Which of these sums is equal to $\mathbb R^4$?

Question

I'm given the following sets: $$U=\{(0,a,b,a-b): a,b \in \mathbb{R}\} \\ V=\{(x,y,z,w): x=y, z=w\} \\ W=\{(x,y,z,w): x=y\}$$

I'm trying to determine which of the following is equal to $\mathbb R^4$: $U+V, U+W, V+W$ where $A+B=\{a+b \ , a\in A, b \in B\}$)

Here's my working:

$U+V=\{x, a+y, b+z, a-b+w \quad : a, b \in \mathbb{R}, x=a+y, b+z=a-b+w\}$

$U+W=\{x, a+y, b+z, a-b+w \quad :a,b \in \mathbb{R}, x=a+y\}=\{x,x, b+z, b+z \}$

$V+W=\{2x,2y,2z,2w \quad : x=y, z=w\}=\{x', x', z', z'\}$

Obviously, $V+W \neq \mathbb R^4$ as $(1,2,3,4) \notin V+W,$ but the other two sums looks very familiar. How would I go about eliminating one of them from being $\mathbb R^4$?

Answer 1

You need to have freely changing variables.

$U$ gives you free second and third terms ($a,b$) while $W$ gives you free first, third and fourth terms.

$U+W=\{x, a+y, b+z, a-b+w \quad :a,b,x,y,z,w \in \mathbb{R}\}=\{x',y', z', w' \quad x',y',z',w' \in \mathbb{R} \}$