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When I was reading a textbook about functional analysis, I notice a theorem:

In metric space, if every continuous functions on a closed set $A$ is bounded, then $A$ must be compact.

In the proof, it says that

Suppose that if $A$ is not compact in metric space, then $\exists \delta>0$ and closed balls $B_n$ in $A$, s.t. $d(B_n, B_m)>\delta, n\not=m$.

My question is how to prove the existence of $B_n$? I have tried the similar method in Hausdorff theorem about completely bounded set, but I couldn’t prove that the distances between $B_n$ have a positive bound below. Any help would be appreciated.

xdyy
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  • What are the sets $B_n$? In any metric space with more than one point, there exists $\delta>0$ and two sets $B_n$, $B_m$ with $d(B_n,B_m)>\delta$ – pancini Mar 10 '24 at 04:34
  • @pancini sorry, I didn't express clearly. What I confused about is just the existence of $B_n$. – xdyy Mar 10 '24 at 06:00
  • Cover $A$ with all possible balls. If these balls $B_n$ don't exist then you can find a finite subcover for $A$. Can you see why? – CyclotomicField Mar 10 '24 at 06:28
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    What about $A=X={1/n: n\in\mathbb N}$ endowed with distance from $\mathbb R$? – Jochen Mar 10 '24 at 07:50
  • The property that all continuous functions are bounded is called pseudocompactness and pseudocompact metric spaces are indeed compact. – Jochen Mar 10 '24 at 07:55
  • @Jochen Thanks for reminding me of this concept. I have read some references mentioned in Pseudocompact’s wiki but I couldn’t find the proof. – xdyy Mar 10 '24 at 09:58
  • @CyclotomicField Sorry, I didn’t know the reason. And what do you mean by “all balls”? All the open balls in the whole metric space $X$? – xdyy Mar 10 '24 at 10:02
  • https://math.stackexchange.com/questions/668905 contains a few proofs. – Jochen Mar 10 '24 at 13:43
  • @Jochen Thanks, I have read about this question 668905, and I know the main goal is the same. But my question is just the detail of the method I mentioned. – xdyy Mar 10 '24 at 14:51

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