I have a sequence of closed, compact sets $\{A_n\}_{n \in \mathbb{N}}$ and $\{B_n\}_{n \in \mathbb{N}}$. I know that both $A_n$ and $B_n$ are decreasing in $n$; i.e.
$n_1 > n_2 \implies A_{n_1}\subseteq A_{n_2} \text{ and } B_{n_1}\subseteq B_{n_2}$
I also know that as $n \rightarrow \infty$, $A_n \rightarrow A$ for some set $A$ and $B_n \rightarrow \emptyset$.
Define the Hausdorff distance as:
\begin{equation}\nonumber
d_H(X,Y) = \inf\{\epsilon \geq 0: X\subseteq (Y)_\epsilon\text{ and }Y\subseteq (X)_\epsilon\}
\end{equation}
where $(Z)_\epsilon$ represents the $\epsilon$-fattening of $Z$.
I want to show the following:
\begin{equation}\nonumber
d_H(A_n\setminus B_n,A) \rightarrow 0 \text{ as } n \rightarrow \infty
\end{equation}
Can this be done? And how?
Best Answer
If $B_n$ goes to a point $p$, then then $ \varepsilon$-ball $B_\varepsilon (p)$ contains $B_n$ and $d_H(A_n,A)<\varepsilon$ for $ n\geq N$ and some $N$.
$ A_n - B_n$ contains $ A_n - B_\varepsilon (p)$ and a closed $\varepsilon$-tubular neighborhood of $A_n - B_\varepsilon (p)$ contains $A_n,\ A_n-B_n$. Hence $ d_H(A_n - B_n,A_n - B_\varepsilon (p)) \leq \varepsilon$.
From triangle inequality $\ast$ of $d_H$, then
\begin{align*} d_H(A_n - B_n, A) & \leq d_H(A_n - B_n,A_n-B_\varepsilon (p)) + d_H( A_n - B_\varepsilon (p),A_n) +d_H(A_n,A) \\&\leq 2\varepsilon + d_H(A_n,A) \\ & \leq 3 \varepsilon \end{align*}
We have a claim $\ast$ that $d_H$ satisfies triangle inequality :
If $d_H(X,Y)=r,\ d_H(Y,Z)=R$, then $(Y)_{r+\epsilon},\ (Y)_{R+\epsilon}$ contains $X,\ Z$ respectively.
Here $(X)_{r+\epsilon}$ contains $Y$ so that $(X)_{R+r+2\epsilon}$ contains $Z$. Similarly $(Z)_{R+r+2\epsilon}$ contains $X$ so that $d_H(X,Z)\leq r+R$.