Understanding sum of subsets of a subspace Before I get to my query, note that throughout I use the definition of the sum of subsets of a vector space as given in "Linear Algebra Done Right":

Suppose $U_1, \dotsc, U_m$ are subsets of V. The sum of $U_1, \dotsc, U_m$ denoted $U_1 + \dotsb + U_m$ is the set of all possible sums of elements of $U_1, \dotsc, U_m$. More precisely $U_1 + \dotsb + U_m= \{u_1 + \dotsb + u_m : u_1 \in U_1, \dotsc, u_m \in U_m \}$.

Further, note that the field $F$ denotes $\mathbb{R}$ or $\mathbb{C}$ and similarly $F^n$ denotes either $\mathbb{R}^n$ or $\mathbb{C}^n$, as is the convention in the aforementioned book.
Now to my problem. Below I give an example from "Linear Algebra Done Right", where frankly I have no clue what is going on.

Suppose that $U= \{(x,x,y,y) \in F^4: x,y\in F \} $ and $W=\{(x,x,x,y)\in F^4:x,y \in F \}$. Then $U+W= \{ (x,x,y,z) \in F^4 : x,y,z \in F \}$.

I would have thought the sum to be $U+W= \{(2x,2x,x+y,2y): x,y\in F \} $, i.e. simply adding the elements of $U$ and $W$.
If any of you would be so kind as to explain to me where I have gone wrong, I would be very grateful. 
 A: 
I would have thought the sum to be $U+W= \{(2x,2x,x+y,2y): x,y\in F \} $, i.e. simply adding the elements of $U$ and $W$.
  If any of you would be so kind as to explain to me where I have gone wrong, I would be very grateful.

The point is that you have to add arbitrary elements from $U$ and $W$, so for two elements $v_1 = (x,x,y,y) \in U$ and $v_2 = (w,w,w,z) \in W$, the sum is $v_1 + v_2 = (x + w, x + w, y + w, y + z) \in U + W$. Your mistake was to take the same parameters for $v_1$ and $v_2$.
Now we see that $v_1 + v_2$ has the same first components, so $U + W \subset \{ (x,x,y,z) \mid x,y,z \in F\}$.
For the other direction, take an arbitrary element $v = (x, x, y, z)\in \{ (x,x,y,z) \mid x,y,z \in F\}$. We now have to find elements $v_1 = (x_1,x_1,y_1,y_1)\in U$ and $v_2=(x_2,x_2,x_2,y_2) \in W$ such that $v = v_1 + v_2$, i.e. we get the following equations:
\begin{align}
x & = x_1 + x_2 \\
y & = y_1 + x_2 \\
z & = y_1 + y_2
\end{align}
So $x_2 = x - x_1$. If we plug this into the second equation we get $y_1 = y - x + x_1$, thus the last equation yields $y_2 = z - y + x - x_1$. Choose any value for $x_1$ and take $x_2$, $y_1$ and $y_2$ according to these equations gives us $v_1$ and $v_2$. you may choose
\begin{align}
x_1 &= 0 \\
x_2 &= x \\
y_1 &= y - x \\
y_2 &= z - y +x.
\end{align}
Now you only have to check that those values really give $v$.
