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Let $(M,d)$ be a metric space and let $H(M)$ denote the set of closed and bounded subset in $M$. Then $(H(M),d_H)$ is a metric space where $d_H$ denotes the Hausdorff distance. Let $\chi$ be the Hausdorff (or ball) measure of non-compactness

How do I prove that the map $\chi\colon H(M)\to\mathbf{R}:B\mapsto\chi(B)$ is Lipschitz?

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Let $A_r$ be the closed $r$-neighborhood of $A$, namely the set of all points whose distance to $A$ is at most $r$. If $d_H(A,B)< r$, then $A\subset B_r$ and $B\subset A_r$, so all we need to do is to prove $\chi(A_r)\le \chi(A)+Cr$ with some constant $C$. To this end, take a finite cover of $A$ by appropriately small sets and replace each of them with its $r$-neighborhood (on increase the radius by $r$ if you are dealing with balls).

You might want to use $2r$ instead of $r$ to avoid any smallest-distance-not-attained issues.

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