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.