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


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 $X$.

  • $\begingroup$ @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. $\endgroup$ – Brian M. Scott Feb 26 '15 at 0:24
  • $\begingroup$ How do you prove that the set of finite subsets of the set of natural numbers is countable? $\endgroup$ – Makoto Kato Feb 26 '15 at 0:27
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    $\begingroup$ @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. $\endgroup$ – Brian M. Scott Feb 26 '15 at 0:32
  • $\begingroup$ How do you prove that a countable union of countable sets is countable? $\endgroup$ – Makoto Kato Mar 5 '15 at 3:22
  • $\begingroup$ @William: You’ll find proofs here, here, and here. $\endgroup$ – Brian M. Scott Mar 5 '15 at 3:50

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