Realizing and embeddability of Gromov Hausdorff convergence

Question

Assume $X$, $Y$ are compact spaces in $\mathbb{R}^n$.

Then does it follow that $d_{\mathrm{GH}}(X,Y) = d_{\mathrm{H}}(X,Y)$?

Clearly, $d_{\mathrm{GH}}(X,Y) \leq d_{\mathrm{H}}(X,Y)$.

However, it seems intuitive to argue that the reverse holds because both live in $\mathbb{R}^n$.

I was wondering moreover if what if $X$, $Y$ are compact metric spaces that can be embedded in an isometric way into $\mathbb{R}^n$. Then does it follow that their Gromov–Hausdorff distance equals their Hausdorff distance (images of spaces identified with isometric copies)?

Answer 1

Let $X$ be any non-empty compact subspace of $ℝ^n$, and let $Y$ be a properly translated version of $X$, i.e., let $Y = X + v$ for some non-zero vector $v$ in $ℝ^n$. The Hausdorff distance of $X$ and $Y$ is non-zero, because $X$ and $Y$ are two distinct non-empty, compact subsets of $ℝ^n$. But $X$ and $Y$ are isometric, whence their Gromov–Hausdorff distance is zero.

Answer 2

Given two isometric spaces we have $d_{GH}(X,Y)=0$.

But for compact (even closed only) subsets of $\mathbb{R}^n$ we have $d_{H}(X,Y)=0$ if and only if $X=Y$. Literally, not only up to isometry.

A concrete counterexample is $X=[0,1]$ and $Y=[1,2]$.

So the answer is: no, these are not equal in general.