every vector space has a basis

This result, trivial in the finite case, is in fact rather surprising when one thinks of infinite dimensionial vector spaces, and the definition of a basis: just try to imagine a basis of the vector space of all continuous mappings $f\colon\mathbb{R}\to\mathbb{R}$ . The theorem is equivalent to the axiom of choice family of axioms and theorems. Here we will only prove that Zorn’s lemma implies that every vector space has a basis.

Theorem.

Let $X$ be any vector space over any field $F$ and assume Zorn’s lemma. Then if $L$ is a linearly independent subset of $X$ , there exists a basis of $X$ containing $L$ . In particular, $X$ does have a basis at all.

Proof.

Let $\mathcal{A}$ be the set of linearly independent subsets of $X$ containing $L$ (in particular, $\mathcal{A}$ is not empty), then $\mathcal{A}$ is partially ordered by inclusion. For each chain $C\subseteq\mathcal{A}$ , define $\widehat{C}=\bigcup C$ . Clearly, $\widehat{C}$ is an upper bound of $C$ . Next we show that $\widehat{C}\in\mathcal{A}$ . Let $V:=\{v_{1},\ldots,v_{n}\}\subseteq\widehat{C}$ be a finite collection of vectors. Then there exist sets $C_{1},\ldots,C_{n}\in C$ such that $v_{i}\in C_{i}$ for all $1\leq i\leq n$ . Since $C$ is a chain, there is a number $k$ with $1\leq k\leq n$ such that $C_{k}=\bigcup\limits_{i=1}^{n}C_{i}$ and thus $V\subseteq C_{k}$ , that is $V$ is linearly independent. Therefore, $\widehat{C}$ is an element of $\mathcal{A}$ .

According to Zorn’s lemma $\mathcal{A}$ has a maximal element, $M$ , which is linearly independent. We show now that $M$ is a basis. Let $\langle M\rangle$ be the span of $M$ . Assume there exists an $x\in X\setminus\langle M\rangle$ . Let $\{x_{1},\ldots,x_{n}\}\subseteq M$ be a finite collection of vectors and $a_{1},\ldots,a_{n+1}\in F$ elements such that

a_{1}x_{1}+\cdots+a_{n}x_{n}-a_{n+1}x=0.

If $a_{n+1}$ was necessarily zero, so would be the other $a_{i}$ , $1\leq i\leq n$ , making $\{x\}\cup M$ linearly independent in contradiction to the maximality of $M$ . If $a_{n+1}\neq 0$ , we would have

x=\frac{a_{1}}{a_{n+1}}x_{1}+\cdots+\frac{a_{n}}{a_{n+1}}x_{n},

contradicting $x\notin\langle M\rangle$ . Thus such an $x$ does not exist and $X=\langle M\rangle$ , so $M$ is a generating set and hence a basis.

Taking $L=\emptyset$ , we see that $X$ does have a basis at all. ∎

Title	every vector space has a basis
Canonical name	EveryVectorSpaceHasABasis
Date of creation	2013-03-22 13:04:48
Last modified on	2013-03-22 13:04:48
Owner	GrafZahl (9234)
Last modified by	GrafZahl (9234)
Numerical id	14
Author	GrafZahl (9234)
Entry type	Theorem
Classification	msc 15A03
Synonym	every vector space has a Hamel basis
Related topic	ZornsLemma
Related topic	AxiomOfChoice
Related topic	ZermelosWellOrderingTheorem
Related topic	HaudorffsMaximumPrinciple
Related topic	KuratowskisLemma