# If a metric space is compact then the set of compact subsets is compact

I am working some exercises from Gerald Edgar's book Measure, Topology, and Fractal Geometry. In section 2.5, which discusses the Hausdorff metric, exercise $$2.5.4$$ asks:

Under what conditions on $$S$$ is $$\mathbb{H}(S)$$ compact, where $$\mathbb{H}(S)=\{K\subseteq S: K \text{ compact}\}$$?

I suspect that for this to hold, $$S$$ must be itself compact. Consider the set of real numbers $$S=\mathbb{R}$$ which is closed but not bounded, and consider the sequence $$([-n,n])_{n\in\mathbb{N}}$$. There is no convergent subsequence (under the Hausdorff metric) since $$\mathbb{R}$$ is unbounded.

So instead let's consider a bounded subset of $$\mathbb{R}$$, say $$S=(-M, M)$$. Now consider the sequence $$([-M+1/n, M-1/n])_{n\in\mathbb{N}}$$. This is again an increasing sequence in $$\mathbb{H}(S)$$, but has no convergent subsequence in $$S$$ since $$S$$ is not closed.

But if we consider the set $$S=[-M,M]$$, I suspect that there must be a convergent subsequence for any sequence of compact subsets of $$S$$. From this I suspect that:

$$S$$ compact $$\implies\mathbb{H}(S)$$ is compact

I don't know where to start to prove this though, if this is after all true.

• Compact in what sense? Is there some topology on the powerset of $S$? Oct 24, 2019 at 11:43
• That wasn't specified, so I'm assuming under any topology Oct 24, 2019 at 11:45
• The discrete topology is never compact on an infinite set, so that seems wrong. Oct 24, 2019 at 11:47
• Section 2.5. is about Hausdorff-metric on sets. That should be mentioned in your question. Oct 24, 2019 at 11:55
• Question updated accordingly, thanks Oct 24, 2019 at 11:58

Part 0, infinitely many empty sets.

If a sequence of subsets of $$X$$ has infinitely many empty members, the empty set is a limiting point. From now on, we will be assuming that the sequence has no empty members.

Part I, Finite subsets.

Proposition a: singletons do have a limit point. Proof: Let $$\{\omega_n\}$$ be a series of singleton sets. Let $$\omega$$ be a limit point for the sequence $$\omega_n$$. This exists since $$X$$ is compact. Then the corresponding sequence of singleton sets, converges to $$\{\omega\}$$. $$\blacksquare$$

Proposition b: If all sequences of subsets of $$X$$ with $$n$$ elements, have a limit point, then so do all sequences of $$n+1$$ element subsets. Proof: For a sequence $$A_i$$ with $$|A_i|=n+1$$, construct the sequence $$B_i=A_i\backslash\{a_i\}$$ for some $$a_i\in A_i$$. (Am i using the axiom of choice here?) The sequence $$B_i$$ has a convergent subsequence $$B_{i_k}$$ with limit $$B$$. The sequence $$a_{i_k}$$ also has a convergent subsequence $$a_{i_{k_l}}$$ with limit $$a$$. This mean $$A_i$$ has the convergent subsequence $$A_{i_{k_l}}$$ with the limit $$B\cup\{a\}$$. $$\blacksquare$$

Proposition c (already proven by induction, not quite but it is so easy to fix that I won't mention the gap): sequences of finite subsets, do have a convergent subsequence.

Part II, Approximation)

Proposition d: For any positive $$\epsilon$$, there exists a finite subset of $$X$$, $$C_\epsilon$$ such that $$\cup_{\omega\in C_\epsilon}B_\epsilon(\omega)=X$$

Proof: Let's assume this is not true, start by a singleton set $$\{\omega_1\}\in X$$. There is at least one point that is at least $$\epsilon$$-far from it. Call it $$\omega_2$$ and add it to the set. This process should never end. The sequence $$\omega_n$$ has no limit point. Contradiction. $$\blacksquare$$

Define the metric (This is not a metric actually, but let us call it one) $$d_\epsilon(A, B)\equiv d_{Hausdorff}(A, B)\times \mathbb{I}[d_{Hausdorff}(A, B)>\epsilon]$$

Proposition d: For any $$\epsilon>0$$, the power set of $$X$$ is compact wrt $$d_\epsilon$$. Proof: For any sequence of sets $$A_n$$, take $$\hat{A}_n$$ to be the finite $$\epsilon$$ approximation of this. Clearly $$d_{Hausdorff}(A_n , \hat{A}_n)\leq\epsilon$$. The finite set sequence has a limit point like $$\hat{A}$$. This is also a limit point for the $$A_n$$ in the $$d_\epsilon$$ sense. $$\blacksquare$$

Part III, the general thing

Proposition e: all sequences of sets have a convergent subsequence.

Proof: Let $$A_n$$ denote the sequence. It has a subsequence that converges under $$d_1$$. Call it $$A^0_n$$. This has a subsequence that converges under $$d_{1/2}$$. Call it $$A^1_n$$. This has a subsequence that converges under $$d_{1/4}$$. Call it $$A^2_n$$. $$\cdots.$$

The subsequence $$A^n_n$$ is convergent under $$d_0=d_{Hausdorff}$$.

$$\blacksquare$$

PS: I wrote this in great hurry and I'm a physicist. So I guess you can find gaps in my proofs. I hope you can fill them up yourself.

• I need an approximation method to jump to the general conclusion. Jan 20, 2020 at 6:37