# Is $\displaystyle \prod_{\alpha \in \mathbb{R}} \{a,b\}$ first countable?

Suppose $\{a,b\}$ is a discrete topological space. Consider $X=\displaystyle \prod_{x \in \mathbb{R}}\{a,b\}$ under the product topology. Is $X$ first countable?

I have taken $x \in X$ and an open set $O$ containing $x$. As $X$ is product topology, there exists a basis element $\displaystyle U=\prod_{\alpha \in R}U_{\alpha}$ where a finite no. of $U_{\alpha}$ are $\{a\}$ or $\{b\}$. But I am unable to produce a countable basis at $x$.

• What makes you think that it is first-countable? – Eric Wofsey Sep 29 '16 at 6:10
• @EricWofsey, I have a hunch it is not. But I am not sure. – user340001 Sep 29 '16 at 6:25
• You can show this is actually the standard Cantor set. – 3-in-441 Sep 30 '16 at 1:11
• @3-in-441. The Cantor set is $2^N.$ Here, we have $2^R.$ – DanielWainfleet Sep 30 '16 at 6:26

Assuming $a\ne b$. Let $p=(p_r)_{r\in \mathbb R}\in \{a,b\}^{\mathbb R}.$

Let $\{U_n:n\in \mathbb N\}$ be a family of nbhds of $p.$ For each $n$ let $p\in \prod_{r\in \mathbb R} C_{n,r} \subset U_n ,$ where $C_{n,r}=\{a,b\}$ for all but finitely many $r.$

For each $n$ let $V_n=\{r: C_{n,r}\ne \{a,b\}\}.$ Then $V_n$ is finite, so $W=\cup_{n\in \mathbb N}V_n$ is countable.

Take $r^*\in \mathbb R$ \ $W$ and let $U^*=\prod_{r\in \mathbb R}C^*_r$ where $C^*_r=\{a,b\}$ when $r\ne r^*,$ and $C^*_{r^*}=\{p_{r^*}\}.$ Let $q=(q_r)_{r\in \mathbb R}$ where $q_r=p_r$ if $r\ne r^*,$ and $q_{r^*}\ne p_{r^*}.$

Now $U^*$ is a nbhd of $p,$ and $q\not \in U^*.$ But $q\in U_n$ for all $n\in \mathbb N.$ So no member of $\{U_n:n\in \mathbb N\}$ is a subset of $U^*.$ So $\{U_n:n\in \mathbb N\}$ is not a local base at $p.$

• Sorry for not responding earlier since I decided to try the question again before taking a look at the answer(and I came up with the same answer). In your solution, $U^*$ is not necessarily open since an uncountable no of $C^*_r$ are not $\{0,1\}$. Instead, we can just change one $C_r^*$ such that $r$ is not in $W$ to $\{p_{r_{*}}\}$. – user340001 Oct 5 '16 at 16:32
• $U^*$ is open because $C^*_r=\{a,b\}$ except for just one $r$, that is ,except for $r=r^*.$ – DanielWainfleet Oct 5 '16 at 23:19
• Ah. You have already fixed an $r*$. I got a little confused by the notation. Thanks! – user340001 Oct 6 '16 at 5:22