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 $\mathcal{S}$ be the collection of all linearly independent supersets of $A$ . First, $\mathcal{S}$ is non-empty since $A\in\mathcal{S}$ . In addition, if $A_{1}\subseteq A_{2}\subseteq\cdots$ 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, $\mathcal{S}$ has a maximal element $B$ . Let $W=\operatorname{span}(B)$ . If $W\neq V$ , pick $b\in V-W$ . If $0=rb+r_{1}b_{1}+\cdots+r_{n}b_{n}$ , where $b_{i}\in B$ , then $-rb=r_{1}b_{1}+\cdots+r_{n}b_{n}$ , so that $-rb\in\operatorname{span}(B)=W$ . But $b\notin W$ , so $b\neq 0$ , which implies $r=0$ . Consequently $r_{1}=\cdots=r_{n}=0$ since $B$ is linearly independent. As a result, $B\cup\{b\}$ is a linearly independent superset of $B$ in $\mathcal{S}$ , contradicting the maximality of $B$ in $\mathcal{S}$ . ∎

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 $\mathcal{S}$ be the collection of all linearly independent subsets of $A$ . $\mathcal{S}$ is non-empty as $\varnothing\in\mathcal{S}$ . Let $A_{1}\subseteq A_{2}\subseteq\cdots$ 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, $\mathcal{S}$ has a maximal element $B$ . In other words, $B$ is a linearly independent subset $A$ . Let $W=\operatorname{span}(B)$ . Suppose $W\neq V$ . Since $A$ spans $V$ , there is an element $b\in A$ 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, $B\cup\{b\}$ is linearly independent, which contradicts the maximality of $B$ in $\mathcal{S}$ . Therefore, $B$ spans $V$ and thus a basis for $V$ . ∎

Corollary 1.

Every vector space has a basis.

Proof.

Either take $\varnothing$ 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 $A\subseteq C$ 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 $A\subseteq B\subseteq C$ . 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