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I am reading "Mathematical Analysis 2nd Edition" by Tom M. Apostol.

Theorem 3.27 $\,\,\,\,$Let $G=\{A_1,A_2,\dots\}$ denote the countable collection of all $n$-balls having rational radii and centers at points with rational coordinates. Assume $x\in\mathbb{R}^n$ and let $S$ be an open set in $\mathbb{R}^n$ which contains $x$. Then at least one of the $n$-balls in $G$ contains $x$ and is contained in $S$. That is, we have $$x\in A_k\subseteq S\,\,\,\,\,\,\,\,\,\text{ for some }A_k\text{ in }G.$$

Theorem 3.28 (Lindelöf covering theorem). $\,\,$ Assume $A\subseteq\mathbb{R}^n$ and let $F$ be an open covering of $A$. Then there is a countable subcollection of $F$ which also covers $A$.

Proof. $\,\,\,$Let $G=\{A_1,A_2,\dots\}$ denote the countable collection of all $n$-balls having rational centers and rational radii. This set $G$ will be used to help us extract a countable subcollection of $F$ which covers $A$. Assume $x\in A$. Then there is an open set $S$ in $F$ such that $x\in S$. By Theorem 3.27 there is an $n$-ball $A_k$ in $G$ such that $x\in A_k\subseteq S$. There are, of course, infinitely many such $A_k$ corresponding to each $S$, but we choose only one of these, for example, the one of smallest index, say $m=m(x)$. Then we have $x\in A_{m(x)}\subseteq S$. The set of all $n$-balls $A_{m(x)}$ obtained as $x$ varies over all elements of $A$ is a countable collection of open sets which covers $A$. To get a countable subcollection of $F$ which covers $A$, we simply correlate to each set $A_{m(x)}$ one of the sets $S$ of $F$ which contained $A_{m(x)}$. This completes the proof.

I cannot understand the following sentence at the end of the proof for 3.28:

"To get a countable subcollection of $F$ which covers $A$, we simply correlate to each set $A_{m(x)}$ one of the sets $S$ of $F$ which contained $A_{m(x)}$."

How to construct a countable subcollection of $F$ which covers $A$ more explicitly?

Does the proof of Theorem 3.28 say the following?

Assume that $x\in A$. Then there is an open set $S_x$ in $F$ such that $x\in S_x$ because $F$ is an open covering of $A$. Then, $\{S_x\mid x\in A\}$ is a subcollection of $F$ which covers $A$. And $G$ helps us to prove $\{S_x\mid x\in A\}$ is countable.

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I'll start with a very rigorous explanation and then I'll give some intuition ($\aleph(A)$ is the cardinality of a set $A$)

(I'll define $m(x)$ rigurusly, skip until I define $\mathscr{M}$ if you don't think it's necessary) For each $x\in A$, let $\mathcal{S}_x\in F$ be an open set containing $x$ and let $m\colon A\to \mathbb{N}$ be defined by $x\mapsto m(x)=\min\{m\in\mathbb{N}\mid x\in A_m\subseteq \mathcal{S}_x\}$. Define $\mathscr{M}=\{m(x)\mid x\in A\}$. For each $m\in\mathscr{M}$, define $S_{m}=\{S\in F\mid A_{m}\subseteq S\}$. We know that for every $m\in\mathscr{M}$, $S_m\neq \emptyset$ ($A_m$ was constructed so that this is satisfied). The sentence you don't understand is saying the following:

Let $c\colon \mathscr{M}\to \bigcup_{m\in\mathscr{M}}S_m$ be a choice function ($\forall m\in\mathscr{M},\; c(m)\in S_m$) then $\text{Im}(c)$ is a countable subcovering.

Indeed, it's countable since $c\colon \mathscr{M}\to \text{Im}(c)$ is a surjection and $\aleph(\mathscr{M})\leq \aleph_0$. $\text{Im}(c)$ consists of elements of $F$ since $\text{Im}(c)\subseteq \bigcup_{m\in\mathscr{M}}S_m\subseteq F$ and is a covering of $A$: let $x\in A$, then $x\in A_{m(x)}\subseteq c(m(x))$ and we are done.

The intuition is the following: For every $x$, you can assign an element of $G$ that is contained in some element of $F$. For each of these select one element of $F$ that contains it and discard the rest. Necessarily, you are left with a countable collection and since this was done for every $x$, it necessarily covers $A$ (do some drawings and it becomes apparent).

For a generalization: Let $(X,\mathscr{T})$ be any space. If $\mathfrak{B}=\{B_\alpha\mid \alpha\in\mathscr{A}\}$ is a collection of open sets that satisfies the property mentioned in 3.27, namely, for every open $G$ in $X$, for every $x\in G$, there is a $B_\alpha$ such that $x\in B_\alpha\subseteq G$, we say that $\mathfrak{B}$ is a basis for the topology of $X$. If $X$ has a countable basis, we say that $X$ is $2^\circ$ countable. Using the exact same proof presented in your question you get that

Proposition. Every $2^\circ$ countable space is Lindelöf.

Just use any countable basis instead of $G$ in the proof.

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    $\begingroup$ Sebastian P. Pincheira, Thank you very much for your answer and edit. $\endgroup$
    – tchappy ha
    Commented Nov 29, 2022 at 5:59

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