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?
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?
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 $(\ast)$.
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.