# Does a nonlinear additive function on R imply a Hamel basis of R?

A function is additive if $f(x+y) = f(x) + f(y)$. Intuitively, it might seem that an additive function from R to R must be linear, specifically of the form $f(x) = kx$. But assuming the axiom of choice, that is wrong, and the proof is rather simple: you just take a Hamel basis of $\mathbb{R}$ as a vector space over $\mathbb{Q}$, and then you define your function f to be different in at least two distinct elements of the basis.

But my question is this: if there is no Hamel basis of $\mathbb{R}$, then must $f$ be linear? To put it another way, does ZF + the existence of a nonlinear additive function imply the existence of Hamel basis of $\mathbb{R}$?

I checked the Consequences of the Axiom of Choice Project, a database of choice axioms and their relationships here, and it said that it didn't know.

• Previously posted on MSE: math.stackexchange.com/questions/366010 – Zhen Lin Apr 20 '13 at 19:22
• Since you copy pasted your question (without bothering to reply to my answer or to my comments), I figured I'll copy-paste my answer here as well. In case you happened to have missed it on MSE somehow. – Asaf Karagila Apr 20 '13 at 19:23
• Asaf, I'm sorry I didn't respond to you. I found your answer, giving reasons why you think it's probably an open problem, to be really helpful. I was just waiting, holding out hope that someone else may be aware of some research that resolved the question. – Keshav Srinivasan Apr 20 '13 at 20:51
• Instead of "certainly" I should have written "it seems clear to me, or at least plausible". I admit I don't have a proof, otherwise I would have written it as an answer. – Goldstern Apr 21 '13 at 14:31
• @Goldstern: This is a naive question. In your BPI argument, how do you ensure that $f$ is real-valued (rather than having values in some hyperreal field)? – Alexander Pruss Jan 11 '14 at 13:48