A property for finite subsets of real numbers Let $B$ be a finite subset of real numbers. Then, is it true
that there exists $A\subseteq \mathbb{R}$ such that

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

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

Not that if $B$ is a singleton, then it is true by choosing $A=\mathbb{R}$,
and if $|B|&gt;1$ then $A-A$ can not be a subgroup (if $A-A$ is a subgroup, then $|\mathbb{R}:A-A|=|B|\geq 2$ that is a contradiction). Here, $B-B=\{ b-\beta:b,\beta\in B\}$, $A+B=\{a+b:a\in A,b\in B\}$.

Also see Two uncountable subsets of real numbers without any interval and two relations .
Take $A=\{0\}$ regardless of $B$, if $B$ is non-empty. $0 \in B-B$ always and $A-A=\{0\}$ trivially, so $(A-A) \cap (B-B) = \{0\}$.

If $A$ is finite and $B$ too, $A-A$ is finite and so is $(A-A) + B$ and so never equals $\Bbb R$.

$\Bbb R$ has no finite subgroups. I see no point in this question..
Take $A=\{0\}$ regardless of $B$, if $B$ is non-empty. $0 \in B-B$ always and $A-A=\{0\}$ trivially, so $(A-A) \cap (B-B) = \{0\}$.

If $A$ is finite and $B$ too, $A-A$ is finite and so is $(A-A) + B$ and so never equals $\Bbb R$.

$\Bbb R$ has no finite subgroups. I see no point in this question..