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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
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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 1

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
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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

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