TITLE: Gromov-Hausdorff convergence for locally finite metric spaces
QUESTION [8 upvotes]: This question might be very easy, but I am little confused by the Gromov-Hausdorff convergence.
My situation is the following: I have a fixed set $X$ which is finite or countable; on it I have locally finite metrics $d_n$ and $d$. For the application that I have in mind, I've found that a good notion of convergence of the sequence of spaces $(X,d_n)$ to the space $(X,d)$ would be described by the uniform convergence of $d_n$ to $d$. Therefore, I am wondering if this convergence is equivalent to Gromov-Hausdorff's convergence.

Question: Are the following statements equivalent:

$d_n\rightarrow d$ uniformly
For any point $x\in X$, the sequence of pointed locally compact metric spaces $(X,d_n,x)$ converges to $(X,d,x)$ in Gromov-Hausdorff sense?


It would sound strange for my intuition if they turn out to be different, but I am in trouble to write down a proof, basically because I am quite new in the definition of Gromov-Hausdorff convergence and I am pretty confused/scared by all these isometric embeddings one should consider.
Thank you in advance for any help,
Valerio

REPLY [10 votes]: 1) does imply 2). An alternative equivalent definition of Hausdorff-Gromov distance is as follows. A correspondence between two sets $X, Y$ is a subset of $X\times Y$ which intersects each horizontal and each vertical fiber. Then, given two metric spaces $(X,d)$, $(Y,e)$, their Hausdorff-Gromov distance is
$$
\frac12 \inf_R \max_{(x,y), (x',y')\in R} |d(x,x')-e(y,y')|,
$$
where the infimum is over all correspondences (This is Theorem 7.3.25 in the book of Burago-Burago-Ivanov).
When the underlying space is the same, the diagonal correspondence gives that the Hausdorff-Gromov distance between $(X,d)$ and $(X,e)$ is at most half the uniform distance between the metrics $d$ and $e$ (likewise for pointed metric spaces). In particular, 1) implies 2).