

and

d

⁢

(

x

3

,

S

2

)

≥

1

2

.

If we proceed inductively, we will find a sequence

{

x

n

}

of norm 1 elements and a sequence of subspaces

S

n

:=

⟨

x

1

,

…

,

x

n

⟩

such that

d

⁢

(

x

n

+

1

,

S

n

)

≥

1

2

.

Under this setting it is easily seen that the sequence

{

x

n

}

is in

B

1

⁢

(

0

)

¯

and satisfies

∥

x

n

-

x

m

∥

≥

1

2

for all

m

≠

n

∈

ℕ

. Therefore,

{

x

n

}

is a sequence in

B

1

⁢

(

0

)

¯

that has no convergent (

http://planetmath.org/ConvergentSequence

)

subsequence

, i.e.,

B

1

⁢

(

0

)

¯

is not compact.

□

Remarks on the proof -

Note that the sequence

{

x

n

}

constructed in the proof does not have a Cauchy subsequence (

http://planetmath.org/CauchySequence

). Thus we have in fact proven the slightly

stronger

result that

X

is finite dimensional if and only if every

bounded

sequence in

X

has a Cauchy subsequence.

For

Hilbert spaces

the proof would be slightly simpler because one could just pick any

orthonormal basis

{

e

n

}

,

and it would

∥

e

n

-

e

m

∥

=

2

for all

m

,

n

∈

ℕ

with

m

≠

n

,

therefore having no convergent subsequence. For general normed spaces we cannot just pick orthonormal elements, since this notion does not exist. Thus, we have to use Riesz Lemma to assure the existence of elements with some .

Title

compactness of closed unit ball in normed spaces

Canonical

name

CompactnessOfClosedUnitBallInNormedSpaces

Date of creation

2013-03-22 17:48:41

Last modified on

2013-03-22 17:48:41

Owner

asteroid (17536)

Last modified by

asteroid (17536)

Numerical id

13

Author

asteroid (17536)

Entry type

Theorem

Classification

msc 46B50

Synonym

closed

unit ball

in a normed space is compact iff the space is finite dimensional

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