If $\{X_j:j\in J\}$ is a family of topological spaces and $A_j\subseteq X_j$, is it true that $\displaystyle\overline{\Pi_{j\in J}A_j}=\displaystyle{\Pi_{j\in J}\overline{A_j}}$? Is there an easy way to prove this?
Of course, we are considering in $\displaystyle{\Pi_{j\in J}X_j}$ the product topology.
Thanks.
If $x\in\overline{\prod_J A_j}$, then $x\in\overline{p_j^{-1}[A_j]}\subseteq p^{-1}_j\left[\overline{A_j}\right]$ for all $j\in J$, so $x\in\prod_J\overline{A_j}$.
For the other inclusion, let $x=(x_j)_J\in\prod_J\overline{A_j}$. This is equivalent to $x_j$ being in $\overline{A_j}$ for each $j\in J$. Now consider a basic neighborhood $U=\prod_J U_j$, where all $U_j$ are open subsets of $X_j$ and almost all $U_j$ are equal to $X_j$. Can you find a $b=(b_j)_J$ in $U\cap\prod_J{A_j}$ ?
Hint: try to prove both inclusions. I am sure that the following characterization turns out to be very useful: $x \in \overline{A}$ if and only if for every open set $U$ containing $x$ we have $U \cap A \neq \emptyset$. Recall also that you need to consider only basic open set.
1.$C=\prod_j\bar{A_j}$ is closed in product topology (hence also closed in box topology): for any $x\in C^c$, there exists $j_0$ such that $x_{j_0}\notin \bar{A_{j_0}}$. $U= \bar{A_{j_0}}\times \prod_{j\neq j_0}X_j$ is open in product topology and $x\in U\subset C^c$. So $C^c$ is open in product topology.
2.Put $A=\prod_jA_j$. Let $A^{cl}$ be the closure of $A$ wrt box topology. Since box topology is finer than product topology, $A^{cl}\subset \bar{A}$. As $A\subset C$, by 1 we have $\bar{A}\subset C$. For any $x\in C$, any $V=\prod_iV_i$ containing $x$, where each $V_i\subset X_i$ is open, we have $x_i\in V_i\cap \bar{A_i}$. There is $y_i\in V_i\cap A_i$. Then $y=(y_i)\in V\cap A$. So $x\in A^{cl}$. Thus we find $A^{cl}=\bar{A}=C$.
