Give two metric space $X,Y$, we denote by $d_{GH}(X,Y)$ the Gromov- Haudorff distance. It seems that it only relies on $X,Y$ not on the ambient metric space $Z$ where we embed $X,Y$ (because in the notation, there is no $Z$). How can we prove this is well-defined?
2026-03-26 02:57:48.1774493868
Does Gromov-Hausdorff distance depends on the choice of the "ambient" metric space?
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$d_{GH}(X, Y)$ is computed by looking at all possible "ambient spaces" $Z$. The phrasings one often sees, like
can make it sound like there is a "special" $Z$. But rather, we're looking at all pairs of embeddings into all (possible) spaces.
EDIT: Specifically, here's a clearer way to define $d_{GH}(X, Y)$.
In particular, we look at all possible embeddings of $X$ and $Y$ into all possible spaces. There is no single $Z$ we're looking at!
Incidentally, it's not hard to prove - and fairly useful - that we can restrict attention to those $Z$s whose underlying set is $X\sqcup Y$ and whose metric restricted to $X$ (resp. $Y$) is just the original $X$-metric (resp. $Y$-metric).
Again, that is:
Note that this is made much clearer if we distinguish between the set $A$ and the metric space $\mathcal{A}=(A,d_A)$. Namely, we have:
Again, though, we're looking at all possible metrics on $X\sqcup Y$ which restrict to $d_X$ and $d_Y$ on $X$ and $Y$, respectively.