Suppose $A \subseteq \mathbb R$ is countable. Show $\exists x\in\Bbb R$ s. t.$A \cap (x+A) =\emptyset$. Suppose $A \subseteq \mathbb R$ is countable. Show $\exists x\in\Bbb R$ s. t. $A \cap (x+A) =\emptyset$. (Here, $x + A = \{ x + a : a ∈ A \}$.
I'm unsure how to proceed.    
I thought about taking the smallest distance between the countable numbers of the set etc but I don't think we can take smallest distance of an infinite set even if it's only countable ie say set were rationals.  I've tried by contrapositive, ie suppose there is no such x but that got me nowhere. 
Any hints?
 A: Consider the set $B$ of differences formed by $A$. Note that $B$ is the image of the countable set $A \times A$ under the difference operation, which makes $B$ countable.
Further, note that if $x \in A \cap (A + r)$, then $x = a_1$ and $x = a_2 + r$ for some $a_1, a_2 \in A$, making $r = a_1 - a_2 \in B$. Since $B$ is countable, there must be uncountably many $r \in \Bbb{R} \setminus B$ for which there is no $x \in A \cap (A + r)$, i.e. $A \cap (A + r) = \emptyset$.
A: HINT: Suppose that $A\cap(x+A)\ne\varnothing$; then there is some $a_0\in A\cap(x+A)$. Since $a_0\in x+A$, $a_0-x\in A$; let $a_1=a_0-x$. We now have two elements of $A$, $a_0$ and $a_1$, such that $a_0-a_1=x$. Conversely, if there are $a_0,a_1\in A$ such that $a_0-a_1=x$, then $x_0\in A\cap(x+A)$. Now, how big is the set $\{a_0-a_1:a_0,a_1\in A\}$ compared with $\Bbb R$?
A: One can show something stronger: let $I$ be uncountable set, then $(f_λ)_{λ∈I}$ a sequence of $I$-many injective functions from uncountable set $X$ to itself such that $f_a(x)=f_b(x)⇒a=b$, then for every countable set $A$, there exists some $i∈I$ such $A\cap f_i[A]=\emptyset$.
Let $I=X=\Bbb R$ and $f_a(x)=a+x$ to get your question.

To prove that let $x∈X$, then if $x\in f_a[A],f_b[A]$ for $a\ne b$, then $f_a^{-1}(x)≠f_b^{-1}(x)$, because $A$ is countable, the set $I_x=\{λ∈I\mid x\in f_λ[A]\}$ is also at most countable, therefore $J=\bigcup_{x∈A}I_x$ is countable union of countable sets, hence countable, so we can take $\kappa\in I\setminus J$, if $A∩f_\kappa[A]≠\emptyset$ there must be $y∈A$ such that $y∈f_\kappa[A]$, so $\kappa\in J$, contradiction.
