TITLE: A union of two subspaces not equal to the vector space.
QUESTION [6 upvotes]: Let $L,M$ two subspaces of the vector space, $V$ such that both $L,M \ne V$.
  Prove: $L\cup M \ne V$.  

I think this is a case of a proof by contradiction.
Lets assume $L \cup M = V$. 
Hence,
$$\dim(V) = \dim(L) + \dim(M) - \dim(L\cap M)$$
How to proceed?

REPLY [3 votes]: Suppose $W \not\subset L $ and $L \not\subset W $, otherwise is trivial.
Let $v \in L-W $ and $u \in W - L $, then if $ L \cup W = V $ we have  $v + u \in L$ or $v + u \in L$, and in both cases we obtain a contradiction.