perhaps this question was asked before, but if not give me a hand for it. My question is how to prove that if $W_1$ and $W_2$ be sub-spaces of vector space $V$ with the union of $W_1$ and $W_2$ is also a subspace, then one of the subspace $W_i$ is contained in other ?
Please give me a hint for start. Where should I look here?
Suppose that $W_1$ has a basis $v_1, \ldots , v_n$, and that $W_2$ has a basis $u_1, \ldots, u_m$.
If neither space is a subspace of the other then there is a $u_i$ not in $W_1$, and a $v_j$ not in $W_2$. But since the union is a subspace that contains $u_i$ and $v_j$ then it must contain $v_j + u_i$. But this implies the sum is in either $W_1$ or $W_2$ and hence that either $u_i \in W_1$ or $v_j \in W_2$, a contradiction.
Hint: What would be happen if for example $W_1$ is not contained in $W_2$ and $W_2$ is not contained in $W_1$? Think of the point that with this assumption the set $W_1\cup W_2$ would not be a subspace of whole space $V$. This is a nice contradicton!
