Compact sets and Kuratowski limit

Question

I have been struggling with the following claim:

Let $A_n$ be a sequence of compact sets and $A$ a compact set. $A=\lim\sup_n A_n=\lim\inf_n A_n$ iff $d_H(A_n,A)\to 0$ where $d_H(.,.)$ is the Hausdorff metric.

$\lim\inf$ and $\lim\sup$ are defined by $\lim\inf_nA_n=\left\{y\in Y:\forall \varepsilon>0,\exists N, n\geq N\quad \implies B_{\varepsilon}(y)\cap A_n\neq\emptyset\quad \right\}$ and, $\lim\sup_n A_n=\left\{y\in Y: \forall\varepsilon>0,\forall N,\exists n\geq N \text{ so that } B_\epsilon(y)\cap A_n\neq\emptyset \right\}$.

Also $d_H (X,Y)=\max\left(\inf\left\{\epsilon>0:Y\subset\bigcup_{x\in X}B_\epsilon(x) \right\}, \inf\left\{\epsilon>0:X\subset\bigcup_{y\in Y}B_\epsilon(y) \right\} \right)$.

$\Rightarrow$: Let $\varepsilon>0$ be given. For each $a\in A$ $\exists N(a,\varepsilon)$ so that $n\geq N$ implies $B_\varepsilon(a)\cap A_n\neq\emptyset$. By compactness of $A$ we can find finite $a_1,a_2\dots a_k$ and associated $N_1,N_2\dots N_k$ so that $A\subset\cup_{i=1}^k B_\varepsilon(a_i)$. Let $N=\max_{1\leq i\leq k}N_i$. I want to show $n\geq N$ implies $A_n\subset A^\varepsilon$ but couldn't manage.

$\Leftarrow$: I want to show $A\subset\lim\inf A_n\subset\lim\sup A_n\subset A$. Let $\varepsilon>0$ be given. There exists $N$ such that $d_H(A_n,A)<\varepsilon$ for $n$ large enough. Pick $a\in A$, so $a\in A_n^{\varepsilon}$ which implies $B_{\varepsilon}(a)\cap A_n\neq\emptyset$ for all such large $n$'s. Thus $a\in\lim\inf A_n$. To finish this side, I need to show $\lim\sup A_n\subset A$ yet again couldn't find the solution. Thanks for any help!

Answer 1

For the direction

$$A = \limsup_n A_n = \liminf_n A_n \implies d_H(A_n,A) \to 0,$$

you need some additional hypothesis on the ambient space, namely that it is compact.

Without that hypothesis, the implication does not hold. An explicit counterexample using compact subsets of $\mathbb{R}$ is $A = [0,1]$, and $A_n = [0,1] \cup \{n\}$ for $n\in \mathbb{N}$. Generally, if the ambient (metric) space is not compact, it contains a sequence $(x_n)_{n\in \mathbb{N}}$ without accumulation point, and then $A = \{x_0\}$ and $A_n = \{x_0,x_n\}$ is a counterexample.

So let's suppose that the ambient space is compact.

What you have so far proves that for large enough $n$, you have $A \subset A_n^{2\varepsilon}$. For the converse inclusion, that $A_n \subset A^\varepsilon$ for all large enough $n$, you use the compactness of the ambient space. Suppose $A_n \setminus A^\varepsilon \neq\varnothing$ for infinitely many $n$. Then

$$F_n = \overline{\bigcup_{k=n}^\infty A_k\setminus A^\varepsilon}$$

is a decreasing sequence of nonempty compact sets, hence $F = \bigcap\limits_{n\in\mathbb{N}} F_n \neq \varnothing$. Now, what can be said about the points in $F$?

For the other direction,

$$d_H(A_n,A) \to 0 \implies A = \limsup_n A_n = \liminf_n A_n,$$

you have correctly shown that $A \subset \liminf_n A_n$. Now let $x\in \limsup_n A_n$. Fix $\varepsilon > 0$. Choose $N$ such that $d_H(A_n,A) < \varepsilon/2$ for $n \geqslant N$. Since $x \in \limsup_n A_n$, there is an $n_\varepsilon > N$ such that $A_{n_\varepsilon} \cap B_{\varepsilon/2}(x) \neq\varnothing$. But, for $y \in A_{n_\varepsilon}$, we also have $B_{\varepsilon/2}(y) \cap A \neq \varnothing$, and hence $B_\varepsilon(x) \cap A \neq \varnothing$. That means $x \in A^\varepsilon$. Since $\varepsilon$ was arbitrary, it follows that

$$x \in \bigcap_{\varepsilon > 0} A^\varepsilon = A.$$

Answer 2

Let's think about this. Some directions don't work. In space $A=[0,1]$, which is compact, let $A_n = \{ k/2^n : k=0,1,\dots,2^n\}$. Then $d_H(A_n,A) \to 0$. But $\limsup A_n = \liminf A_n = \bigcup A_n$ is not compact at all.