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 $$\gamma:=\{x\colon d(x,A)=d_h(B,C)+d_h(A,B)<d_h(A,C)\}.$$ 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 $$d(x,x_1)<\varepsilon/2+\varepsilon/2^2+\dots=\varepsilon.$$

  • $\begingroup$ 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 $\bigcup\limits_{k=n}^\infty C_k$. $\endgroup$ – Bach Jun 14 at 15:40
  • $\begingroup$ I've expanded that paragraph. $\endgroup$ – user10354138 Jun 14 at 16:08
  • $\begingroup$ 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 $X\not \in K(X) .$ $\endgroup$ – DanielWainfleet Jun 15 at 2:48

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