every vector space has a basis

This result, trivial in the finite case, is in fact rather surprisingwhen one thinks of infinite dimensionial vector spaces, and thedefinition of a basis: just try to imagine a basis of the vector spaceof all continuous mappings $f:ℝ\to ℝ$. The theorem isequivalent to the axiom of choice family of axioms and theorems. Herewe will only prove that Zorn’s lemma implies that every vector spacehas a basis.

Theorem.

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

Proof.

Let $𝒜$ be the set of linearly independent subsets of $X$containing $L$ (in particular, $𝒜$ is not empty), then $𝒜$is partially ordered by inclusion. For each chain $C\subseteq 𝒜$,define $\widehat{C}=\bigcup C$. Clearly, $\widehat{C}$ is an upper bound of $C$. Next weshow that $\widehat{C}\in 𝒜$. Let $V:=\{v_1,\mathrm{\dots },v_n\}\subseteq \widehat{C}$be a finite collection of vectors. Then there exist sets $C_1,\mathrm{\dots },C_n\in C$ such that $v_i\in C_i$ for all $1\le i\le n$. Since $C$ isa chain, there is a number $k$ with $1\le k\le n$ such that$C_k=\bigcup _{i=1}^n{C}_i$ and thus $V\subseteq C_k$, that is $V$ islinearly independent. Therefore, $\widehat{C}$ is an element of $𝒜$.

According to Zorn’s lemma $𝒜$ has a maximal element, $M$, whichis linearly independent. We show now that $M$ is a basis. Let $⟨M⟩$be the span of $M$. Assume there exists an $x\in X\setminus ⟨M⟩$. Let$\{x_1,\mathrm{\dots },x_n\}\subseteq M$ be a finite collection of vectors and$a_1,\mathrm{\dots },a_{n+1}\in F$ elements such that

$$a_1{x}_1+\mathrm{\cdots }+a_n{x}_n-a_{n+1}x=0.$$

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

$$x=\frac{a_1}{a_{n+1}}{x}_1+\mathrm{\cdots }+\frac{a_n}{a_{n+1}}{x}_n,$$

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

Taking $L=\mathrm{\varnothing }$, we see that $X$ does have a basis at all.∎