Normally, the Euclidean space is introduced as $\mathbb R^n$. However, I've now been thinking about how one might define the $n$-dimensional Euclidean space only from the properties of the metric. I've come up with the following conjecture:

A metric space $(M,d)$ is an $n$-dimensional Euclidean space iff it has the following properties:

Line segment (L): For any two points $A,B\in M$ and any number $\lambda\in [0,1]$, there exists exactly one point $C\in M$ so that $d(A,C)=\lambda\,d(A,B)$ and $d(C,B)=(1-\lambda)\,d(A,B)$.

Uniqueness of extension (U): If for any points $A,B,C,D\in M$ with $A\ne B$ we have $d(A,C)=d(A,B)+d(B,C)=d(A,D)=d(A,B)+d(B,D)$ then $C=D$.

Homogeneity (H): For any four points $A,B,C,D\in M$ with $d(A,B)=d(C,D)$ there exists an isometry $\phi$ of $M$ so that $\phi(A)=C$ and $\phi(B)=D$.

Scale invariance (S): For any $\lambda>0$ there exists a function $s\colon M\to M$ so that for any two points $A,B\in M$ we have $d(s(A),s(B)) = \lambda\,d(A,B)$.

Dimension (D): The maximal number of different points $P_1,\ldots,P_k$ so that each pair of them has the same distance is $n+1$.

Now my question: Is this correct? That is, do those conditions already guarantee that the metric space is an $n$-dimensional Euclidean space? If not, what would be an example of a metric space which is not Euclidean, but fulfils all the conditions above?

What I already found (unless I've done an error, in that case, please correct):

It is easy to see that it contains a full line for each pair of points: Given the points $A$ and $B$, the condition (L) already gives the points in between $A$ and $B$. Now for any $r>0$, (S) tells us that there exist two points $C,D$ so that $d(C,D) = (r+1)\,d(A,B)$. Then (L) guarantees the existence of a point $E$ with $d(C,E)=1$ and $d(E,D)=r$. And (H) guarantees us an isometry $\phi(C)=A$ and $\phi(E)=B$. Then the line segment from $A$ to $\phi(D)$ extends the line segment in the direction of $B$. (U) guarantees us that this extension is unique.

If we define a straight line $l$ as a set of points so that for any three points $A, B, C\in l$ the largest of their distances is the sum of the other two distances, then from we also get immediately that two lines can intersect at most in one point (because if they have two points in common, then (L) guarantees that all points in between are also common, and I just showed that the extension is also unique).

I can also use the law of cosines to define the angle $\phi = \angle ABC$ as $\cos\phi = \frac{d(A,C)^2-d(A,B)^2-d(B,C)^2}{2\,d(A,B)\,d(B,C)}$ (of course the law of the cosine assumes Euclidean geometry, but since I'm defining the angle, this just means that if the space is not Euclidean, the angle I just defined is not the usual angle). It is obvious that this angle is independent of scaling (because a common factor just cancels out).

I also think that with the definition of the angle above, I should get that the sum of angles in the triangle is always $\pi$ (because I can just map the three points individually on three points with the same distance onto a known Euclidean plane, and there I know that the angles add up to $\pi$).

However is that already sufficient to show that it is an Euclidean space? Or could there be some strange metric space where all this is true without it being an Euclidean space?

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    $\begingroup$ @ClementC.: According top your link, it is "page not found" ;-) But anyway, my point was that I wanted to use only the metric. $\endgroup$ – celtschk Jul 6 '13 at 21:48
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    $\begingroup$ @ChrisEagle: Could you please give me a concrete counter example? $\endgroup$ – celtschk Jul 6 '13 at 21:49
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    $\begingroup$ @ClementC.: Strange, now your first link works. Must have been a temporary server problem. Anyway, even your second link is not what I want because it also refers to an additional structure, namely the Euclidean vector space defined in your first link. $\endgroup$ – celtschk Jul 6 '13 at 21:58
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    $\begingroup$ @user254665: According to the triangle inequality, $d(A,C)\le d(A,B)+d(B,C)$, therefore $$d(A,C)^2-d(A,B)^2-d(B,C)^2\le((d(A,B)+d(B,C))^2-d(A,B)^2-d(B,C)^2 = 2d(A,B)d(B,C)$$ and thus $\cos\phi \le 1$. Further be wlog $d(A,B)\ge d(B,C)$. Again due to the triangle inequality $d(A,B) \le d(A,C)+d(C,B)$, thus $d(A,C) \ge d(A,B) - d(B,C)$, thus $d(A,C)^2 - d(A,B)^2 - d(B,C)^2 \ge -2 d(A,B) d(B,C)$, thus $\cos\phi\ge-1$. $\endgroup$ – celtschk Sep 1 '15 at 12:16
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    $\begingroup$ Stepping back to $\mathbb R^2$, found a metric space approach - math.stackexchange.com/a/1991372/432081 $\endgroup$ – CopyPasteIt Jan 20 at 2:43

Maybe this has some relevance: Cayley–Menger determinants.

(Most of this Wikipdia article was destroyed on November 11th by a user called "Toninowiki". I've restored much of what was destroyed. The original poster in this present thread has commented below that the article does not deal with higher dimensions. That is wrong. If you look at it and don't see anything on higher dimensions, then look at the version of the article that was there before November 11th. Or at the one I left there a few minutes ago.)

  • $\begingroup$ Thank you for the link. It allows to show that my straight lines indeed are straight. Unfortunately the article doesn't give an extension to higher dimension. $\endgroup$ – celtschk Jan 25 '14 at 10:41
  • $\begingroup$ @celtschk : Sorry, you're wrong. It does go into higher dimensions. Just look at the article before the edits done by "Toninowiki" on November 11th. $\endgroup$ – Michael Hardy Jan 25 '14 at 21:09
  • $\begingroup$ @celtschk : I've restored the parts of the article that were destroyed by "Toninowiki" on November 11. $\endgroup$ – Michael Hardy Jan 25 '14 at 21:19
  • $\begingroup$ Ah, now it's much more useful. Thank you. $\endgroup$ – celtschk Jan 26 '14 at 0:02

I think the issue here is that there is some confusion between your use of the terms metric space, Euclidean and $\mathbb{R}^{n}$.

The most general object in the list above is that of a metric space.

The axioms given in the question are satisfied by any complete homogenous metric space -- for example the hyperbolic metric on the unit $n$-ball.

So if you start with the underlying space $\mathbb{R}^n$, and give $\mathbb{R}^n$ the standard metric, then the axioms are satisfied.

If you take a different underlying set, say the unit ball, and put a hyperbolic metric on it, again the axioms are satisfied!

So your axioms do not really distinguish between different metrics. They just give properties satisfied by many metrics on different spaces.

If you are asking if $\mathbb{R}^n$ can be given a metric that satisfies the axioms but where the cosine law say is different, then the answer is no.

Hope this helps!

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    $\begingroup$ Does the hyperbolic metric on the unit ball really satisfy the scale invariance axiom (S)? $\endgroup$ – celtschk Jan 25 '14 at 10:11

In order to show that your axioms are correct, it suffices to show that they are equivalent to Tarski's axioms of Euclidean geometry, which have been proven to be correct. Below I have written them out in the context of a metric space, to make the equivalence easier to verify.

Tarski's axioms only have one primitive notion, that of points, and only two relations, congruence and betweenness.

Restricting to metric spaces, it is very straightforward to define an appropriate notion of congruence to substitute into Tarski's axioms: points $p_1, p_2, p_3, p_4$ are congruent (written $p_1 p_2 \equiv p_3 p_4$) if and only if $d(p_1, p_2) = d(p_3, p_4)$.

We can also define an appropriate notion of betweenness in entirely metric terms. (The inspiration for this definition comes from convex metric spaces, of which Euclidean space is an example. However, see also 1 or 2.) We say that the point $p_2$ is between the points $p_1$ and $p_3$ if and only if equality is achieved in the triangle inequality: $d(p_1, p_2) + d(p_2, p_3) = d(p_1,p_3)$.

Note that, with these definitions of congruence and betweenness, axioms 1,2,3, and 6 in this numbering become redundant, so we don't need to mention them. Thus, in the case of metric space models for Euclidean geometry, the number of axioms reduce automatically from 11 to 7. This leads to the following metric space axiom list for Euclidean geometry (with new numbering):

  1. Segment Extension: Given points $p_1,p_2,p_3,p_4$: $$\begin{array}{l}\text{There exists a point }p_5\text{ such that: }\\ d(p_1,p_2) + d(p_2,p_5) = d(p_1,p_5)\text{ and }d(p_2,p_5) = d(p_3,p_4) \,. \end{array} $$
  2. Five Segment Axiom: Given points $p_1,p_2,p_3,p_4$ and $q_1,q_2,q_3,q_4$, with $d(p_1,p_2) > 0$: $$ \begin{array}{l} \text{If:} \\ \begin{array}{ll}d(p_1,p_2) = d(q_1,q_2) \,, & d(p_2,p_3) = d(q_2,q_3) \,, \\ d(p_1,p_4) = d(q_1,q_4) \,, & d(p_2,p_4) = d(q_2,q_4) \,, \\ d(p_1,p_2)+d(p_2,p_3) = d(p_1,p_3) \,, & d(q_1,q_2)+d(q_2,q_3) = d(q_1,q_3) \end{array} \\ \text{Then } d(p_3,p_4) = d(q_3,q_4) \,. \end{array} $$
  3. (Inner) Pasch Axiom: Given points $a,b,c$ and $p,q$: $$ \begin{array}{l} \text{If } d(a, p)+d(p, c) = d(a,c)\text{ and }d(b, q)+d(q, c) = d(b,c) \\ \text{Then there exists a point }x\text{ such that: }\\ d(p,x)+d(x,b) = d(p,b)\text{ and }d(q,x)+d(x,a) = d(q,a) \,. \end{array} $$
  4. Lower Dimension Axiom ($n \ge 2$): There exist points $a,b,c$, and $n-1$ distinct points $p_1, \dots, p_{n-1}$, such that: $$ \begin{array}{l} d(a,b)+d(b,c) < d(a,c) \,, d(b,c) + d(c,a) < d(b,a) \,, d(c, a)+ d(c, b) < d(c,b) \,, \text{ and} \\ d(a, p_1) = d( a, p_2) = \dots = d( a, p_{n-1})\,, d(b, p_1) = d( b, p_2) = \dots = d( b, p_{n-1})\,, \\ \text{and }d(c, p_1) = d( c, p_2) = \dots = d( c, p_{n-1}) \,. \end{array} $$
  5. Upper Dimension Axiom ($n \ge 2$): There exist points $a,b,c$, and $n$ distinct points $p_1, \dots, p_{n-1}, p_n$, such that: $$ \begin{array}{l} \text{If } d(a, p_1) = d(a,p_2) = \dots = d(a, p_n)\,, d(b, p_1) = d(b, p_2) = \dots = d(b, p_n) \,, \\ \text{and }d(c, p_1) = d(c, p_2) = \dots = d(c, p_n) \\ \text{Then at least one of }d(a,b)+d(b,c) = d(a,c)\,, d(b,c)+d(c,a) = d(b,a) \,, \\ \text{ or }d(c,a)+d(a,b) = d(c,b) \text{ is true.} \end{array} $$
  6. Euclid Axiom: Given points $p_1, p_2, p_3, p_4$ and $t$: $$ \begin{array}{l} \text{There exist points }x,y\text{ such that: } \\ \text{If }d(p_1,p_4) >0\text{, }d(p_1,p_4)+d(p_4,t) = d(p_1,t)\text{, and }d(p_2,p_4)+d(p_4,p_3) = d(p_2,p_3) \\ \text{Then } d(p_1,p_2)+d(p_2,x) = d(p_1,x)\text{, }d(p_1,p_3)+d(p_3,y) = d(p_1,y)\,,\\ \text{and }d(x,t)+d(t,y) = d(x,y) \,. \end{array} $$
  7. Continuity Schema

Note: The Five Segment Axiom is a "rewording" of the Side-Angle-Side postulate. Taxicab geometry is an example of a geometry satisfying most (if not all) of the axioms of Euclidean geometry except for the Side-Angle-Side postulate. The Five Segment Axiom therefore as a result is necessary to make the notion of angle in Euclidean geometry a useful (isotropic) one.

The Euclid axiom is a "rewording" of the Parallel Postulate. Thus it has the same geometric significance as the Parallel Postulate, which I assume is well-known to you.

The Inner Pasch Axiom is equivalent to the Pasch Axiom (also called sometimes the Outer Pasch Axiom) which is also equivalent to the Plane Separation Axiom. It is necessary to prevent really degenerate spaces from also being models of Euclidean geometry. The interpretation of the axiom's geometric significance is easiest to understand when we assume additionally that the metric space is complete (see below, this case also makes Axiom 7 redundant): see here or here.

If we assume additionally that the metric space is (Cauchy) complete, we get the full second-order logic continuity axiom (a la Hilbert) and a clean relationship with the real numbers. As a result, Axiom 7 becomes redundant when we assume a complete metric space. See (1)(2)(3).

Note also that the lower dimension axiom, in the case that $n=2$, simplifies to the following:

Lower Dimension ($n = 2$): There exist points $a,b,c$ such that: $$ \begin{array}{l} d(a,b)+d(b,c) < d(a,c) \,, \\ d(b,c)+d(c,a) < d(b,a) \, \\ d(c,a)+d(a,b)<d(c,b) \,. \end{array} $$

Since three points are collinear if and only if one of the points is between the other two, this is equivalent to the statement "there exist three non-collinear points".

In the case of $n = 1$ (which amounts to an essentially coordinate-less characterization of a space homeomorphic to a real closed subfield of $\mathbb{R}$, or of a space homeomorphic to $\mathbb{R}$ if we demand a complete metric space) one has the following simpler versions of both the upper and lower dimension axioms:

Lower Dimension Axiom ($n = 1$): There exist two distinct points $a,b$.

Upper Dimension Axiom ($n = 1$): Given three points $a,b,c$, at least one of the following is true: $$ \begin{array}{c} d(a,b)+d(b,c) = d(a,c)\\ d(b,c)+d(c,a) = d(b,a) \\ d(c, a)+d(a,b) = d(c,b) \end{array} $$ In other words, any three points $a,b,c$ are collinear.

For $n= 0$, all of the axioms can be replaced by a single axiom:

Upper Dimension Axiom ($n = 0$): For any points $a,b$, one has that $a = b$.

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    $\begingroup$ Thank you, that is definitely useful. If I'm not mistaken, Segment Extension is effectively what I've proved in my question when showing that there's a complete line for each two points. I initially thought Cauchy-completeness should follow from my axioms, but I now notice that there's nothing preventing extra infinitesimally close points (I demand the existence of points based on real numbers, thus preventing e.g. $\mathbb Q^n$, but I don't demand the non-existence of points that cannot be reached that way. $\endgroup$ – celtschk Nov 12 '17 at 12:16

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