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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.

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

1 Answer 1

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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.

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  • $\begingroup$ I need an approximation method to jump to the general conclusion. $\endgroup$
    – K. Sadri
    Jan 20, 2020 at 6:37

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