A union of two subspaces not equal to the vector space. [duplicate]

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?

Solution 1:

Let's break down the proof in two cases:

• If $L \subset M$ (resp. $M \subset L$) then $L \cup M = M \neq V$ (resp. $L \cup M = L \neq V$);
• Otherwise choose $x \in L \setminus M$, $y \in M \setminus L$. Then $x + y$ is neither in $L$ (for then $y$ would be in $L$) nor in $M$ (same reason). Therefore $x + y \not\in L \cup M$.

Solution 2:

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.