If $S$ spans a vector space $V$, does $S$ contain a basis?

Question

As the title states, if $V$ is a vector space and $S \subseteq V$ spans $V$, does $S$ contain a basis?

This is true if $V$ is finite dimensional. However, what if $V$ is infinite dimensional?

Answer 1

As others have said, assuming the existence of a basis for every vector space is equivalent to the axiom of choice. So, let's assume it.

By Zorn's lemma, you can find a maximal linearly independent subset of $S$, because the union of a chain (with respect to set inclusion) of linearly independent sets is still linearly independent, since this condition deals with a finite subset at a time.

Call this maximal independent subset $T$; then the span of $T$ is a subspace of the span of $S$. If it is different, then there is an element $v\in S$ that is not in the span of $T$. This is a contradiction, because $T\cup\{v\}$ would be linearly independent, against the maximality of $T$.

If the span of $S$ is finitely generated, one can give a constructive method for finding a basis out of it: this is the main difference. No constructive method can exist in general, because of the equivalence with the axiom of choice (the span of the vector space itself is, of course, the whole vector space).

Answer 2

If you assume the axiom of choice, in the same way one 'sifts' a finite set of spanning vectors, you can find a linearly independent subset of $S$ that still spans $V$.

Answer 3

In modern mathematics we often appeal to an axiom known as "the axiom of choice", which allows us to control (to some degree) the behavior of infinitary objects, such as infinite dimensional vector spaces. This axiom is equivalent to Zorn's lemma which you may have met before, in particular in the proof that every vector space has a basis.

Using the axiom of choice we can prove that every spanning set contains a basis. This can be easily done with Zorn's lemma, simply consider all the linearly independent sets contained in the spanning set, and order them by inclusion. Zorn's lemma holds for this partial order, and therefore a maximal element exists, it is not hard to show that this maximal element must be a basis.

But in fact the assertion "Every spanning set of a vector space contains a basis" is equivalent to the axiom of choice.

This means that if the axiom of choice fails, then we can use counterexamples for this failures and construct a vector space that some generating set does not contain a basis (in fact, we can generate a space that has no basis to begin with).