Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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

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 $\gamma(A) \le \gamma(A+K) \le \gamma(A) + \gamma(K) = \gamma(A)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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