@JonasTeuwen Hehehe OK. Remember the problem that went "Suppose that $f(x+y)=f(x)+f(y)$ and $f$ is continuous at $0$. Show that $f$ is continuous for all $x$."
@JohnSenior I thought it was because he figured he was too sadistic giving all the diseases and such and this is like to make it up a bit. To make it bearable.
@JonasTeuwen PROOF First note that $f(0)=f(0)+f(0)$ so that $f(0)=0$. Then we have that $$\lim_{x\to 0}f(x)=0$$. Now, this is equivalent to $$\lim_{x\to 0}(f(x)+f(p))=f(p)$$ for any fixed $p$. Then we obtain $$\lim_{x\to 0}f(x+p)=f(p)$$, but that is equivalent to $$\lim_{x\to p}f(x)=f(p)$$ $\blacktriangle$
maybe get a discontinuous solution by something like this: define $f$ to be some random real for each $b_\alpha$ in the Hamel basis, and extend to all of $\mathbb{R}$ by the functional equation
I think the Hamel basis we are talking about here is a base for $\mathbb{R}$ as a vector space over $\mathbb{Q}$
