1. ## Unions of subspaces

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

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

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

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

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

Thanks!

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

3. Damn, owned by a one liner.

Thanks a bunch.