TITLE: Any countable $A \subseteq \mathbb{R}$ satisfies $(x+A) \cap A = \emptyset$ for some $x$
QUESTION [7 upvotes]: I have to prove that 

if $A \subseteq \mathbb{R}$ is countable, then 
  $\exists x \in \mathbb{R}\, (x+A) \cap A = \emptyset, $

where $x+A$ denotes the set $\{x + a \mid a \in A\}$.
I can see why this is true for some specific subsets (like the set of rationals or the set of algebraic numbers), but the general approach eludes me. Any hints would be appreciated.

REPLY [11 votes]: Note that $x+A$ and $A$ meet if and only if there are $y,z\in A$ with $x+y=z$, so $x=z-y$. This means that $(x+A)\cap A\ne\emptyset$ iff $x\in A-A$, which is a countable set, as it is the image of $A\times A$ under the map $(y,z)\mapsto y-z$. 
So, since $\mathbb R$ is uncountable, we can find values of $x$ not in $A-A$, and any such $x$ works.

REPLY [2 votes]: Suppose the conclusion does not hold: i.e. for every $x$ in $\mathbb{R}$, there are elements $a_x$ and $b_x$ of $A$ such that $x+a_x=b_x$. This defines an injection from $\mathbb{R}$ to $A^2$: just map $x$ to $(a_x,b_x)$. Thus $|\mathbb{R}|\le |A^2|$, so $|A|\ge |\mathbb{R}|$, and in particular $A$ is uncountable.