every vector space has a basis
This result, trivial in the finite case, is in fact rather surprising when one thinks of infinite dimensionial vector spaces, and the definition of a basis: just try to imagine a basis of the vector space of all continuous mappings . The theorem is equivalent to the axiom of choice family of axioms and theorems. Here we will only prove that Zorn’s lemma implies that every vector space has a basis.
Let be any vector space over any field and assume Zorn’s lemma. Then if is a linearly independent subset of , there exists a basis of containing . In particular, does have a basis at all.
Let be the set of linearly independent subsets of containing (in particular, is not empty), then is partially ordered by inclusion. For each chain , define . Clearly, is an upper bound of . Next we show that . Let be a finite collection of vectors. Then there exist sets such that for all . Since is a chain, there is a number with such that and thus , that is is linearly independent. Therefore, is an element of .
According to Zorn’s lemma has a maximal element, , which is linearly independent. We show now that is a basis. Let be the span of . Assume there exists an . Let be a finite collection of vectors and elements such that
If was necessarily zero, so would be the other , , making linearly independent in contradiction to the maximality of . If , we would have
contradicting . Thus such an does not exist and , so is a generating set and hence a basis.
Taking , we see that does have a basis at all. ∎
|Title||every vector space has a basis|
|Date of creation||2013-03-22 13:04:48|
|Last modified on||2013-03-22 13:04:48|
|Last modified by||GrafZahl (9234)|
|Synonym||every vector space has a Hamel basis|