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The following statement is present in "Elementos de Topología General" by Angel Tamariz and Fidel Casarrubias.

Statement

Let be $X$ a metrizable countably compact space: so for any $\delta>0$ there exist a finite subset $A_\delta\subseteq X$ such that $X=\bigcup_{x\in A_\delta}B(x,\delta)$.

Proof. Previously we remember that any infinite countable subset of a contably compact space have a limit point. So if the statement was false then there exist $\delta>0$ and a countable subset $F=\{x_n:n\in\Bbb{N}\}$ such that $d(x_n,x_m)\ge \delta$ for any $m,n\in\Bbb{N}:n\neq m$, but this means that $F$ is an infinite contable subset such that has not limit points and this would be inconsisten with the hypothesis of the countable compactness of $X$.

So I don't understeand the following things:

  1. if the statement was false then there exist $\delta>0$ and a countable subset $F=\{x_n:n\in\Bbb{N}\}$ such that $d(x_n,x_m)\ge \delta$ for any $m,n\in\Bbb{N}:n\neq m$;
  2. If 1. was true then $F$ has not limit points.

As reference I say that "A space is countably compact if every countable cover has a finite subcover".

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

enter image description here

Could someone help me, please?

Henno Brandsma
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Antonio Maria Di Mauro
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1 Answers1

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Suppose the assertion is false. Then there must be some $\delta>0$ for which it fails, so there cannot be a finite $A_\delta$ as advertised.

We can construct the set by recursion again, just like in your previous construction: we pick $x_1\in X$ arbitrarily. Then $B(x_1, \delta) \neq X$, or else $A_\delta = \{x_1\}$ would have worked. So pick $x_2 \notin B(x_1, \delta)$, and note that $d(x_1, x_2) \ge \delta$. Having chosen at stage $n$ points $F_n = \{x_1,x_2,\ldots, x_n\}$ such that for any $1\le i < j \le n$ we have $ d(x_i, x_j) \ge \delta$ we again note that this finite set is not $A_\delta$ so we pick $$x_{n+1} \notin \bigcup_{x \in F_n} B(x, \delta)$$ and this ensures that we can add $x_{n+1}$ to our set, and that all distances remain $\ge \delta$ between them.

Suppose that the set of all $x_n$ had a limit point $p$. Then any ball of radius $\frac{\delta}{2}$ around $p$ contains more than one point from the set ( even infinitely many) but that forces their distances to be $< \delta$, contrary to construction. So there is no limit point.

Henno Brandsma
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