Metric spaces where each point sees each distance exactly once

Question

Let $M$ be a metric space, and $S=\{d(x,y):x,y\in M\}$ be the set of all distances between points in $M$. Let's call $M$ a unique distance space if for all $x\in M$ and all $r\in S$, there exists a unique $y\in M$ with $d(x,y)=r$. As an example, the vertices of a generic box in $\mathbb{R}^n$ will be a unique distance space with $|S|=2^n$. On the other hand, $\mathbb{R}$ with the usual metric is not a unique distance space, because given $x\in \mathbb{R}$ and $r>0$, there are two points of distance $r$ from $x$.

Main Question: Does there exist a unique distance space with $S=\mathbb{R}_{\geq 0}$?

One attempt

Let $M$ be a vector space over $\mathbb{F}_2$ with basis $B$, and let $f:B\to\mathbb{R}_{>0}$. If for $x,y\in M$ we define $d(x,y)$ to be the sum of $f(b)$ over all $b\in B$ where the coefficient of $b$ in $x$ and $y$ are different, then $d$ is a metric on $M$. If we choose $f$ so that the values of $\sum_{b\in T}f(b)$ are distinct for all finite subsets $T\subseteq B$, then $M$ is a unique distance space.

Some examples of this construction:

• If $M$ is finite-dimensional, then $M$ can be isometrically embedded as a box in $\mathbb{R}^n$ under the $L^1$ ("Manhattan") metric.
• If $B=\{e_n:n\in\mathbb{Z}\}$ and $f(e_n):=2^{n}$, then we can map $M$ bijectively onto the non-negative dyadic rationals $\mathbb{Z}[\frac12]_{\geq 0}$ by $e_n\mapsto 2^n$, and $d$ will correspond to the bitwise XOR operator. The fact that $M$ is a unique distance space can be seen directly by noting that $r=x\text{ XOR }y$ if and only if $y=x\text{ XOR }r$.

In order to use this technique to solve the original problem, we would need the following question to have a positive answer:

Question 2: Does there exist a set $A\subseteq\mathbb{R}_{\geq 0}$ such that every positive real number can be expressed uniquely as a sum of finitely many distinct elements of $A$?

If such a set $A$ existed, we could use it as a basis for $M$ with $f(a)=a$, giving an affirmative answer to the Main Question. At first this seems like a straightforward transfinite induction argument (well-order $\mathbb{R}_{>0}$, then successively include in $A$ any values that haven't yet occurred as a finite sum), but the problem is that $x$ can't be included in $A$ if it equals a difference of two previously occurring finite sums. If this happens, we will need to include other elements in $A$ that sum to $x$ (while simultaneously worrying about the fact that each of those elements may not be allowed in $A$, etc). I don't see a straightforward way to handle this.

Other constructions?

We could take the metric space completion of the dyadic rational example above. This comes extremely close to working, except for the fact that binary expansions of real numbers are not unique. Specifically, if $x,r\in \mathbb{R}_{\geq 0}-\mathbb{Z}[\frac12]$, then there is a unique point of distance $r$ from $x$ (and likewise if either $x$ or $r$ is zero). But if $x$ and $r$ are positive real numbers and at least one of them is in $\mathbb{Z}[\frac12]$, then there are two points of distance $r$ from $x$.

Question 3: Are there any other general methods for constructing unique distance spaces besides the ones mentioned above?

While I would be happy to hear any thoughts on this more open-ended question, I won't accept such a response as an official answer unless significant time passes without any answers to the Main Question.

Even for finite metric spaces this appears to be a fairly subtle combinatorial problem. Any finite unique distance space $M$ with $|M|>1$ must have $|M|$ even (given a nonzero $r\in S$, the set of pairs $\{x,y\}$ with $d(x,y)=r$ partitions $M$), but can every even number occur as a cardinality of a unique distance space? The box construction shows that powers of $2$ are possible, and I worked out an example with $|M|=6$ by hand, but I don't know of any other ways to generate examples.

Answer 1

Theorem. The complex plane $\mathbb C$ contains a unique distance subspace $X$ such that $\{|x-y|:x,y\in X\}=\mathbb R_{\ge 0}$.

Proof. Let $\mathbb R_+:=\{r\in\mathbb R:r>0\}$. Using the Axiom of Choice, choose a well-order $\preceq$ on the set $\mathbb C\times\mathbb R_+$ such that for every pair $(z,r)\in\mathbb C\times \mathbb R_+$ the set ${\downarrow}(z,r):=\{(z',r')\in\mathbb C\times\mathbb R_+:(z',r')\preceq (z,r)\}$ has cardinality $|{\downarrow}(z,r)|<|\mathbb C\times \mathbb R_+|=\mathfrak c$. For a nonempty subset $A\subseteq \mathbb C\times\mathbb R_+$ by $\min_\preceq A$ we denote the smallest element of $A$ with respect to the well-order $\preceq$.

By transfinite induction of length $\mathfrak c$, we can construct transfinite sequences $(x_\alpha)_{\alpha\in\mathfrak c}$, $(y_\alpha)_{\alpha\in\mathfrak c}$, $(Z_\alpha)_{\alpha\in\mathfrak c}$ such that for every ordinal $\alpha\in\mathfrak c$, the following conditions are satisfied:

(i) $Z_\alpha=\{x_\beta,y_\beta:\beta<\alpha\}\subseteq\mathbb C$;

(ii) $(x_\alpha,|x_\alpha-y_\alpha|)=\min_{\preceq}((\mathbb C\times\mathbb R_+)\setminus (E_\alpha\cup F_\alpha\cup G_\alpha))$ where

$E_\alpha:=\{(x,|x-y|):x,y\in Z_\alpha\}$,

$F_\alpha:=\bigcup_{x,y\in Z_\alpha}\{(z,r)\in\mathbb C\times \mathbb R_+: x\ne y\;\wedge\;|z-x|=|z-y|\}$,

$G_\alpha:=\bigcup_{x,y\in Z_\alpha}\{(z,r):z\ne y\ne x\;\wedge\;|z-x|=|y-x|\}$;

(iii) for every distinct point $x,y\in Z_\alpha\cup\{x_\alpha\}$ with $x\ne x_\alpha$ we have $y_\alpha\notin \{z\in\mathbb C:|z-x_\alpha|=|z-x|\}\cup\{z\in\mathbb C:|z-x|=|y-x|\}\cup\{z\in\mathbb C:|z-x|=|z-y|\}$.

The condition (ii) uniquely determines the point $x_\alpha$ and the real number $r_\alpha=|x_\alpha-y_\alpha|$. Then we choose a point $y_\alpha$ on the circle $\{z\in\mathbb C:|z-x_\alpha|=r_\alpha\}$ so that the condition (iii) is satisfied. The existence of $y_\alpha$ follows from the fact that the intersection of two distinct circles contains at most two points and the intersection of a circle and a line contains at most two points.

After completing the inductive construction, we will obtain the desired unique distance subspace $X:=\{x_\alpha,y_\alpha:\alpha\in\mathfrak c\}$ of the complex plane such that $\{|x-y|:x,y\in X\}=\mathbb R_{\ge 0}$.

Let us explain the last equality. Given any $z\in X$ and $r\in\mathbb R_+$, we should find a point $y\in X$ such that $|z-y|=r$. To derive a contradiction, assume that $|z-y|\ne r$ for every point $y\in X$. Then $(z,r)\notin E_\alpha$ for all $\alpha\in\mathfrak c$.

Next, we show that $(z,r)\notin F_\alpha\cup G_\alpha$ for every ordinal $\alpha<\mathfrak c$. In the opposite case, we can find points $x,y\in X$ such that $z\ne x\ne y$ and $|z-x|=|z-y|$ or $|z-y|=|x-y|$. Then also $z\ne y$, so $x,y,z$ are pairwise distinct points in $X$. Choose ordinals $\alpha,\beta,\gamma<\mathfrak c$ such that $x\in\{x_\alpha,y_\alpha\}$, $y=\{x_\beta,y_\beta\}$, $z=\{x_\gamma,y_\gamma\}$.

Since the points $x,y,z$ are pairwise distinct, the ordinals $\alpha,\beta,\gamma$ cannot be equal. So, only 12 case are possible:

1. $\alpha=\beta<\gamma$. If $z=x_\gamma$, then the equalities $|z-x|=|z-y|$ or $|z-y|=|x-y|$ contradict the choice of $x_\gamma$ in the conditionn (ii). If $z=y_\gamma$, then $|z-x|=|z-y|$ or $|z-y|=|x-y|$ contradict the choice of $y_\gamma$ in the condition (iii).

2. $\gamma<\alpha=\beta$. In this case $\{x,y\}=\{x_\alpha,y_\alpha\}$ and the equalities $|z-x|=|z-y|$ or $|z-y|=|x-y|$ contradict the choice of $y_\alpha$ in the condition (iii).

3. $\alpha=\gamma<\beta$. If $y=x_\beta$, then then the equalities $|z-x|=|z-y|$ or $|z-y|=|x-y|$ contradict the choice of $x_\beta$ in the conditionn (ii). If $y=y_\beta$, then $|z-x|=|z-y|$ or $|z-y|=|x-y|$ contradict the choice of $y_\beta$ in the condition (iii).

4. $\beta<\alpha=\gamma$. In this case $\{x,z\}=\{x_\gamma,y_\gamma\}$ and the equalities $|z-x|=|z-y|$ or $|z-y|=|x-y|$ contradict the choice of $y_\gamma$ in the condition (iii).

5. $\beta=\gamma<\alpha$. If $x=x_\alpha$, then then the equalities $|z-x|=|z-y|$ or $|z-y|=|x-y|$ contradict the choice of $x_\alpha$ in the conditionn (ii). If $x=y_\alpha$, then $|z-x|=|z-y|$ or $|z-y|=|x-y|$ contradict the choice of $y_\alpha$ in the condition (iii).

6. $\alpha<\beta=\gamma$. This case is analogous to the case 2, 4.

7. $\alpha<\beta<\gamma$. This case is analogous to the cases 1, 3, 5.

8. $\alpha<\gamma<\beta$. This case is analogous to the cases 1, 3, 5, 7.

9. $\beta<\alpha<\gamma$. This case is analogous to the cases 1, 3, 5, 7,8

10. $\beta<\gamma<\alpha$. This case is analogous to the cases 1, 3, 5, 7, 8, 9.

11. $\gamma<\alpha<\beta$. This case is analogous to the cases 1, 3, 5, 7, 8, 9, 10.

12. $\gamma<\beta<\alpha$. This case is analogous to the cases 1, 3, 5, 7-11.

In all 12 cases, we derive contradictions, which show that $(x,r)\notin E_\alpha\cup F_\alpha\cup G_\alpha$ for all $\alpha\in\mathfrak c$. The condition (ii) ensures that $\{(x_\alpha,|x_\alpha-y_\alpha|):\alpha\in\mathfrak c\}\subseteq{\downarrow}(x,r)$ and hence the set $\{(x_\alpha,|x_\alpha-y_\alpha|):\alpha\in\mathfrak c\}$ has cardinality strictly less than the continuum. The Pigeonhole Pronciple ensures that $(x_\alpha,|x_\alpha-y_\alpha|)=(x_\beta,|x_\beta-y_\beta|)$ for some ordinals $\alpha<\beta$, which implies $(x_\beta,|x_\beta-y_\beta|)\in E_\beta$ and contradicts the inductive condition (ii). This contradiction completes the proof.

$\quad\square$

Answer 2

This is by no means close to an answer, but I find these structures which you have introduced very appealing, here is a way to take the product of two finite unique distance spaces:

Suppose $(M_1,d_1)$ and $(M_2,d_2)$ are finite unique distance spaces, and let $\delta>0$ be the smallest difference between distinct numbers in $Im(d_2)$, we can then scale $(M_1,d_1)$ appropriately and assume w.l.o.g. that $$Im(d_1)\subseteq [0,\delta) $$

We then define the metric

$$ d: (M_1\times M_2)^2 \to \mathbb{R} \\ ((x_1,x_2),(y_1,y_2)) \mapsto d_1(x_1,y_1) + d_2(x_2,y_2) $$

To prove that this is again a unique distance space is fairly straightforward, if we suppose that

$$ d((x_1,x_2),(s_1,s_2)) = d((x_1,x_2),(t_1,t_2)) $$

Then we get

$$ d_1(x_1,s_1) + d_2(x_2,s_2) \\ = d_1(x_1,t_1) + d_2(x_2,t_2) $$

Since $d_2(x_2,s_2)$ and $d_2(x_2,t_2)$ differ by more than $\delta$ if $s_2\neq t_2$, the only way the above equality can be true is if $s_2=t_2$. Thus

$$ d_1(x_1,s_1) = d_1(x_1,t_1)$$

And since $M_1$ is a unique distance space, this means $s_1 = t_1$, so their product is also a unique distance space.

This technique can probably extend at least partially to infinite metric spaces, but not in a very useful way, since one would need to know that $ M_1$ is bounded and that there is a minimum difference between distinct real numbers in $Im(M_2)$