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Let $X$ be a metric space and let $A\subset X$ be a bounded subset of $X$. I read on Wikipedia that the Hausdorff- and Kuratowski measures of non-compactness ($\alpha$, resp. $\beta$) satisfy the inequalities $$\alpha(A)\leq \beta(A)\leq 2\cdot\alpha(A)$$ How can I prove these inequalities?

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2 Answers 2

Lemma 1. Let $x\in X$ and $r\geq 0$, then $\mathrm{diam}(B(x,r))\leq 2r$.

Proof. Indeed for all $p,q\in b(x,r)$ we have $$ \rho(p,q)\leq \rho(p,x)+\rho(x,q)\leq r+r=2r $$ Hence, $$ \mathrm{diam}(B(x,r))=\sup\{\rho(p,q):p,q\in B(x,r)\}\leq2r $$

Lemma 2. Let $S\subset X$ and $d=\mathrm{diam}(S)<+\infty$, then for all $p\in S$ we have $$ S\subset B(p,d) $$

Proof. Take arbitrary $p\in S$, then for all $s\in S$ we have $\rho(p,s)\leq\mathrm{diam}(S)=d$. This is exactly means that $s\in B(p,d)$. Since $s\in S$ is arbitrary $S\subset B(p,d)$. $$ $$

Theorem. Let $A\subset X$, then $$ \alpha(A)\leq\beta(A)\leq 2\alpha(A) $$

Proof. Denote $$ \mathrm{Ball}(X,r)=\{B(x,r):x\in X\} $$ $$ \mathrm{Bounded}(X,d)=\{S\subset X:\mathrm{diam}(S)\leq d\} $$ $$ \mathcal{F}(M)=\{\mathcal{S}\subset 2^M:\mathrm{Card}(\mathcal{S})<\aleph_0\} $$ then $$ \alpha(A)=\inf\{r>0:\quad \mathcal{B}\in\mathcal{F}(\mathrm{Ball}(X,r)),\quad A\subset\cup\mathcal{B}\} $$ $$ \beta(A)=\inf\{d>0:\quad \mathcal{S}\in\mathcal{F}(\mathrm{Bounded}(X,d)),\quad A\subset\cup\mathcal{S}\} $$

Fix $\varepsilon>0$. Then there exist a family $\mathcal{B}=\{B_i:i\in\{1,\ldots,n\}\}\in\mathcal{F}(\mathrm{Ball}(X,r))$ of balls of radius $r$ such that $A\subset\cup\mathcal{B}$ and $r<\alpha(A)+\varepsilon$. Each ball of radius $r$ have diameter at most $2r$ (see lemma 1), so for all $i\in\{1,\ldots,n\}$ we have $\mathrm{diam}(B_i)\leq 2r$. Hence $\mathcal{B}\in\mathcal{F}(\mathrm{Bounded}(X,2r))$. Since $\mathcal{B}\in\mathcal{F}(\mathrm{Bounded}(X,2r))$ and $A\subset\cup\mathcal{B}$, then $$ \beta(A)\leq 2r\leq 2\alpha(A)+2\varepsilon $$ Since $\varepsilon>0$ is arbitrary, then $$ \beta(A)\leq 2\alpha(A)\tag{1} $$

Again fix $\varepsilon>0$. Then there exist a family $\mathcal{S}=\{S_i:i\in\{1,\ldots,n\}\}\in\mathcal{F}(\mathrm{Bounded}(X,d))$ such that $A\subset\cup\mathcal{F}$ and $d<\beta(A)+\varepsilon$. Since every set of diameter $d$ is contained in some ball of radius $d$ (see lemma 2), then for each $i\in\{1,\ldots,n\}$ we have a ball $B_i$ of radius $r$ such that $S_i\subset B_i$. Consider $\mathcal{B}=\{B_i:i\in\{1,\ldots,n\}\}$, obviously $\mathcal{B}\in\mathcal{F}(\mathrm{Ball}(X,d))$. Moreover $$ A\subset\cup\mathcal{S}=\bigcup\limits_{i=1}^n S_i\subset\bigcup\limits_{i=1}^n B_i=\cup\mathcal{B} $$ Since $A\subset\cup\mathcal{B}$ and $\mathcal{B}\in\mathcal{F}(\mathrm{Ball}(X,d))$, then $$ \alpha(A)\leq d<\beta(A)+\varepsilon. $$ Since $\varepsilon>0$ is arbitrary, then $$ \alpha(A)\leq \beta(A)\tag{2} $$ The result follows from $(1)$ and $(2)$.

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These relationships come from the inequalities between the radius and the diameter of a ball. In the following, $(X,d)$ is a metric space. Remark that, in the definition of the Hausdorff measure (is that how it is called?), taking the balls open or closed does not change anything, so we might as well assume that they are closed.

Inequality $\alpha \leq \beta$

Let $K \subset X$. Let $\varepsilon >0$. By the definition of the Kuratowski measure, we can find a finite covering $U_1$, $\cdots$, $U_n$ of $K$ such that $\text{Diam} (U_i) \leq \beta (K) + \varepsilon$ for all $i$. Now, let us take for all $i$ some $x_i \in U_i$. Then, for all $y \in U_i$, one has $d(x_i, y) \leq \text{Diam} (U_i)$, so that $U_i \subset \overline{B} (x_i, \text{Diam} (U_i)) \subset \overline{B} (x_i, \beta (K) + \varepsilon)$. Thus, $K$ is covered by the balls $\overline{B}(x_1, \beta (K) + \varepsilon)$, $\cdots$, $\overline{B} (x_n, \beta (K) + \varepsilon)$. By the definition of the Hausdorff measure, this implies that $\alpha (K) \leq \beta (K) + \varepsilon$. Since this holds for all $\varepsilon >0$, we get finally $\alpha (K) \leq \beta (K)$.

Inequality $\beta \leq 2 \alpha$

The diameter of a ball $B (x,r)$ of radius $r$ is at most $2r$. Indeed, for all $y$ and $z$ in $B(x,r)$, one has $d(y,z) \leq d(y,x) + d(x,z) \leq 2r$, so that $\text{Diam} (B(x,r)) \leq 2r$. Let $K \subset X$. Any covering of $K$ by balls of radius $r$ is a covering of $K$ by subsets of diameter at most $2r$. Let $\varepsilon >0$, and consider a finite covering of $K$ by balls of radius at most $\alpha (K) + \varepsilon$. Then this is a covering of $K$ by subsets of diameter at most $2r + 2\varepsilon$. Hence, $\beta (K) \leq 2 \alpha (K) + 2\varepsilon$. Since this is true for any $\varepsilon >0$, we get $\beta (K) \leq 2 \alpha (K)$.

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