Let be $S$ a metric space. We define the hyperspace $H(S)$ as the metric spaces consisting of every no empty compact subset of $S$ and the Hausdorff metric.
I want that $H(S)$ be compact imposing condition on $S$. But I get a little confuse with the conditions, I mean if we suppose that $S$ is compact I do not figure out what to do with that condition.
I believe that the easiest way is taking a infinite sequence $A_1, A_2,..., A_n,$ in $H(S)$-we can suppose that every element in the sequence is different from the others- and try to find a convergent sub sequence. But I do not fin anything.
Any hint?
Of course $H(S))$ had better be the class of nonempty compact subsets of $S$...
Say $S$ is compact. Two proofs:
First proof: Since $S$ is compact, $S$ is totally bounded: Let $\epsilon>0$. There exist $x_1,\dots,x_N\in S$ with $$S=\bigcup_{n=1}^NB(x_n,\epsilon).$$
Now for every $F\subset\{1,\dots\,N\}$ let $H_F$ be the set of all $K\in H(S)$ such that $K\cap B(x_n,\epsilon)\ne\emptyset$ for all $n\in F$ while $K\cap B(x_n,\epsilon)=\emptyset$ for all $n\notin F$. We certainly have $$H(S)=\bigcup_F H_F,$$and you can show that if $K_1,K_2\in H_F$ then the Hausdorff distance from $K_1$ to $K_2$ is less than $\epsilon$. So $H(S)$ is totally bounded.
And $H(S)$ is complete. Leaving out a lot of details: Say $(K_n)$ is a Cauchy sequence in $H(S)$. Let $K$ be the set of all $x\in S$ such that there exist $k_j\in K_j$ with $k_j\to x$. Then (details, details) you can show that $K_n\to K$.
So $H(S)$ is compact, being complete and totally bounded.
Second proof: This one is more fun. For $K\in H(S)$ define $f_K\in C(K)$ by $$f_K(x)=d(x,K)=\min_{y\in K}d(x,y).$$
Take the standard metric on $C(K)$, defined by $d(f,g)=\max_{x\in S}|f(x)-y(x)|$. You can show that $$d(K_1,K_2)=d(f_{K_1},f_{K_2}).$$
So by Arzela-Ascoli it's enough to show that the set of all $f_K$ is closed, uniformly bounded, end equicontinuous. The last two conditions are trivial. To show that set of all $f_K$ is closed: Say $f_{K_n}\to f$ uniformly. Let $K$ be the set of $x$ where $f(x)=0$. Show that $f=f_K$.