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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 $y$?
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No, and in fact, most of your conditions are not needed.

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.

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  • $\begingroup$ sorry, I forgot to mention that both $S_1$ and $S_2$ are finite sets (I modified the question) $\endgroup$ – Victor Jun 29 '16 at 6:33
  • $\begingroup$ I think you should remove the fa.functional-analysis tag, and rather add co.combinatorics. Probably there is still some simple counterexample, I would start by reading about unit disk graphs. $\endgroup$ – domotorp Jun 29 '16 at 13:54
  • $\begingroup$ In fact, this question came from a conjecture that the class of co-biparite unit disk graphs is 'self-complementary', which was posed in this paper arxiv.org/abs/1602.08148 . Formally, the conjecture says that if one has a unit disk graph with a fixed partition of its vertices into two cliques, then by complementing the edges between the cliques we get a unit disk graph again. $\endgroup$ – Victor Jun 29 '16 at 15:09
  • $\begingroup$ The conjecture was proved for a special case (when the edges between the cliques form a C4-free bipartite graph), and we tend to believe that it is true in general. I just thought that ideas from other areas of mathematics may be helpful here. $\endgroup$ – Victor Jun 29 '16 at 15:09
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    $\begingroup$ Yeah, I was thinking about the same conjecture (and I thought that it must fail), in fact, this conjecture seems more interesting than the original question. I think in general it's a good idea to give as much motivation as possible when you pose a question on MO, then people might get more interested - like I did just now! $\endgroup$ – domotorp Jun 29 '16 at 16:30

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