# Name of a metric space concept

I have a metric space with the following property (a bit like having unique geodesics): for any points $a,b,x,y$ with $d(a,b)=d(a,x)+d(x,b)=d(a,y)+d(y,b)$ and $d(a,x)=d(a,y)$, we have $x=y$. Is there an established name for this?

(UPDATE: the condition $d(a,x)=d(a,y)$ was omitted by mistake in the original question.)

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Uhm. One-point set? –  darij grinberg Jun 30 '11 at 17:01
To elucidate Darij's comment: you probably want to add the condition $d(a,x) = d(a,y)$. –  André Henriques Jun 30 '11 at 17:06
No time to check, but it looks related to Definition 2.6 in this paper: home.lu.lv/~ibula/lv/petnieciba/raksti/moravica.pdf –  gowers Jun 30 '11 at 18:50
Not the same though: any subset of $\mathbb{R}^d$ with the Euclidean metric satisfies this property but not the one in the Bula paper. –  Anthony Quas Jun 30 '11 at 19:18
Oh yes -- it's like the uniqueness part of strict convexity but not the existence part. It's probably too much to hope that a metric space with Neil's property can be isometrically embedded into a strictly convex metric space in Bula's sense. –  gowers Jun 30 '11 at 19:46

## 1 Answer

It is easy to prove that the completion of your space can be any separable metric space where metric spheres are nowhere dense.

Does not it scare you?

Answer to the comment: Not all of your spaces can be embedded into a metric tree.

BTW, there is a nice characterization of subsets in a metric tree: $$| x - y | + | a - b | \le \max\{|x-a| + |y-b|,|x-b|+|y-a|\}$$ for all points $x,y,a,b$ (here $|x-y|$ denotes the distance from $x$ to $y$).

In other words, the values $$X=|x-a|+|y-b|,$$ $$Y=|x-b|+|y-a|,$$ $$Z=|x-y|+|a-b|$$ satisfy ulrtatriangle inequality. This inequality can be also thought as a discrete analog of CAT[−∞] inequality.

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Not really. The particular class of spaces that I am thinking about arise in a very combinatorial way, and I hope to show that each of them is the vertex set of a tree with the obvious edge-counting metric. (In particular, my spaces have only finitely many points.) I have proved the property in my question as a step towards that. –  Neil Strickland Jul 1 '11 at 7:31
I will answer above. –  Anton Petrunin Jul 1 '11 at 8:02
Doesn't a 4-cycle satisfy that inequality? –  Gjergji Zaimi Jul 1 '11 at 9:06
Ups, now it is correct. –  Anton Petrunin Jul 1 '11 at 9:15