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Let $(X_i,p_i)$, $(X,p)$ be pointed connected proper metric spaces (i.e. the closures of balls are compact). Are the following two statements equivalent?

  1. $\forall r > 0: \bar{B}_r(p_i) \stackrel{GH}{\to} \bar{B}_r(p)$.
  2. $\forall r > 0: (\bar{B}_r(p_i),p_i) \stackrel{GH}{\to} (\bar{B}_r(p),p)$.

In 2, the pointed Gromov-Hausdorff distance is defined as usual but with respect to the pointed Hausdorff distance $d_H((A,a),(B,b)) := d_H(A,B) + d(a,b)$.

Obviously, 2 implies 1, so the question is whether or not/under which conditions the other implication holds.

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    $\begingroup$ Note that there are a few jokes: e.g. although $((1+1/n)\mathbf{Z},0)\to (\mathbf{Z},0)$, the closed 1-balls does not converge. $\endgroup$
    – YCor
    Commented Oct 9, 2014 at 13:43
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    $\begingroup$ I added the hypothesis of connectedness of the metric spaces. $\endgroup$
    – dg.jan
    Commented Oct 13, 2014 at 13:17
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    $\begingroup$ Anyway the same "joke" can hold in a connected space (e.g. use the union of the line $Im(z)=1$ and the segments $[n,n+i]$ for $n\in\mathbf{Z}$ in the complex plane, instead of $\mathbf{Z}$). $\endgroup$
    – YCor
    Commented Oct 13, 2014 at 20:21
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    $\begingroup$ Here is a non-connected counterexample: Consider the example given in the second answer to this question: mathoverflow.net/questions/182719/…. Then let $X_i = X$ be the space defined there with $p_i = x$ and $p = y$. It is compact, but can easily be turned into a noncompact one by adding an infinite ray starting at $c$. $\endgroup$
    – Wolfgang Spindeler
    Commented Oct 14, 2014 at 7:16

1 Answer 1

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Here is a counterexample constructed from the $5$-point example given by Włodzimierz Holsztyński here.

Consider the set $Z = \{x,y,a,b,c\}$ made into a metric space via $$d(x\ y) = d(a\ b) = 1,$$ $$d(x\ a) = d(y\ b) = 2,$$

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

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

First we construct a connected metric space $Z'$ simply by connecting each two points by edges of lengths given by the respective distances and taking the metric on $Z'$ the induced intrinsic one (i hope this is clear). Now we form a noncompact space $Z''$ by adding a ray (that is the interval $[0,\infty[$) starting at $c$. Finally consider the subset $X \subset Z''$ given by $Z''$ without the edge connecting $a$ to $c$ and the one connecting $b$ to $c$. Equip $X$ with the induced restricted metric $d$ (not the intrinsic one). Then for all $r > 0$ the balls $B_r(x) \subset X$ and $B_r(y) \subset X$ are isometric. Therefore, taking $X_i = X$ and $p_i = x$ we find that $B_r(p_i) \to B_r(p)$ in Gromov-Hausdorff sense for all $r > 0$, but $(B_r(p_i),p_i) = (B_r(x),x)$ does not converge to $(B_r(p),p) = (B_r(y),y)$ for all $r \geq 6$ since there is no isometry of $X$ taking $x$ to $y$.

Maybe a sufficient condition might be that the metrics of the $X_i$ are intrinsic (and complete), i.e. distances are given by infimal (minimal) lengths of continuous curves connecting points.

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