# Can you construct a basis for an infinite dimensional vector space from a set of vectors that span that space?

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.

• Since the Span operation only consider finite combinations of vectors, then yes, the usual algorithm you use in the finite case is still valid in any case. May 12, 2016 at 12:55
• Consider the set of all families of linearly independent vectors, partially ordered with the inclusion, and use Zorn's lemma. May 12, 2016 at 13:06
• proofwiki.org/wiki/Vector_Space_has_Basis
– Ian
May 12, 2016 at 13:06
• The question is a duplicate of math.stackexchange.com/questions/654089/… but that question has no satisfactory answer for the infinite case. May 12, 2016 at 13:21

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.