# Proof that$(a+A)\cap A=\varnothing$

I finished my test and there is a question I completely failed but that my teacher did not go over, so I was hoping someone could post a correction of it, so that I understand what I was supposed to do for next time.

Suppose that A is a countable set of real numbers. Show that there exists a real number a such that $(a+A)\cap A=\varnothing$. (Note: By definition, $a+A= \{a+r\mid r\in A\}$.)

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Let $B = \{ x \mid (x + A)\cap A \neq \varnothing \} \subseteq \{ x-y \mid x,y \in A \} = C$. Of course $C$ is countable, hence is $B$, but $\mathbb{R}$ in uncountable and therefore $\mathbb{R} \backslash B$ is not empty.
Let $a$ be such that $(a+A) \cap A$ is nonempty, i.e. there are $b,c \in A$ such that $a + b = c$.
Then $a = b-c$ lies in the set $\{b - c : b,c \in A\}$.
Since this set is countable with $A$ being countable, it is not all of $\mathbb{R}$.