Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be an additive function. We have that $f(n)=nf(1)$ for $n\in\mathbb{Z}$ by induction, and we can extend to the rationals by setting $nf\left(\cfrac{m}{n}x\right)=f(mx)=mf(x)\Rightarrow f\left(\cfrac{m}{n}x\right)=\cfrac{m}{n}f(x)$. Hence, $f(q)=qf(1)$ for all $q\in\mathbb{Q}$.
But let's say we were given a little more information. Let's say there was some set $S=\{ r_{i}\}$ such that $f(r_{i})=r_{i}f(1)$. My question is this: What properties must $S$ have for us to conclude that $f$ is linear as opposed to one of the continuous nowhere, unbounded, pathological solutions? Besides, of course, things like $S$ being the set of reals from $0$ to $1$ (correct me if I'm wrong, but that would mean $f$ is continuous, right?), or being the set of reals in any interval for that matter.
One thing I noticed is that $S=H$ for some Hamel basis $H$ of $\mathbb{R}$ over $\mathbb{Q}$ would satisfy the condition, since if $x=\sum_{i=1}^{n}r_{i}h_{i}$ for $r_{i}\in\mathbb{Q}$ and $h_{i}\in H$, then we would have \begin{align*} f(x)&=f\left(\sum_{i=1}^{n}r_{i}h_{i}\right)\\ &=\sum_{i=1}^{n}f(r_{i}h_{i})\\ &=\sum_{i=1}^{n}r_{i}f(h_{i})\\ &=\sum_{i=1}^{n}f(1)r_{i}h_{i}\\ &=f(1)\sum_{i=1}^{n}r_{i}h_{i}\\ &=f(1)x. \end{align*}
But are there any other sets that work that aren't either things like intervals or supersets of Hamel bases? Also, what would the minimum cardinality of $S$ need to be? Thanks.
Assuming the axiom of choice, $r_i$ must contain a Hamel Basis. The reason is that if the finite $\mathbb{Q}$-linear combinations of the $r_i$ do not generate all of $\mathbb{R}$, then we can extend it to a Hamel Basis and define $f$ to be linear on the the subspace generated by the $r_i$ and then however we like on the other basis elements.
Since the set of finite $\mathbb{Q}$-linear combinations of an infinite set $S$ has the same cardinality as $S$, the minimum possible cardinality of $S$ is the cardinality of the continuum.
I will add the note that if we do not assume the axiom of choice, it is possible that $S=\mathbb{Q}$. The reason for this is that if you reject the axiom of choice, it is possible that all sets are measureable, but constructing a nonlinear function that obeys the addition rule would allow you to construct a nonmeasureable set (the idea is to do something vaguely similar to the Vitali Set; I'll see if I can find a reference)