# The Class of Non-empty Compact Subsets of a Compact Metric Space is Compact

This is a question from my homework for a real analysis course. Please hint only.

Let $M$ be a compact metric space. Let $\mathbb{K}$ be the class of non-empty compact subsets of $M$. The $r$-neighbourhood of $A \in \mathbb{K}$ is

$$M_r A = \lbrace x \in M : \exists a \in A \text{ and } d(x,a) < r \rbrace = \bigcup_{a \in A} M_r a.$$

For $A$, $B \in \mathbb{K}$ define

$$D(A,B) = \inf \lbrace r > 0 : A \subset M_r B \text{ and } B \subset M_r A \rbrace.$$

Show that $\mathbb{K}$ is compact.

(Another part of the question is that if $M$ is connected, then so is $\mathbb{K}$, but this is not assigned).

Many thanks.

• You need to show that every open cover of $\mathbb{K}$ has a finite subcover. So it seems that $M_{r}(A)$ is an open cover of $A$. – PrimeNumber Jan 26 '11 at 16:33

## 3 Answers

It is probably easier to appeal to the Heine-Borel Theorem

Step (1): Show that $D$ is a metric on $\mathbb{K}$.

Now that $(\mathbb{K},D)$ is a metric space, by the Heine-Borel theorem, it suffices to show that it is complete and totally bounded.

Step (2): Let $A_i$ be a Cauchy sequence in $\mathbb{K}$, show that it converges. (In fact, you can show that as long as $M$ is a complete metric space, then so is $\mathbb{K}$ with the metric you just wrote down.)

Step (3): Show that for every $\epsilon > 0$ there exists a finite open cover of $\mathbb{K}$ by $\epsilon$-balls. (In fact, as long as $M$ is totally bounded, so will $\mathbb{K}$.)

Step (1) is obvious, so I won't give a hint.

For Step (2), let $A_i$ be your Cauchy sequence. Consider the set $A$ of points $a$ such that for any $\epsilon > 0$, $B(a,\epsilon)$ intersects all but finitely many $A_i$.

For Step (3), start with a finite cover of $M$ by $\epsilon$ balls, let $S$ be the set of the center points of those balls. Consider the power-set $P(S)$ (the set of all subsets of $S$) as a set of points in $\mathbb{K}$.

• What was your last sentence? "Consider the power-set $P(S)$ as a set of points in $M$"? – PrimeNumber Jan 26 '11 at 18:44
• @PEV: I meant what I wrote. The elements of $P(S)$ are sets consisting of finitely many points in $M$, and hence are compact sets, and hence they are points in $\mathbb{K}$. – Willie Wong Jan 26 '11 at 20:51
• I see. I think there there is a word that the end. – PrimeNumber Jan 26 '11 at 20:53
• @PEV: I see. The "consider..." is a statement in imperative mood: it is the hint. – Willie Wong Jan 26 '11 at 23:20
• Thanks a lot Willie! This is a much better anwser then I expected to get! Many thanks! – milcak Jan 27 '11 at 5:32

Hints:

1. It suffices to show that $\mathbb{K}$ is complete and totally bounded.
2. For every $\varepsilon > 0$ there is a constant $C(\varepsilon)$ such that every $K \in \mathbb{K}$ can be covered by at most $C(\varepsilon)$ balls of radius $\varepsilon$ around points in $K$. To see this, cover $M$ with finitely many balls of radius $\varepsilon/2$, let $K \in \mathbb{K}$ be arbitrary and pick $x_{n}$ in the intersection of $K$ with those balls that intersect $K$. The balls with radius $\varepsilon$ around these balls will cover $K$.
3. Given a Cauchy sequence in $\mathbb{K}$, show that it converges (this doesn't need compactness!).
• ah, the problem with giving "hints". Between your hint 2 and my hint 3 we have basically the entire proof of the proposition. :) – Willie Wong Jan 27 '11 at 12:18

As an aside (I do not suggest you prove it this way): this $\mathbb{K}$ is just the so-called hyperspace of $M$, also denoted $H(M)$ (all non-empty closed sets of $M$ in general), and an alternative, non-metric description of its topology is by describing a subbase for it, consisting of all $[U] = \{A \in \mathbb{K}:\, A \cap U \neq \emptyset \}$ and $<U> = \{ A \in \mathbb{K}:\, A \subset U \}$, where $U$ ranges over all non-empty subsets of $M$. If $M$ is compact so is $H(M)$, by a simple application of the Alexander subbase lemma. This is a more general topological way of viewing this space (it's a theorem that for compact metric spaces $M$ the space $H(M)$ is compact metrizable as well, and one of the metrics is the one descibed in the question).

• Thanks for the anwser! I haven't been introduced to these topological concepts before, I'll be sure to read up. – milcak Jan 27 '11 at 5:34