Theorem 1   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 $\mc{A}$ be the set of linearly independent subsets of $X$ containing $L$ (in particular, $\mc{A}$ is not empty), then $\mc{A}$ is partially ordered by inclusion. For each chain $C\subseteq\mc{A}$ , define $\h{C}=\cup C$ . Clearly, $\h{C}$ is an upper bound of $C$ . Next we show that $\h{C}\in\mc{A}$ . Let $V:=\{v_1,\ldots,v_n\}\subseteq\h{C}$ 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 $C_k=\Bigcup_{i=1}^nC_i$ and thus $V\subseteq C_k$ , that is $V$ is linearly independent. Therefore, $\h{C}$ is an element of $\mc{A}$ .

According to Zorn's lemma $\mc{A}$ has a maximal element, $M$ , which is 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 $\{x_1,\ldots,x_n\}\subseteq M$ be a finite collection of vectors and $a_1,\ldots,a_{n+1}\in F$ elements such that \begin{equation*} a_1x_1+\cdots+a_nx_n-a_{n+1}x=0. \end{equation*}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 \begin{equation*} x=\frac{a_1}{a_{n+1}}x_1+\cdots+\frac{a_n}{a_{n+1}}x_n, \end{equation*}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 $L=\emptyset$ , we see that $X$ does have a basis at all.