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I am reading this post where George expresses some worries he has concerning a proof in Munkres' Topology book. I myself have a different worry, although I am sure it is related. Here is the relevant passage:

Let ${B_n}$ be a countable basis and $\mathcal{A}$ an open cover of $X$. For each positive integer $n$ for which it is possible, choose an element $A_n$ of $\mathcal{A}$ containing the basis element $B_n$. The collection $\mathcal{A'}$ of the sets $A_n$ is countable, since it is indexed with a subset $J$ of the positive integers. Furthermore, it covers X: given a point $x \in X$, we can chosse an element $A$ of $\mathcal{A}$ containing $x$. Since $A$ is open, there is a basis element $B_n$ such that $x \in B_n \subset A$. Because $B_n$ lies in an element of $\mathcal{A}$, the index $n$ belong to the set $J$, so $A_n$ is defined; since $A_n$ contains $B_n$, it contains $x$. Thus $\mathcal{A'}$ is a countable subcollection of $\mathcal{A}$ that covers $X$.

I take it that the sentence "For each positive integer $n$ for which it is possible, choose an element $A_n$ of $\mathcal{A}$ containing the basis element $B_n$" means: Given $B_n \in \mathcal{B}$, where $n \in \Bbb{N}$, if there exists an $A \in \mathcal{A}$ such that $B_n \subseteq A$, then $A_n := A$; and let $\mathcal{A}'$ be the collection of all such $A_n$. I think this is an accurate reformulation (I try to avoid using modal terms in mathematical discourse). My worry is, what if it's never 'possible'; i.e., what if the antecedent is never true?

user193319
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  • Since we have a basis of the topology, it is true sufficiently often. – Daniel Fischer Oct 22 '17 at 15:45
  • @DanielFischer Okay. So the idea is that, since $A \in \mathcal{A}$ is open, $A$ can be written as $\bigcup_{k \in I} B_k$ for some indexing set $I \subseteq \Bbb{N}$; thus $B_k \subseteq A$ for every $k \in I$, and so we have this set inclusion at least for those basis elements indexed by $I$? Thus, if it occurs for $B_n$, where $n$ is not necessarily in $I$, then define $A_n := A$ and put it in $\mathcal{A}'$; and the previous sentence shows, as you put it, that the antecedent is true sufficiently often. Does this sound right? – user193319 Oct 22 '17 at 15:55
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    Not quite. Leaving aside $X = \varnothing$ (in which case the empty subfamily of $\mathcal{A}$ covers $X$), and assuming all elements of $\mathcal{A}$ nonempty, it is true that we can arrange it so that some specific element of $\mathcal{A}$ is chosen. But the point is that we don't need to do that, whatever choices we make, we get a countable subcover. Let $J = {n \in \mathbb{N} : (\exists A\in \mathcal{A})(B_n \subseteq A)}$. Then $J$ is countable (as a subset of $\mathbb{N}$), and $\mathcal{B}' = { B_n : n \in J}$ is an open cover of $X$. – Daniel Fischer Oct 22 '17 at 16:19
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    Then for every $f \colon J \to \mathcal{A}$ with $B_n \subseteq f(n)$ for all $n \in J$, it follows that $$X = \bigcup_{n \in J} B_n = \bigcup_{n \in J} f(n).$$ – Daniel Fischer Oct 22 '17 at 16:19

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Look at it this way: Take $x \in X$. As $\mathcal{A}$ is an open cover there is some $A_x \in \mathcal{A}$ that contains $x$. As $\{ B_n: n \in \mathbb{N}\}$ is a base we have that there is some $n_x$ such that $x \in B_{n_x} \subseteq A_x$. So at least for that $n_x$ we have such a member of $\mathcal{A}$ that contains $B_{n_x}$, so being a base forces that this condition will hold often, for many $n$.

The construction just efficiently chooses just $1$ such set for each $n$ for which it is possible, because we want a "small" subcover. We do use the axiom of choice in a strong way (at least countable choice, which most people don't have that much of an issue with).

Henno Brandsma
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  • How come there are uncountably many $x$'s but countably many $A_x$'s covering $X$? Is this because some $B_nx$ contain uncountably many $x$? – Tedebbur Jan 08 '19 at 17:50
  • @Serpenche it will certainly be the case that some $B_{n_x}$ has uncountably many points if $X$ is uncountable and the base is countable. – Henno Brandsma Jan 08 '19 at 17:54
  • So, this is another place where we use $Axiom$ $of$ $Choice$, right? We pick a representative among uncountably many $x$'s? – Tedebbur Jan 08 '19 at 18:02
  • I was talking about the $x$ of $n_x$, however, that $x$ was chosen at the beginning and it does not represent $B_{n_x}$. Sorry. – Tedebbur Jan 08 '19 at 18:33
  • @Serpenche for every $x$ we can pick a unique $n_x$, namely the minimal one that works. So we don’t need choice here. – Henno Brandsma Jan 09 '19 at 00:25
  • But there are uncountably many $x$'s, how do we choose a "unique" $n_x$ for each $x$? – Tedebbur Jan 10 '19 at 09:30
  • @Serpenche I just defined $n_x$ by a logical formula: $n_x = \min {n \in \mathbb{N}: \exists A \in \mathcal{A}: x \in B_n \subseteq A}$ which is well-defined as the set of integers is non-empty and $\mathbb{N}$ is well-ordered. The fact that $X$ is uncountable is irrelevant: for each real $x$ I can define a unique real number $2x$ and so the function $f(x)=2x$ is well-defined. The same holds for the function $x \to n_x$. – Henno Brandsma Jan 10 '19 at 13:48