Suppose I have an infinite dimension vector space V (not necessarily countably infinite). Suppose a have a set S that spans the space. If V is finite dimensional, it is trivial to construct a basis using elements in S that form a basis for V. Is this also valid when V has a non-countable number of dimensions?
EDIT: Wikipedia claims: "Theorem 3: Let V be a finite-dimensional vector space. Any set of vectors that spans V can be reduced to a basis for V by discarding vectors if necessary (i.e. if there are linearly dependent vectors in the set). If the axiom of choice holds, this is true without the assumption that V has finite dimension."
However, it doesn't provide a reference nor explain how the proof works.
Best Answer
Yes, any spannning set contains a basis.
There is a difference between the finite case and the infinite case. Suppose throughout that $S$ spans some vector space $V$.
If $B$ is a maximal independent subset of $S$ then $B$ is a basis for $V$: It's sufficient to show that $B$ spans $V$, and hence it's sufficient to show that $S$ is contained in the span of $B$. But if $x\in S$ is not in the span of $B$ then $B\cup\{x\}$ is independent, contradicting the maximality of $B$.
If $B$ is a minimal spanning subset of $S$ then $B$ is a basis for $V$: If $B$ is not independent then some $x\in B$ is a linear combination of the other elements of $B$, so $B\setminus\{x\}$ spans $V$, contradicting the minimality of $B$.
So if $S$ is finite you can obtain a basis $B\subset S$ by letting $B$ be a maximal independent subset of $S$ or a minimal spannning subset of $S$. On the other hand, if $S$ is infinite then Zorn's lemma shows easily that $S$ has a maximal independent subset, but the obvious "direct" proof by Zorn's lemma that $S$ contains a minimal spanning subset doesn't work; we leave it as an exercise for the reader to see why not.