TITLE: Example of an additive but not linear map beetween real topological linear spaces
QUESTION [2 upvotes]: Let $X$ and $Y$ be real topological linear spaces. It's easy to prove that if $f\colon X \to Y$ is additive and continuous, then it's linear. 
I'm looking for a counterexample if the continuity request is omitted. More specifically, consider the following cases: 

$X$ and $Y$ are finite dimensional and the topology is given by a norm (in particular $X=Y=\mathbb R$). Then “continuity” and “linearity” are, in fact, equivalent [modulo additivity]. 
$X$ and $Y$ are normed but not necessarily finite dimensional.
What about a generalization to the complex case, i.e. $X$ and $Y$ are linear spaces over $\mathbb C$?

I don't have any particularly useful idea, apart from the observation that, for the first point, continuity at a point implies continuity everywhere. This is so because if $f$ is continuous at $x_0$, and $y_n\to y$, then $f(y_n)=f(y_n-y+x_0)+f(y-x_0)$, and the first term converges to $f(x_0)$ by continuity in $x_0$. So, for the $X=Y=\mathbb R $ case, we seek for a function $f$ such that $f(\frac{p}{q})=\frac{p}{q}f(1)$, for $\frac{p}{q}\in \mathbb Q$, and $f$ is discontinuous everywhere.
Any hint, also for only one of the above points, is highly appreciated, thank you.

REPLY [1 votes]: Since $\mathbb R$ is a $\mathbb Q$-vector space, the automorphism of the subspace $\mathbb Q[\sqrt 2]$ that maps $a+b\sqrt 2\mapsto a-b\sqrt 2$ can be extended to an automorphism of $\mathbb R$.
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