Is the following metric characterization of the real line true (and known)?



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> A nonempty complete metric space $(X,d)$ is isometric to the real line if and only if for every $c\in X$ and positive real number $r$ there exist two points $a,b\in X$ such that $d(a,b)=2r$ and $\{a,b\}=\{x\in X:d(x,c)=r\}$.
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**Added in Edit:** Without the completeness of $(X,d)$ this hypothetical characterization of the real line is not true: using the idea from the answer of @PietroMajer, by transfinite induction of length $\mathfrak c$, one can construct a dense $\mathbb Q$-linear subspace $L$ of the Euclidean plane $\mathbb R^2$ such that $|\{x\in L:\|x\|=r\}|=2$ for every positive real number $r$. More details can be found in [this preprint](https://arxiv.org/abs/2305.07354).



A metric characterization of the real line
Yes: in the new version of the question, with the word "complete" added, this is indeed a characterization of the real line.


In order to generate as much confusion as possible, but also for convenience, let's give the name "Banakh space" to any metric space satisfying the condition in your question with the word "complete'' deleted.


**Theorem:** Every complete Banakh space is isometric to $\mathbb R$ (with the usual metric).


This follows fairly easily from:


**Lemma:** Every Banakh space (whether complete or not) contains an isometric copy of a dense subset of $\mathbb R$.


To prove the theorem from the lemma, suppose $(X,d)$ is a complete Banakh space. Assuming the lemma, there is an isometric copy of some dense $Q \subseteq \mathbb R$ in $X$. Because $(X,d)$ is complete, the closure of this copy of $Q$ in $X$ is an isometric copy of the Cauchy completion of $Q$, which is $\mathbb R$. Let the map $r \mapsto \bar r$ be an isometric embedding of $\mathbb R$ in $X$. We are done if we can show that $X = \{ \bar r :\, r \in \mathbb R \}$. Aiming for a contradiction, suppose there is some $x \in X \setminus \{ \bar r :\, r \in \mathbb R \}$, and let $a = d(x,\bar 0)$. But then there are (at least) three points in $X$ at distance $a$ from $\bar 0$: $x$, $\bar a$, and $- \bar a$, contradicting that $X$ is a Banakh space.


Now to prove the lemma. We'll build up an embedding of a dense subset of $\mathbb R$ into $X$, one piece at a time. The first step is to find an isometric copy of $A = \{0\} \cup \{2z+1 :\, z \in \mathbb Z\}$ (the odd integers plus $0$) inside $X$.


To start, let $\bar 0$ be any point of $X$. There are exactly two points at distance $1$ from $\bar 0$: let's call these $\bar 1$ and $-\bar 1$ (it doesn't matter which point gets which label). By the Banakh property, $d(-\bar 1,\bar 1) = 2$.


Next, observe that there are exactly two points in $X$ with distance $2$ from $\bar 1$. We already have a name for one of these points, $-\bar 1$; let's call the other one $\bar 3$. By the Banakh property, $d(-\bar 1,\bar 3) = 4$. So we have:
$$d(-\bar 1,0) = d(\bar 0,\bar 1) = 1,\ d(-\bar 1,\bar 1) = d(\bar 1,\bar 3) = 2,\ d(-\bar 1,\bar 3) = 4.$$
To compute $d(\bar 0,\bar 3)$, we use the triangle inequality twice:
$$d(\bar 0,\bar 3) \leq d(\bar 0,\bar 1)+d(\bar 1,\bar 3) = 3,$$
$$d(\bar 0,\bar 3) \geq d(-\bar 1,\bar 3)-d(-\bar 1,\bar 0) = 3.$$
Hence $d(\bar 0,\bar 3) = 3$, and we have an isometric embedding of $\{-1,0,1,3\}$ into $X$.


For the next step, we can add in $-3$ the same way we did $3$. That is, observe that there are exactly two points in $X$ with distance $2$ from $-\bar 1$. One of these points is $\bar 1$, and we call the other one $-\bar 3$. By the Banakh property, $d(-\bar 3,\bar 1) = 4$. Using the triangle inequality as above, we can get $d(-\bar 3,\bar 0) = 3$. Lastly, because $d(-\bar 3,\bar 1) = 4 \neq 2 = d(\bar 3,\bar 1)$, we have $-\bar 3 \neq \bar 3$, but both of these are at distance three from $\bar 0$. Thus by the Banakh property, $d(-\bar 3,\bar 3) = 6$.
Thus we obtain an isometric embedding of $\{-3,-1,0,1,3\}$ into $X$.


Next we add in $5$ and $-5$ in a similar fashion. (I'll go through the details rather than just leaving it at "similar" though.) There are two points at distance $2$ from $\bar 3$. One of them is $\bar 1$; let's call the other $\bar 5$. The Banakh property gives $d(\bar 1,\bar 5) = 4$. But we also know already that $d(-\bar 3,\bar 1) = 4$, so another use of the Banakh property gives $d(-\bar 3,\bar 5) = 8$. Once we know both $d(-\bar 3,\bar 5)$ and $d(\bar 3,\bar 5)$ (i.e., the distance from $\bar 5$ to the least and greatest of our previously constructed points) the triangle inequality fills in all the other distances from previously constructed points to $\bar 5$. For example,
$$d(\bar 0,\bar 5) \leq d(\bar 0,\bar 3)+d(\bar 3,\bar 5) = 5,$$
$$d(\bar 0,\bar 5) \geq d(-\bar 3,\bar 5)-d(-\bar 3,\bar 0) = 5.$$
Similar computations give $d(-\bar 1,\bar 5) = 6$, and so we have an isometric embedding of $\{-3,-1,0,1,3,5\}$ into $X$.


Next add in $-5$ the same way we did $5$. That is, observe that there are exactly two points in $X$ with distance $2$ from $-\bar 3$. One of these points is $-\bar 1$, and we call the other one $-\bar 5$. The distances from $-\bar 5$ to $-\bar 1, \bar 0, \bar 1, \bar 3$ are computed just as they were for $\bar 5$. Then we observe that $-\bar 5 \neq \bar 5$ (because their distances to $\bar 1$ are different) but they are both distance $5$ from $\bar 0$, and this implies $d(-\bar 5,\bar 5) = 10$.
Thus we obtain an isometric embedding of $\{-5,-3,-1,0,1,3,5\}$ into $X$.


Continuing in this way, we can, two points at a time, build up an isometric embedding of $A$ into $X$, denoted by the map $z \mapsto \bar z$.


Once this is done . . . do it again! By the same exact method, we can find an isometric embedding of $\frac{1}{3} A$ (all odd integer multiples of $\frac{1}{3}$, plus $0$) in $X$, beginning with the same base point $\bar 0$ as before. Let's denote this new embedding by $z \mapsto \bar{\bar z}$ (so $\bar{\bar 0} = \bar 0$). But then notice that $A \subseteq \frac{1}{3}A$ and, by the Banakh property, for each $a \in A \setminus \{0\}$ there can only be two points of $X$ at distance $|a|$ from $\bar 0$. Thus $\{-\bar a,\bar a\} = \{-\bar{\bar a},\bar{\bar a}\}$. In this way we see that $\{ \bar z :\, z \in A \}$ is naturally included in $\{ \bar{\bar z} :\, z \in \frac{1}{3}A \}$.


The same works with $\frac{1}{3^k}$ in place of $\frac{1}{3}$ for any $k$. Doing this for every $k$, and gluing things together in the obvious way, we get our isometric embedding of a dense subset of $\mathbb R$ in $X$. (Namely, the set of all fractions with a power of $3$ in the denominator.)


