Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 2, Problem 26

Problem

Let $X$ be a metric space in which every infinite subset has a limit point. Prove that $X$ is compact.

Answer

By Problem 23 and 24, $X$ has a countable base. Consider an arbitrary open cover $\{G_\alpha\}_{\alpha \in A}$ of $X$. For any $\alpha \in A$ and $x \in G_\alpha$, $x$ is contained in some base $V_{\alpha,x} \subset G_\alpha$. Since the base element $V_{\alpha,x}$ is contained in countable family, $G_\alpha = \bigcup_{x \in G_\alpha}V_{\alpha,x}$ has a countable subcover, denoted by $\{G_n\}_{n=1}^\infty$. Suppose there is no finite subcollection of $\{G_n\}$ covers $X$. Then $F_n:=(G_1 \cup \cdots \cup G_n)^c$ is nonempty for each $n$. Let $E$ be a set which contains a point from each $F_n$. If $E$ is a finite set, then $E$ is covered by finite subcover of $\{G_n\}$, say $G_{n_1},\cdots,G_{n_N}$. However, it contradicts to the fact that $E$ has to contain point in $(G_1 \cup \cdots \cup G_{n_\ast})^c$ where $n_\ast = \max(n_1,\cdots,n_N)$. Hence, $E$ is infinite and has a limit point $y$. Let $y \in G_k$ for some $k$. Since $G_k$ is open, there exists $r>0$ such that $N_r(y) \subset G_k$. Since $y$ is a limit point, $N_r(y)$ contains infinitely many points. However, this contradicts to the fact that at most $k$ points of $E$ are contained in $G_k$.

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