# If all balls at $x$ and $y$ are isometric is there an isometry sending $x$ to $y$?

Let $(X,d)$ be a metric space and $x,y \in X$. Assume that for all $r > 0$ the balls $B_r(x)$ and $B_r(y)$ are isometric.

Is it true that there exists an isometry of $X$ sending $x$ to $y$?

No. Let $x$ and $y$ be connected by an edge and let's use the graph distance as our metric. At $x$, connect paths of length $n$ for each $n\in\mathbb N$. At $y$, do the same, but also connect an infinite path.

• Thank you. I guess i need to figure out the appropiate conditions i might have in my situation. – Wolfgang Spindeler Oct 6 '14 at 17:20
• I upvoted your answer but you could do a bit more to help to interpret it; I mean to replace interpreting your answer with a clear and complete description. – Włodzimierz Holsztyński Oct 6 '14 at 18:01
• For my part, I found this answer to be elegant and easy to understand. While thinking about it, I got a "model-theoretic flash": it reminds me of issues of "local isomorphism" which arise in that subject. I wonder if one can make a real connection there... – Pete L. Clark Oct 6 '14 at 20:59
• I can't turn this example into a (Riemannian, say) manifold example. Is it me, or is this worth a new question? – Marco Golla Oct 6 '14 at 21:03
• Follow-up:mathoverflow.net/questions/182741/… – Bjørn Kjos-Hanssen Oct 7 '14 at 3:42

There is a 5-point example $$\ X := \{x\ y\ a\ b\ c\},\$$ with (symmetric) metrics $$\ d\$$ as follows:

$$d(x\ y) = d(a\ b) = 1$$

$$d(x\ a) = d(y\ b) = 2$$

$$d(x\ b) = d(y\ a) = 3$$

$$d(x\ c) = d(y\ c) = 6$$ $$d(a\ c) = 5\qquad\qquad d(b\ c) = 4$$

Oviously, $$\ B_r(x)\$$ and $$\ B_r(y)\$$ are isometric for every $$\ r>0,\$$ while there is no isometry $$\ f:X\rightarrow X\$$ for which $$\ f(x)=y$$.

REMARK 1  This example is ironic because while the respective balls are isometric, the isometry of the balls doesn't respect the centers. In this sense every bounded (and especially--finite) required example would be ironic.

REMARK 2  Number $$\ 5\$$ is minimal.

• Excellent. Sometimes the number 5 is more complicated than the number $\infty$. – Bjørn Kjos-Hanssen Oct 6 '14 at 22:31
• @Bjørn, thank you. You have produced a nice quotation (proverb, wisdom, ...). – Włodzimierz Holsztyński Oct 6 '14 at 22:35