Definition of Gromov-Hausdorff distance

Question

I am working with the following definition of the Gromov-Hausdorff distance for two metric spaces $X$ and $Y$:

$d_{G-H}(X, Y) = \inf\{d_H(X, Y) : d \text{ is admissible on } X \sqcup Y \}$, where admissible on the disjoint union $X \sqcup Y$ means that $d$ extends the metrics of $X$ and $Y$.

I read the following example with the independent metric spaces $X =[0,1]$ and $Y = [1,2]$. $d_{G-H}(X, Y) = 0$, as they are isometric. But if I consider $X \sqcup Y$ = $\{(0, x), (1, y) : x \in X, y \in Y\}$, then according to this metric we would have, for example, $d((0,0), (1,1)) = 0$ right? But then this would not be a metric because $(0,0) \neq (1,1)$?

Answer 1

I think the $inf$ in the definition is not reachable by a distance but you can make a sequence of distance $d^n$ such that $d^n_H(X,Y) \to 0$.

Take $d_n(a,b)$ in the following way:

$$ d_n(a,b)=\left\{\begin{array}{ll} |a-b|, & a \in X , b\in X \text{ or } a \in Y , b\in Y\\ |a-(b-1)| + \frac{1}{n}, & a \in X, b \in Y\\ |(a-1)-b| + \frac{1}{n}, & a \in Y, b \in X\end{array}\right. $$

This is an admissible distance and $d^n_H(X,Y)= \frac{1}{n}$, taking $n \to 0$ we have that $d_{G-H}(X,Y)=0$.