Every linearly independent subset of a vector space is contained in a basis

Let F be a field, let V be a nonzero F-vector space, and let E \subseteq V be a linearly independent subset. Prove that there exists a basis for V over F which contains E.

Let S \subseteq \mathcal{P}(V) be the set of all linearly independent subsets of V which contain E. Note that S is partially ordered by \subseteq, and is nonempty since E \in S.

Let C \subseteq S be a chain; that is, a linearly ordered subset. We claim that \bigcup C \in S. First, certainly E \subseteq \bigcup C. To see that \bigcup C is linearly independent, let D = \{d_i\} \subseteq \bigcup C be a finite subset and suppose \sum a_id_i = 0. Since D is finite and C is linearly ordered, there exists an element C_t \in C such that D \subseteq C_t. Since C_t is linearly independent, a_i = 0 for all i. Thus \bigcup C is linearly independent. So \bigcup C \in S, and in fact is an upper bound for C. By Zorn’s lemma, there exists a maximal element B \in S.

We claim that V = \mathsf{span}\ B. To see this, suppose there exists an element v \in V \setminus \mathsf{span}\ B. Suppose we have r,s_i \in F and b_i \in B (finitely many) such that rv + \sum s_ib_i = 0. If r \neq 0, then v = \sum -s_ir^{-1}b_i \in \mathsf{span}\ B, a contradiction. Thus r = 0, and we have \sum s_ib_i = 0. Since B is linearly independent, s_i = 0 for all i. Thus B \cup \{v\} is linearly independent and contains E, violating the maximalness of B in S. Thus we have V = \mathsf{span}\ B.

So B is a linearly independent generating set (i.e. basis) for V which contains E.

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