How do I prove that a given set is an affine space? 
Let $A$ be an affine space over a vector space $V$. Let $B$ be a nonempty subset of $A$ such that $W:=\{a-b\in V:a,b\in B\}$ is a subspace of $V$. In this case, how do I prove that for each $b\in B$ and $y\in W$, $b+y\in B$?
Thank you in advance.
 A: This is wrong, hence you can't prove it :-)
As an example choose
$A=V=\mathbb{R}^2$
and $B:=\{(x,y) \ | \ x=0 \lor y=0\}$
as the union of the two axes. You can easily see, that $W=\mathbb{R}^2$ holds (for $v=(x,y)\in V$ take $a=(x,0)$ and $b=(0,-y)$). But obviously $B$ is not an affine space. E.g. $(0,0)+(1,1) =(1,1)\notin B$ but $(0,0) \in B$ and $(1,1) \in W$.
A: If $B$ is itself an affine space of $V$ and a subset of $A$, then we get the desired conclusion.
Since $A$ is an affine space of $V$, there exists a subspace $U$ of $V$ and a vector $v$ in $V$ such that $A = v + U = \{v + u:u \in U\}.$ Since $B$ is also affine and a subset of $A$, there exists a subspace $U'$ of $U$ such that $B = v + U' = \{v + u':u \in U'\}$. Let $a, b \in B$. Then, there exist $x_a,x_b \in U'$ such that
$$
a = v + x_a \\
b = v + x_b.
$$
If $y \in W$, there exist $a,b \in B$ such that $y = a - b = (v + x_a) - (v + x_b) = x_a - x_b \in U'$. Let $b = v + x_b' \in B$. Then,
$$
b + y = v + x_b' + (x_a - x_b) = v + (x_b' + x_a - x_b).
$$
$U'$ is a subspace of $V$, so $U'$ is closed under addition and scalar multiplication. Furthermore, $x_b',x_a,x_b\in U'$. Hence, $(x_b'+x_a-x_b) \in U'$, so $v + (x_b' + x_a - x_b)\in B$.
