Let $X$ be a Banach space, $\dim X<\infty$, $A$ be a bounded subset of $X$ and $\alpha$ be Kuratowski measure of noncompactness. How to prove that $\alpha (A) =0$ if and only if $A$ is relatively compact?
Measure of noncompactness
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real-analysis
functional-analysis
compactness
1 Answers
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By definition of ball mesure of non-compactness $ \beta(A)=\inf\left\{r>0:\quad\exists\{B(x_i,r)\}_{i=1}^n\quad A\subset\bigcup\limits_{i=1}^nB(x_i,r)\right\} $ then $ \begin{align} \beta(A)=0 &\;\overset{\text{definition of infimum}}{\Longleftrightarrow}\; \forall\varepsilon>0\;\;\exists\{B(x_i,\varepsilon)\}_{i=1}^n\;\;A\subset\bigcup\limits_{i=1}^n B(x_i,\varepsilon)\\ &\;\overset{\begin{array}{cc}\text{definition of relative}\\ \text{compactness}\end{array}}{\Longleftrightarrow}\; A \quad\text{is relatively compact} \end{align} $ It remains to recall that $\beta(A)\leq\alpha(A)\leq 2\beta(A)$. Note that condition $\mathrm{dim}\; X<\infty$ is redundant here.
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1@Mokata, you should read this [article](https://en.wikipedia.org/wiki/Totally_bounded_space) – 2018-04-04