Prove that $U+W = \{u+w\mid u\in U, w\in W\}$ is a finite-dimensional subspace of $V$ Let $V$ be a vector space over a field $k$ and let $U,W$ be finite-dimensional subspaces of $V$. Prove that $U+W = \{u+w\mid u\in U, w\in W\}$ is a finite-dimensional subspace of $V$.
I know how to prove that $U\cap W$ is a subspace of $V$ but I'm having a hard time grasping how to prove $U+W$
This is what I was doing in regards to the first part of proving closed under addition:
Let $u,w\in U+W$ and $u\in U, w\in W$
Since $U$ is a space, it is closed under addition and $u+w\in U$. Also, $W$ is a space so it is closed under addition as well and so $u+w \in W$. So, $u+w\in W$ and $u+w\in W$ shows that $u+w\in U+W$. Thus, $U+W$ is closed under addition. 
But this is almost exactly how I've proved $U\cap W$ is a subspace and I have a feeling I've made a mistake somewhere. 
 A: To show that $U+W$ is closed under addition, you need to assume two elements in $U+W$, and show their addition is also in $U+W$.
So let $v\in U+W$ and $v'\in U+W$. Suppose $v=u+w, v'=u'+w'$, where $u,u'\in U$ and $w,w'\in W$.
Then $v+v'=(u+u')+(w+w')$. This is in $U+W$ since $U$ and $W$ are closed under addition, thus $u+u'\in U$ and $w+w'\in W$.
A: To show $U + W$ is a subspace of $V$ it must be shown that $U + W$ contains the the zero vector, is closed under addition and is closed under scalar multiplication. Showing the zero vector is in $U + W$ is very easy. Since $U, W$ are subspaces of V, $0 \in U,V.$ Thus, $0 + 0 = 0 \in U + W.$ Now let $x, y \in U + W.$ This means $ x = a + b$ for some $a \in U$ and $b \in W$ and $y = c + d$ for some $c \in U$ and $d \in W.$ Now, $x + y = (a + b) + (c + d) = (a + c) + (b + d).$ Since a,c $\in U$ and $U$ is a subspace of $V$ $a + c \in U.$ The same argument can be made for $b + d \in W.$ Thus, $x + y \in U + W$ and $U + W$ is closed under addition. Showing $U + W$ is closed under scalar multiplication is similar and I leave it to you to work out.   
A: "Closed under addition" means that $u+w \in U$ would be true if $u,w \in U.$ Written more concisely, $$U + U \subseteq U.$$ It's not enough when $u \in U$ and $w$ isn't. This is why your proof doesn't work.
You can make a proof around the computation $$(U+W) + (U+W) = (U+U) + (W+W) \subseteq U + W.$$
A: There's no reason for $u+w$ to be in $U$ since $w$ doesn't necessarily belong to $U$. 
What you have to prove is that, if $u+w\in U+W$, $u'+w'\in U+W$ ($u,u'\in U$, $w,w'\in W$) then  $(u+w)+(u'+w')$ has the form $(u''+w'')$ for some $u''\in U , w''\in W$, and similarly for $\lambda (u+w)$.
