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 $y$?


2 Answers 2


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.

  • 1
    $\begingroup$ Thank you. I guess i need to figure out the appropiate conditions i might have in my situation. $\endgroup$ Oct 6, 2014 at 17:20
  • 1
    $\begingroup$ 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. $\endgroup$ Oct 6, 2014 at 18:01
  • 6
    $\begingroup$ 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... $\endgroup$ Oct 6, 2014 at 20:59
  • 3
    $\begingroup$ I can't turn this example into a (Riemannian, say) manifold example. Is it me, or is this worth a new question? $\endgroup$ Oct 6, 2014 at 21:03
  • 2
    $\begingroup$ Follow-up:mathoverflow.net/questions/182741/… $\endgroup$ Oct 7, 2014 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.

  • 10
    $\begingroup$ Excellent. Sometimes the number 5 is more complicated than the number $\infty$. $\endgroup$ Oct 6, 2014 at 22:31
  • $\begingroup$ @Bjørn, thank you. You have produced a nice quotation (proverb, wisdom, ...). $\endgroup$ Oct 6, 2014 at 22:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.