# Set of points with unique distances

Is there a set of points in $\mathbb{R}^n$ such that every positive distance is realized by exactly one pair of points in the set?

I can see that if it exists, the set must be uncountable and unbounded, and cannot contain any line segment.

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 Wrote a number of comments, but realized that I did not understand the question. Sorry about that. – JBeardz Dec 1 '11 at 23:43

Say that a subset $X$ of $\mathbb{R}$ is metrically rigid if each positive distance is realized by at most one pair of points in $X$.

Let $\{d_\xi:\xi<2^\omega\}$ be an enumeration of the positive reals. Let $X_0=\varnothing$. Suppose that $\eta<2^\omega$ and for each $\xi<\eta$ we have a set $X_\xi\subseteq\mathbb{R}$ such that for each $\xi<\eta$ the following hold:

$\qquad(1)_\xi: X_\zeta\subseteq X_\xi$ whenever $\zeta\le\xi$;
$\qquad(2)_\xi: |X_\xi|\le\max \{|\xi|,\omega \}$;
$\qquad(3)_\xi: X_\xi$ is metrically rigid; and
$\qquad(4)_\xi: d_\xi$ is realized by some pair of points of $X_\xi$.

Let $Y_\eta=\bigcup\limits_{\xi<\eta}X_\xi$; it’s easy to see that $Y_\eta$ is metrically rigid and that $|Y_\eta|\le\max\{|\eta|,\omega\}$.

If $d_\eta$ is realized by some pair of points of $Y_\eta$, let $X_\eta=Y_\eta$, and observe that $(1)_\eta-(4)_\eta$ hold.

Begin Edit:

Otherwise, let

\begin{align*}S_\eta&=\{x\in\mathbb{R}:\exists \xi<\eta\;\exists\zeta\le\eta\;\big(|x-x_\xi|=d_\zeta\big)\}\;,\\ M_\eta&=\left\{\frac12(x_\xi+x_\zeta):\xi,\zeta<\eta\right\},\\ C_\eta&=\left\{x\in\mathbb{R}:\exists y\in M_\eta\left(|x-y|=\frac{d_\eta}2\right)\right\},\text{ and}\\ Z_\eta&=S_\eta\cup M_\eta\cup C_\eta\;. \end{align*}

$|Z_\eta|\le\max\{2|\eta|^2,\omega\}<2^\omega$, so we can choose $x_\eta,y_\eta\in\mathbb{R}\setminus Z_\eta$ so that $|x_\eta-y_\eta|=d_\eta$. Set $X_\eta=Y_\eta\cup\{x_\eta,y_\eta\}$, and again observe that $(1)_\eta-(4)_\eta$ hold.

Only the metric rigidity of $Y_\eta$ might be in doubt. Clearly any violation must involve at least one of the two new points and a distance not in $\{d_\xi:\xi<\eta\}$. This could happen in one of two ways. First, $x_\eta$ or $y_\eta$ could be the midpoint of two points of $Y_\eta$, but this is excluded by the inclusion of $M_\eta$ in $Z_\eta$. Secondly, there might be $x,y\in Y_\eta$ such that $|x-x_\eta|=|y-y_\eta|$. But in that case the pairs $\{x,y\}$ and $\{x_\eta,y_\eta\}$ have the same midpoint, so this possibility is excluded by the inclusion of $C_\eta$ in $Z_\eta$.

My thanks to Arthur Fischer for catching an oversight in the original version of this section.

End Edit.

Thus, the construction goes through to $2^\omega$.

Finally, let $X=\bigcup\limits_{\xi<2^\omega}X_\xi$; then $X$ is a metrically rigid subset of $\mathbb{R}$ (and by abuse of terminology of $\mathbb{R}^n$ for any $n$) that realized every positive distance.

Added: Some may find it interesting that the continuum hypothesis is equivalent to the statement that $\mathbb{R}$ is the union of countably many metrically rigid sets; see Corollary 7 of this old paper. ($\mathbb{R}$ is not the union of finitely many such sets.)

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 There appears to be a small gap in your justification. When moving from $Y_\eta$ to $X_\eta$ (in the case where points must be added) you must also ensure that (1) neither $x_\eta$ nor $y_\eta$ lie mid-way between two (distinct) points of $Y_\eta$, and (2) $|x-x_\eta| \neq |y-y_\eta|$ for all $x,y \in Y_\eta$. (1) can be taken care of by a simple modification of the set $Z_\eta$, while (2) seems to require looking at the set of all pairs from $\mathbb{R} \setminus Z_\eta$ which are $d_\eta$ apart. The induction does seems to go through, however. – Arthur Fischer♦ Dec 2 '11 at 13:33 @Arthur: You’re absolutely right; I was trying to simplify the argument and overdid it. It should be okay now, but I’d appreciate it if you cast an independent eye over it. – Brian M. Scott Dec 2 '11 at 17:39 Looks perfect to me! – Arthur Fischer♦ Dec 2 '11 at 19:25 @Arthur: Thanks! – Brian M. Scott Dec 2 '11 at 19:36