Let $(E, d)$ be a metric space and $\mathcal C$ the set of all non-empty compact subsets of $E$. We define a metric $d_H$ on $\mathcal C$ by $$ d_H (A, B) := \max \left \{ \max_{x \in B} d(x, A), \max_{y \in A} d (y, B) \right \}\quad \forall A,B \in \mathcal C, $$ with $d(x, A) := \inf_{y \in A} d(x, y)$. I'm trying to prove below property which I have seen from this Wikipedia page.
Theorem: If $E$ is totally bounded, then so is $\mathcal C$.
Could you please have a check on my attempt?
My attempt: Fix $\varepsilon > 0$. Because $E$ is totally bounded, there are $x_1, x_2, \ldots, x_m \in E$ such that $$ E = \bigcup_{i=1}^m B(x_i, \varepsilon). $$
Let $\mathcal D$ be the collections of all non-empty subsets of $X := \{x_1, \ldots, x_m \}$. Clearly, $\mathcal D$ is finite and $\mathcal D \subset \mathcal C$. We pick a map $\varphi:E \to X$ such that $d(x, \varphi (x)) < \varepsilon$ for all $x \in E$.
Fix $A \in \mathcal C$ and let $C := \{ \varphi (a) \mid a \in A\} \in \mathcal D$. We pick a "reverse" map $\psi:C \to A$ such that $d(c, \psi (c)) < \varepsilon$ for all $c \in C$. We have
- $$\max_{a \in A} d(a, C) = \max_{a \in A} \min_{c \in C} d(a,c) \le \max_{a \in A} d(a, \varphi (a)) < \varepsilon.$$
- $$\max_{c \in C} d(c, A) = \max_{c \in C} \min_{a \in A} d(c, a) \le \max_{c \in C} d(c, \psi (c)) < \varepsilon.$$
It follows that $$ d_H (A, C) = \max \left \{ \max_{a \in A} d(a, C), \max_{c \in C} d(c, A) \right \} < \varepsilon. $$
This completes the proof.
Remark: In a previous thread, I proved that
Theorem: If $E$ is complete, then so is $\mathcal C$.
By this characterization of compactness in metric spaces, we obtain the following corollary.
Corollary: If $E$ is compact, then so is $\mathcal C$.
