Here's the problem I'm dealing with:

Let $(X,d)$ be a compact metric space and let $(U_{\lambda})_{\lambda \in \Lambda}$ be an open cover of $X$. Show that there exist $\delta >0$ such that for all $A \subset X$ with diam$A < \delta$, there exist $\lambda \in \Lambda$ such that $A \subset U_{\lambda}$.

I am not sure how to solve this one. I tried to take an open ball of radius $r= \delta + \epsilon$, for a small non-negative number $\epsilon$ which contains $A$, i.e. $A \subset B(x,r)$ for some $x \in A$. Then I consider a finite sub cover of $X$ : $(U_{\lambda})_{\lambda =1}^{n}$. Now this is also a cover of $A$, so I guess that I can find some open sets in my finite sub cover such that my ball is the union of these elements, i.e. $B(x,r)= \cup_{i=1}^k U_i$.

Is it clear that $\cup_{i=1}^k U_i$ is again contained in one of the $U_{\lambda}$ of the cover ?

Thanks for your help.

  • 3
    $\begingroup$ See en.wikipedia.org/wiki/Lebesgue%27s_number_lemma $\endgroup$ – Kavi Rama Murthy Apr 15 at 11:55
  • $\begingroup$ Oh I see. Thank you. $\endgroup$ – Alain Apr 15 at 11:57
  • $\begingroup$ @KaviRamaMurthy Reading the proof, I don't really understand why the function is defined like that. Also do you understand why the minimum has to be strictly positive ? Thanks again $\endgroup$ – Alain Apr 15 at 12:28
  • $\begingroup$ If $f(x)=0$ then $d(x,C_i)=0$ for all $i$ which implies $x \in C_i=A_i^{c}$ for each $i$. This means $x $ does not belong to any $A_i$, a contradiction. $\endgroup$ – Kavi Rama Murthy Apr 15 at 12:35
  • $\begingroup$ thank you for your help. $\endgroup$ – Alain Apr 15 at 12:45

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