5

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:

Added descriptive set theory tag after looking over links in Noah's answer.

CopyPasteIt
  • 11,870
  • You put all this effort, but not into searching... shame. – Asaf Karagila Oct 22 '18 at 07:00
  • 1
    https://math.stackexchange.com/questions/110125/is-there-a-non-trivial-example-of-a-mathbb-q-endomorphism-of-mathbb-r https://math.stackexchange.com/questions/115486/what-is-operatornameaut-mathbbr https://math.stackexchange.com/questions/302257/is-ex-the-only-isomorphism-between-the-groups-mathbbr-and-mathbb https://math.stackexchange.com/questions/166176/a-question-concerning-on-the-axiom-of-choice-and-cauchy-functional-equation https://math.stackexchange.com/questions/1591691/is-there-a-nice-discontinuous-bijective-homomorphism-f-mathbbr-to – Asaf Karagila Oct 22 '18 at 07:01
  • @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 08:34
  • @AsafKaragila Also, I get a question in my head and pursue it without always searching first. I like exercising my neurons oblivious of established theory (some of which, of course, I can't master). – CopyPasteIt Oct 22 '18 at 08:36
  • (1) Don't trust on the community to "always close duplicates", even I get sometimes lazy and don't do it. (2) Part of asking the question should be searching for it, that is part of the research you're expected to make. Just saying "Oh, I'm lazy, and I will probably just go ahead and ask" is borderline insulting to the people who wrote all those other answers that answer your question. And if you want to actually exercise your neurons, finding an almost-duplicate and thinking about how that actually answered your question is a far superior exercise. – Asaf Karagila Oct 22 '18 at 08:39
  • @AsafKaragila So what keyword strategy am I missing? I arrived at the exact question I asked following my path. What words are you assuming should pop right in my head. – CopyPasteIt Oct 22 '18 at 08:43
  • 1
    Within the tag [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 09:15
  • @AsafKaragila thank you sir. I never searched through a tag before. learn something new everyday! – CopyPasteIt Oct 22 '18 at 09:17
  • 1
    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". – hmakholm left over Monica Oct 22 '18 at 11:49
  • @Henning: Perhaps you're right, and I am too quick to anger. But the OP did in fact prove my point, claiming that they haven't spent any time searching for the answer on their own. – Asaf Karagila Oct 22 '18 at 11:58
  • @AsafKaragila I guess when you are angry you don't carefully read comments - see 2nd comment here. going offline from this --- good day! – CopyPasteIt Oct 22 '18 at 12:02
  • CopyPasteIt, perhaps. But that makes https://math.stackexchange.com/questions/2965382/automorphisms-on-mathbb-r-and-the-axiom-of-choice?noredirect=1#comment6122445_2965418 all the more confusing. – Asaf Karagila Oct 22 '18 at 12:03

1 Answers1

5

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.

Noah Schweber
  • 260,658
  • Pretty cool to see a reference link to Math.StackExchange in the wikipedia https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation article. (the web - not your grandfather's research tool). – CopyPasteIt Oct 22 '18 at 02:04
  • 1
    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 07:02
  • @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 08:51
  • 1
    @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 09:11
  • @AsafKaragila But reading background material, comments, etc, can be instructive to neophytes. But enuf - I will google and search this site for at least 5 minutes before asking a question. Does that sound fair? (I do this part time as a hobby). – CopyPasteIt Oct 22 '18 at 09:16
  • 1
    @CopyPasteIt: Yes, that would be a good start. – Asaf Karagila Oct 22 '18 at 09:16
  • @AsafKaragila You're absolutely right, of course. I mentioned AD just because it's natural on its own, and knowing about it might preemptively answer other similar questions by the OP. – Noah Schweber Oct 22 '18 at 11:45
  • Noah, just as a point to be made, AD is not like a nuke for a mosquito, it's like destroying the planet to cure one case of the sniffles. :) – Asaf Karagila Oct 22 '18 at 16:18
  • Hey, little Timmy's not sniffling anymore. :) – Noah Schweber Oct 22 '18 at 16:27