Weird partitions of the real line Partition $\mathbb{R}$ into subsets $S_r$ indexed by $r\in\mathbb{R}$ with the following property: For all $r₁, r_2\in\mathbb{R}$:
$$\forall x\in S_{r_1}\forall y\in S_{r_2}:\; x+y\in S_{r_1+r_2}$$
An obvious family of examples is given by $S_r=\{a(r)\}$, where $a$: $\mathbb{R}\longrightarrow\mathbb{R}$ is an additive injection.
Are there any other (weird) examples? In particular, must the $S_r$ be singletons?
Note: If $r(x)$ denotes the unique $r\in\mathbb{R}$ with $x\in S_{r}$, then $r$ is an additive map $\mathbb{R}\longrightarrow\mathbb{R}$.
 A: Note that $G := S_0$ is an additive subgroup of $\mathbf R$, as: If $0 \in S_r$, then $0 + 0  =0 \in S_{r+r} = S_{2r}$, hence $r = 0$, that is $r = 0$. If $x,y\in S_0$, then $x+y \in S_0$, and finally, if $x \in S_0$ and $-x \in S_r$, than $x+(-x) \in S_{r+0} = S_r$, but $0 \in S_0$, hence $r = 0$. That is $-x \in S_0$. 
Moreover, each $S_r$ is a translate of $S_0$: If $x \in S_r$, then 
$$ y \in S_r \iff y-x \in S_0 $$
On one side, let $y-x\in S_0$, then $y = x+(y-x) \in S_{r+0} = S_r$, on the other side, let $y\in S_r$, then $-y\in S_{-r}$ due to $0 \in S_0$, hence $y-x \in S_{r-r} = S_0$. 
That is, if we consider the quotient group $\mathbf R/S_0 = \{S_r \mid r \in \mathbf R\}$ as a group, then this group must be isomorphic to $\mathbf R$, as we want to have $S_r + S_t = S_{r+t}$. On the otherhand, subgroups $S_0 \le \mathbf R$ with $\mathbf R/ S_0 \cong \mathbf R$ correspond to onto group homomorphisms, $a \colon \mathbf R \to \mathbf R$ with kernel $S_0$. To have a weird example, let $B$ your favourite $\mathbf Q$-base of $\mathbf R$, then $B$ is in bijection with $\mathbf R$, choose any bijection $\phi$. Then define $a\colon \mathbf R \to \mathbf R$ by $a(\sum q_b b) = \sum q_b \phi(b)$. Then $a$ is onto but not one-to-one.
