Hausdorff metric and compactness

compactnessmetric-spacesreal-analysis

Exercise 7.4.15 (Introduction to Real Analysis by Jiri Lebl): Let $X$ be a metric space and $C \subset \mathscr{P} (X)$ the set of nonempty compact subsets of $X$. Using the Hausdorff metric from Exercise 7.1.8, show that $(C, d_H)$ is a metric space. That is, show that if $L$ and $K$ are nonempty compact subsets then $d_H(L, K) =0$ if and only if $L=K$.

I know from Exercise 7.1.8 that the Hausdorff metric is a pesudo metric because $d_H(X,Y)$ can be zero for $X \not= Y$. For example, in Wikipedia it shows that $d_H(\{3,6\}, \{1,3,6,7\}) =0$. I also know that if $L$ is compact, every sequence in $L$ has a subsequence convergent in $L$. But, I do not know how to approach this question. I appreciate if you give some help.

Best Answer

Let $L, K \subset X$ be nonempty compact subsets.

Suppose $d_H(L, K)=0$. Then, by definition of $d_H$, we have $\sup_{x \in L} d(x, K)=\sup_{x \in K} d(x, L)=0$. Since $K$ is closed, $d(x, K)=0 \iff x \in K$, for all $x$, in particular, $x \in L$. So $L \subset K$. Symmetrically, $K \subset L$, too, so $K=L$.

The converse should be clear enough. Otherwise, let me know.

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