Let $A,B \subset \mathbb R^n$ be non empty compact sets and $d_H$ be Hausdorff distance. I'm thinking that if we know the distance between two sets, the difference between their diameters is bounded. How to (dis)prove that if $d_H(A,B)=r$ then $diam(A)+2r \ge diam(B) \ge diam(A)-2r$?

up vote 3 down vote accepted

I like to use the definition of Hausdorff distance which looks like this: $d_H(A,B)$ is the infimum of all $d\in\mathbb{R}$ such that $A$ is in the $d$-neighborhood of $B$ and $B$ is in the $d$-neighborhood of $A$. (This definition is the easiest to draw a picture of, and is equivalent to any other definition you might know.)

So if $d_H(A,B)=r$, then for every $\epsilon>0$, each set $A,B$ is in the $(r+\epsilon)$-neighborhood of the other.

Okay, so take any two points $x,y\in B$. Then each has a corresponding point $x', y'$ in $A$ at distance at most $r+\epsilon$, i.e. $$ d(x,x')<r+\epsilon, $$ $$ d(y,y')< r+\epsilon. $$

Since $x',y'\in A$, we have $$d(x',y')\leq diam(A).$$

Apply the triangle inequality to get $$ d(x,y) \leq diam(A) + 2r+2\epsilon.$$

Since $x$ and $y$ were arbitrary points in $B$, and $\epsilon>0$ was arbitrary, we get $$ diam(B) \leq diam(A) + 2r. $$

Of course the other inequality follows the same way.

It might be a good idea to draw pictures convincing yourself that these bounds are sharp (that is, you can find sets where the inequalities become equalities).

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