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I'm currently studying about metric spaces, specifically on topic about compactness, here's the exercise in my lecture notes

In usual metric space $\mathbb R$, Prove that the set $$A = \left\{\frac{2n+2}{2n+1} : n \in \mathbb{N}\right\} \cup \left\{0,1\right\}$$ is compact.

Now, I know that the intuition to prove that A is compact by proving that every open cover of $A$ has a finite subcover, but some things that I'm not sure are :

  1. How do I define the open cover that covers $A$? At glance, I know that A is set that contains points near $1$, I'm not sure about what to do with $0$.
  2. I've read that to prove compactness, one might start by investigating the limit points of the set, is this correct? If so, then should I try to construct finite subcover that revolving around the point $1$?

any insight would really help, thanks beforehand.

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  • $\begingroup$ I would say: it's closed and bounded. $\endgroup$ – MJD May 29 at 13:26
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Let $(A_\lambda)_{\lambda\in\Lambda}$ be an open cover of $A$. There are $\lambda_0,\lambda_1\in\Lambda$ such that $0\in A_{\lambda_0}$ and that $1\in A_{\lambda_1}$. Since $\lim_{n\in\Bbb N}\frac{2n+2}{2n+1}=1$, there is some $N\in\Bbb N$ such that $n\geqslant N\implies\frac{2n+2}{2n+1}\in A_{\lambda_1}$. For each $n\in\{1,2,\ldots,N-1\}$, take $\lambda_{n+1}\in\Lambda$ such that $\frac{2n+2}{2n+1}\in A_{\lambda_{n+1}}$. So$$A\subset\bigcup_{n=0}^{N+1}A_{\lambda_n}.$$

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  • $\begingroup$ Second to last sentence. Shouldn't it be "such that $$\frac{2n+2}{2n+1} \in A_{\lambda_{n+1}}?$$" $\endgroup$ – fwd May 29 at 11:36
  • $\begingroup$ @fwd Yes. I shall edit my answer. $\endgroup$ – José Carlos Santos May 29 at 11:55
  • $\begingroup$ thanks for the input, I'll try to continue from this! $\endgroup$ – singularity May 29 at 12:30
  • $\begingroup$ To continue what? What is missing in my answer? $\endgroup$ – José Carlos Santos May 29 at 13:04
  • $\begingroup$ @JoséCarlosSantos sorry I mean continuing my work, yours already outlined the part that I'm confused, thanks again :) $\endgroup$ – singularity May 29 at 14:42

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