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This is a question I have come across while studying for a topology Ph.D. qualifying exam. I believe my proof is correct, but I would like feedback to make sure I have a correct understanding of metric spaces and compactness, as well as on the clarity of my proof. The question is as follows. Let $\beta$ be an open cover of the metric space $(X,d)$. Show tat if $X$ is compact, then $\exists \epsilon > 0$ such that $\forall A \subset X$ such that the diameter of $A \leq \epsilon$, $\exists B \in \beta$ that contains $A$.

Here is my proof. Since $X$ is compact, $\beta$ has a finite subcover. Let, $B_1 \cup . . . \cup B_n$ be a finite subcover. Note, there exist $B_i , B_j$ such that $B_i \cap B_j \neq \emptyset$. If this were not the. case, then $X$ would not contain all its limit points and would not be closed, and therefore not compact. Since$B_1 \cup . . . \cup B_n$ is finite, we can consider the diameter of $B_i \cap B_j \forall i,j \leq n$. Then let $\epsilon$ be the minimum of such diameters. Then, consider some subset $A$ with a diameter less than or equal to $\epsilon$. If $A$ lies entirely in some $B_i$, we are done. If it lies in multiple $B_i$, we know it must lie in the intersection of all the $B_i$'s since the diameter of $A$ is less than the minimum diameter of the intersection of any $B_i, B_j$. Hence, since it lies in the intersection, $A$ must be fully contained in some $B_i$.

Any feedback is helpful. Thank you.

Kenneth Winters
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    "Note, there exist $B_i,B_j$ such that $B_i\cap B_j\neq \varnothing$. If this were not the. case, then $X$ would not contain all its limit points and would not be closed, and therefore not compact." does not make sense. There are 1) logical leaps in this sentence and more importantly 2) $X$ is the entire space and thus closed – Andrew Aug 08 '23 at 17:06
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    $\varepsilon$ is called Lebesgue number of the cover $\beta$, by the way – Jakobian Aug 08 '23 at 17:06
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    Your $B_j$'s might very well be pairwise dijoint. E.g. a finite discrete space is compact. – Anne Bauval Aug 08 '23 at 17:06
  • It's funny because the exact same question (i.e. about the same problem) appeared few days ago here. – Jakobian Aug 08 '23 at 17:08
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    Have a look at Lebesgue Number Lemma, it is your exact question. @Jakobian sorry your comment loaded late for me! – Igor Aug 08 '23 at 17:08
  • Related – Anne Bauval Aug 08 '23 at 17:12
  • Thank you for your comments. I did not realize this was called the Lebesgue Number Lemma, hence why I did not see the previous questions on it. I see the mistake now with my claim that $B_i, B_j$ are disjoint. In my head, I was thinking this had to be true for $X$ to be closed because each $B_i$ is open and therefore does not contain its boundary points. So I thought there would have to be an overlap between the open balls. However, as @AndrewZhang pointed out, X is obviously closed because it is an entire space. So the fact that $B_i$ doesn't contain its boundary points is not helpful. – Kenneth Winters Aug 08 '23 at 17:36

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