TITLE: Prove that the union of three subspaces of $V$ is a subspace iff one of the subspaces contains the other two.
QUESTION [16 upvotes]: Prove that the union of three subspaces of V is a subspace iff one of the subspaces contains the other two.
I can do this problem when I am working in only two subspaces of $V$ but I don't know how to do it with three. 
What I tried is:
If one of the subspaces contains the other two, Then their union is obviously a subspace because the subspace that contains them is a subspace. (Is this sufficient??).
If the union of three subspaces is a subspace.....
How do I prove that one of the subspaces must contain the other two from here?
*When proving this for two I said that there is an element in one of the subspaces that is not the other and proved by contradiction that one of the subspaces must be contained in the other. How would I do this for three?

REPLY [2 votes]: With regard to Gina and Jeff's answers, I believe the simplification at the start can be made easier (at least in my opinion).
Suppose that subspaces $U_1,U_2,U_3$ union to form a subspace.
We have 2 cases.
Case 1: Suppose $U_i\subseteq \cup_{j\neq i}U_j$ for some $i\in \{1,2,3\}$.
WLOG, let $i=1$. Then the problem is reduced to the case of 2 subspaces $U_j, j=2,3$.
WLOG again, we have $U_2\subseteq U_3$, then $U_1\subseteq U_2 \cup U_3 = U_3$.
Hence $U_3$ contains the other 2 sets.
Case 2: $\forall i\in \{1,2,3\}, U_i\not\subseteq \cup_{j\neq i}U_j$.
Then for each $i$, there exists $\mathbf{u}_i\in U_i$ such that $u_i\notin \cup_{j\neq i}U_j$, i.e.  $u_i\notin U_j$ for $j\neq i$.
After this, we can make a similar argument to that of Gina's. For simplicity, I assume the field in concern to be the real/complex field like in Axler's book.
$\mathbf{u}_1 + \mathbf{u}_2, 2\mathbf{u}_1 + \mathbf{u}_2, \mathbf{u}_1 + 2\mathbf{u}_2$ must all be in $U_3\setminus (U_1 \cup U_2)$. (Note that all these vectors are in $\cup_{1\leq i\leq 3} U_i$.) In particular, these vectors lie in $U_3$. Hence, $\mathbf{u}_1, \mathbf{u}_2 \in U_3$, a contradiction.
Hence only case 1 is possible and we are done.