Under what conditions are there only a finite number of distinct generating subsets of a vector space?

Let $V$ be a vector space and $W$ be a subspace of $V$.

Define $I\triangleq \{S\subset W: \text{span} S= W\}$.

Since $\text{span} W=W$, $I$ is nonempty.

Under what conditions, is $I$ finite?

Moreover, i don't understand why subspace $W$ is in the hypothesis, since it seems to me that this hypothesis does not generalize anything more than just saying vector space $V$. Is there anything more general by saying "Let $W$ be a subspace of $V$" rather than not mentioning $W$ in this problem?

-
@Tobias To be honest, i cannot think of any approach to attack this problem when $W$ is infinite. Would you give me a hint? – Jj- Feb 14 '13 at 13:55
You're right, $V$ is useless here. – 1015 Feb 14 '13 at 13:58
Hint: If $v$ appears in some spanning set, then you can replace $v$ by $av$ for any non-zero scalar $a$ and not change the fact that the set spans the entire space. – Tobias Kildetoft Feb 14 '13 at 13:59
Consider the cases where $W$ has finite/infinite dimension over the field $K$. Consider $K$ finite/infinite. – 1015 Feb 14 '13 at 13:59

Hint: Try to prove this (using the comments): $|I|<\infty \iff |W|<\infty$.
($W$ can be infinite either if the base field is infinite or if the dimension is infinite.)
If $W$ is finite, how could $I$ be infinite? And if $W$ is infinite, what happens if you just remove a singleton from $W$? How many singletons are there?