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Can an additive function be nonlinear without AC?

  1. Apr 18, 2013 #1
    A function is additive if f(x+y) = f(x) + f(y). Intuitively, you might think that an additive function on R is necessarily 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 R as a vector space over Q, and then you define your function f to be different in at least two distinct elements of the basis.

    But my question is, does the same conclusion hold in the absence of the axiom of choice? Or does the negation of choice force the function to be linear?

    Any help would be greatly appreciated.
    Thank You in Advance.
  2. jcsd
  3. Apr 18, 2013 #2
    The answer is a bit unsatisfying. In the absence of AC, you work in the ZF axioms. It turns out that in the ZF axioms, the question you ask cannot be answered. Both the question (there is a nonlinear additive function) as its negation turn out to be consist with ZF set theory. So if you want, you can accept an axiom that says "all additive functions are linear". In ZF, this is perfectly ok. It will not turn out to be consist with the axiom of choice. However, if you want to accept an axiom that says "there is a nonlinear additive function", then this will be ok in ZF too (and in ZFC, where it is a theorem instead of an axiom).

    If you want to know why those things are consistent with ZF set theory, then there is no easy answer. You'll need to study the technique of forcing.
  4. Apr 18, 2013 #3
    Thanks micromass. But does ZF + no Hamel basis of R imply that all additiive functions are linear? In other words, does the existence of a nonlinear additive function imply the existence of a Hamel basis of R?
  5. Apr 18, 2013 #4
  6. Apr 18, 2013 #5
    micromass, a working version of the website is here. I just tried using it. In its notation, 366 refers to the existence of a discontinuous additive function from ℝ to ℝ, and 367 refers to the existence of a Hamel basis of R. When I made it give the implication table of 366 and 367, it gave the result that 367 implies 366, as expected, and then it said 366 does not imply 367. But then when I asked it to give a list of models satisfying 366 but not 367, it said no models. So what's going on? Does it knows that there exists such a model, but it doesn't know of any examples?

    It just seems really counterintuitive that there can be a nonlinear additive function in the absence of a Hamel basis of R.

    EDIT: I misread what the implication table was saying. It was not saying 366 does not imply 367. It was saying that it does not know whether 366 implies 367.
    Last edited: Apr 18, 2013
  7. Apr 21, 2013 #6
    I've been told by various people on the internet that in order for there to be a nonlinear additive function on R, R must be able to be decomposed into a direct sum of nontrivial subspaces over Q. First of all, is that true? It certainly seems intuitively plausible. And if it is true, is that condition strictly weaker than R having a Hamel basis over Q? It would seem like it would be weaker.

    By the way, that other link to the Consequences of the Axiom of Choice Project seems to be down now.
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