Gromov-Hausdorff convergence for locally finite metric spaces - MathOverflow

Gromov-Hausdorff convergence for locally finite metric spaces

2011-12-05T10:09:29Z

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

Answer by Tom Leinster for Gromov-Hausdorff convergence for locally finite metric spaces

2011-12-05T10:45:34Z

No: 2 does not imply 1.

Let $X$ be the set $\mathbb{Z} \times \mathbb{Z}$. Define a metric $d$ on $X$ by $$ d((x, y), (x', y')) = 2|x-x'| + |y-y'| $$ and define a metric $e$ on $X$ by $$ e((x, y), (x', y')) = |x-x'| + 2|y-y'|. $$ Then $d \neq e$, but for each $x \in X$, the pointed metric spaces $(X, d, x)$ and $(X, e, x)$ are isometric in a basepoint-preserving way (rotate by 90 degrees about $x$).

Now consider the sequence $$ d, e, d, e, \ldots $$ of metrics on $X$. It does not converge uniformly (or even pointwise). However, the sequence $$ (X, d, x), (X, e, x), (X, d, x), (X, e, x), \ldots $$ of pointed metric spaces does converge in the Gromov-Hausdorff sense: since they're all pointedly isometric, the distance between each element of the sequence and the next is always zero.

Answer by Pablo Shmerkin for Gromov-Hausdorff convergence for locally finite metric spaces

2011-12-05T13:07:17Z

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).