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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.

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  • $\begingroup$ $A, B \subseteq \mathbb R$ I imagine? $\endgroup$
    – mathcounterexamples.net
    Commented May 23, 2022 at 7:04
  • $\begingroup$ @mathcounterexamples.net Yes $\endgroup$
    – Eufisky
    Commented May 23, 2022 at 7:04
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    $\begingroup$ I do not see a connection to topology. $\endgroup$
    – Paul Frost
    Commented May 23, 2022 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
    Commented May 24, 2022 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 $\endgroup$
    – Alexander
    Commented May 24, 2022 at 15:06

1 Answer 1

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$A$ can be $ \ldots \cup [-2,-1) \cup [1,2) \cup [4,5) \cup \ldots$ and
$B$ will then be $ \ldots \cup [-1,1) \cup [2,4) \cup [5,7) \cup \ldots$

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