Gromov-Hausdorff convergence for non-compact metric spaces

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.

• 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.
– YCor
Oct 9 '14 at 13:43
• I added the hypothesis of connectedness of the metric spaces. Oct 13 '14 at 13:17
• 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 $\mathbf{Z}$).
– YCor
Oct 13 '14 at 20:21
• 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 $c$. Oct 14 '14 at 7:16

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.