# About Lindeloef covering theorem ("Mathematical Analysis 2nd Edition" by Tom M. Apostol)

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.

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.

• Sebastian P. Pincheira, Thank you very much for your answer and edit. Commented Nov 29, 2022 at 5:59