Special subset of an Euclidean space This is a nice problem that I found it somewhere and thought to share it with everyone!
Does there exist a subset $S \subset \mathbb{R}^n$ s.t. for every non-zero $t \in \mathbb{R}^n\;, \; S \cap (S+t)$ has precisely one element?
 A: Edit: The original version of this assumed the Continuum Hypothesis. I have no idea why I thought that it was necessary: the same argument works perfectly well without it. I’ve made the necessary minor adjustments below.
Such a set can be constructed (if we assume the axiom of choice, but I take that for granted). Note first that the condition on $S$ is equivalent to the property $(\ast)$: for each non-zero $t \in \mathbb{R}^n$ there is a unique pair of points $x_t,y_t \in S$ such that $y_t = x_t + t$: $y_t$ is the unique point common to $S$ and $S+t$. I’ll construct $S$ to have this equivalent property.
Let $P$ be a subset of $\mathbb{R}^n \setminus \{0\}$ containing exactly one member of each pair $\{t,-t\}$ with $t \ne 0$. For $A \subseteq \mathbb{R}^n$ let $D(A) = P \cap (A-A)$; the $\pm t$ with $t \in D(A)$ are the translations already realized between points of $A$. Let $A^* = A \cup (A+D(A)) \cup (A-D(A))$; $A^*$ is $A$ together with the set of points in $\mathbb{R}^n$ that can reached from $A$ by translations already realized in $A$. Note that $A^*$ is countable whenever $A$ is. (In fact $A \subseteq (A+D(A)) \cup (A-D(A))$ if $A$ has at least two points.)
We can enumerate $P$ as $\{t_\xi:\xi < 2^\omega\}$. Suppose that $\eta < 2^\omega$, and for each $\xi < \eta$ we’ve constructed a set $S_\xi \subseteq \mathbb{R}^n$ such that:
$\qquad(a)_\xi$ $\vert S_\xi \vert < 2^\omega$;
$\qquad(b)_\xi$ $S_\xi \subseteq S_\zeta$ whenever $\xi < \zeta < \eta$; and
$\qquad(c)_\xi$ $t_\xi \in D(S_\xi)$.
Let $T_\eta = \bigcup\limits_{\xi<\eta} S_\xi$. If $t_\eta \in D(T_\eta)$, let $S_\eta = T_\eta$. Otherwise, $\vert T_\eta \vert < 2^\omega$, so $\vert T_\eta^* \vert < 2^\omega$, and we can choose $s_\eta,s_\eta' \in \mathbb{R}^n \setminus T_\eta^*$ such that $s_\eta' = s_\eta + t_\eta$ and $\frac12(s_\eta+s_\eta') \notin \{\frac12(x+y):x,y \in T_\eta\}$. Let $S_\eta = T_\eta \cup \{s_\eta,s_\eta'\}$; clearly $(a)_\eta - (c)_\eta$ are satisfied, and the construction goes through to $2^\omega$.
Now let $S = \bigcup\limits_{\xi< 2^\omega} S_\xi$; clearly $D(S) = P$. Suppose that for some $\eta < 2^\omega$ there are distinct pairs $x,x'$ and $y,y'$ in $S$ such that $x'=x+t_\eta$ and $y'=y+t_\eta$. No pair is added at stage $\eta$ if $t_\eta$ is already realized in $T_\eta$, and every point added after stage $\eta$ avoids $S_\eta \pm t_\eta$, so $x,x',y,y' \in T_\eta$. Let $\xi < \eta$ be minimal such that $x,x',y,y' \in S_\xi$. Then at least one of $x,x',y,y'$ must belong to $\{s_\xi,s_\xi'\}$, and since $\{s_\xi,s_\xi'\} \cap D(T_\xi) = \varnothing$, $\{s_\xi,s_\xi'\}$ must contain one point from each pair $\{x,x'\}$ and $\{y,y'\}$.
If $s_\xi = x$ and $s_\xi' = y$, then $y'-x' = (y+t_\eta)-(x+t_\eta) = y-x = t_\xi$. But in this case $x',y' \in T_\xi$, so $t_\xi \in D(T_\xi)$, and nothing would have been added at stage $\xi$. The case $s_\xi = x', s_\xi' = y'$ is obviously similar.
If $s_\xi = x$ and $s_\xi' = y'$, then $\frac12(s_\xi+s_\xi') = \frac12(x+y') = \frac12((x'-t_\eta) + (y + t_\eta)) = \frac12(x'+y)$; but in this case $x',y \in T_\xi$, so $s_\xi$ and $x_\xi'$ were chosen in such a way that $\frac12(s_\xi+s_\xi') \ne \frac12(x'+y)$. The case $s_\xi = x', s_\xi' = y$ is obviously similar.
It follows that $S$ satisfies $(\ast)$.
A: If I write formally your problem, it states that for any $t\neq 0$ there exists a unique $s\in S$ such that $s+t\in S$. Let us denote such $s$ through $f(t)$, so $t\mapsto f(t)$ is uniquely defined for any $t\neq 0$. 
Let $t',t''\neq 0$ be such that $t'+t''\neq 0$. What do we have:
$$
f(t')+t'\in S
$$
$$
f(t'')+t''\in S
$$
so
$$
f(t')+f(t'')+(t'+t'')\in S
$$
hence 
$$
f(t')+f(t'') = f(t'+t'')
$$
for all $t',t''\neq 0,t'+t''\neq 0$. Now you can see that $f(\mathbb R^n\setminus \{0\})\subseteq S$ for some additive function $f:\mathbb R^n\setminus\{0\}\to\mathbb R^n$.
But now, let pick up $t^*:f(t^*)\neq 0$. Then $f(t)+f(t^*) = f(t+t^*)\in S$ as well as $f(t-t^*)+f(t^*) = f(t)\in S$ which violates the uniqueness.
