Proof that any set with less elements than the dimension of vector space is not spanning I need help proving this. I understand that any set with elements more than the dimension of vector space is dependent.
 A: If your vector space is $V$, I will assume that $\dim V<\infty$. Let $S$ be your set and assume that $S$ spans $V$ and that $\#S<\dim V$. Let $S^\star$ be a maximal subset of $S$ among those subsets of $S$ which are linearly independent. Then $\operatorname{span}S^\star=\operatorname{span}S=V$. Therefore, $S^\star$ is a basis of $V$, which is impossible, since $\#S^\star\leqslant\#S<\dim V$.
A: Suppose that $S$ spans $V$ (some arbitrary vector space)
Recall that a set, $S = \{u_i\}_{i=1}^k \subseteq V$, to span a vector space $V$ means that, $\forall v \in V,$  $\exists \{\mu_i\}_{i=1}^k \in F$ so that $$v = \sum_{i}\mu_iu_i $$
We will have that $\dim V \leq k = |S|$.  Why?  
The dimension of $V$ is the cardinality of a basis set for $V$.  Recall that a basis is a set, with the smallest number of elements possible, which spans the space.  So if we supposed, by contradiction, that $\dim V > k$, then we would have some vector $v$ that could not be completely written as a sum of elements of $S$.  It immediately follows that $S$ does not span $V$.  This contradicts the hypothesis that $S$ spans $V$.  $\hspace{1.9in}$$\square$
So we have established that if a set spans a vector space it follows that it has at least as many elements as the dimension of that vector space.  The contrapositive to this statement is your desired result.
