Let $X$ and $Y$ be two complete proper length spaces, $x \in X$ and $y \in Y$. Assume for every $r>0$ the closed balls $\overline{B_r(x)}$ and $\overline{B_r(y)}$ are isometric.
Does there exist an isometry $X \to Y$?

Remark: If you add that the isometries between the balls all map $x$ to $y$, you get a pointed isometry between $(X,x)$ and $(Y,y)$.

Definitions: A metric space is called proper if all closed balls are compact. A metric space is called length space if the distance of two points equals the infimum of the length of curves connecting these points, i.e. $d(x,y) = \inf \{ L(c) \mid c(0) = x, c(1) = y\}$.

In fact, for a complete length space, this infimum is a minimum.

  • $\begingroup$ I feel that it would not hurt to add a definition (rather than a link) of "proper length". $\endgroup$ Mar 8, 2016 at 7:41

1 Answer 1


The answer is NO.

To get an idea look at the following picture. Imagine it is extended down infinitely and each segment exchanged to a thin rectangle with $\ell_\infty$-metric.

enter image description here

The center of the ball has to be at one of the ends.

There are ways to extend it down to get non-isometric spaces. At one step down you have two choices --- extend it to the right $(+1)$ or to the left $(-1)$. The space can be encoded by an infinite sequence of $\pm1$. Two such spaces are isometric if the sequences coincide starting from some moment or they have opposite sign starting from some moment. In particular the space for $$(+1,+1,+1,+1,\dots)\ \ \text{and}\ \ (+1,-1,+1,-1,\dots)$$ are not isometric.

  • $\begingroup$ Thanks for your respond. I spent some time thinking about this, but do not get it: Could you please explain how you mean to do the extensions? $\endgroup$
    – dg.jan
    Mar 7, 2016 at 9:36
  • $\begingroup$ @dg.jan The answer is updated. $\endgroup$ Mar 10, 2016 at 14:27

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