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

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?

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.

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 $\prod_n X_n$.