# Hausdorff distance between a sequence of sets and a limiting set

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:

$$$$\nonumber d_H(X,Y) = \inf\{\epsilon \geq 0: X\subseteq (Y)_\epsilon\text{ and }Y\subseteq (X)_\epsilon\}$$$$ where $$(Z)_\epsilon$$ represents the $$\epsilon$$-fattening of $$Z$$.

I want to show the following:

$$$$\nonumber d_H(A_n\setminus B_n,A) \rightarrow 0 \text{ as } n \rightarrow \infty$$$$

Can this be done? And how?

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

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

• Ah I see! Thank you! Yes, this is what I was looking for! Commented Feb 12, 2021 at 15:56
• 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. Commented Feb 27, 2021 at 19:08
• 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! Commented Feb 28, 2021 at 14:28