False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-04 (gpt-5.6-sol-codex-subscription)

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FALSE: $\prod_{i} U_{i}$ is open in the product topology whenever every $U_{i}$ is open

Statement

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 $R^{N}$ with no choice principle at all.

The refutation uses $R^{N} = \prod_{k \in N} R$ with the usual topology on each factor (The absolute value makes $R$ 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 $R$ .

Facts & Assumptions

Given: The product $P : = \prod_{k \in N} R$ with the product topology, the set $C : = \prod_{k \in N} (0, 1) \subseteq P$ , and the point $c \in P$ with $c_{k} = 1 / 2$ for every $k$ , where $1 / 2$ is the inverse of $ι (2)$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[A1]

A basis for the product topology on $P$ is the family of boxes $\prod_{k} O_{k}$ with every $O_{k}$ open in $R$ and $O_{k} = R$ off a list $j_{0}, \dots, j_{n - 1}$ with $n \in N$ (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).

[A2]

$(0, 1) = {t \in R : 0 < t < 1}$ ; it is nonempty and $0 < 1 / 2 < 1$ , since $a < (a + b) / 2 < b$ whenever $a < b$ ; and $(0, 1)$ is open in the usual topology of $R$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, The absolute value makes $R$ 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).

[L1]

$0 < 1$ , so $1 < 1 + 1 = 2$ (The multiplicative identity is positive).

[L2]

$ι$ is strictly increasing on $N$ , hence injective (Canonical naturals are positive and strictly increasing, The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[L3]

For every natural $n \geq 1$ and reals $a_{0}, \dots, a_{n - 1}$ the set ${a_{0}, \dots, a_{n - 1}}$ has a maximum (Every nonempty finite set of reals has a maximum and a minimum).

[L4]

A topology is a family of subsets of the underlying set, and every member of a basis of it is open (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

Refutation

technique · direct

1.1

$c \in C$ , since $c_{k} = 1 / 2 \in (0, 1)$ for every $k$ by [A2].

1.2

$2 \notin (0, 1)$ , since $1 < 2$ by [L1] and membership of $(0, 1)$ requires $t < 1$ by [A2].

A2L1

1.3

Suppose $C$ were open in the product topology. Then by [A1] there is a basic product-open $O = \prod_{k} O_{k}$ with $c \in O \subseteq C$ , and $O_{k} = R$ for every $k$ outside a list $j_{0}, \dots, j_{n - 1}$ with $n \in N$ .

A1L4assume-hyp

2.1

There is $j \in N$ with $O_{j} = R$ : for $n = 0$ the list is empty and $j : = 0$ serves; for $n \geq 1$ the set ${ι (j_{0}), \dots, ι (j_{n - 1})}$ has a maximum by [L3], attained at some $m_{0} < n$ , and $j : = j_{m_{0}} + 1$ satisfies $ι (j) > ι (j_{m})$ , hence $j \neq j_{m}$ , for every $m < n$ by [L2].

step 1.3L2L3

3.1

Let $y \in P$ be the point with $y_{j} : = 2$ and $y_{k} : = c_{k} = 1 / 2$ for $k \neq j$ . Then $y \in O$ , since $y_{j} \in R = O_{j}$ and $y_{k} = c_{k} \in O_{k}$ for $k \neq j$ , using $c \in O$ .

step 1.3step 2.1

4.1

$y \notin C$ , since $y_{j} = 2 \notin (0, 1)$ by step 1.2.

step 1.2step 3.1

5.1

Steps 3.1 and 4.1 contradict $O \subseteq C$ from step 1.3, so $C$ is not open in the product topology although every factor $(0, 1)$ is open in $R$ ; the claim is therefore false.

step 1.1step 1.3step 3.1step 4.1∎

Remarks

The correct statement, and why the finiteness is there. The basic open sets of a product are the finite intersections of the sets $π_{i}^{- 1} [U]$ , 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 $R^{N}$ , 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.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

55 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Product topology (Wikipedia) (standard reference, not scraped)
Box topology (Wikipedia) (standard reference, not scraped)

FALSE: ∏iUi∏i​Ui​ is open in the product topology whenever every UiUi​ is open