# A clarification on pointed Gromov-Hausdorff convergence

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 $p$.

• 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. – YCor May 23 '16 at 13:31

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.
• 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 $X$ ? – Thomas Richard May 23 '16 at 19:08
• 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. – Thomas Richard May 23 '16 at 20:25