Let $(X,d)$ be a metric space such that for all distinct $x_1, x_2 \in X$:

  • For $r \in [0,d(x_1,x_2)]$, there exists a unique $y_r \in X$ such that $d(x_1,y_r)=r$ and $d(x_1,y_r)+d(y_r,x_2)= d(x_1,x_2)$.
  • For $r \in (d(x_1,x_2),\infty)$, there exists a unique $y_r \in X$ such that $d(x_1,y_r)=r$ and $d(x_1,x_2)+ d(x_2,y_r)=d(x_1,y_r)$.

For those familiar with the terms, this should be equivalent to an externally convex M-space. As far as I understand, any normed vector space over the reals should satisfy this property.

Now let's say we have five points $O, A_1, A_2, B_1, B_2$ such that for $i=1,2$, $\rho >1$, $d(O,B_i)=\rho \cdot d(O,A_i)$ and $d(O,B_i)=d(O,A_i)+d(A_i,B_i)$.

Question: Then do we have that $d(B_1,B_2)= \rho \cdot d(A_1,A_2)$?

Attempt: At first I assumed that the answer to this question would obviously be yes, and that fiddling around with the given information would immediately lead to a proof.

However, I quickly realized that the triangle inequality does not give a tight upper bound for either side of the equation, and I was unable to manipulate it (e.g. using minus signs) to also give me any lower bound. All I found was a common upper bound for $d(B_1,B_2)$ and $\rho \cdot d(A_1,A_2)$: $$d(B_1,B_2) \le \rho \cdot (d(O,A_1) + d(O,A_2))\,, \quad\quad \rho \cdot d(A_1,A_2) \le \rho \cdot(d(O,A_1)+d(O,A_2)) \,. $$ But since this upper bound isn't tight, even if I find a common lower bound as well, it won't be enough to force both quantities to have the same value.

My next thought was to use the hypothesis to identify a unique midpoint $A_3$ between $A_1$ and $A_2$ and a unique midpoint $B_3$ between $B_1$ and $B_2$, then somehow show that $d(O,B_3)=\rho \cdot d(O,A_3)$ and $d(O,B_3)=d(O,A_3)+d(A_3,B_3)$, by using at some point an appeal to uniqueness. Then somehow this would show that $\rho\cdot d(A_1,A_3) = d(B_1,B_3)$ and thus that $d(B_1,B_2) = \rho \cdot d(A_1,A_2)$.

But for any of this argument to work, it seems like I would need the following equivalent formulation of the parallel postulate/Euclid's axiom:

Let a line segment join the midpoint of two sides of a given triangle. That line segment will be half as long as the third side.

I was worried that this might be equivalent to the SAS postulate and thus I would have to assume what I wanted to prove. However, that seems not to be the case.

Nevertheless, I am still confused about whether this could work, since the parallel postulate is supposed to be equivalent to the pythagorean theorem, which obviously does not hold in arbitrary normed vector spaces over the reals, whereas I thought my assumptions were weaker than a normed vector space over the reals.

So it seems like neither of these obvious approaches would work, which means that instead I should try to find a counterexample. However, all of my intuition for this comes from $\mathbb{R}^n$ with either the Euclidean or the taxicab metric -- cases for which the result is true. Thus I am not sure what direction to look in order to find a counterexample.


Consider $\mathbb{H}^2$ Then $$d(B_1,B_2) > \rho \cdot d(A_1,A_2)$$

  • $\begingroup$ I'm sure this is completely obvious, but I don't follow. Does $\mathbb{H}$ mean the quaternions? What is the metric for $\mathbb{H}$? Are $A_i, B_i$ completely arbitrary? $\endgroup$ – Chill2Macht Apr 14 '17 at 10:15
  • 1
    $\begingroup$ $\mathbb{H}^2=(\{(x,y)|y>0\},\frac{dx^2+dy^2}{y^2} )$ is 2-dimensional hyperbolic space. And it holds for distinct $A_i,\ B_i$ $\endgroup$ – HK Lee Apr 14 '17 at 11:15
  • $\begingroup$ Oh OK that makes a lot more sense than what I was thinking of. I really appreciate you fixing this for me. $\endgroup$ – Chill2Macht Apr 14 '17 at 12:05

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