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Let $\gamma$ denote the Hausdorff/Kuratowski measure of noncompactness defined on a Banach space $(X,\|\cdot\|)$. I was wondering whether $\gamma(A)=\gamma(A+K)$ holds for $A\subset X$ is bounded and $K\subset X$ is compact.

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Yes, since $\gamma(A) \le \gamma(A + K)$ by monotonicity and $\gamma(A+K) \le \gamma(A) + \gamma(K) = \gamma(A)$ as for example mentioned in the wiki article you linked. –  martini May 15 '12 at 20:28

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I'll try and give a proof for $\gamma(A + B) \le \gamma(A) + \gamma(B)$ for bounded sets $A,B \subseteq X$. So let $\epsilon > 0$, then there is a finite covering $U_i$, $i\in I$, of $A$ with balls of radius at most $\gamma(A) + \epsilon$ resp. sets with diameter at most $\gamma(A) + \epsilon$. Also we choose a finite cover $V_j$, $j \in J$ of $B$ with the $V_j$ balls of radius at more $\gamma(B) + \epsilon$ resp. sets with diameter at most $\gamma(B) + \epsilon$. Then $A + B$ is covered by $U_i + V_j$, $(i,j) \in I \times J$, which is a finite cover consisting of balls of radius at most $\gamma(A) + \gamma(B) + 2\epsilon$ resp. of sets with diameter at most this. Therefore $\gamma(A + B) \le \gamma(A) + \gamma(B) + 2\epsilon$. As $\epsilon$ was arbitrary, the result follows.

In your case you have, as mentioned above $\gamma(A) \le \gamma(A+K) \le \gamma(A) + \gamma(K) = \gamma(A)$.

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