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Prove that the product of a countable number of separable spaces is separable space.

If each $(X_i,T_i)$ is separable, let $A_i \subseteq X_i$ be a countably dense subset.

Then $ cl(A_1 \times A_2 \times ...) =cl(A_1) \times cl(A_2) \times ... = X_1 \times X_2 \times ... = \prod_{i=1}^{\infty}(X_i,T_i)$. This shows it's dense but I have a feeling $\prod A_i$ is not countable because isn't an infinite product of countably infinite spaces an uncountable space?

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Yes, that infinite product is uncountable, unless all but finitely many of the factors are singletons.

Hint. For each $i$ pick a "base point" $b_i\in A_i.$ Let $S$ be the subset of $A_1\times A_2\times\cdots$ consisting of points $(x_1,x_2,\dots)$ such that $x_i=b_i$ for all but finitely many $i.$ Show that $S$ is countable and dense in $X_1\times X_2\times\cdots.$

Trickier but true: the product of $2^{\aleph_0}$ separable spaces is separable.

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Recall that a basic open set of $\prod_n X_n$ is of the form $u=u_1\times u_2\times\cdots u_n\times X_{n+1}\times X_{n+2}\times\cdots$ where $u_k$ is an open set in $X_k$.

For each $k>1$ pick some fixed element $o_k\in X_k$.

Let $A=\{(a_1,a_2,\cdots a_n,o_{n+1},o_{n+1},\cdots)\vert a_1\in A_1,\cdots,a_n\in A_n\}$

Then $A$ is a countable dense subset of $\prod_n X_n$.

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