I am trying to prove (or disprove) this statement:
Let $V$ be a vector space, and let $L,S\subseteq V$. If $L$ is linearly independent and $\text{span}(S)=V$, then $|L|\leq |S|$.
Is this statement true? I have consulted many linear algebra books, but they only provide the proof under the additional assumption that $L$ and $S$ are finite. The proof of the finite case goes like this: we assume for the sake of contradiction that $|L|>|S|$, then we repeatedly substitute elements of $S$ by elements of $L$ until $S$ is exhausted before $L$ and thus we arrive at a contradiction.
I think, in essence, they prove it by using Recursion theorem. I have been trying to generalize this by using Transfinite Recursion, but there are some difficulties, like how to recursively define the operation to iterate which is a bit different from the finite case which does not involve infinite ordinals.
I have also read this:
Infinite dimensional vector space. Linearly independent subsets and spanning subsets
but it's mainly focused on a special case and dependency on Axiom of Choice. Can anyone hint me this statement's truth value? or where to find them?
Thanks!
Best Answer
It is true. It is a consequence of this reference: Bourbaki, Algebra, Ch. II, Linear Algebra, §7, Vector Spaces, n°1, theorem 2.
The proof is rather short and relies on Zorn's lemma.
With your hypotheses, $L$ is not contained in $S$, but consider $S'=S\cup L$ and apply the above theorem.