# A metric characterization of the real line

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

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 $$\{a,b\}=\{x\in X:d(x,c)=r\}$$.

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 $$r$$.

• Just in case, papers by Lelek+Nitka may be nice to consult. Mar 15 at 17:49
• @WlodAA Thank you for the suggestion. I have found some paper of Lelek and Nitka (matwbn.icm.edu.pl/ksiazki/fm/fm49/fm49117.pdf), but I do not see there an answer to my question, maybe I had in mind some other paper of Lelek and Nitka? Mar 15 at 18:09
• I take it you are explicitly not assuming that $(X,d)$ is complete? Mar 16 at 4:50
• Is there an easy proof from this property that any two points have a unique point equidistant from them? Mar 16 at 10:18
• Will it be easier if we verify that the space under consideration is a 1-dimensional Busemann G-space en.wikipedia.org/wiki/Busemann_G-space? Mar 22 at 14:52

## 3 Answers

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.)

• But isn’t “complete Banakh space” redundant? :) Mar 16 at 19:49
• I had the same construction in mind but I got stuck in the next to last paragraph because I was inadvertantly using a slightly weaker version of Banakh's property. The weaker property states that for every $x$ and every $r > 0$ there is a unique unordered pair $\{a,b\}$ with $d(a,b) = 2r$ and $d(a,x) = d(b,x) = r$. This does not exclude the existence of another pair $\{a',b'\}$ with $d(a',x) = d(b',x) = r$ but $d(a',b') < 2r$. I still wonder if this weaker property is sufficient. Mar 16 at 19:49
• @FrançoisG.Dorais I hope that for the weaker property there is a counterexample of the form $K\times \mathbb R\subseteq \mathbb R^2$, where $K$ is a suitable curve, for example a circle or parabola. Mar 17 at 13:36
• @FrançoisG.Dorais: Here's an idea for a counterexample. Take two parallel lines in the plane, say $y=0$ and $y=1$, and connect them with a line segment, let's say $\{0\} \times [0,1]$. Let $d$ denote the taxicab metric on this set, and then let $X$ denote the subspace consisting of just the two lines (without the connecting segment). Mar 17 at 13:43

A tentative construction of a counterexample would be: take $$X$$ to be an additive subgroup of a real normed space $$(E,\|\cdot\|)$$ such that for any $$r>0$$ there there is exactly one pair $$\pm x$$ of elements of $$X$$ of norm $$r$$, and such that $$X$$ is not a real line. Then by translation invariance, $$X$$ verifies the stated metric property.

Construction of $$X$$ (Failed) Let $$V\subset\mathbb R$$ be a complementary subspace of $$\mathbb Q$$ in $$\mathbb R$$ as $$\mathbb Q$$-linear space (so $$V$$ is a version of the Vitali set). Consider $$X:=\mathbb Q\times V$$ as a metric subspace of the real normed space $$\big(\mathbb R^2,\|\cdot\|_1\big)$$. This does not work. Maybe another norm does ?

• Thank you very much for your answer. Very nice idea! But I am not sure that the equality $\|x\|_1=r=q+v$ has just two solutions. Why not four? $x=(\pm q,\pm v)$? Mar 15 at 18:59
• It can happen that your construction indeed works (at least under CH) when one can construct such a group by transfinite induction. Mar 15 at 22:48
• It seems that the Euclidean plane does contain a dense $\mathbb Q$-linear subspace $L\subseteq \mathbb R^2$ such that for every $r>0$ the set $\{x\in L:\|x\|=r\}$ is a doubleton. The subspace $L$ can be constructed by transfinite induction of length $\mathfrak c$. It remains to resolve the complete case, which was suggested by Will Brian. Mar 15 at 23:39
• A proof of $2 \neq -2$: Let $r_x(y)$ denote the reflection of $y$ on $x$ (so $d(x,y) = d(x,r_x(y))$). Let $0$ and $1$ be two distinct points with $d(0,1) = 1$, and let $2 = r_1(0)$, $-1 = r_0(1)$, $-2 = r_{-1}(0)$. Suppose that $2 = -2$. Let $P = r_0(2)$. Then $d(P,0) = 2$, $d(P,2) = 4$, $d(1,2) = 1$, so $d(P,1) \ge 3$ by the $P12$ triangle, but also $d(0,1) = 1$, so $d(P,1) \le 3$ by the $P01$ triangle, so $d(P,1) = 3$. Similarly, $d(P,-1) = 3$, so $2 = d(1,-1) = 6$. So indeed $2 \neq -2$. Mar 16 at 10:04
• It seems that the argument of WIll Brian enhanced by the argument of @user42355 indeed yields an isometric embedding of dyadic rationals to that space, which would resolve the "complete" question affirmatively. Mar 16 at 11:08

This long comment is a modest contribution to OP's initial question.

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

Question. Let $$(X, d)$$ be a metric space satisfying $$(B)$$. Is $$(X, d)$$ isometric to the Euclidean line $$(\mathbb{R}, \vert \cdot \vert)$$?

Claim 1. Let $$(X, d)$$ be a metric space satisfying $$(B)$$. Then for every $$(x, y) \in X^2$$, there is a unique $$z \in X$$ such that $$d(x, z) = d(y, z)$$.

Claim 1 answers a question asked in a comment by Matt F.

Claim 2. Let $$(X, d)$$ be a metric space satisfying $$(B)$$ and let $$x \in X$$. Then $$X = \bigcup_{\rho} Q_{x, \rho} = \{x \} \sqcup \bigsqcup_{\rho} (Q_{x, \rho} \setminus \{x\})$$ where $$\rho$$ ranges in a complete set of positive representatives of $$\mathbb{R}^{\times} / \mathbb{Q}^{\times}$$ and $$Q_{x, \rho}$$ is a subspace of $$X$$ containing $$x$$ and is endowed with an isometry onto $$(\rho \mathbb{Q}, \vert \cdot \vert)$$ which maps $$x$$ to $$0$$.

Claim 3. Let $$X$$ be a subset of the $$n$$-dimensional Euclidean space $$(\mathbb{R}^n, \Vert \cdot \Vert_2)$$ containing $$0$$ and such that $$(X, \Vert \cdot \Vert_2)$$ satisfies $$(B)$$. Then $$(X, +)$$ is a divisible subgroup of $$(\mathbb{R}^n, +)$$.

The example given by our last claim fails to satisfy $$(B)$$ as its distance set $$d(X \times X)$$ is a strict subset of $$\mathbb{R}_{> 0}$$.

Claim 4. Let $$K$$ be a subfield of $$\mathbb{R}$$ and let $$x, y \in \mathbb{R}$$ be such that $$1, x$$ and $$x^2 + y^2$$ are linearily independent over $$K$$. Let $$X = K \cdot (1, 0) \oplus K \cdot (x, y) \subseteq \mathbb{R}^2$$. Let $$D = \{ \Vert v \Vert_2 \, \vert \, v \in X\}$$. Then for every $$\rho \in D$$, there are $$w, w' \in X$$ such that $$\Vert w \Vert_2 = \Vert w' \Vert_2 = \rho$$, $$\Vert w - w' \Vert_2 = 2 \rho$$ and $$\{w, w'\} = \{v \in X \, \vert \, \Vert v \Vert_2 = \rho\}$$.

To prove Claim 1, we shall use the following two lemmas:

Lemma 1. Let $$(X, d)$$ be a metric space satisfying $$(B)$$. Let $$x \in X, \rho \in \mathbb{R}_{> 0}$$ and $$\{a, b\} = \{y \in X \, \vert \, d(x, y) = \rho\}$$. Then there is a subspace $$Z_{x, a}$$ of $$X$$ containing $$\{x, a\}$$ and an isometry from $$Z_{x, a}$$ onto $$(\rho \mathbb{Z}, \vert \cdot \vert)$$ mapping $$x$$ to $$0$$ and $$a$$ to $$\rho$$. The latter conditions define $$Z_{x, a}$$ uniquely, and we have $$Z_{x, a} = Z_{x, b}$$. We denote this set by $$Z_{x, \rho}$$ when the choice of a specific isometry is irrelevant.

Lemma 2. Let $$(X, d)$$ be a metric space satisfying $$(B)$$. Let $$x \in X, \rho \in \mathbb{R}_{> 0}$$ and $$\{a, b\} = \{y \in X \, \vert \, d(x, y) = \rho\}$$. Then there is a subspace $$Q_{x, a}$$ of $$X$$ containing $$\{x, a\}$$ and an isometry from $$Q_{x, a}$$ onto $$(\rho \mathbb{Q}, \vert \cdot \vert)$$ mapping $$x$$ to $$0$$ and $$a$$ to $$\rho$$. The latter conditions define $$Q_{x, a}$$ uniquely, and we have $$Q_{x, a} = Q_{x, b}$$. We denote this set by $$Q_{x, \rho}$$ when the choice of a specific isometry is irrelevant.

Notation (from @user42355). Let $$(X, d)$$ be a metric space satisfying $$(B)$$. Let $$x \in X, \rho \in \mathbb{R}_{> 0}$$ and $$\{a, b\} = \{y \in X \, \vert \, d(x, y) = \rho\}$$. We define $$\sigma_{x, \rho}: \{a, b\} \rightarrow \{a, b\}$$ by $$\sigma_{x, \rho}(a) = b$$ and $$\sigma_{x, \rho}(b) = a$$.

Proof of Lemma 1. This is essentially Will Brian's proof. Let $$\{a, b\} = \{y \in X \, \vert \, d(x, y) = \rho\}$$, let $$x_0 = x$$ and $$x_1 = a$$. Then define inductively $$x_{i + 1} = \sigma_{x_i, \rho}(x_{i - 1})$$ for every $$i \ge 1$$. By assumption, we have $$d(x_i, x_j) = \vert i - j \vert \rho \text{ if }\vert i - j \vert = 1 \text{ or if } i - j \text{ is even } \textbf{(1)}$$ It follows from $$(1)$$ and the triangle inequality that $$d(x_0, x_{2k + 1}) \le (2k + 1) \rho$$ and $$d(x_0, x_{2k + 1}) \ge d(x_0, x_{2k + 2}) - d(x_{2k + 1}, d_{2k + 2}) = (2k + 1) \rho$$ for every $$k \ge 0$$. Thus $$d(x_0, x_{2k + 1}) = (2k + 1) \rho$$ for every $$k \ge 0$$. Using again the identities $$(1)$$ and the triangle inequality, we obtain that $$d(x_i, x_j) \le \vert i - j \vert \rho$$ and $$d(x_i, x_j) \ge \vert d(x_0, x_i) - d(x_0, x_j) \vert = \vert i - j \vert \rho$$. Thus $$d(x_i, x_j) = \vert i - j \vert \rho$$ for every $$i,j \ge 0$$, which shows that $$x_i \mapsto i \rho$$ is an isometry from $$\{x_i\}_{i \ge 0}$$ into $$(\rho \mathbb{Z}, \vert \cdot \vert)$$ which maps $$x$$ to $$0$$ and $$a$$ to $$\rho$$. Now, define inductively $$x_{- i - 1} = \sigma_{x_{-i}, \rho}(x_{- i + 1})$$ for every $$i \ge 0$$. Applying our above reasoning to $$\{x_i\}_{i \ge -1}$$, we see that the previous isometry can be extended by mapping $$x_{-1}$$ to $$- \rho$$. By a straightforward induction, we deduce that $$x_i \mapsto i \rho$$ extends to an isometry from $$\{x_i\}_{i \in \mathbb{Z}}$$ onto $$(\rho \mathbb{Z}, \vert \cdot \vert)$$. Let us set $$Z_{x, a} = \{x_i\}_{i \in \mathbb{Z}}$$. Starting anew with $$x_0 = x$$ and $$x_1 = b$$, we obtain in the same way a subspace $$Z_{x, b}$$. It is trivial to check that $$Z_{x, a} = Z_{x, b}$$ from property $$(B)$$.

Proof of Lemma 2. Set $$Q_{x, \rho} = \bigcup_{q \in \mathbb{Q}_{> 0}} Z_{x, q \rho}$$ with $$Z_{x, q \rho}$$ as in Lemma 1. Observing that $$Z_{x, nq\rho} \subset Z_{x, q \rho}$$ for every $$n \in \mathbb{N}_{> 0}$$, the result follows.

The proofs of Claim 1 and 2 are now immediate:

Proof of Claim 1. Let $$Q_{x, y} = Q_{x, d(x, y)}$$ be as in Lemma 2. We define then $$z$$ as the unique point of $$X$$ which maps to $$d(x, y)/2 \in d(x,y)\mathbb{Q}$$.

Proof of Claim 2. Apply Lemma 2.

Proof of Claim 3. For $$x \in \mathbb{R}^n$$, denote by $$s_x$$ the symmetry with respect to $$x$$, i.e, the isometry defined by $$s_x(y) = 2x - y$$. Since $$X$$ is invariant under $$s_x$$ for every $$x \in X$$, it is invariant under the translation $$s_x \circ s_y$$ of vector $$2(y - x)$$ for every $$x, y \in X$$. As $$0 \in X$$, by assumption, the space $$X$$ is invariant under translation by $$2y$$ for every $$y \in X$$. We infer from Claim 1 that $$X$$ is actually invariant under translation by $$y$$ for every $$y \in X$$. Therefore $$(X, +)$$ is an additive subgroup of $$(\mathbb{R}^n, +)$$. The fact that $$(X, +)$$ is divisible is a direct consequence of Lemma 2.

Proof of Claim 4. It suffices to show that for every pair $$(w, w') \in X^2$$, the condition $$\Vert w \Vert_2 = \Vert w' \Vert_2$$ implies $$\det(w, w') = 0$$. This readily follows from the $$K$$-linear independence assumption on $$1, x$$ and $$x^2 + y^2$$.