The product of a sequence of separable topological spaces is separable.

*Proof.*

Let be a sequence of separable spaces. For each , let denote a countable dense subset. Let have the product topology, . Let , . Let , which is clearly countable, since it is a countable union of countable sets.

Now, I will show that is dense in . Let , be a basis element of , such that . Then, , where is basis element of , or . Furthermore, there exists , such that implies . Now, for each , with , since is dense in . Then, . Thus, and so .

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This solution is good. You might say fix some element x \in X