Let $W$ be a vector space and let $U$ and $V$ be finite-dimensional subspaces. a) Show that $U ∩ V$ is a subspace of $W$.
b) Show that $U + V = \{u + v : u \in U, v \in V\}$ is a subspace of $W$.
c) Show that $\dim(U+V) = \dim(U) + \dim(V) - \dim(U ∩ V)$.
I have no idea where to start.  Any help?
 A: For part a) Use the definition of what a subspace means. It is first closed under vector addition, has an identity, and is closed under scalars.  First show closure under vector addition which is basically proving this:  If $x,y\in U\cap V\implies x+y\in U+V$. Now the identity is trivial to show. Since $U$ and $V$ are finite subspaces it follows by definition that... And for scalar multiplication. Let $r\in\mathbb{R}$ where $r$ will be a scalar. Let $x\in U\cap V\implies x\in U$ and $x\in V$. It follows that since $U$ and $V$ are subspaces then $rx\in U$ and $rx\in V \implies rx\in U\cap V$. Hence we conclude... 
For part b) Same thing. Let $x\in U+V$. First show closure. That is if $x,y\in U+V$ where $x=a+b$ and $y=c+d$ where $a,c\in U$ and $b,d\in V$ then we see $x+y=(a+b)+(c+d)$ which is still an element of the set $U+V$. For the identity its easy. I will denote $0$ to be the identity both in $U$ and $V$ since it follows by definition of a subspace. Thus $0=0+0\in U+V$ And lastly scalar multiplication which I will leave up to you.
For part c) Recall what dimension means and remember $U+V$ is a set.
I forgot to put arrows on top of the elements but they are all vectors. So you can place arrows for clarity. 
