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Are there two uncountable subsets $A, B$ of real numbers such that:

(1) $(A-A)\cap (B-B)=\{ 0\}$,

(2) $(A-A)+B=\mathbb{R}$ or $(B-B)+A=\mathbb{R}$ ?

We know that if one of them contains an interval, then the condition (1) is impossible, since every uncountable subset has an accumulation point. Also, $B=\mathbb{Z}$ and $A=(0,1)$ satisfy (1) and (2) but the condition (1) does not hold and $\mathbb{Z}$ is countable (see Subsets of real numbers satisfying the two conditions). Note that $B-B=\{ b-\beta:b,\beta\in B\}$, $A+B=\{a+b:a\in A,b\in B\}$.

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Sure. For instance, pick a basis $S$ for $\mathbb{R}$ as a vector space over $\mathbb{Q}$ and split it as the union of two disjoint uncountable sets $T$ and $U$, and let $A$ and $B$ be the $\mathbb{Q}$-spans of $T$ and $U$, respectively. Then $A-A=A$ and $B-B=B$ so your two conditions amount to the fact that $\mathbb{R}$ is the direct sum of $A$ and $B$.

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  • $\begingroup$ Thanks for your answer. Do you have any example which both $A$ and $B$ are not subgroups? $\endgroup$
    – M.H.Hooshmand
    Commented Dec 14, 2019 at 7:04
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    $\begingroup$ Replace $A$ with $A\setminus\{0\}$, say. $\endgroup$
    – Eric Wofsey
    Commented Dec 14, 2019 at 13:23

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