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Following a reference from "Elementos de Topología General" by Angel Tamariz and Fidel Casarrubias.

Theorem Let be $X$ a $T_1$ space such that every infinite and numerable subset have a limit point: so $X$ is countably compact.

proof. Well we suppose that $X$ is not countably compact and so there exist a open numerable cover $\mathcal{U}=\{U_n:n\in\mathbb{N}\}$ such that it have not a finite subcover.

So we choose $x_1\in X$ and $U_{n_1}\in\mathcal{U}$ such that $x\in U_{n_1}$. So since $\mathcal{U}$ have not a finite subcover it is $X\setminus(\bigcup_{i=1}^{n_1}U_i)\neq\varnothing$ and so we choose $x_2\in X\setminus(\bigcup_{i=1}^{n_1}U_i)$. Then since $\mathcal{U}$ is an open cover there exist $n_2\in\mathbb{N}$ such that $x_2\in U_{n_2}$ (observe that $n_1<n_2$). So we suppose that we made $n_1,...,n_k\in\mathbb{N}$ and $x_1,...,x_k\in X$ such thaat $n_1<...<n_k$ and $x_1\in U_{n_1}$ and $x_j\in U_{n_j}\setminus(\bigcup_{i=1}^{n_j-1}U_i)$ for any $j\in\{2,...,k\}$. Since $\bigcup_{i=1}^{n_k}U_i\neq X$ there are exist $x_{n_{k+1}}\in X\setminus(\bigcup_{i=1}^{n_k}U_i)$ and $n_{k+1}\in\mathbb{N}\setminus\{1,2,...,k\}$ such that $x_{n_{k+1}}\in U_{n_{k+1}}$. So in this way we recursively made the point $x_{k+1}\in U_{n_{k+1}}\setminus(\bigcup_{i=1}^{n_k}U_i)$.

So using the previous recursive process we can define the infinite numerable set $F=\{x_k:k\in\mathbb{N}\}$ and the sequence $\{U_k:k\in\mathbb{N}\}$ in $\mathcal{U}$. So we prove that $F$ have not limit point in $X$. Indeed for any $x\in X$ there exist $n\in\mathbb{N}$ such that $x\in U_n$ and $U_n$ contains at most a finite collection $G$ of points of $F$. So $(U_n\setminus G)\cup\{x\}$ is a neighborhood of $x$ that not contains point of $X$ that are different from $x$.

Well I don't understand why $U_n$ contains at most a finite collection $G$ of points of $F$. Could someone help me, please?

Here the original proof in Spanish (I hope mine was a good translation).

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Antonio Maria Di Mauro
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1 Answers1

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So having $x \in X$ and some $U_n$ ($n$ fixed for this argument) containing it, let $n_k$ be the first of the indices used in the recursive construction that is larger or equal to $n$. By construction then $x_{k+1}$ and higher indexed ones, like $x_{k+2}$ etc. ) are not in $U_n$ as $n \le n_k < n_{k+1}$, and $x_{k+1}$ was chosen to lie outside $\bigcup_{i=1}^{n_k} U_i$ in the recursive step, so only those $x_l$ with $l \le k$ can be in $U_n$, so at most finitely many, as claimed.

Henno Brandsma
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