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).

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.