5
$\begingroup$

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.

Improve this question
$\endgroup$
12
  • 2
    $\begingroup$ Previously posted on MSE: math.stackexchange.com/questions/366010 $\endgroup$
    – Zhen Lin
    Commented Apr 20, 2013 at 19:22
  • 3
    $\begingroup$ 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. $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 20, 2013 at 19:23
  • 1
    $\begingroup$ 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. $\endgroup$
    – Keshav Srinivasan
    Commented Apr 20, 2013 at 20:51
  • 1
    $\begingroup$ 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. $\endgroup$
    – Goldstern
    Commented Apr 21, 2013 at 14:31
  • 1
    $\begingroup$ @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)? $\endgroup$
    – Alexander Pruss
    Commented Jan 11, 2014 at 13:48

1 Answer 1

Reset to default
12
$\begingroup$

To my knowledge this is an open problem.

If one looks at Herrlich The Axiom of Choice, there is a diagram (7.23, p. 156) of implications related to non-measurable sets (which include discontinuous solution to the Cauchy functional equation problem), one can see that this is pretty far down below the existence of a Hamel basis.

Had it been known to be equivalent, an arrow back would be there -- and the book is not that old.

I doubt Herrlich would have omitted a reference to such fact, had it been known, and the last time I looked around, I couldn't find anything newer which proved anything related to that (and I did look around several times when answering a couple of questions).

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .