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It's obviously not injective in the case of $f(x)=0$. I'm wondering if it's injective in all other cases. The other linear solutions of the form $f(x)=c\cdot x$ where $c$ is some constant are injective, what about the "wild" additive functions?

Sil
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    The usual sort of discontinuous example is constructed, say, by taking a basis for $\mathbb R$ as a vector space over $\mathbb Q$ and then projecting onto a basis vector. functions constructed in that manner are clearly not injective. – lulu Feb 15 '16 at 20:43
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    @lulu That's not true. An uncountable family of well-known examples are the additive involutions, which can be defined like this: partition a Hamel basis to two sets $ I $ and $ J $, set $ f ( x ) = x $ for $ x \in I $ and $ f ( x ) = - x $ for $ x \in J $, and extend linearly (over $ \mathbb Q $) to the whole $ \mathbb R $. Then you get a solution of Cauchy's functional equation that satisfies $ f \big( f ( x ) \big) = x $. This implies that $ f $ is bijective. And there are $ 2 ^ { 2 ^ { \aleph _ 0 } } $ of these functions, more than the $ 2 ^ { \aleph _ 0 } $ linear functions. – Mohsen Shahriari May 03 '21 at 05:10
  • @MohsenShahriari if you or lulu could summarize this and also talk about surjectivity, this post could be the target of a "close as duplicate" of the following new post: https://math.stackexchange.com/questions/4701162/is-every-non-zero-additive-function-injective-surjective?noredirect=1&lq=1 – Anne Bauval May 17 '23 at 15:55
  • @lulu same message to you – Anne Bauval May 17 '23 at 15:55
  • Or maybe the general answer (about the 4 cardinalities of inj-surj, noninj-surj, inj-notsurj, bij, as a function of the dimensionS and the cardinality of the field) has already been written somewhere on this site? – Anne Bauval May 17 '23 at 17:42

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