Let $(X_i)$ be a countable collection of bounded metric spaces, pre-compact in the Gromov-Hausdorff metric. It is well-known that for any choice of non-principal ultrafilter $U$ on $\mathbb N$, the metric ultra-limit of $(X_i)$-s with respect to $U$ is a Gromov-Hausdorff limit of some subsequence $(X_{i_k})$.

Let $Y$ be a Gromov-Hausdorff limit of a subsequnce $(X_{i_j})$. Is it isometric to the metric ultralimit of $(X_i)$ for some choice of ultrafilter on $\mathbb N$?

  • $\begingroup$ In your first paragraph 1) you need a uniform precompactness condition in order to have convergence of a subsequence 2) if the metric spaces are bounded but not uniformly, the ultralimit is not defined: you need to choose base-points. $\endgroup$
    – YCor
    Dec 5, 2016 at 17:52
  • 1
    $\begingroup$ On the other hand no uniform precompactness assumption is required for the (trivial) converse: if $(X_n,x_n,d_n)$ GH-converges to $(X,x,d)$, then every ultralimit of $(X_n,x_n,d_n)$ is (basepoint-preserving) isometric to $(X,x,d)$. $\endgroup$
    – YCor
    Dec 5, 2016 at 18:54

1 Answer 1


Yes, any ultrafilter which contains $\{i_j\}_j$ will do.


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