Prove that if $A$ is a countable set, and is a subset of $\Bbb R$, then there exists a real number $x$ such that $A$ intersect $A + x$ is disjoint. I am struggling with this question, and any help would be appreciated. I hope I'm clear with the terminology though. 
 A: Suppose not. Consider $x \in [0,1]$. If $A + x$ intersects $A$, then $x = a - b$ where $a,b \in A$. Thus, since $[0,1]$ is uncountable, the set $X = \{a-b \, | \, a,b \in A\}$ is uncountable. However $A \times A$ is countable and you can define a surjection $f:A \times A \to X$ onto $X$ by $f(a,b) = a - b$. Thus by theorem, the cardinality of $X$ is no more than the cardinality of $A \times A$, which is countable, contradiction.
A: Here's one way you could do it. Let $A = \{a_1,a_2,a_3,\ldots\}$ and for a fixed integer $k\in \Bbb{N}$ define $C_k = \{\pm|a_k-a_j|:j > k,\space j \in \Bbb{N}\}$. $C_k$ is the set of distances between $a_k$ and every other element in $A$ with an index greater than $k$. Hence, $$\bigcup_{k=1}^\infty C_k$$ is the set of distances between any two elements of $A$. Each $C_k$ is countable, and a union of countably many countable sets is countable, so $\bigcup_{k=1}^\infty C_k$ is a countable set. As such, this means $\Bbb{R} \setminus \bigcup_{k=1}^\infty C_k$ is nonempty. Now what if we choose an $x$ in $\Bbb{R} \setminus \bigcup_{k=1}^\infty C_k$?
A: For every element in $A$ define $X_a=\{y\in R/ a\in (A+y)\}$ it's clear that $X_a$ is countable for all elements $a\in A$ hence:
$$\bigcup_{a\in A}X_a $$
is countable, and  $\mathbb{R}$ is not countable there exists $x\in \mathbb{R}\backslash\bigcup_{a\in A}X_a$ which is equivalent to $A\cap (A+x)=\emptyset$
