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Let $\{A_\alpha : \alpha \in I\}$ a family of closed subsets of a compact metric space $(X,d)$ such that $\cap_{\alpha \in I} A_\alpha =\emptyset$. Prove that $\exists \varepsilon>0$ such that if $B \subseteq X$ with $\delta(B)<\varepsilon$ then $\exists \beta \in I$ such that $B \cap A_\beta = \emptyset$. Where $$\delta(B)=\sup \{d(x,y):x,y \in B\}.$$

My idea. Since $\cap_{\alpha \in I} A_\alpha =\emptyset$ we have $X-\cap_{\alpha \in I} A_\alpha = X$ then $\cup_{\alpha \in I} (X-A_\alpha) =X$, so $\{X-A_\alpha : \alpha \in I\}$ is an open cover of $X$ and since $X$ is compact we have that $X \subset \cup_{i=1}^{n} (X-A_{\alpha_i})$. This is what I have at the moment.

Alejandra Benítez
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1 Answers1

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I will assume that each $A_\alpha$ is not empty; otherwise, the problem is trivial.

It follows from what you wrote that there are $\alpha_1,\alpha_2,\ldots,\alpha_n\in I$ such that $\bigcap_{k=1}^nA_{\alpha_k}=\emptyset$. Since each $A_\alpha$ is not empty, $n>1$. Take a number $\varepsilon>0$ which is smaller than the distance from $A_{\alpha_i}$ to $A_{\alpha_j}$ for any two distinct numbers $i,j\in\{1,2,\ldots,n\}$. If $B\subseteq X$ is such that $\delta(B)<\varepsilon$, then $B\cap A_{\alpha_i}\ne\emptyset$ for, at most, one $i\in\{1,2,\ldots,n\}$. In fact, if $B\cap A_{\alpha_i},B\cap A_{\alpha_j}\ne\emptyset$ with $i\ne j$, then take $b_i\in B\cap A_{\alpha_i}$ and $b_j\in B\cap A_{\alpha_j}$. Then $d(b_i,b_j)\leqslant\delta(B)<\varepsilon$. But this is impossible, since $b_i\in A_{\alpha_j}$, $b_j\in A_{\alpha_j}$, and $\varepsilon$ is smaller than the distance from $A_{\alpha_i}$ to $A_{\alpha_j}$. So, it is proved that $j\ne i\implies B\cap A_{\alpha_j}=\emptyset$.

José Carlos Santos
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