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I know this question is frequently asked on this website, but I'm not sure if my proof is correct:

Question. Prove that if each factor space $X_n$, $n=1,2,\cdots$, is separable, then so is the product space $X=\prod_{n\in\mathbb N} X_n$.

My Attempt. Let $\{X_n:n\in\mathbb N\}$ be a countable set of separable spaces, and for each $n\in\mathbb N$ let $A_n$ be a countable dense subset of $X_n$. Then $A:=\prod_{n\in\mathbb N}A_n$ is clearly countable. Now for any $x\in X$, let $O$ be an open set containing $x$. Then there exists a subset $U:=\prod_{n\in\mathbb N} U_n$ of $O$ with $U_n$ being open in $X_n$ for each $n\in\mathbb N$. Now since $A_n$ is dense in $X_n$ for any $n\in\mathbb N$, the intersection $U_n\cap A_n$ is nonempty. Then taking $y_n\in U_n\cap A_n$ for each $n\in\mathbb N$, we have that $y=(y_1,y_2,\cdots)$ is a member of $A\cap O$ and hence $A$ is dense in $X$. Therefore $X$ is separable.

What I want to ask is that in my attempt, I haven't used the fact that $U_n=X_n$ for all but finitely many $n$. Is this proof still correct, or did I miss something? All help would be appreciated.

WishofStar
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  • Hint: your set $A_n$ is not countable unless all but finitely many of the $X_n$ comprise exactly one point. See https://math.stackexchange.com/questions/244427/proving-separability-of-the-countable-product-of-separable-spaces-using-density to see how you exploit the fact that $U_n = X_n$ for all but finitely many $n$. – Rob Arthan Dec 06 '20 at 15:29
  • @RobArthan Do you mean that $A$ is not countable? But $A$ is a countable product of countable spaces $A_n$, so I think it must be countable. – WishofStar Dec 06 '20 at 17:12
  • Not so. E.g., the set of all infinite sequences of 0s and 1s is uncountable and it's a countable product of finite sets. – Rob Arthan Dec 06 '20 at 18:35

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