This exercise is given in Munkres, but it is not stated in which topology that it holds. Do I take the product topology or box topology?
Additionally, I have a question on how to solve this problem. Let $A = A_1 \times A_2 \dots$ be the space, and let $B_n$ be the countable dense subset of $A_n$. Can I take $B = B_1 \times B_2 \times \cdots$ to be the countable dense subset, or do I have to be more clever than that?
Best Answer
It’s the product topology, and you have to be a bit cleverer than that, because $\prod_nB_n$ is in general not countable.
HINT: For each $n$ fix a point $x_n\in B_n$, and consider the set of points of your $B$ that are equal to $x_n$ for all but at most finitely many indices.
In fact a stronger result is true: the product of $2^\omega=\mathfrak{c}=|\Bbb R|$ separable spaces is separable, but this is significantly harder to prove. See this question for more information.