I'm working on a proof of $\prod_{\alpha\in\Lambda}\overline{A_\alpha}=\overline{\prod_{\alpha\in\Lambda}A_\alpha}$ in the product topology. This has been asked before, i.e. Closure in a product of topological spaces, The closure of a product is the product of closures? but they aren't explicit on the parts which are giving me trouble.
If $(C_\alpha)_{\alpha\in\Lambda}$ is a collection of closed sets, the product $\prod_{\alpha\in\Lambda}C_\alpha$ can be written as $\bigcap_{\alpha\in\Lambda}\pi_\alpha^{-1}(C_\alpha)$, and continuity of $\pi_\alpha$ implies $\pi_\alpha^{-1}(C_\alpha)$ is closed, so $\prod_{\alpha\in\Lambda}C_\alpha$ is also closed. Setting $C_\alpha=\overline{A_\alpha}$ this proves $\prod_{\alpha\in\Lambda}\overline{A_\alpha}\supseteq\overline{\prod_{\alpha\in\Lambda}A_\alpha}$.
For the reverse inclusion, suppose $x=(x_\alpha)_{\alpha\in\Lambda}\in\prod_{\alpha\in\Lambda}\overline{A_\alpha}$, and let $U=\prod_{\alpha\in\Lambda}U_\alpha$ be a basic open set containing $x$, with $I\subseteq \Lambda$ finite such that $U_\alpha=X_\alpha$ for $\alpha\notin I$. Since $x_\alpha\in \overline{A_\alpha}\cap U_\alpha$, there is a $y_\alpha\in A_\alpha\cap U_\alpha$, so an application of the axiom of choice would give a $y=(y_\alpha)_{\alpha\in\Lambda}\in$ $\prod_{\alpha\in\Lambda}A_\alpha\cap U$ as desired.
Is AC necessary here? So far I haven't even been able to show that $\prod_{\alpha\in\Lambda}\overline{A_\alpha}\ne\emptyset$ implies $\prod_{\alpha\in\Lambda}A_\alpha\ne\emptyset$, and surely the theorem can't be proven without establishing this.
The theorem is equivalent of the axiom of choice. Suppose the theorem is true. First, we can recover the weak form mentioned at the end:
Proof: If $\prod_{\alpha\in\Lambda}A_\alpha =\emptyset$, then $\prod_{\alpha\in\Lambda}\overline{A_\alpha}=\overline{\prod_{\alpha\in\Lambda}A_\alpha }=\overline{\emptyset}=\emptyset$.
Now let us show that $\forall{\alpha\in\Lambda},A_\alpha\ne\emptyset\implies \prod_{\alpha\in\Lambda}A_\alpha\ne\emptyset$ (where the $A_\alpha$ are just sets, not members of a topological space), which is equivalent to the axiom of choice.
Let $X_\alpha=A_\alpha\sqcup\{a_\alpha\}$ (the disjoint union of $A_\alpha$ and a new point), and topologize $X_\alpha$ using the excluded point topology, in which a set is open iff it is either $X_\alpha$ or a subset of $A_\alpha$. Then given any open set containing $a_\alpha$, the definition of the topology implies that the set must be all of $X_\alpha$, and hence must meet $A_\alpha$ (which we are assuming is non-empty), so $a_\alpha\in\overline{A_\alpha}$ for this topology. Thus $(a_\alpha)_{\alpha\in\Lambda}$ is an element of $\prod_{\alpha\in\Lambda}\overline{A_\alpha}$, and applying the theorem above we get $\prod_{\alpha\in\Lambda}A_\alpha\ne\emptyset$ as desired.