TITLE: A vector space over $R$ is not a countable union of proper subspaces
QUESTION [13 upvotes]: I was looking for alternate proofs of the theorem that "a vector space $V$ of dimension greater than $1$ over an infinite field $\mathbf{F}$ is not a union of fewer than $|\mathbf{F}|$ proper subspaces" and possible generalizations.
A simple measure-theoretic proof over $\mathbb{R}$ is as follows: By countable additivity the sum of the measures of any collection of subspaces is zero since the measure of each subspace is zero, which is a contradiction.
I would like to look at proofs over arbitrary infinite fields and would like to know if similar statements hold for say modules (finitely generated or otherwise) over infinite rings.

REPLY [8 votes]: I wrote a short note on exactly this problem.  It will appear in the Monthly one of these years.
Note in particular that your quoted statement is not necessarily true for infinite-dimensional vector spaces: an infinite-dimensional vector space over any field can be covered by $\aleph_0$-many proper subspaces.
You ask also about modules over rings.  Yes, there has been work on that.  I believe that Apoorva Khare, a mathematician who independently proved most of the results in my note, also has some work on the case of modules over various rings.  If you google for "covering numbers", you'll find that many, many papers on this topic have been written.