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For two subsets $X,Y$ of a metric space $(M,d)$ the Hausdorff distance is defined as $$ d_{\mathrm H}(X,Y) = \max\{\,\sup_{x \in X} \inf_{y \in Y} d(x,y),\, \sup_{y \in Y} \inf_{x \in X} d(x,y)\,\},$$ or, equivalently, $$d_H(X,Y) = \inf\{\epsilon \geq 0\,;\ X \subseteq Y_\epsilon \ \mbox{and}\ Y \subseteq X_\epsilon\},$$ where $X_\epsilon:=\{x\in M\mid d(x,X)\leq\epsilon\}$. My questions regards the behavior of the Hausdorff distance in the following situation. Let $A,B\subseteq M$ and define for $v\in M$ the sets $$C_v:=(v+A)\cap B$$ and assume $C_v\neq\emptyset$ for all $v$ under consideration. Does the following continuity property hold for every $\varepsilon>0$ there exists $\delta>0$ s.t. $$d(v,w)<\delta\Rightarrow d_H(C_w,C_v)<\varepsilon.$$ I may give you some thoughts on my question:

  • If we take two sets $A,B$ s.t. $d_H(A,B)\leq \varepsilon$ and intersect with some set $C$, we, in generally, do not have $d_H(A\cap C,B\cap C)\leq\varepsilon,$ even if the intersections are non-empty. (This would have implied the implication above)
  • Considering translations of the same set seems more restrictive, s.t. I think that the implication might hold, even though I am not entirely convinced.

I would appreciate some input or a counterexample.

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    $\begingroup$ What does $v+A$ stand for? If $M$ is just a metric space there is no operation of addition in it. $\endgroup$ Jun 15, 2018 at 9:54
  • $\begingroup$ This is right, I was more specifically thinking about a vector space, thus $v+A=\{x\in M\mid x=v+a, a\in A\}$. $\endgroup$
    – Tsuyoi
    Jun 15, 2018 at 10:06

1 Answer 1

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No.

Let $M=\Bbb R^2$, $A=B=[0,1]\times [0,1]\cup [1,2]\times\{0\}$. Then $$C_{(0,t)}=\begin{cases}[0,1]\times[\max\{0,t\},\min\{1,1+t\}]&0<|t|<1\\ A&t=0\end{cases}$$ and therefore $$ d_H(C_{(0,t)},C_{(0,0)})=\begin{cases}\sqrt{1+t^2}&0<|t|<1\\0&t=0\end{cases}$$

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  • $\begingroup$ Nice example. But what if $A$ and $B$ are assumed to be closed and convex? $\endgroup$
    – amsmath
    Feb 26, 2021 at 1:58

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