In my book, it is stated that the following are equivalent:
(a) $v-w \in U$
(b) $v+U = w + U$
(c) $(v+U) \cap (w+U) \neq \varnothing$
The proof for (a) $\Rightarrow$ (b) is the following:
$v-w \in U$, if $u \in U$, then $v+u = w + ((v - w) + u) \in w + U$, same for the other direction.
While I get this proof, I don't see how it applies to this example:
Say $U = \{(x+3,2x) \in \mathbb R^2:x \in \mathbb R\}$, $v = (2,0), w = (5,0)$. Then $w-v = (3,0) \in U$, so (b) has to follow. However, when drawing $v+U$ and $w+U$, you get two parallel lines, not equal lines, so $v+U$ and $w+U$ are disjoint. What am I missing?
Thanks.
These equivalences only hold when $U$ is a vector subspace of a vector space $V$ (or more generally a subgroup of an abelian group, since this doesn't involve the scalars): indeed in the proof you use the fact that since $v-w$ and $u\in U$ their sum is also in $U$.
But your $U$ is not a vector subspace, as it doesn't contain the zero vector, and in fact it is not closed under addition.