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 $\mathbb N$?



can every Gromov-Hausdorff limit be realized as a metric ultralimit?
Yes, any ultrafilter which contains $\{i\_j\}\_j$ will do.


