# Any countable $A \subseteq \mathbb{R}$ satisfies $(x+A) \cap A = \emptyset$ for some $x$

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.

• Proof by condradiction if for every $x$ there exist $b, a \in A$ so that $x + b = a$ then $\mathbb R \subset B = \{a \pm b|a,b \in A\}$. Why is that impossible? – fleablood Mar 16 '16 at 22:22

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.
• The same argument shows that there is such an $x$ as long as $A$ has size strictly smaller than $\mathbb R$. (Admittedly, this may be just the same as countable, depending on whether CH holds.) – Andrés E. Caicedo Oct 23 '12 at 22:22
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.