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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, where the completeness assumption was quickly added. Will Brian proved that any complete Banakh space is isometric to the Euclidean line.

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.

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.

Luc Guyot
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    Why do you say that the question remains unanswered? Do you doubt that the claim added into Taras Banakh's original question is true? (IIRC I convinced myself that it works, it's not that difficult.) Or is the question subtly different in a way I failed to notice? – Emil Jeřábek Apr 01 '23 at 19:13
  • The question I am asking here is exactly the same as the original question of Taras Banakh. I have failed to prove the proposed claim. This is why I would be glad to read a detailed proof. – Luc Guyot Apr 01 '23 at 20:40
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    Why the transcendental-number-theory tag? – Gerry Myerson Apr 01 '23 at 22:40
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    @GerryMyerson Dear Gerry, thanks for asking. While I was looking for a counter-example, I felt that reasoning with a transcendance basis of $\mathbb{R}$ over $\mathbb{Q}$, in combination with my Claim 4 (see my "answer" to the original post), or something similar, could be successful. This is quite speculative at the moment, but I still hope that some number theoretic tools could come in handy. – Luc Guyot Apr 01 '23 at 23:27
  • @LucGuyot What is wrong with the counterexample that Taras Banakh "added in edit"? – Timothy Chow Apr 01 '23 at 23:52
  • @TimothyChow Dear Timothy, I am certain that you had no intention to be offensive. But this is the second time that I am asked the same question, and I do feel offended. The claim has no proof. I cannot produce a proof myself. The original question remains unanswered because it has no MO answer. I am asking for a detailed proof. – Luc Guyot Apr 02 '23 at 00:14
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    @LucGuyot: I think one issue causing confusion for people commenting on your question is that in the main text of your question you do not point out Taras Banakh offered a solution (albeit without proof) in the previous question. In fact, you emphasize that the previous question contains failed counter-examples, which gives the opposite impression. – Sam Hopkins Apr 02 '23 at 00:53
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    @LucGuyot I agree with Sam Hopkins that the wording of your question is unclear. I recommend that you rephrase your question by explicitly referring to the proposed counterexample, and then either explaining in detail why the counterexample is a "failed counterexample," or asking that someone spell out more details about how the counterexample is supposed to work. As it stands, it's not clear whether you have identified an explicit problem with the proposed counterexample, or are just unable to see why it works. – Timothy Chow Apr 02 '23 at 01:08
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    Also, "property (B)" should be expanded to "Banakh Property" or something else to make the origin obvious and to avoid confusion with notations from early authors where "property (B)" is sometimes used for "Baire property" or "Banach property". – François G. Dorais Apr 02 '23 at 01:20
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    @FrançoisG.Dorais, re, for maximal clarity, spaces satisfying property (B) should be called Banakh spaces. – LSpice Apr 02 '23 at 01:24
  • @SamHopkins Thanks for pointing out the defects of my wording. I have done my best to improve it. I hope you understand that I feel uncomfortable with questions on my opinion about a solution that is not articulated yet. – Luc Guyot Apr 02 '23 at 01:29
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    Just do it; don’t try to use a transcendence basis. Enumerate the positive reals as ${r_\alpha:\alpha<2^\omega}$, and by transfinite recursion, build a sequence of elements $x_\alpha\in\mathbb R^2$ such that $|x_\alpha|=r_\alpha$, and such that $|a|=|b|\implies a=\pm b$ for all $a,b$ in the $\mathbb Q$-linear span $\mathbb Q{x_\alpha:\alpha<2^\omega}$. When looking for $x_\alpha$, you have $2^\omega$ potential choices for $|x_\alpha|=r_\alpha$, and $<2^\omega$ obstructions $(a,b)$. Check that each obstruction defines a line ... – Emil Jeřábek Apr 02 '23 at 06:51
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    ... or a circle (this is the only thing that requires you to compute something), thus it only invalidates at most $2$ points on the circle ${x:|x|=r_\alpha}$. It follows that there are $<2^\omega$ bad points, and $2^\omega$ valid choices for $x_\alpha$ remain. (Unless $\mathbb Q{x_\beta:\beta<\alpha}$ already contains a point with norm $r_\alpha$, in which case you declare it to be $x_\alpha$.) – Emil Jeřábek Apr 02 '23 at 06:53
  • @EmilJeřábek Dear Emil, thank you very much. I see it now. The nice cardinality argument of the second part (removing i tersections of less than $2^{\omega}$ lines and circles from a continuous circle) had eluded me, I'll wait a couple of days to see if someone, possibly, wants to turn your comments into answer. If not, I'll write it. (Removing the number theory tag.) – Luc Guyot Apr 02 '23 at 11:25

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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 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 $C_\alpha$.

Emil Jeřábek
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