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.

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