Let $X$ be a complete metric space. Define $H$ to be the set of non-empty compact subsets of $X$. Now let $A_n$ be a cauchy sequence in $H$. (Considering the Hausdorff metric on $H$). Then can it be proved that $\cup A_n$ is compact?
I came across this question in an attempt to prove that H is complete. So I'd like a proof that does not use the completeness of H, if possible.
This is not true. Let $X=\mathbb{R}$ and $A_n=\{1/n\}$. This is a Cauchy sequence (it converges to $\{0\}$ in the Hausdorff metric) but $\bigcup_n A_n$ is not compact.
However, it is true that $\overline{\bigcup_n A_n}$ is compact. Since it is a closed subset of a complete space, it is also complete. So it remains to prove that it's totally bounded. And for that it suffices to show that the union $A = \bigcup_n A_n $ is totally bounded. For every $\epsilon>0$ there is $N$ such that $A$ is contained in the $\epsilon/2$-neighborhood of the set $B_N=\bigcup_{n=1}^N A_n $. Since $B_N$ is compact, it has a finite $\epsilon/2$-net. This gives a finite $\epsilon$-net for $A$.