Verifying an example of sum of subsets Book: Linear Algebra Done Right 3E
Here's an example from the textbook that I cannot make sense of:
Suppose that $U = \{(x,x,y,y) \in \mathbf{F^4}: x,y\in\mathbf{F}\}$ and $W = \{(x,x,x,y) \in \mathbf{F^4}: x,y\in \mathbf{F} \}$. Then $$U+W = \{ (x,x,y,z)\in \mathbf{F^4}: x,y,z\in \mathbf{F} \}$$
To me it makes sense that $$U+W = \{ (2x,2x,x+y,2y)\in \mathbf{F^4}: x,y\in \mathbf{F} \}$$
 A: You seem to be a little confused about set-builder notation. $U$ is the set of all four-tuples where the first two and last two things are the same. $W$ is the set of all four-tuples where the first three things are the same. Your answer is wrong because of examples like $(1,1,2,2)\in U$, $(4,4,4,5)\in W$, so $(5,5,6,7)\in U+W$. Note that $5$ isn't even since the "first two things" in the element of $U$ and the "first three things" in the element of $W$ can be different.

A useful thing to keep in mind is that the variables in set-builder notation are "dummy variables", a.k.a. "bound variables": the notation means the same thing no matter what variables you have. A way to get at the right answer is to make these variables all different to make it easier to think about.
$U=\left\{(a,a,b,b)\in\mathbf{F}^4:a,b\in\mathbf{F}\right\}$ and $V=\left\{(c,c,c,d)\in\mathbf{F}^4:c,d\in\mathbf{F}\right\}$. Then $U+W=\left\{(a+c,a+c,b+c,b+d)\in\mathbf{F}^4:a,b,c,d\in\mathbf{F}\right\}$. Clearly this is a subset of $\left\{(x,x,y,z)\in\mathbf{F}^4:x,y,z\in\mathbf{F}\right\}$. To show that it's a superset, set $a=x-y,b=0_{\mathbf{F}},c=y,d=z$.
