3
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $$d(x,x_1)<\varepsilon/2+\varepsilon/2^2+\dots=\varepsilon.$$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.