Every vector space over a field has a basis, assuming the axiom of choice. More generally, any linearly independent subset extends to a basis. This is an algebraic Hamel basis: each representation uses a finite linear combination, even when the basis is infinite.[ref]
Use the poset of linearly independent subsets containing :It is a set, being a subset of the power set of , and is nonempty because .
A short proof outline supplies both essential steps:
- For a nonempty chain in a partial order , put . Any finite linear relation in lies in one member of : among finitely many containing members, one contains the rest. Its linear independence makes every coefficient zero. Thus is an upper bound. For the empty chain in a partial order, use ; its empty union need not contain .
- Zorn's lemma gives a maximal element . If a vector lies outside its linear span, then is linearly independent: a linear relation with nonzero coefficient on would express in that linear span. This contradicts maximality, so is a spanning set and hence a basis.[ref]
Taking proves existence, including , whose basis is the empty set. The chain in a partial order need not be indexed by the natural numbers: only the finitely many members needed for a particular linear relation are compared. An arbitrary union of linearly independent sets can fail: and for a nonzero vector over the real numbers are each linearly independent, but their union is linearly dependent. The nesting of a chain in a partial order is what prevents this failure.
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