Cartesian Product Dense Let $\left\{ \left(X_\alpha, \tau_\alpha\right)\right\}$ be a family of topological spaces, and let $ X=\prod_{\alpha}^{}{X_\alpha}$ , prove $\prod_{\alpha}^{}{A_\alpha}$ is dense in $\prod_{\alpha}^{}{X_\alpha}$ if and only if each $A_\alpha \subseteq X_\alpha$ is dense. 
 A: ($\rightarrow$)  $cl(\prod A_\alpha)=cl(\bigcap \pi_\alpha^{-1}(A_\alpha))\subseteq \bigcap cl(\pi_\alpha^{-1}(A_\alpha))\subseteq $
$\subseteq \bigcap \pi_\alpha^{-1}(cl(A_\alpha))=\bigcap \pi_\alpha^{-1}(X_\alpha)=\prod X_\alpha$
because $\pi_\beta$ is continuos and for each continuos function $f: X\to Y$ you have that $ cl(f^{-1}(A))\subseteq f^{-1}(cl(A))$
($\leftarrow$) Let $A$ an open set of $ X_\beta$, then $\pi_\beta^{-1}(A)$ is an open set of $\prod X_\alpha$ but $\prod A_\alpha$ is dense so 
$(\prod A_\alpha)\cap \pi_\beta^{-1}(A)\neq \emptyset$
then 
$A_\beta\cap A\neq \emptyset$
A: Suppose each $A_\alpha$ is dense in its $X_\alpha$. Then let $O=\prod_\alpha O_\alpha$ be a basic non-empty open subset of the product, so all $O_\alpha$ are non-empty open in $X_\alpha$ and all but finitely many are equal to $X_\alpha$ itself.
For those finitely many non-trivial $O_\alpha$ we known that they intersect $A_\alpha$ by that set being dense, and for the other $\alpha$ that’s trivial. So any non-empty basic open subset of the product intersects $\prod_\alpha A_\alpha$ and this implies that latter set is dense in the product.
If, on the other hand, $\prod_\alpha A_\alpha$ is given to be dense in the product, fix any arbitrary $\alpha_0$ and consider $O$ open and non-empty in $X_{\alpha_0}$. Then $\pi_{\alpha_0}^{-1}[O]$ is (basic) open in the product and so intersects $\prod_\alpha A_\alpha$ in some point $(a_\alpha)_\alpha$. Then clearly, $a_{\alpha_0} \in A_{\alpha_0} \cap O$ and so $A_{\alpha_0}$ is dense in $X_{\alpha_0}$ and as the index was arbitrary, all $A_\alpha$ are dense in $X_\alpha$. 
