# Prove that the set $A = \left\{\frac{2n+2}{2n+1} : n \in \mathbb{N}\right\} \cup \left\{0,1\right\}$ is compact

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.

• I would say: it's closed and bounded. – MJD May 29 at 13:26

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$$A\subset\bigcup_{n=0}^{N+1}A_{\lambda_n}.$$
• Second to last sentence. Shouldn't it be "such that $$\frac{2n+2}{2n+1} \in A_{\lambda_{n+1}}?$$" – fwd May 29 at 11:36