# Additive groups $\mathbb R$ and $\mathbb C$

Are the additive groups $$(\mathbb R,+)$$ and $$(\mathbb C,+)$$ isomorphic in Zermelo–Fraenkel set theory with the negation of AC?

Added remark: I was told at a lecture that the groups are isomorphic while assuming the axiom of choice. A natural (above) question arose, but I was not able to sort that out. I did not realize before, that the negation of AC is a kind of useless axiom...

• Are you asking whether the axiom of choice is need to define $\mathbb R$, $\mathbb C$, and the addition on both of them? – José Carlos Santos Sep 25 '18 at 12:25
• @JoséCarlosSantos More likely whether it's needed to define an isomorphism between them. – Arthur Sep 25 '18 at 12:25
• @Arthur You are probably right. – José Carlos Santos Sep 25 '18 at 12:26
• Note that for a specific statement like this, ZF with the negation of AC is functionally equivalent to just ZF. All that the negation of AC does is guarantee that some set of non-empty sets, somewhere, doesn't have a choice function. We have no way of knowing what that set is, and therefore it would be very difficult to leverage that existence into proving anything about the existence of certain functions between $\Bbb R$ and $\Bbb C$. – Arthur Sep 25 '18 at 12:29
• Try not to just post questions, as people won't know where to start to be able to help you (and you will get downvotes and close votes, as happened here). Try to explain what you have tried and where and why you got stuck/need help. It would also help if you said where you found this question (week X of course Y, where we covered topic Z). – user1729 Sep 25 '18 at 15:21

Without the axiom of choice (or a suitable similar assumption), you cannot prove that the two groups are isomorphic. On the other hand, you cannot prove that they aren't isomorphic either.

Assuming ZF is consistent, ZF is consistent both with axioms which let you prove the isomorphism (like AC), and with axioms which let you disprove it (see the answer by Asaf Karagila). That means that isoomorphism cannot be shown either way from ZF alone.

• Got it. A natural question arises – are the groups not isomorphic in some ZF + a suitable axiom? – byk7 Sep 25 '18 at 12:38
• @byk7 No common axioms that I can think of right now. Although, you could probably add this non-isomorphicness specifically as an axiom. It would seem kindof forced, but I believe you would get a consistent system (as consistent as ZF) out of it. – Arthur Sep 25 '18 at 12:42
• @byk7: Every set of reals has the Baire Property/Lebesgue measurable. The former does not require stronger axioms. – Asaf Karagila Sep 25 '18 at 13:05
• Why can't one prove this?! – user1729 Sep 25 '18 at 15:20
• @user1729 Because (assuming ZF is consistent), ZF is consistent both with axioms which let you prove it (like AC), and with axioms which let you disprove it (see the answer by Asaf Karagila). That means that it cannot be shown either way from ZF alone. – Arthur Sep 25 '18 at 15:38

As mentioned in the other answer, you cannot prove in $$\sf ZF+\lnot AC$$ that $$\Bbb R$$ and $$\Bbb C$$ are isomorphic as additive groups.

On the one hand, the failure of choice is not a localized statement. It tells you nothing about where it fails, it is consistent that any "reasonable set" and its power set are all well-orderable, so all the theorems with the axiom of choice follow immediately. In particular, it is perfectly reasonable that choice fails but $$\Bbb R$$ can be well-ordered, in which case there are Hamel bases for $$\Bbb R$$ and $$\Bbb C$$ over $$\Bbb Q$$, and the groups are isomorphic.

On the other hand, assumptions such as "Every set of reals has the Baire property" , "Every set of reals is Lebesgue measurable", or The Axiom of Determinacy (which itself implies the two former statements) imply—at least in the presence of Dependent Choice—that any homomorphism between Polish groups is continuous, so in particular any homomorphism between $$\Bbb R$$ and $$\Bbb C$$ is continuous. But it is fairly easy to show that no isomorphism can be continuous.

One should remark, however, about the consistency strength of such failures. If all sets are Lebesgue measurable (again, assuming Dependent Choice), then a theory stronger than just $$\sf ZF$$ is consistent. Even more in the case of Determinacy. However, the assumption that every set has the Baire property does not increase the consistency strength.