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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?

Martin Sleziak
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Schmidt
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1 Answers1

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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.

Community
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Brian M. Scott
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  • Is it true that a set $E$ is dense in $X$ if every open basis element of $X$ contains points in $E$? Because if so, then this set is easily seen to be a countable dense subset. – Schmidt Mar 03 '16 at 00:39
  • @Schmidt: Yes, it’s true. If $\mathscr{B}$ is a base for the topology of a space $X$, a set $D\subseteq X$ is dense in $X$ if and only $B\cap D\ne\varnothing$ for each non-empty $B\in\mathscr{B}$. This isn’t hard to prove. – Brian M. Scott Mar 03 '16 at 00:41