We say that a metric space $(X, d)$ is a *Banakh space* if for every $\rho \in \mathbb{R}\_{> 0}$ and every $x \in X$, there are
$a,b \in X$ such that $\{y \in X \, \vert \, d(x, y) = \rho\} = \{a, b\}$ and $d(a, b) = 2 \rho$.


**Question on Banakh spaces**. Let $(X, d)$ be a Banakh space. Is $(X, d)$ isometric to the Euclidean line $(\mathbb{R}, \vert \cdot \vert)$?


This question was first asked by Taras Banakh in his post [A metric characterization of the real line](https://mathoverflow.net/q/442772/84349), where the completeness assumption was quickly added.
Will Brian proved that [any **complete** Banakh space is isometric to the Euclidean line](https://mathoverflow.net/a/442872/84349).


Taras Banakh also claimed, with a sketch of proof, that the Euclidean plane $(\mathbb{R}^2, \Vert \cdot \Vert\_2)$ contains a dense subspace which is a Banakh space; see **Added in Edit** in the [original post](https://mathoverflow.net/q/442772/84349).


This would answer the above **Question on Banakh spaces** in the negative, but I do not see how to turn Taras Banakh's sketch into a complete proof.


This question is, I hope, an incentive to write a complete answer for those who know how to do it.


Detailed proofs are welcome.



An incomplete characterisation of the Euclidean line?
Taras Banakh writes in the original question that *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 $r$*. This is indeed rather straightforward to do; it’s a [just-do-it](https://gowers.wordpress.com/2008/08/16/just-do-it-proofs/) proof.


First, observe that if $L$ has the properties above, the translation invariance of $L$ implies that it is a Banakh space, but $L$ is not isometric to $\mathbb R$ as it is not complete. Thus, it answers the question in the negative.


To construct $L$, let us fix an enumeration $\mathbb R\_{>0}=\{r\_\alpha:\alpha<\mathfrak c\}$. We will build by transfinite recursion a sequence $X=\{x\_\alpha:\alpha<\mathfrak c\}\subseteq\mathbb R^2$ such that $\|x\_\alpha\|=r\_\alpha$, and such that $\|a\|=\|b\|\implies a=\pm b$ for all $a,b\in L$, where $L$ is the $\mathbb Q$-linear span $\mathbb QX$. Let $\alpha<\mathfrak c$, and assume that $\{x\_\beta:\beta<\alpha\}$ has been already defined such that $\|x\_\beta\|=r\_\beta$, and $\|a\|=\|b\|\implies a=\pm b$ for all $a,b\in L\_\alpha=\mathbb Q\{x\_\beta:\beta<\alpha\}$; we will define $x\_\alpha$.


If $\|a\|=r\_\alpha$ for some $a\in L\_\alpha$, we can define $x\_\alpha$ as one of the two such elements $a$. Thus, we may assume $L\_\alpha$ is disjoint from the circle $C\_\alpha=\{x\in\mathbb R^2:\|x\|=r\_\alpha\}$.


We pick $x\_\alpha$ as a suitable element $x\in C\_\alpha$. We need to ensure that for each $a,b\in L\_\alpha$ and $q,r\in\mathbb Q$ such that $(a,q)\ne\pm(b,r)$, $\|a+qx\|\ne\|b+rx\|$. This is a collection of $|\alpha|+\aleph\_0<\mathfrak c$ possible obstructions; we will show that each obstruction set is a line or a circle different from $C\_\alpha$, and therefore intersects $C\_\alpha$ in at most $2$ points. Thus, only $|\alpha|+\aleph\_0<\mathfrak c$ points of $C\_\alpha$ violate some obstruction, and we can choose $x\_\alpha$ as any of the remaining $\mathfrak c$ points. (Moreover, if $\alpha=1$, we can make sure $x\_1$ is not a scalar multiple of $x\_0$, thus the resulting set $L$ will not be contained in a line; being a $\mathbb Q$-linear set, this will make it dense in $\mathbb R^2$.)


The obstructions $\|a+qx\|=\|b+rx\|$ are of the following kind:


* $q=r=0$: Then $b\ne\pm a$, thus $\|a\|\ne\|b\|$ by the induction hypothesis, i.e., the obstruction set is empty.
* $q=\pm r\ne0$: Scaling everything by $q^{-1}$, and possibly negating $(b,r)$, we may assume $q=r=1$ and $a\ne b$. Then the obstruction set $\{x:\|x+a\|=\|x+b\|\}$ is a line (the perpendicular bisector of $-a$ and $-b$).
* $q\ne\pm r$: If you do the algebra, it is easy to see that the obstruction set $\{x:\|qx+a\|=\|rx+b\|\}$ is a circle. Moreover, it contains the point $x=(a-b)/(r-q)\in L\_\alpha$, hence the circle is different from $C\_\alpha$.


