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Define two real numbers to be rationally equivalent provided their difference is a rational number.

from Royden Real Analysis

flag	asked May 3 2011 at 15:21 z0q0vk 3●1

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Are you saying that this is a problem in Royden's book? – Pete L. Clark May 3 2011 at 17:02

Andreas Blass · Accepted Answer · 2011-05-03 15:30:22Z UTC

If there were a choice function for this equivalence relation, there would be a set of real numbers that is not Lebesgue measurable. But the existence of such a set is not provable in ZF (i.e., without the axiom of choice). Even with the axiom of choice, i.e., in ZFC, it's consistent that no such set is definable. (On the other hand, with additional assumptions, like Gödel's axiom of constructibility, one can have a definable, in fact $\Delta^1_2$ , choice function.)

Find a explicit choice function of the “rationally equivalence class”

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