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According to Burago-Burago-Ivanov, one says that the sequence of pointed metric spaces $(X_n,d_n,p_n)$ GH-converges to $(X,d,p)$ if for every $R>0,\varepsilon>0$, there exists a $N$ such that for $n\geq N$ one can find $f:B_{X_n}(p_n,R)\to X$ such that :

  • $f(p_n)=p$,

  • for every $x,y\in B_{X_n}(p_n,R)$, $|d_{X_n}(x,y)-d_X(f(x),f(y))|\leq\varepsilon$,

  • the $\varepsilon$-neighborhood of the image of $f$ contains $B_X(p,R-\varepsilon)$.

Before stating this definition, the authors give an informal one :

"Roughly speaking, a sequence $\{X_n\}$ of metric spaces converges to a space $X$ if for every $r > 0$ the balls of radius $r$ in $X_n$ centered at some fixed points converge (as compact metric spaces) to a ball of radius $r$ in $X$."

Then they go on to say :

The actual definition (Definition 8.1.1 below) is more complicated, but in most cases it is equivalent to this description.

My question is :

  • Under which assumptions is the informal definition equivalent to the rigorous one ?

I know that the metric on the limit space being intrinsic should play a role, as exemplified by the sequence $(1-\tfrac{1}{n})\mathbb{Z}$, which GH-converges to $\mathbb{Z}$ according to the formal definition, but fails to do so for the informal one.

Actually Ex 8.1.3 in Burago-Burago-Ivanov show that the formal definition implies that for every $R>0$, $B_{X_n}(p_n,R)$ GH-converges to $B_X(p,R)$ if $X$ is a length space.

I tried to prove a converse to this statement but was stuck at one point, building maps $f$ which satisfy conditions (2) and (3) of the definition is not complicated when $(X,d,p)$ is a length space. However I am not sure if one can the points $f(p_n)$ from staying too far away from $p$.

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  • $\begingroup$ It's maybe just that the "informal definition" is awkwardly stated, and should be "for every $r$-ball of $X_n$ converges to the $r$-ball of $X$ except possibly on the $r$-sphere", which better matches the reality. $\endgroup$ – YCor May 23 '16 at 13:31
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In other words you want to remove the condition $f(p_n)=p$.

In this case, instead of sequence $X_n$ one can take a fixed space $Y$ which is not isometric to $X$, but such that for some points $p\in X$ and $q\in Y$ the balls $B(p,R)_X$ and $B(q,R)_Y$ are isometric for any radius $R<\infty$.

One example is given here, let me describe an other way to think about it. Imagine that you cut the vertical side of right half plane in the $\ell_\infty$-plane using the following pattern. Equip the obtained paper with induced intrinsic metric; this way you get space $X$. Now you do the same for the positive quadrant to get space $Y$. You can take the origin as the marked points in both spaces.

enter image description here

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    $\begingroup$ I am having some trouble understanding your construction, I'll think about it and accept the answer once I understand it. $\endgroup$ – Thomas Richard May 23 '16 at 17:21
  • $\begingroup$ @ThomasRichard I updated the answer, maybe it is more readabale now. $\endgroup$ – Anton Petrunin May 23 '16 at 18:52
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    $\begingroup$ I am still not sure what is going on : does one start with the right half plane and remove the "dyadic comb" from it to get the space $X$ ? $\endgroup$ – Thomas Richard May 23 '16 at 19:08
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    $\begingroup$ What freedom does one have in the choice of the comb and how can that give isometric balls for all $r>0$ ? I'm sorry to be slow on that one. $\endgroup$ – Thomas Richard May 23 '16 at 20:25
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    $\begingroup$ @ThomasRichard, for each size, we have to decide if the closest tooth of that size is above or below the origin. The obtained spaces are isometric if the sequence of these signs coincide (or opposite) starting from some size. $\endgroup$ – Anton Petrunin May 23 '16 at 20:31

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