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Let us assume $A,B\subseteq \Bbb R^+ $ are disjoint sets and $A\cup B=\Bbb R^+ $. Furthermore, $ \forall_{x,y\in A} \ x+y \in B \ \text{and} \ \ \forall_ {x,y \in B} \ x+y \in A.$

Is it possible to provide an example for such $A$ and $B$?

If not, I would like to know why, and also prove the existence of such $A$, $B$ using Zorn's lemma. However, I couldn't figure out how to do it.

Please do not provide full proof as I am only looking for hints.

Asaf Karagila
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Akira
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  • Really, it's the only obvious thing that you can do with Zorn's lemma. – Asaf Karagila Apr 02 '17 at 20:00
  • I tried many sets as the partial order set of the lemma but none led me to the solution so far. Any hints would be appreciated.

    For some reason I missed the other existing question similar to mine but I would still like to know why we can't provide an example for such A and B.

     – Akira Apr 02 '17 at 20:08
  • Would an answer like "because there are models where such partitions are not definable without parameters, and there are models without choice where such partitions do not exist" satisfy you? Or would you prefer something along the lines of "such sets cannot be measurable/have the Baire property/etc. and we know that it is consistent without choice that every set of reals is measurable/has the Baire property/etc."? I don't mind digging into that, but what type of answer would satisfy you? – Asaf Karagila Apr 02 '17 at 20:17
  • The first one. How can this explanation be applied in this case? – Akira Apr 02 '17 at 20:24
  • Since there are models without such partitions, we cannot prove their existence without appealing to the axiom of choice. Since the axiom of choice does not give us definitions for sets, just means to prove their existence, it means that we cannot specify such sets. – Asaf Karagila Apr 02 '17 at 20:27
  • What do you mean by saying 'such partitions?'. There is a partition between A and B but we cannot provide it, am I right? I presume my question is 'why can't we?'. – Akira Apr 02 '17 at 20:46
  • If by "provide" you mean write down an explicit formula, then the answer is in my previous comment. If you mean something else, then you should perhaps clarify your use of language. – Asaf Karagila Apr 02 '17 at 20:47
  • We cannot write down an explicit formula, that I understand, and I also understand how it leads us to not being able to specify such sets. I do not understand however, how do we know if our model has or doesn't have partition we could provide an explicit formula for. If we are entering some kind of circular argument, I beg your pardon, but there is still something I believe I am missing here. – Akira Apr 02 '17 at 20:55
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    We can prove that such partitions will imply the existence of something that does not exist in our model. For example, a non-measurable set, or a set without the Baire property, or a discontinuous homomorphism from $\Bbb R$ to itself. – Asaf Karagila Apr 03 '17 at 03:58

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