Automorphisms on $(\mathbb R,+)$ and the Axiom of Choice

We know that the algebraic automorphisms of the real numbers under addition is not in $$\text{1:1}$$ correpondence with $$\mathbb R \setminus \{0\}$$; see here.

The argument uses the AOC.

Suppose we drop the AOC from $$\text{ZFC}$$ replacing it with

Axiom (GR):

The injective mapping

$$\quad \Phi: \mathbb R \setminus \{0\} \to \text{AutomorphismGroup(} \mathbb R ,+ \text{)}$$

is surjective.

Has this $$\text{ZF+GR}$$ been tried and/or does this lead to $$1 = 0$$?

Update:

• You put all this effort, but not into searching... shame. – Asaf Karagila Oct 22 '18 at 7:00
• @AsafKaragila Well I did do some searching. And if it was an exact duplicate I am sure it would have been marked so. Again, I can get animated and even dizzy working on this! Thanks for the extra links! – CopyPasteIt Oct 22 '18 at 8:34
• Within the tag axiom-of-choice searching for "automorphisms real" gives a quick answer. This is why so much effort has been put into the tag system. – Asaf Karagila Oct 22 '18 at 9:15
• FWIW, I did some searching after first seeing this question, also with no clear result. I think it is rather uncharitable of @AsafKaragila to consistently assume that everybody who knows less than him is "too lazy to search". – Henning Makholm Oct 22 '18 at 11:49

It is indeed consistent, and in fact is a consequence of the extremely powerful axiom of determinacy.

Specifically, AD implies that every homomorphism from $$(\mathbb{R},+)$$ to itself is continuous, and in particular of the form $$a\mapsto ar$$ for some $$r\in\mathbb{R}$$. See here for some discussion of how nasty any other endomorphism would have to be; AD rules out such sets (e.g. implies that every set of reals is measurable).

Of course, as Asaf observes below, AD is truly massive overkill (like, nuking a mosquito); I'm mentioning it because AD is a natural alternative to AC which you may independently want to know about.

Now AD isn't actually cheap: the theory ZF+AD proves the consistency of ZF, that is, the axiom determinacy is of high consistency strength. We can prove the consistency of ZF+GR relative to ZF alone; however, this is a bit more technical.

• Pretty cool to see a reference link to Math.StackExchange in the wikipedia en.wikipedia.org/wiki/Cauchy%27s_functional_equation article. (the web - not your grandfather's research tool). – CopyPasteIt Oct 22 '18 at 2:04
• I'm confused as to the role of AD here. You just need "Every set of reals has the Baire property". Heck, even "Every set is measurable" is enough, and that has far less in terms of consistency. Also, how do you prove the consistency of ZF+AD without using forcing and large cardinals, and how is it less technical? Yes, the proof of BP is hard and technical, but the proof of AD is also very hard and technical. Oh, and neither can be proved consistent with just forcing. – Asaf Karagila Oct 22 '18 at 7:02
• @AsafKaragila See how great it is I asked my question! Look,, this is esoteric stuff but even the specialists can bounce it around to see what falls out. If I did not ask my question you and Noah would have missed an opportunity for an exchange. – CopyPasteIt Oct 22 '18 at 8:51
• @CopyPasteIt: I don't think we would have missed that opportunity. I think we had it before, and it's not the first time we talk about this sort of stuff, and it won't be the last time either. – Asaf Karagila Oct 22 '18 at 9:11
• @CopyPasteIt: Yes, that would be a good start. – Asaf Karagila Oct 22 '18 at 9:16