Show the relationship between a compact(non empty) set $A$ and a sequence $A_j$ in $R^n$: Let $A_j$ $(j\in N)$ be a sequence of non-empty compact subsets of $R^n$, and $A$ is a non-empty compact set, when $A_j$ converges in the Hausdorff metric to $A$, I need to show that:
$$
A=\overline {A_1 \cup A_2 \cup \cdots} \cap \overline{A_2 \cup A_3 \cup \cdots} \cap \overline {A_3 \cup A_4 \cup \cdots} \ldots
$$
Have no clue about this, please give me some details about how to prove it. Thanks a lot！
 A: HINT: Let $x\in A$ and $\epsilon>0$; there is an $m\in\Bbb Z^+$ such that $d_H(A_n,A)<\epsilon$ whenever $n\ge m_\epsilon$. Suppose that $n\ge m_\epsilon$. Since
$$d_H(A_n,A)=\max\left\{\sup_{y\in A_n}\inf_{z\in A}\|y-z\|,\sup_{z\in A}\inf_{y\in A_n}\|y-z\|\right\}\;,$$
we have 
$$\sup_{y\in A_n}\|y-x\|\le d_H(A_n,A)<\epsilon\;.$$
$A_n$ is compact, so there is a $y_{n,\epsilon}\in A_n$ such that $\|y_{n,\epsilon}-x\|=\sup_{y\in A_n}\|y-x\|<\epsilon$. 


*

*Use this to show that for each $\epsilon>0$ and $m\in\Bbb Z^+$ there is a $z_{\epsilon,m}\in\bigcup_{n\ge m}A_n$ such that $\|z_{\epsilon,m}-x\|<\epsilon$.  

*Deduce that $x\in\operatorname{cl}\bigcup_{n\ge m}A_n$ for each $m\in\Bbb Z^+$.  

*Conclude that $A\subseteq\bigcap_{m\in\Bbb Z^+}\operatorname{cl}\bigcup_{n\ge m}A_n$.


This gives you half of the desired result. To prove the opposite inclusion, I suggest starting with an arbitrary $x\in\Bbb R^n\setminus A$ and showing that $x\notin\bigcap_{m\in\Bbb Z^+}\operatorname{cl}\bigcup_{n\ge m}A_n$. You can do this by showing that if $m$ is large enough, $x\notin\operatorname{cl}\bigcup_{n\ge m}A_n$; you’ll want to use the fact that $\inf_{y\in A}\|x-y\|>0$.
