Products commute with subspaces; for infinite nonempty families, the closure identity $\overline{\prod A_i}=\prod \overline{A_i}$ uses the Axiom of Choice

Let $(X_i,{\mathcal{T}}_i{)}_{i\in I}$ be topological spaces, let $A_i\subseteq X_i$ for each $i$, and give ${\prod }_i{X}_i$ the product topology (The product set ${\prod }_{i\in I}{X}_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space). Then:

1. Subspaces commute with products. The product of the subspace topologies on the $A_i$ (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace) is exactly the subspace topology that ${\prod }_i{A}_i$ inherits from ${\prod }_i{X}_i$. So the phrase "the product of the subspaces $A_i$" names one topology, whichever of the two routes is taken.
2. Closure of a product is the product of the closures. In ${\prod }_i{X}_i$, $\overline{{\prod }_{i\in I}{A}_i}={\prod }_{i\in I}\overline{A_i},$ closures being taken in ${\prod }_i{X}_i$ and in $X_i$ respectively (Interior, closure, boundary, exterior, derived set and isolated point in a topological space). In particular ${\prod }_i{A}_i$ is closed in the product whenever every $A_i$ is closed in $X_i$.

No hypothesis of nonemptiness is imposed in claim 2: if some $A_{i_0}$ is empty then both sides are empty, since $\bar{\mathrm{\varnothing }}=\mathrm{\varnothing }$. When every $A_i$ is nonempty the inclusion $\supseteq$ of claim 2 uses the Axiom of Choice for infinite $I$ (The Axiom of Choice) and Every natural-number-indexed list of nonempty sets has a choice function on its family of values for $I$ a natural number, and that is the only place in the item where a choice principle appears.

• Claim 1 is what lets "$\prod A_i$" be written without a warning. Every later item that forms a product of subspaces, the Hilbert cube and the Cantor set among them, silently uses it: the topology on $[0,1{]}^{\mathbb{N}}$ obtained by taking the product of the subspaces $[0,1]\subseteq \mathbb{R}$ is the topology it inherits as a subset of ${\mathbb{R}}^{\mathbb{N}}$.

• Claim 2 fails for the box topology in the direction one might expect it to hold. Nothing above is claimed for ${\mathcal{T}}^{\mathrm{\square }}$. The inclusion $\bar{A}\subseteq \prod \overline{A_i}$ survives there without any choice principle, since it uses only continuity of the projections, which holds for the box topology as well. The reverse inclusion also holds for the box topology, but only with the Axiom of Choice (The Axiom of Choice): given $x\in \prod \overline{A_i}$ and a box ${\prod }_i{U}_i$ around $x$, every $U_i\cap A_i$ is nonempty, and a choice function picks a point of $\left({\prod }_i{U}_i\right)\cap {\prod }_i{A}_i$. What is genuinely not claimed here is a choice-free proof of that half.

• The choice is spent on the coordinates that the basic open set leaves unrestricted. Those are all but finitely many, and for each of them the point $y$ of step 2.2 needs some member of $A_i$; the finitely many restricted coordinates are handled by Every natural-number-indexed list of nonempty sets has a choice function on its family of values alone. That split is exactly why the finite case of claim 2 is a theorem of ZF.