# If any infinite numerable subset of the $T_1$ space $X$ have a limit point then $X$ is countably compact

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 $$n_1). So we suppose that we made $$n_1,...,n_k\in\mathbb{N}$$ and $$x_1,...,x_k\in X$$ such thaat $$n_1<... 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).

• As per the comments on my answer: the sequence of elements from $\mathcal{U}$ should probably say $\{U_{n_k}: k \in \Bbb N\}$ to make it clearer. Mar 21 '20 at 22:54
• Hi professor Brandsma, could I ask your assistance to solve the problem that I explain here? Mar 22 '20 at 18:25
• I’ll type something later. Busy now. Mar 22 '20 at 18:28
• Okay, don't worry; thanks!!! Mar 22 '20 at 18:29

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.
• Okay, this it is clear if $U_n$is an element of the sequence $\{U_k:k\in\mathbb{N}\}$; but if $U_n$ is an element of $\mathcal{U}$ such that is not in $\{U_k:k\in\mathbb{N}\}$, why the result holds? Mar 21 '20 at 18:15
• @AntonioMariaDiMauro you can always choose a $U_n$ because it’s a cover. Mar 21 '20 at 18:17
• So the made sequence $\{U_k:\in\mathbb{N}\}$ of element of $\mathcal{U}$ is a permutation of the element of $\mathcal{U}$? Mar 21 '20 at 18:18
• Sorry, but I don't understan what you wrote. Perhaps do you want say that $\{U_k:k\in\mathbb{N}\}$ is a subsequence of $\mathcal{U}$ such that is a cover of $X$? Mar 21 '20 at 18:22