The closure of a product is the product of closures?

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.

Solution 1:

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.

Solution 2:

If $x\in\overline{\prod_J A_j}$, then $x\in\overline{p_j^{-1}[A_j]}\subseteq p^{-1}\left[\overline{A_j}\right]$ 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 $U\cap\prod_J{A_j}$ ?