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Here is the question: Let $\{U_i\}_{i \in A}$ be an open cover of a compact metric space $X$. Show that there exists $\epsilon \gt 0$ such that for each $x \in X$, the open ball $B(x;\epsilon)$ is contained in $U_i$ for some $i \in A$.

Here is my attempt:

Since $X$ is compact, it is totally bounded. This means for each $\epsilon \gt 0$, there exists a finite number of open balls of radius $\epsilon$ that cover $X$.

This means that the family of open balls $\{B_i(x;\epsilon)\}_{i \in A}$ is an open cover for $X$.

I am stuck here and have no idea whatsoever after this. Am I going the right way? Any help would be appreciated.

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  • $\begingroup$ I think looking at totally bounded might be on the wrong track: you're trying to limit how small things can get, not how large they can get. You should use the fact that in a compact space, you can refine any open cover to a finite subcover. $\endgroup$ – Joppy Apr 24 '17 at 14:39
  • $\begingroup$ I don't know what it is that you're trying to prove. Everything follows by definition. If $x\in X$, then there is some $U_i$ such that $x \in U_i$. And then by definition of open sets in metric spaces, there is an $\epsilon > 0$ such that $B(x, \epsilon) \subseteq U_i$. This would mean the family of all such open balls is an open cover of $X$. $\endgroup$ – Mark Apr 24 '17 at 14:48
  • $\begingroup$ @user264885 The $\epsilon>0$ you mention depends on $x$. It must be shown to be true for some independent $\epsilon>0$. $\endgroup$ – drhab Apr 24 '17 at 14:57
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    $\begingroup$ en.m.wikipedia.org/wiki/Lebesgue's_number_lemma $\endgroup$ – Henno Brandsma Apr 24 '17 at 15:23
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Suppose it is not true.

Then a sequence $(x_n)_n$ exists in $X$ such that $B(x_n,\frac1{n})$ is not contained in any of the $U_i$.

A compact metric space is sequentially compact so there is a convergent subsequence $(x_{n_k})_k$.

Let $x$ denotes its limit and let $x\in U_i$.

Then it can be shown that for $k$ large enough $B(x_{n_k},\frac1{n_k})\subseteq U_i$ and a contradiction is found.

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  • $\begingroup$ Thank you for the answer. But is there another way to prove it without using the sequentially compact part? Because I haven't learnt that term yet.. $\endgroup$ – Cruso James Apr 24 '17 at 15:10
  • $\begingroup$ @CrusoJames Might be, but I have nothing else (yet), and I am a slow thinker. I suggest you do not accept this answer. That increases the chance that others will take a look at your question. Good luck. $\endgroup$ – drhab Apr 24 '17 at 15:14
  • $\begingroup$ This might be interesting for you. $\endgroup$ – drhab Apr 24 '17 at 15:17

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