Zorn’s lemma and bases for vector spaces

Zorn’s lemma and bases for vector spaces

In this entry, we illustrate how Zorn’s lemma can be applied in proving the

existence

of a basis for a

vector space

. Let

V

be a vector space over a field

k

.

Proposition 1

.

Every linearly independent subset of

V

can be extended to a basis for

V

.

This has already been proved in this entry (

http://planetmath.org/EveryVectorSpaceHasABasis

). We reprove it here for completion.

Proof.

Let

A

be a linearly independent subset of

V

. Let

𝒮

be the

collection

of all

linearly independent

supersets

of

A

. First,

𝒮

is non-empty since

A

∈

𝒮

. In

addition

, if

A

1

⊆

A

2

⊆

⋯

is a chain of linearly independent supersets of

A

, then their union is again a linearly independent superset of

A

(for a proof of this, see here (

http://planetmath.org/PropertiesOfLinearIndependence

)). So by Zorn’s Lemma,

𝒮

has a

maximal element

B

. Let

W

=

span

⁡

(

B

)

. If

W

≠

V

, pick

b

∈

V

-

W

. If

0

=

r

⁢

b

+

r

1

⁢

b

1

+

⋯

+

r

n

⁢

b

n

, where

b

i

∈

B

, then

-

r

⁢

b

=

r

1

⁢

b

1

+

⋯

+

r

n

⁢

b

n

, so that

-

r

⁢

b

∈

span

⁡

(

B

)

=

W

. But

b

∉

W

, so

b

≠

0

, which implies

r

=

0

. Consequently

r

1

=

⋯

=

r

n

=

0

since

B

is linearly independent. As a result,

B

∪

{

b

}

is a linearly independent superset of

B

in

𝒮

, contradicting the maximality of

B

in

𝒮

.
∎

Proposition 2

.

Every

spanning set

of

V

has a subset that is a basis for

V

.

Proof.

Let

A

be a spanning set of

V

. Let

𝒮

be the collection of all linearly independent subsets of

A

.

𝒮

is non-empty as

∅

∈

𝒮

. Let

A

1

⊆

A

2

⊆

⋯

be a chain of linearly independent subsets of

A

. Then the union of these sets is again a linearly independent subset of

A

. Therefore, by Zorn’s lemma,

𝒮

has a maximal element

B

. In other words,

B

is a linearly independent subset

A

. Let

W

=

span

⁡

(

B

)

. Suppose

W

≠

V

. Since

A

spans

V

, there is an

element

b

∈

A

not in

W

(for otherwise the span of

A

must lie in

W

, which would imply

W

=

V

). Then, using the same

argument

as in the previous

proposition

,

B

∪

{

b

}

is linearly independent, which contradicts the maximality of

B

in

𝒮

. Therefore,

B

spans

V

and thus a basis for

V

.
∎

Corollary 1

.

Every vector space has a basis.

Proof.

Either take

∅

to be the linearly independent subset of

V

and apply proposition 1, or take

V

to be the spanning subset of

V

and apply proposition 2. ∎

Remark

. The two propositions above can be combined into one: If

A

⊆

C

are two subsets of a vector space

V

such that

A

is linearly independent and

C

spans

V

, then there exists a basis

B

for

V

, with

A

⊆

B

⊆

C

. The proof again relies on Zorn’s Lemma and is left to the reader to try.

Title

Zorn’s lemma and bases for vector spaces

Canonical

name

ZornsLemmaAndBasesForVectorSpaces

Date of creation

2013-03-22 18:06:49

Last modified on

2013-03-22 18:06:49

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

9

Author

CWoo (3771)

Entry type

Result

Classification

msc 16D40

Classification

msc 13C05

Classification

msc 15A03

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