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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.

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    $\begingroup$ 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? $\endgroup$ Apr 1, 2023 at 19:13
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    $\begingroup$ Why the transcendental-number-theory tag? $\endgroup$ Apr 1, 2023 at 22:40
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    $\begingroup$ @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. $\endgroup$ Apr 2, 2023 at 0:53
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    $\begingroup$ 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 ... $\endgroup$ Apr 2, 2023 at 6:51
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    $\begingroup$ ... 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 $x_\alpha$.) $\endgroup$ Apr 2, 2023 at 6:53

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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 $C_\alpha$.

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