# 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