Vector basis

Assuming choice, every vector space has a basis

Let be a vector space (or more generally, a module over a division ring), be a linearly independent set in and be a spanning set in containing . Then there exists a basis for for which .1 #m/thm/linalg Hence any linearly independent set belongs to some basis, and every spanning set contains a basis.

Proof

Consider the set of all linearly independent subsets of containing , which is inhabited since . Clearly forms a complete lattice. If is a chain, then the union is linearly independent and satisfies . Hence the hypothesis of Zorn's lemma is satisfied so assuming choice has a maximal element which is linearly independent.

This proof relies on Zorn's lemma and hence the axiom of choice.

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Footnotes

  1. 2008. Advanced Linear Algebra ↩