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Does a countable product of topological spaces each of which has a countable basis have a countable basis?

user26857
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Makoto Kato
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1 Answers1

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Yes. Let $X=\prod_{n\in\Bbb N}X_n$, and for $n\in\Bbb N$ let $\mathscr{B}_n$ be a countable base for $X_n$. For each finite $F\subseteq\Bbb N$ let

$$\mathscr{B}_F=\left\{\prod_{n\in\Bbb N}U_n:U_n\in\mathscr{B}_n\text{ if }n\in F\text{ and }U_n=X_n\text{ otherwise}\right\}\;;$$

$\mathscr{B}_F$ is countable. Let

$$\mathscr{B}=\bigcup\{\mathscr{B}_F:F\subseteq\Bbb N\text{ is finite}\}\;;$$

$\mathscr{B}$ is the union of countably many countable sets and is a base for $X$.

Brian M. Scott
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  • @William: There’s an obvious bijection between $\mathscr{B}F$ and $\prod{n\in F}\mathscr{B}_n$. The latter is the product of finitely many countable sets and therefore is countable. – Brian M. Scott Feb 26 '15 at 00:24
  • How do you prove that the set of finite subsets of the set of natural numbers is countable? – Makoto Kato Feb 26 '15 at 00:27
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    @William: For each $n\in\Bbb N$ let $\mathscr{F}_n$ be the family of $n$-element subsets of $\Bbb N$; clearly every finite subset of $\Bbb N$ is in one of the $\mathscr{F}_n$, and there are only countably many of them, so it suffices to see that each $\mathscr{F}_n$ is countable. For each $n$ there is an easy injection from $\mathscr{F}_n$ into $\Bbb N^n$; the latter, as the product of finitely many countable sets, is countable, so $\mathscr{F}_n$ is countable as well. – Brian M. Scott Feb 26 '15 at 00:32
  • How do you prove that a countable union of countable sets is countable? – Makoto Kato Mar 05 '15 at 03:22
  • @William: You’ll find proofs here, here, and here. – Brian M. Scott Mar 05 '15 at 03:50
  • So we need the axiom choice to prove that the question is affirmative. Right? – Makoto Kato Mar 07 '15 at 04:54
  • @William: Only the weaker axiom of countable choice, and no choice at all if each of the spaces comes equipped with a counted base, i.e., one with a specified indexing by $\Bbb N$ or a subset thereof. (It’s not really something to worry about, considering that one of the most important and useful theorems in topology is actually equivalent to $\mathsf{AC}$.) – Brian M. Scott Mar 07 '15 at 04:59
  • "one of the most important and useful theorems in topology is actually equivalent to AC." What is it? – Makoto Kato Mar 07 '15 at 05:02
  • @William: The Tikhonov product theorem, that says that an arbitrary product of compact spaces is compact. – Brian M. Scott Mar 07 '15 at 05:03
  • We can prove that a coutable product of metric compact spaces is compact without AC. No? – Makoto Kato Mar 07 '15 at 05:06
  • @William: What do you mean by usual? – Brian M. Scott Mar 07 '15 at 05:08
  • Metric compact spaces – Makoto Kato Mar 07 '15 at 05:08
  • @William: No: it requires some choice just to prove that all products of countably infinitely many finite discrete spaces are compact. – Brian M. Scott Mar 07 '15 at 05:10
  • That's interesting. Could you elaborate? – Makoto Kato Mar 07 '15 at 05:12
  • @William: This PDF has the standard proof that the full Tikhonov theorem implies $\mathsf{AC}$. You can easily adapt it to show that if countable products of finite discrete spaces are compact, then arbitrary products $\prod_{n\in\Bbb N}X_n$ of non-empty finite sets $X_n$ are non-empty, i.e., the axiom of choice for countable families of finite sets. This is known to be independent of $\mathsf{ZF}$. – Brian M. Scott Mar 07 '15 at 05:19
  • In most pracatical examples, a countable product of metric compact spaces can be trivially proved to be non-empty without AC. For example, the countable product of the interval $[0, 1]$. – Makoto Kato Mar 07 '15 at 05:41
  • @William: Speaking as a set-theoretic topologist, I really have no idea what you mean by practical: that’s not a distinction that I make. – Brian M. Scott Mar 07 '15 at 05:43
  • I mean a case in which a countable family of compact metric spaces is given concretely or constructibly. – Makoto Kato Mar 08 '15 at 03:34
  • @William: That’s simply not a distinction that I find useful or very interesting. – Brian M. Scott Mar 08 '15 at 07:54