Hausdorff's distance of some sets 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 !  
 A: 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?
A: 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:


*

*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.

*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.
