FALSE: $\prod_i U_i$ is open in the product topology whenever every $U_i$ is open

False claim: if $U_i$ is open in $X_i$ for every $i \in I$, then $\prod_{i \in I} U_i$ is open in $\prod_{i \in I} X_i$ with 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).

What is true is the version with a restriction on how many factors may be cut down: $\prod_i U_i$ is open in the product topology when every $U_i$ is open and $U_i = X_i$ for all but finitely many $i$, those being exactly the basic product-open sets (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, Basis and subbasis for a topology, and the topology generated by a family of sets). The unrestricted claim is the definition of the box topology, which is finer, and is strictly finer under the hypotheses of claim 3 of The box topology is finer than the product topology, the two agree for a finite index set in ZF, and, assuming the Axiom of Choice for nonempty factors, the box topology is strictly finer whenever infinitely many factors have a nonempty proper open subset, which assumes the Axiom of Choice (The Axiom of Choice); the witness written out below exhibits the strictness in $\mathbb{R}^{\mathbb{N}}$ with no choice principle at all.

The refutation uses $\mathbb{R}^{\mathbb{N}} = \prod_{k \in \mathbb{N}}\mathbb{R}$ with the usual topology on each factor (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$, and it is unbounded, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not) and the single open set $U_k := (0,1)$ in every factor: $\prod_{k} (0,1)$ is not open in the product topology, although $(0,1)$ is open in $\mathbb{R}$.

• The correct statement, and why the finiteness is there. The basic open sets of a product are the finite intersections of the sets $\pi_i^{-1}[U]$, and each of those constrains one coordinate only; a finite intersection therefore constrains finitely many coordinates. Constraining all of them at once, as $C$ does, is a box, and a box need not be a union of such finite intersections.

• Nothing is wrong with $\prod_k (0,1)$ as a set or as a space. It is a perfectly good subspace of $\mathbb{R}^{\mathbb{N}}$, and by claim 1 of Products commute with subspaces; for infinite nonempty families, the closure identity $\overline{\prod A_i}=\prod \overline{A_i}$ uses the Axiom of Choice its subspace topology is the product of the subspace topologies of the factors. What fails is only that it is not an open subset of the ambient product.

• The same computation with shrinking intervals gives the sharper failure. Replacing $(0,1)$ by $(-1/(k+1),\ 1/(k+1))$ produces a box whose only product-interior point would have to have all but finitely many coordinates unrestricted, and that box separates the two topologies outright; that is the false statement immediately before this one.