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?

  • 1
    $\begingroup$ What have you tried? Are $\{A_{n}\}_{n \in \mathbb{N}}$ and/or $\{B_{n}\}_{n \in \mathbb{N}}$ assumed to be compact? $\endgroup$
    – user711689
    Feb 11, 2021 at 16:30
  • 1
    $\begingroup$ $B_n\rightarrow \emptyset$ so that $\emptyset=(\emptyset)_\epsilon$ contains $B_n$ for some large $n$. Hence $B_n$ is empty set ? $\endgroup$
    – HK Lee
    Feb 11, 2021 at 16:44
  • $\begingroup$ They are assumed to be compact, yes. $\endgroup$
    – JDoe2
    Feb 11, 2021 at 18:24
  • $\begingroup$ @HKLee . I have the same trouble.... & if $X\ne \emptyset =Y$ then $d_H(X,Y)=\inf \,\emptyset .$ $\endgroup$ Feb 12, 2021 at 14:33
  • $\begingroup$ DanielWainfleet : I believe that OP wanted to say something like measure 0 set $\endgroup$
    – HK Lee
    Feb 12, 2021 at 14:37

1 Answer 1


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 $d_H(X,Z)\leq r+R$.

  • $\begingroup$ Ah I see! Thank you! Yes, this is what I was looking for! $\endgroup$
    – JDoe2
    Feb 12, 2021 at 15:56
  • 1
    $\begingroup$ Just a quick question; $A_n-B_n$ is not necessarily compact so how can the inequality be used? $d_H$ is only a metric on compact sets. $\endgroup$
    – JDoe2
    Feb 27, 2021 at 19:08
  • $\begingroup$ Ah okay I think I see... so the final step follows as $d_H(X,Z)\leq r+R+2\epsilon$ for all $\epsilon>0$, it is therefore true that $d_H(X,Z)\leq r+R$? Thank you! $\endgroup$
    – JDoe2
    Feb 28, 2021 at 14:28

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