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Munkres in his book states that:

Theorem 30.3 Suppose that $X$ has countable basis, then every open covering of $X$ contains a countable subcollection covering $X$.

$\textbf{Proof.}$

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 $X$.

My first doubt is when he states $A_n$ is indexed with $J \subset \mathbb{Z}^+$. Why is this true? My second doubt is about the construction of $\mathcal{A'}$: he states that $\mathcal{A'}$ is the collection of the sets $A_n$, but could have $A, A^* \in \mathcal{A'}$ such that $B_n \subset A \ \cap \ A^*$. In this case, I think that we need have one of these sets in $\mathcal{A'}$ to ensure that $\mathcal{A'}$ is countable, but how exactly do I do this?

Thanks in advance!

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  • $\begingroup$ @Vim Not for each $n$. Only for the $n$ such that some $A_n\in\mathcal A$ satisfying $B_n\subseteq A_n$ indeed exists. $\endgroup$ – drhab Apr 29 '17 at 17:39
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Addressing your first doubt, the value of $n$ ranges over the set $J$ of all $n$ such that $B_n \subset A$ for some $A \in \mathcal{A}$. Since the list of $B_n$ is a countable enumeration of the basis elements, $n$ is always a positive integer. Some integers $n$ may not be in $J$, if there is no element of $\mathcal{A}$ containing the basis element $B_n$. But in any case the set $J$ is a subset of the positive integers.

As for your second doubt, what is happening is that for every $B_n$, we choose one $A \in \mathcal{A}$ containing it (if possible, as per the first point), and label it $A_n$. By definition of $J$ this means $n \in J$. We have a surjection from $\mathbb{Z}^+$ to $J$, so $J$ must be countable. Alternately, we have a surjection from the set of all $B_n$ to the set of all $A_n$. Since the former set is countable, so must the latter. Multiple different $A_i$ may contain the same basis element $B_n$, but $B_n$ will have been associated to only one of them . The others will be chosen for different basis elements.

EDIT:


I will add more detail about the construction of the set $\mathcal{A}'$.

First, for every $n \in \mathbb{Z}^+$, define the set $\mathcal{S_n} = \{A \in \mathcal{A}: B_n \subset A\}$. Then define the set $J = \{n \in \mathbb{Z}^+: \mathcal{S_n} \neq \emptyset\}$. Now, we have the countable collection $\mathcal{S} = \{\mathcal{S_n}: n \in J\}$ consisting of nonempty sets.

By the axiom of (countable) choice, there exists a choice function, $f: J \rightarrow \bigcup\limits_{n \in J}\mathcal{S_n}$ such that $f(n) \in \mathcal{S_n}$ for every $n \in J$.

Then define $A_n = f(n)$. Since $A_n \in \mathcal{S_n}$, we have that $B_n \subset A_n$ (by definition of $\mathcal{S_n}$). This is what is meant by "choosing" an element $A_n$ containing $B_n$.

We then have the set $\mathcal{A}' = \{A_n: n \in J\}$.


When I said that "$B_n$ is associated to one of the $A_i$", that was really just a vague way of saying that there is a function $g$ from the set $\{B_n: n \in J\}$ to the set $\mathcal{A}'$. This function would just be defined by $g(B_n) = f(n) = A_n$.

By definition, a function can map $B_n$ to only one value. So only one $A_n$ is "associated" to a given $B_n$, even if there might be multiple $A_i$ containing $B_n$.

But since every $A \in \mathcal{A}'$ is (by definition) equal to $f(i)$ for some $i \in J$, this means $A = f(i) = g(B_i)$ and $g$ is surjective. So every $A$ is "associated" to some $B_i$.

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  • $\begingroup$ Thanks for the explanation of my first doubt, I understood well this now, but my second doubt remains: how exactly can I ensure that "$B_n$ will have been associated to only one of them. The others will be chosen for different basis elements."? $\endgroup$ – George Apr 29 '17 at 16:22
  • $\begingroup$ @George that is the axiom of choice at work. You choose exactly one, and the axiom of countable choice garantuees that this can be done. $\endgroup$ – Henno Brandsma Apr 29 '17 at 18:56
  • $\begingroup$ @Tob, I understood now, thanks! $\endgroup$ – George Apr 29 '17 at 19:15
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Starting with base $\mathcal B=\{B_n\mid n=1,2,\dots\}$ we arrive at $\mathcal A':=\{A_n\mid n\in J\}\subseteq\mathcal A$.

Here $J\subseteq\{1,2,\cdots\}$ with: $$n\in J\iff B_n\subseteq A\text{ for some }A\in\mathcal A$$ One of sets of $\mathcal A$ that cover $B_n$ is simply picked out and labeled as $A_n$.

It is evident that $J$ as a subset of $\{1,2,\dots\}$ is countable and that consequently $\mathcal A'$ is countable.

Does this makes things more clear to you?

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  • $\begingroup$ Yes, but, this lead us to my second doubt: $\mathcal{A'}$ is the collection of the sets $A_n$, but could have $A, A^* \in \mathcal{A'}$ such that $B_n \subset A \ \cap \ A^*$. In this case, I think that we need have one of these sets in $\mathcal{A'}$ to ensure that $\mathcal{A'}$ is countable, but how exactly do I do this? $\endgroup$ – George Apr 29 '17 at 16:12
  • $\begingroup$ I think it is a problem, because we can define $n \rightarrow A_n$ for $n \in J$, but if $B_n \subset A \ \cap \ A^*$, so $n \rightarrow A$ and $n \rightarrow A^*$, but this would imply that the function $f: J \longrightarrow \mathcal{A'}$ defined by $n \rightarrow A_n$ is not well defined $\endgroup$ – George Apr 29 '17 at 16:17
  • $\begingroup$ There can indeed be several sets $A\in\mathcal A'$ with $B_n\subseteq A$, but choosing is no problem. Just take $A_n$ (i.e. the one with equal index as $B_n$), which has been chosen on forehand. As I said: "One of the sets of $\mathcal A$ that cover $B_n$ is simply picked out and labeled as $A_n$" $\endgroup$ – drhab Apr 29 '17 at 17:22
  • $\begingroup$ We have $n\to B_n$ for every $n$ and in the special case that $n\in J$ we have $n\to B_n\to A_n$ (well-defined). $\endgroup$ – drhab Apr 29 '17 at 17:28
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Let the base be $\{B_n: n \in \mathbb{N}\}$. Let $\mathcal{A}$ be an open cover of $X$.

Then as others already said: define $J \subseteq \mathbb{N}$ by $n \in J$ iff $\exists A \in \mathcal{A}: B_n \subseteq A$.

Then $J$ is a countable set (as a subset of $\mathbb{N}$) and for every $n \in J$ the set $\mathcal{A}_n = \{A \in \mathcal{A}: B_n \subseteq A\}$ is by definition of $J$ a non-empty subcollection of $\mathcal{A}$. (your doubt is that it can have more than one element).

The Axiom of Countable Choice says that there is a function $f :J \rightarrow \cup_n \mathcal{A}_n$ such that $f(n) \in \mathcal{A}_n$ for all $n \in J$.

Then $\mathcal{A}' = \{f(n): n \in J \}$ is as required:

Suppose $x \in X$, then $\exists A_x: x \in A_x \in \mathcal{A}$, as we have a cover. So $x \in B_{n_x} \subseteq A_x$, for some $B_{n_x}$ as $A_x$ is open and the $B_n$ form a base. Then by definition, $n_x \in J$, so $f(n_x) \in \mathcal{A}_{n_x}$, so $x \in B_{n_x} \subseteq f(n_x) \in \mathcal{A}'$, so $\mathcal{A}'$ is a cover of $X$, indexed by the countable set $J$.

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