Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 2, Problem 12

Problem

Let $K \subset \mathbb{R}^1$ consist of $0$ and the numbers $1/n$, for $n=1,2,3,\cdots$. Prove that $K$ is compact directly from the definition (without using Heine-Borel theorem).

Answer

Let $\{U_\alpha\}_\alpha \in A$ be a open cover of $K$. Then there exists $\alpha_0 \in A$ such that $0 \in U_{\alpha_0}$. Since $U_{\alpha_0}$ is open, there exists $r>0$ such that $N_r(0) \subset U_{\alpha_0}$. If we choose $N \in \mathbb{N}$ such that $1/N < r$, then $1/n \in N_r(0) \subset U_{\alpha_0}$ for $n \geq N$. For $n=1,\cdots,N-1$, there exist $\alpha_n \in A$ such that $1/n \in U_{\alpha_n}$. Then $K$ can be covered by $U_{\alpha_0},U_{\alpha_1},\cdots,U_{\alpha_{N-1}}$.

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