1. ## Vector Space proof

Hey,

Can you check to see if this problem is correct? Thank you!

Prove that if, If $\displaystyle U$ and $\displaystyle V$ are both subspaces of the vector space $\displaystyle W$, then $\displaystyle U \cup V$ is not a vector space unless one of the subspaces $\displaystyle U$ or $\displaystyle V$ are contained in the other.

$\displaystyle \text{\emph{Proof.}}$

Let $\displaystyle U$ and $\displaystyle V$ be vector spaces. Clearly if either $\displaystyle U$ or $\displaystyle V$ is
contained in the other, then $\displaystyle V \cup U$ is a vector space because
their union would simply be the space that the other one in
contained in. Now, suppose for the sake of contradiction that $\displaystyle U \cup V$ is a vector space and neither $\displaystyle U$ nor $\displaystyle V$ are contained
within the other. Then obviously, since neither is a subset of the
other, there exist elements $\displaystyle \mathbf{v} \in V$ and $\displaystyle \mathbf{u} \in U$ such that
$\displaystyle \mathbf{v} \notin U$ and $\displaystyle \mathbf{u} \notin V$. Consider the vector $\displaystyle \mathbf{v} + \mathbf{u}$. Since we are assuming that $\displaystyle U \cup V$ is a vector space, $\displaystyle \mathbf{v} + \mathbf{u} \in U \cup V$ so that $\displaystyle \mathbf{v} + \mathbf{u} \in U$ or $\displaystyle \mathbf{v} + \mathbf{u} \in V$.
If $\displaystyle \mathbf{v} + \mathbf{u} \in V$, then by closure $\displaystyle (\mathbf{v} + \mathbf{u}) + (-\mathbf{v})$ $\displaystyle = \mathbf{u}$
$\displaystyle \in V$ since $\displaystyle (-\mathbf{v}) \in V$. This obviously contradicts that $\displaystyle \mathbf{u}$
$\displaystyle \notin V$.A similar argument can be made if $\displaystyle \mathbf{v} + \mathbf{u} \in U$, so $\displaystyle U$
$\displaystyle \cup V$ is not a vector space. Therefore, for $\displaystyle U \cup V$ to be a
vector space, one of the spaces must be contained within the other.

2. looks good to me.