Could any one tell me how to prove this one?

The product topology has a countable base if and only if the topology of each coordinate space has a countable base and all but a countable number of coordinate spaces are indiscrete

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    $\begingroup$ If one of the coordinate spaces is empty, it won't matter how many others are indiscrete. $\endgroup$ – GEdgar Aug 25 '12 at 23:37
  • $\begingroup$ Why would you expect that to be true? You can improve your question by giving a fuller context for where you encountered the question and what you have tried. $\endgroup$ – Carl Mummert Aug 26 '12 at 2:54

The harder direction is to show that the product space has a countable base, then each of the factors also has a countable base, and all but countably many of them have the indiscrete topology.

Let $X=\prod_{i\in I}X_i$, and let $\mathscr{B}$ be a countable base for $X$. Fix $x=\langle x_i:i\in I\rangle\in X$. For $i\in I$ let $$Y_i=\Big\{\langle y_j:j\in I\rangle\in X:y_j=x_j\text{ for all }j\in I\setminus\{i\}\Big\}\;;$$ if $\pi_i:X\to X_i$ is the projection map, it’s a standard elementary result that $\pi_i\upharpoonright Y_i$ is a homeomorphism of $Y_i$ onto $X_i$. In particular, for each $B\in\mathscr{B}$ and $i\in I$, $\pi_i[B\cap Y_i]$ is open in $X_i$. For $i\in I$ let $\mathscr{B}_i=\{\pi_i[B\cap Y_i]:B\in\mathscr{B}\}$; clearly each $\mathscr{B}_i$ is countable, and I claim that $\mathscr{B}_i$ is a base for $X_i$.

Suppose that $p\in U\subseteq X_i$, where $U$ is open. Let $y$ be the unique point of $Y_i$ such that $\pi_i(y)=p$. Then $\pi_i^{-1}[U]$ is an open nbhd of $y$ in $X$, so there is some $B\in\mathscr{B}$ such that $y\in B\subseteq \pi_i^{-1}[U]$. It follows that $p\in\pi_i[B\cap Y_i]\subseteq U$, where $\pi_i[B\cap Y_i]\in\mathscr{B}_i$; this establishes the claim.

Now let $I_0$ be the set of $i\in I$ such that the topology on $X_i$ is not indiscrete, and suppose that $I_0$ is uncountable. For each $i\in I_0$ let $U_i$ be a non-empty proper open subset of $X_i$. Without loss of generality assume that the point $x$ fixed at the beginning of the proof is such that $x_i\in U_i$ for all $i\in I_0$. Then $x\in\pi_i^{-1}[U_i]$ for each $i\in I_0$, so for each $i\in I_0$ there is a $B(i)\in\mathscr{B}$ such that $x\in B(i)\subseteq\pi_i^{-1}[U_i]$. $\mathscr{B}$ is countable, and $I_0$ is uncountable, so there are a $B\in\mathscr{B}$ and an uncountable $I_1\subseteq I_0$ such that $B(i)=B$ for every $i\in I_1$. Clearly $x\in B\subseteq\prod_{i\in I_1}U_i\times\prod_{i\in I\setminus I_1}X_i$.

On the other hand, $B$ is open in the product topology on $X$, so so there are a finite $F\subseteq I$ and open sets $V_i$ in $X_i$ for $i\in F$ such that

$$x\in\prod_{i\in F}V_i\times\prod_{i\in I\setminus F}X_i\subseteq B\subseteq\prod_{i\in I_1}U_i\times\prod_{i\in I\setminus I_1}X_i\;.$$

Pick any $i\in I_1\setminus F$; then $X_i\subseteq\pi_i[B]\subseteq U_i\subsetneqq X_i$, which is absurd. Hence $I_0$ must be countable.

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$(\Rightarrow)$: Suppose $\{X_i\}_{i\in I}$ is a collection of spaces with all but countably many indiscrete, let $\{B_{ij}\}_{j\in \mathbb N}$ be a countable basis for each $X_i$, and let $X$ be their product. The sets of the form $\prod_{i\in I} B_{ij_i}$ with all but finitely many $B_{ij_i}$ equal to the whole set $X_i$ form a basis for $X$. Note that when constructing $\prod_{i\in I} B_{ij_i}$ there are only countably many $i$ for which $B_{ij_i}$ could be not the whole set. Thus there are only countably many ways to choose a finite number of $i$ to make $B_{ij_i}$ not the whole set (since $B_{ij_i}=X_i$ for $X_i$ indiscrete). Furthermore, for any fixed choice of finitely many values of $i$ to make $B_{ij_i}$ not the whole set, there are only countably many ways to choose the $B_{ij_i}$. Thus we have a countable basis.

$(\Leftarrow)$: Suppose some $\{X_i\}_{i\in I}$ is a collection of spaces with product $X$ which has a countable basis $\{B_j\}_{j\in\mathbb N}$. Let $\pi_i:X\to X_i$ denote projection onto the $i^{th}$ component. Then each $X_i$ has a countable basis $\{X_i\cap \pi_i[B_j]\}_{j\in\mathbb N}$. Furthermore, if uncountably many of the $X_i$ are indiscrete then uncountably many of the sets $\{X_i\cap \pi_i[B_j]\}_{j\in \mathbb N}$ contain sets other than $X_i$, yet for fixed $j$ this can only be true for finitely many sets (for any open set $U$ in $X$, $\pi_i(U)=X_i$ for all but finitely many $i$) and there are only countably many choices for $j$, thus we have a contradiction.

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  • $\begingroup$ In the first sentence you want with all but countably many indiscrete. In the second paragraph $X_i$ isn’t a subset of $X$, so $X_i\cap B_j$ doesn’t make sense; you have to work a little harder than that. $\endgroup$ – Brian M. Scott Aug 25 '12 at 23:14
  • $\begingroup$ @BrianM.Scott Thanks for the correction. For $X_i\cap B_j$, the correct notation is rather torturous and the inclusion $X_i\to X$ is obvious, so I was under the impression that this was a standard abuse of notation. $\endgroup$ – Alex Becker Aug 25 '12 at 23:18
  • $\begingroup$ I would never use it, and I’d take off for it at least in an undergraduate or first-year graduate course. (I’d comment negatively on it at any level.) At the very least it’s likely to be confusing for someone who is just learning to work with products. $\endgroup$ – Brian M. Scott Aug 25 '12 at 23:23
  • $\begingroup$ @BrianM.Scott I've revised the answer. $\endgroup$ – Alex Becker Aug 25 '12 at 23:28
  • $\begingroup$ $\pi_i[B_j]\subseteq X_i$, so you don’t need to intersect it with $X_i$. However, there’s still a bit of work to do to show that $\{\pi_i[B_j]:j\in\Bbb N\}$ is a base for $X_i$: you can’t assume that the $B_j$ belong to the standard base for the product. I’ve gone ahead and written up my version of that direction. $\endgroup$ – Brian M. Scott Aug 26 '12 at 0:24

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