Hausdorff distance: Prove that if $(E,d)$ is complete then, $(\mathcal{K}(E), \mathcal{H})$ is also complete

Question

Let $(E,d)$ be metric space. We denote by $\mathcal{K}(E)$ the set of compact subsets of $E$

that is $$ \mathcal{K}(E)=\{K\subset E: K\text{ is compact in } E\}$$

Prove that if $(E,d)$ is complete then, $(\mathcal{K}(E), \mathcal{H})$ is also complete. where for $A\subset E$ and $B\subset E$ one has $$ \mathcal{H}(A,B)= \max\{\sup_{y\in B}\inf_{x\in A} d(x,y) ,~~~\sup_{x\in A} \inf_{y\in B} d(x,y)\}$$ What about the converse ?

I was only able to prove that $\mathcal H$ is a distance. But for the completeness I don't know how to do it. Please help.

Answer 1

The converse ($\mathcal K(E)$ complete $\implies E$ complete) readily follows from the fact that for any Cauchy sequence $x_n$ in $E$, the sequence of one-point subsets $\{x_n\}$ is Cauchy in $\mathcal K(E)$, and therefore converges. The limit $\lim \{x_n\}$ then turns out to be a one-point set whose sole element is $\lim x_n$.

For the direct implication, I give an "atypical" proof, which may be entertaining. Let $\{A_n\}$ be a Cauchy sequence in $\mathcal K(E)$. As was shown elsewhere, the set $F:=\overline{\bigcup_n A_n}$ is compact, so we may forget $E$ and restrict our attention to $F$.

Next, we'll forget the Cauchy sequence and prove more: $\mathcal K(F)$ is compact, i.e., every sequence has a convergent subsequence. (Recall that compact spaces are complete.)

To every nonempty compact subset $A\subset F$ associate its distance function $d_A(x) = \operatorname{dist}(x, A)$, where $x\in F$. The following facts are easy to check, and are worth knowing anyway:

1. The functions $d_A$ are equicontinuous (they are Lipschitz with constant 1) and are bounded by $\operatorname{diam}F$.
2. The Hausdorff distance between $A, B$ is exactly $\sup_F |d_A-d_B|$.

By the Arzelà-Ascoli theorem, any sequence $d_n=d_{A_n}$ has a uniformly convergent subsequence, say $d_{n_k}\to f$. It remains to show that $f$ is the distance function of its zeros set $f^{-1}(0)$. To this end, note that $$f = \liminf d_n = \sup_{m} \inf_{n\ge m} d_n $$ hence $$ f^{-1}(0) = \bigcap_m \{\inf_{n\ge m} d_n =0 \} = \bigcap_m B_m\quad \text{where } B_m = \overline{\bigcup_{n\ge m} A_n} $$ Being the intersection of nested compact sets, $f^{-1}(0)$ is nonempty. Using the nested compact sets lemma again, one can show that $$ \operatorname{dist}\left( x, \, \bigcap_m B_m\right) = \lim_m \operatorname{dist}(x, B_m) $$ where the right hand side is precisely $\liminf d_m$, which is $f$.

Answer 2

I think you can try following these steps:

1) If $(D_n)\subset \mathcal{K}(E)$ is a decreasing sequence, then $$ \lim_n D_n = \bigcap_n D_n. $$

Now let $(K_n) \subset\mathcal{K}(E)$ be a Cauchy sequence.

2) Since $(K_n)$ is Cauchy, the sets $B_m := \bigcup_{n=m}^\infty K_n$ are totally bounded, hence the sets $D_m := \overline{B_m}$ are compact and the sequence $(D_n)$ is decreasing.

3) By Step 1, $D_n \to D := \bigcap_m D_m$.

4) Use again the assumption that $(K_n)$ is Cauchy to conclude that $K_n \to D$.