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 spaceMathworldPlanetmath. 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 collectionMathworldPlanetmath of all linearly independentMathworldPlanetmath supersetsMathworldPlanetmath of A. First, 𝒮 is non-empty since A∈𝒮. In additionPlanetmathPlanetmath, if A1⊆A2⊆⋯ 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 elementMathworldPlanetmath B. Let W=span⁡(B). If W≠V, pick b∈V-W. If 0=r⁢b+r1⁢b1+⋯+rn⁢bn, where bi∈B, then -r⁢b=r1⁢b1+⋯+rn⁢bn, so that -r⁢b∈span⁡(B)=W. But b∉W, so b≠0, which implies r=0. Consequently r1=⋯=rn=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 A1⊆A2⊆⋯ 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 propositionPlanetmathPlanetmathPlanetmath, 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