# Transformation inverting distances between two sets of diameter 1

Let $S_1, S_2 \subseteq \mathbb{R}^2$ be two finite disjoint sets of points in the plane with $\texttt{diam}(S_1) \leq 1$ and $\texttt{diam}(S_2) \leq 1$.

Does there always exist a transformation $f: S_1 \cup S_2 \rightarrow \mathbb{R}^2$ such that:

1. $\texttt{diam}(f(S_1)) \leq 1$, $\texttt{diam}(f(S_2)) \leq 1$; and
2. for any pair of points $s_1 \in S_1$ and $s_2 \in S_2$ we have $d(s_1,s_2) \leq 1 \iff d(f(s_1), f(s_2)) > 1$, where $d(x,y)$ is the Eucledean distance between $x$ and $y$?

Take a unit square, and let $S_1$ be one of its sides, and $S_2$ another side parallel to it. This example has the property that for every $s_1\in S_1$ there is exactly one $s_2\in S_2$ for which $d(s_1,s_2)\le 1$. Consider the uncountably many open sets defined as $H_{s_1}=\{x\mid d(f(s_1),x)>1\}$ for $s_1\in S_1$. Each of these must contain exactly one point of $S_2$. But using that the plane is a Lindelöf space, already countably many of the $H_{s_1}$ must cover the whole plane. This contradicts that $S_2$ has uncountably many points.
• sorry, I forgot to mention that both $S_1$ and $S_2$ are finite sets (I modified the question) – Victor Jun 29 '16 at 6:33