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We define $H^{n}$ for the set of all compact subsets of $\mathbb{R}^n$. Define the metric $\Delta$ in $H^{n}$ as following.Let $A,B \in H^{n}$ then define

$d(x,B):= \min \lbrace d(x,y): y \in B \rbrace$

$d^{*}(A,B):=\max\lbrace d(x,B); x \in A \rbrace$

$\Delta(A,B):=\max \lbrace d^*(A,B),d^*(B,A) \rbrace$

We can check that $(H^n, \Delta)$ is a complete metric space.

My question is

  • what is the distance $d^*(A,B)$ and $d^*(B,A)$ in which $A$ is the unit disk and $B$ is the unit square.
  • What are $d^*(A,B)$ and $d^*(B,A)$ if $A$ is the unit circle and $B$ is the unit disk?

My calculations always lead to confusion since I think $A$ and $B$ are symmetric. Please feel freely helping me solve this problems.

Thank for reading!


Update: I edited my post since I remembered the wrong definition. This problem is in fact problem 6 page 259 in the book "Invitation to Dynamical System". In the solution guide(I have just found it), the author gave the result, but I do not understand it clearly. I still get confusion. Please help me. Sorry for the wrong definition. Thank you very much !

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Hint: there are points of the unit circle whose distance from the unit square is 1. –  Robert Israel Dec 11 '11 at 8:53
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I guess you meant $\Delta(A,B):=\max \lbrace d^*(A,B),d^*(B,A) \rbrace$. –  Christian Blatter Dec 11 '11 at 12:45
    
Thank, Christian Blatter. It's my mistake. It should be $\Delta(A,B):=\max \lbrace d^*(A,B),d^*(B,A) \rbrace$ –  Arsenaler Dec 11 '11 at 16:30
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2 Answers 2

Something is wrong here. Observe that it $A\subseteq B$, then $d^*(A,B)=0$. It follows that $\Delta(A,B)=0$. Hence $\Delta$ is not even a metric.

Isn't the Hausdorff metric usually the diameter of the symmetric difference of two sets? Probably you mean $\max$ in the definition of $\Delta$.

Anyhow. Solve the following two questions: Which point on the disk has the maximal distance to the circle? Which points on the square have the maximal distance to the disk?

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I am so sorry for it, I fixed it. Thanks. –  Arsenaler Dec 11 '11 at 16:33
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When $(M,d)$ is a complete metric space then the distance function $d:M\times M\to {\mathbb R}_{\geq0}$ is symmetric, i.e., $d(x,y)=d(y,x)$ for all $x$, $y\in M$.

Given such an $(M,d)$ and a closed set $B\subset M$ the function

$$d(x,B):=\min_{y\in B}\ d(x,y)$$

is a new function, again denoted by $d(\cdot,\cdot)$, with domain the pairs $(x,B)$ of a point $x\in M$ and a closed set $B\subset M$. Intuitively $d(x,B)$ is the distance from $x$ to the nearest point in $B$; it is continuous as a function of $x$.

When $B$ is fixed and $x$ is allowed to run over a compact set $A\subset M$ then $d(x,B)$ will assume a maximum on $A$, and the value of this maximum is denoted by $d^*(A,B)$. Intuitively $d^*(A,B)$ denotes the maximal distance that a point of $A$ can have from $B$. This new function $d^*(\cdot,\cdot)$ is defined, e.g., on the set of pairs of compact subsets of $M$, but it is not symmetric in its two variables. By symmetry of the construction, however, the Hausdorff distance $\Delta(A,B):=\max\{d^*(A,B),d^*(B,A)\}$ is symmetric in $A$ and $B$.

Now I'm turning to your two examples:

  1. Assume that $A=\{(x,y)\ |\ x^2+y^2\leq 1\}$ and $B=[0,1]^2$. Then the points on the lower left quarter of $\partial A$ have a distance $1$ from $B$, and all other pointa of $A$ have a distance $\leq1$ from $B$. Therefore $d^*(A,B)=1$. On the other hand the point $(1,1)\in B$ has a distance $\sqrt{2}-1$ from $A$, and all other points of $B$ have a smaller distance from $A$. Therefore $d^*(B,A)=\sqrt{2}-1<1$. It follows that $\Delta(A,B)=1$ in this case.

  2. Since $A$ is contained in $B$, all points of $A$ have distance $0$ from $B$; so $d^*(A,B)=0$. On the other hand the point $(0,0)\in B$ has distance $1$ from $A$, and all other points of $B$ have a distance $ <1$ from $A$. Therefore $d^*(B,A)=1$ and $\Delta(A,B)=1$ also.

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