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Assume ZF+ not AC. Then how many solutions are there for Cauchy functional equation?

Thank you

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  • $\begingroup$ Maybe only the continuous ones. It is consistent that all sets of reals are Lebesgue measurable. And possibly many, since one can deny Choice in many ways. For example we can have a Hamel basis but not full Choice. $\endgroup$
    – André Nicolas
    Apr 22, 2013 at 21:21

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The answer is undecidable. We know it could be $2^{\aleph_0}$ and it could be $2^{2^{\aleph_0}}$. I am unaware of results that it could be an intermediate cardinality, though.

It is true that there are always the continuous ones (and there are $2^{\aleph_0}$ of those), but it is consistent that there are only the continuous ones. For example in Solovay's model or in Shelah's model of $\sf ZF+DC+BP$ (the last one denotes "all sets of reals have the Baire property").

Assuming only that the axiom of choice fails is not enough to conclude in what manner it fails, and whether or not the real numbers are even well-orderable or not.

It is consistent that the axiom of choice fails and the real numbers are well-orderable, in which case the usual proof as if the axiom of choice holds shows that there are $2^{2^{\aleph_0}}$ solutions.

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