# Proving that a certain space of compact sets is a complete metric space

Let $$(X,d)$$ be a metric space, and let $$K(X)$$ denote the collection of all non-empty compact subsets of $$X$$. Define a function, $$d_h\colon K(x)\times K(x)\to\mathbb R$$ by letting $$d_h(A,B)=\inf\{\varepsilon\colon A\subseteq U_\varepsilon(B) \text{ and } B\subseteq U_\varepsilon(A)\},$$ where $$U_\varepsilon(S)=\{x\in X\colon d(x, S)<\varepsilon\}$$, and the distance $$d(x, S)$$ from a point $$x\in X$$ to a non-empty subset $$S$$ of $$X$$ is defined by $$d(x,S)=\inf\{d(x,s)\colon s\in S\}$$.

Prove that $$(K(X), d_h)$$ is a complete metric space.

My attempt:

First, we need to show that $$d_h$$ is a metric. It is trivial to show that $$d_h\ge 0$$ and $$d_h=0$$ if and only if $$d_h=0$$. Also, it is clear that $$d_h(A,B)=d_h(B,A)$$.

To see that for every $$A, B, C\in K(X)$$, we have $$d_h(A,C)\le d_h(A,B)+d_h(B,C)$$, we consider $$\gamma_{A}:=\{x\colon d(x,A)=d_h(A,B)\}$$ and $$\gamma_{B}:=\{x\colon d(x,B)=d_h(B,C)\}$$.

If $$d_h(A,C)\le d_h(A,B)$$ or $$d_h(A,C)\le d_h(B,C)$$ then we are through, otherwise we claim that $$d_h(B,C)\ge d_h(A,C)-d_h(A,B).$$ Suppose the converse, $$\gamma_B$$ would be completely lying in the interior of $$\gamma_A$$. By the definition of $$d_h(B,C)$$, $$C$$ would be lying in $$U_{d_{h}(B,C)}$$ and in the interior of $$\gamma:=\{x\colon d(x,A)=d_h(B,C)+d_h(A,B) Interchanging the roles played by $$A$$ and $$C$$, we will have a contradiction with the definition of $$d_h(A,C)$$. Hence $$d_h$$ is a metric.

However, I got stuck in proving completeness. Suppose we have a sequence $$\{C_n\}_{n=1}^\infty$$ with $$C_n\in K(X)$$ and $$d_h(C_n,C_m)\to 0$$ as $$n,m\to\infty$$, I failed to find out the limit set. I think it should be $$\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty C_k$$ but I am not even sure whether it is compact or not. Any help here? Thank you!

Edit: As user10354138 pointed out, we should assume $$X$$ is a complete metric space.

You need to assume $$X$$ is complete, otherwise any Cauchy sequence $$x_n\in X$$ not converging to a limit gives you a corresponding sequence $$\{x_n\}\in K(X)$$ that does not converge to a limit.
After that, you need to stick a closure (again, see the Cauchy sequence example) $$C_\infty=\bigcap_{n=1}^\infty \overline{\bigcup_{k=n}^\infty C_k}$$ The first nonobvious point is $$\overline{\bigcup_{k\geq n}C_k}$$ is compact: it is a closed subset of the complete space $$X$$, so it suffices to show total boundedness. So we need to show we can cover with finitely many $$2\varepsilon$$-balls, which using $$(C_k)$$ is Cauchy we reduce to covering the finitely many initial $$C_k$$ by $$2\varepsilon$$-balls, plus covering $$C_m$$ with $$\varepsilon$$-balls where all $$C_n$$s, $$n>m$$, lie within $$\varepsilon/2$$ of $$C_m$$. Now enlarge the $$\varepsilon$$-balls to $$2\varepsilon$$-balls and we covered the lot.
So $$C_\infty$$ is nested intersection of compacts so is nonempty compact. We also know $$d_h(C_n,\overline{\bigcup_{k\geq m} C_k}) \leq\sup_{k\geq m} d_h(C_n, C_k)\xrightarrow{\min(m,n)\to\infty} 0$$ So we may assume, by replacing $$C_n$$ with $$\overline{\bigcup_{k\geq n} C_k}$$, that $$(C_n)$$ are nested. In this case the proof of $$d_h(C_n,C_\infty)\to 0$$ just need to bound $$\sup_{x\in C_n} d(x,C_\infty)$$ (the $$\sup_{x\in C_\infty} d(x,C_n)=0$$ come from nested): for $$\varepsilon>0$$, pick sequence $$n_k$$ such that $$d_h(C_m,C_n)<\varepsilon/2^k$$ for all $$m,n\geq n_k$$. Then every point $$x_1\in C_{n_1}$$ is less than distance $$\varepsilon/2$$ from a point $$x_2\in C_{n_2}$$, which in turn is less than distance $$\varepsilon/2^2$$ from $$x_3\in C_{n_3}$$, etc. and $$x_n$$ is Cauchy, so $$x_n\to x\in C_\infty$$, with $$d(x,x_1)<\varepsilon/2+\varepsilon/2^2+\dots=\varepsilon.$$
• Thank you! But still, I am confused about the last paragraph. Can you clarify a little bit why we have $d_h(C_\infty, C_n)\to 0$? Can we prove this rigorously? Especially now we are required to take the closure of $\bigcup\limits_{k=n}^\infty C_k$. – Bach Jun 14 at 15:40
• An example where $d$ is not a complete metric and $d_h$ is also not complete: Let $X=(0,1]$ with $d(x,y)=|x-y|.$ Let $A_n=[1/n,1]$ for $n\in \Bbb N.$ Then $(A_n)_n$ is a $d_h$- Cauchy sequence. Suppose it had a limit $B\in K(X).$ Suppose $x\in X$ \ $B$. Now $B$ is compact so $B$ is closed in $\Bbb R.$ So for some $r\in (0,x)$ the set $(x-r,x+r)$ is disjoint from $B$. But then $d_h(B,A_n)\ge (x-r)$ whenever $1/n<(x-r),$ contrary to $A_n\to B.$ Therefore by contradiction we have $B=X$. This is absurd because $X\not \in K(X) .$ – DanielWainfleet Jun 15 at 2:48