1. ## Vector Space proof

Hey,

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

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

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

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

2. looks good to me.