$\mathbb R$ is an uncountably dimensional vector space over $\mathbb Q.$ We can define as many endomorphisms of this vector space as we want by picking their values on the elements of the basis. However, to have a basis we need to use the axiom of choice, so this way is non-constructive. Any $\mathbb R-$ endomorphism of $\mathbb R$ is also a $\mathbb Q-$endomorphism. But can we give a concrete example of a $\mathbb Q-$endomorphism of $\mathbb R$ that is not an $\mathbb R-$ endomorphism?

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    $\begingroup$ No. Other people can give more details, but it's consistent with ZF that the only such endomorphisms are given by scalar multiplication. (I believe one approach to proving this is to show that any non-trivial such endomorphism can be used to construct a non-measurable subset of $\mathbb{R}$, and there's a model of ZF in which no such sets exist.) $\endgroup$ Feb 16, 2012 at 20:57
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    $\begingroup$ See also en.wikipedia.org/wiki/Cauchy's_functional_equation . $\endgroup$ Feb 16, 2012 at 21:02
  • $\begingroup$ @QiaochuYuan Thanks! Now I remember proving the density of the graphs of such functions during my first year! $\endgroup$
    – user23211
    Feb 16, 2012 at 21:05

1 Answer 1


Using the axiom of choice this is trivial. Choose a Hamel basis, and take any permutation of it.

However without the axiom of choice it is perfectly feasible to have a model in which there is no Hamel basis, that is $\mathbb R$ as a vector space over $\mathbb Q$ has no basis.

Of course that one does not need a basis to have endomorphisms, however we can make sure that indeed there are no endomorphisms of $\mathbb R$ over $\mathbb Q$. From such automorphism we can generate non-measurable sets, so if we happen to live in a model of ZF in which every set of real numbers is Lebesgue measurable there can be no such endomorphism.

Natural examples for such models are models of ZF when assuming The Axiom of Determinacy instead of The Axiom of Choice; or Solovay's model.


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