# Thread: Unions of subspaces

1. ## Unions of subspaces

Question: Let $W_1, W_2$ be subspaces of vector space $V$. Prove that $W = W_1 \cup W_2$ is a subspace of $V$ if and only if $W_1 \subseteq W_2$ or $W_2 \subseteq W_1$.

I managed to prove it one way. I managed to show that if $W_1 \subseteq W_2$ or $W_2 \subseteq W_1$, then $W_1 \cup W_2$ is a subspace of $V$.

However, I cannot go the other way. Surely the zero vector is contained in $W$ because it is contained in $W_1,W_2$ and thus their union. So that's not the problem; the problem seems to be whether or not the structure $W$ is closed under addition.

So I imagine that if neither $W_1$ nor $W_2$ are subsets of each other, then there exists $x \in W_1$ such that $x \notin W_2$ and there exists $y \in W_2$ such that $y \notin W_1$. And somehow, I need to show that there exists $x,y \in W$ such that $x+y \notin W$ from that. This would show that if neither $W_1 \subseteq W_2$ nor $W_2 \subseteq W_1$ hold, then $W=W_1 \cup W_2$ fails to be a subspace of $V$

How do I do so? Am I even thinking on the right track?

Thanks!

2. Let $x \in W_1 - W_2$ and let $y \in W_2-W_1$ consider $x+y=z$ then if $z \in W_1$ then $y=z-x \in W_1$ which can't be, and if $z \in W_2$ then $x=z-y \in W_2$ and so $W_1 \cup W_2$ is not a subspace.

3. Damn, owned by a one liner.

Thanks a bunch.