Let V be a vector space (over a field), and let \mathcal{S} \subset \mathcal{P}(V) be the set of all sets of linearly independent vectors, partially ordered by set inclusion.

Any chain \mathcal{C} \subseteq \mathcal{S} is bounded above by S_{\text{max}} = \bigcup_{S \in \mathcal{C}} S which is a linearly independent set of vectors because any finite subset of S_{\text{max}} must be contained in some element of \mathcal{C}. Thus, by Zorn’s Lemma there exists at least one maximal element M \in \mathcal{S}.

Suppose for contradiction that M does not span. Then there is some v \in V which cannot be expressed as a finite linear combination of elements of M. But this implies M' = M \cup \{v\} \in \mathcal{S} and M \subset M', contradicting the maximality of M with respect to inclusion.

So M must span, and since it is a linearly independent set, this makes it a basis.

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