# $A,B$ such that $A\cap B=\emptyset$ and $A\cup B=\mathbb{R}$ and $B=\{x+y : x,y\in A\}$?

If set $$A,B$$ satisfy $$A\cap B=\emptyset,A\cup B=I$$, and $$B=\{x+y : x,y\in A\}$$, can $$I$$ be real number set $$\mathbb{R}$$?

I think the answer is yes, but I can't construct it. If $$A$$ is odd number set, $$B$$ is even number set, then $$I$$ is integer set $$\mathbb{Z}$$. But how can the set of irrational numbers and the set of other rational numbers be put into the set $$A$$? I don't understand.

• $A, B \subseteq \mathbb R$ I imagine? May 23 at 7:04
• @mathcounterexamples.net Yes May 23 at 7:04
• I do not see a connection to topology. May 23 at 8:56
• I'm not certain this is possible for the rational numbers, but can't see a proof either way. If it is possible for the rationals, proving it for the reals will probably require the Axiom of Choice. May 24 at 8:22
• I don't think we can construct the real number, because B is forming by the help of A. So it's not possible to write a irrational number in the sum of two rational number May 24 at 15:06