## Every vector space has a basis

Let $F$ be a field. Prove that every $F$-vector space has a basis.

If $V = 0$, then we say $\emptyset$ is a basis for $V$. Henceforth we assume that $V$ is nonzero.

Let $S \subseteq \mathcal{P}(V)$ be the set of all linearly independent subsets of $V$. Note that $S$ is partially ordered by $\subseteq$ and is nonempty since any nonzero singleton set is linearly independent.

Let $C \subseteq S$ be a chain; that is, if $C_1, C_2 \in C$, then either $C_1 \subseteq C_2$ or $C_2 \subseteq C_1$. We claim that $\bigcup C$ is linearly independent. To see this, let $D = \{d_i\} \subseteq \bigcup C$ be a finite subset and suppose $\sum a_id_i = 0$. Now each $d_i$ is in some $C_i \in C$. Since $D$ is finite, there is some index $t$ such that $D \subseteq C_t$. Since $C_t$ is linearly independent, $a_i = 0$ for all $i$. Thus $\bigcup C$ is linearly independent, and thus $\bigcup C \in S$. That is, the chain $C$ has an upper bound in $S$. By Zorn’s lemma, there exists a maximal element $B \in S$.

We claim that $B$ is a basis for $V$. To see this, suppose there exists an element $v \in V \setminus \mathsf{span}\ B$. Now let $D \subseteq B$ be a finite subset, and suppose $av + \sum b_id_i = 0$. If $a \neq 0$, then we have $v = \sum -b_ia^{-1}d_i \in \mathsf{span}\ B$, a contradiction. Thus $a = 0$. But since $B$ is linearly independent, $b_i = 0$ for all $i$. Thus $B \cup \{v\}$ is linearly independent, violating the maximalness of $B$ in $S$. Thus in fact $V = \mathsf{span}\ B$. Thus $B$ is a linearly independent generating set (i.e. basis) for $V$.