Hausdorff distance and isometry

Question

Let $(M,d)$ be a metric space and $\Phi (M) $ be the collection of all bounded and closed subsets $X \subset M$.

For all $X,Y \in \Phi(M)$, we define their Hausdorff distance by:

$$ \rho (X,Y) = \max \{ \underset{x \in X}{\sup }d(x,Y),\underset{y \in Y}{\sup } d(y,X) \} $$

Consider the function $\xi : \Phi(M) \rightarrow \mathcal{F}(M;\mathbb{R})$, given by: $\xi (X) = d_X$ and $d_X(z) = d(z,X)$, $\forall z\in M$.

1) Show that the distance between $d_X$ and $d_Y$ is $< +\infty$

2) Show that $\xi$ is an isometry.

My attempt:

1)

I think I should consider $\mathcal{F}(M;\mathbb{R})$ with the metric $\beta (f,g) = \underset{z \in M}{\sup }|f(z)-g(z)|,$ if $f$ and $g$ are bounded functions.

So, I proved the following:

$|d_X(z) - d_Y(z)| \leq \sup \{d(x,y): x \in X, y \in Y\} = \alpha(X,Y)$.

We have $\alpha(X,Y) < + \infty$, since $X$ and $Y$ bounded $\Rightarrow$ $X \cup Y$ bounded.

Then we can take the sup and we have $\beta (d_X,d_Y)< + \infty$, thus the distance between $d_X$ and $d_Y$ is $< +\infty$.

Am I right? I'm not sure because we have that $|d_X-d_Y|$ is bounded but $d_X$ and $d_Y$ are not necessarily bounded, right? Doesn't it cause any problem?

2) I have no idea to prove it using the metric $\beta$.

Is this statement true with the metric $\beta$?

How could I prove it?

Thanks in advance.

Answer 1

1. I would consider in $\mathcal{F}(M,\mathbb{R})$ the supremum norm instead of the metric $\beta$ you defined. That is, for every bounded function $f$, I would consider $\Vert f \Vert_{\infty}= \sup_{z\in M} \vert f(z)\vert$. With this, what you did is better justified (since $d_X-d_Y$ is bounded) and you avoid the problems of using the metric $\beta$ with functions that are not bounded.

2. This can be proved using the norm $\Vert \cdot \Vert_{\infty}$. We have to prove $$ \Vert\xi(X)-\xi(Y)\Vert_{\infty}=\Vert d_X-d_Y\Vert_{\infty}=\rho(X,Y) $$ We can show, using triangle inequality, for all $z\in M$, and all $(x,y)\in X\times Y$, $$ \vert d(z,x)-d(z,y)\vert\leq d(x,y)\leq \rho(X,Y) $$ Taking sequences $(x_n)\subset X$ and $(y_n)\subset Y$ s.t. $d(x_n,z)\to d_X(z)$ and $d(y_n,z)\to d_Y(z)$, we deduce, for all $z$, $$ \vert d_X(z)-d_Y(z)\vert\leq\rho(X,Y) $$ which implies $$ \Vert d_X-d_Y\Vert_{\infty}\leq\rho(X,Y) $$ For the other inequality, notice, for all $(x,y)\in X\times Y$, $$ d(x,Y)=\vert d(x,X)-d(x,Y)\vert \leq \Vert d_X-d_Y\Vert_{\infty}\\ d(y,X)=\vert d(y,X)-d(y,Y)\vert \leq \Vert d_X-d_Y\Vert_{\infty} $$ which implies $$ \sup_{x}d(x,Y)\leq \Vert d_X-d_Y\Vert_{\infty}\\ \sup_{y}d(y,X)\leq \Vert d_X-d_Y\Vert_{\infty} $$ which implies $$ \rho(X,Y)\leq \Vert d_X-d_Y\Vert_{\infty} $$