Problem : If $ M $ and $N$ are two subspaces of the vector space $V$ such that $\forall v \in V $ , $ v \in M $ or $ v \in $ (or both) . Prove that at least one of the is equal to $ V $
My Approach Assume that $ M \neq V $ , we are required to prove $ N = V $ .
Assume contrary, let $\dim N < \dim V$.
We assume that $ M \cap N = \phi $ , in case the intersection is not null, we delete the elements from $ N $ . Now, $$\dim(M+N) + \dim (M\cap N) = \dim M + \dim N$$
Since $ \dim(M\cap N) = 0 $ and $M+N = V$. Let $\dim M =m , \dim N = n . \dim V = v$. Therefore we get $n+m = v$ We know there is an isomorphism between $V$ and $\mathbb F^v$.
Consider the vectors in $\mathbb V$ corresponding to $(1,0,0,\dots) , (0,1,0,0,\dots), \dots$ in $\mathbb F^v$.
Exactly $n$ of these belong to $N$ (which form a basis) and exactly m of these belong to $M$ (which again form a basis).WLOG we can assume that
$(1,0,0,\dots) , (0,1,0,\dots) , \dots , (0,0,\dots,1, 0,\dots) $ are the vectors present in $N $ ...$(*)$
Now consider the vector corresponding to $(1,1,1,\dots)$. It must belong to either $ N $ or $ M$ . WLOG assume it belongs to $N$ . Then we can get a set of $n+1$ linearly independent vectors in $N$. $n$ vectors as described at $(*)$ and one of the form $(0,0,0,\dots , 1,1,1,\dots)$ But since we know that number of elements in basis is greater or equal to the number of elements in an independent set, we arrive at a contradiction since $n+1 > n$
My proof inherently assumes that the vector space given is a finite dimensional vector space. And also I feel that my proof is really clumsy. Is there a better way to prove it ?