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.

  • $\begingroup$ $A, B \subseteq \mathbb R$ I imagine? $\endgroup$ May 23 at 7:04
  • $\begingroup$ @mathcounterexamples.net Yes $\endgroup$
    – Eufisky
    May 23 at 7:04
  • 1
    $\begingroup$ I do not see a connection to topology. $\endgroup$
    – Paul Frost
    May 23 at 8:56
  • $\begingroup$ 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. $\endgroup$
    – aschepler
    May 24 at 8:22
  • $\begingroup$ 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 $\endgroup$
    – Epsilon101
    May 24 at 15:06


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