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Let $A=B(a,r)$ and $B=B(b,s)$ tow balls in a metric space $(E,d)$ such that $r\leq s$ we define the Hausdorff Distance by : $$H(A,B)=\max\{\sup_{x\in A } d(x,B),\sup_{y\in B} d(y,A)\}$$ That is true ? $$H(A,B)= d(a,b)+s-r$$

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    No. Consider the space $E$ consisting of just two points. – 2017-02-14
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    i don't understand your idea – 2017-02-14
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    What did you try to solve this problem? – 2017-02-14

1 Answers 1

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Let $d$ be the discrete metric, i.e., $d(x,y)=1$ if and only if $x\neq y$, otherwise $0$. Then if $0 \leq r < s < 1$, each ball contains precisely $1$ point as $(E,d)$ is a metric space.

Clearly for the Hausdorff distance $d(A,B) = 1$, but $d(a,b) + s - r = 1 + s - r > 1$.