I know that given two metric spaces $X$ and $Y$, one can realize the Gromov-Hausdorff metric between the spaces as the Hausdorff distance between $\tilde{X},\tilde{Y}$ as subsets of $X\sqcup Y$. Given a sequence metric spaces $(X_n,d_n)$ such that $X_n\overset{GH}{\to}X_\infty$ (converge in the Gromov-Hausdorff sense), can we embed all the sequence isometrically $\{ X_n \}$ in a metric space $Z$ such that $$d_H(\tilde{X}_n,\tilde{X}_m)=d_{GH}(X_n,X_m)?,$$ where $\tilde{X}_n$ is the realization of $X_n$ in $Z$ and $d_H(A,B)$ is the Hausdorff distance induced in $Z$.
I have a feeling that this is true, but I haven't found a place where this is implied or explicitly stated. I'm assuming that if there is a counter-example, it should be something relatively simple I haven't considered. I'd also appreciate partial insights on the matter.
Later edit:
Since no one has answered so far, I thought maybe I should add more thoughts I had on the matter since posting this, as it could maybe lead to a constructful answer from someone.
I know that if a space $(Z,d)$ is second countable and complete, that the Hausdorff distance on compact subsets is complete. So if $X_n$ are compact metric subsets, then it seems reasonable that if I can embed them appropriately in some complete $(Z,d)$, then $X$ would be embedded in $Z$ by the completeness of $d_H$ (Hausdorff-distance). So I would like to ask three easier questions:
- Given $N$ compact metric spaces, $\{(X_n,d_n) \}_{n=1}^N$, can we find by an iterative process embed them in $Z_N:=\underset{n=1}{ \overset{N}{\bigsqcup} }X_n$ such that their differnces from one another in the Gromov-Hausdorff distance agree with their Hausdorff distance of subsets of $Z_N$?
- If 1 is true for every $N$, can we take the direct limits of the metric spaces, obtain a metric space $Z$ in which all embedding agree with the Gromov Hausdorff distance?
- Can we take the completion of $Z$ and then ensure that $X_\infty$ be isometric to a subset of $Z$?