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

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FALSE: the product topology and the box topology agree on every product

Statement

False claim: for every family of topological spaces $(X_{i})_{i \in I}$ the product topology and the box topology on $\prod_{i \in I} X_{i}$ are the same 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).

The claim is correct for a finite index set. It fails when the index set is infinite and the factors have enough open sets, 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 needs no choice principle at all. The refutation below writes down the standard witness explicitly, in $R^{N} = \prod_{k \in N} R$ with every factor carrying the usual topology (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): the shrinking box

$B : = \prod_{k \in N} (- \frac{1}{k + 1}, \frac{1}{k + 1})$

is open in the box topology and is not open in the product topology. No choice principle is used, the factors of $B$ being given by a formula.

Facts & Assumptions

Given: The index set $N$ , the product $P : = \prod_{k \in N} R$ with each factor carrying the usual topology, the box $B$ of the statement, and the point $z \in P$ with $z_{k} = 0$ for every $k$ . Here $1 / (k + 1)$ abbreviates $1 / ι (k + 1)$ , the inverse of the canonical natural (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[A1]

A basis for the box topology is the family of all boxes $\prod_{k} U_{k}$ with every $U_{k}$ open; a basis for the product topology is the family of boxes with $U_{k} = R$ off a set listed as ${j_{0}, \dots, j_{n - 1}}$ for some natural $n$ (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, 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).

[A2]

$(a, b) = {t \in R : a < t < b}$ , and a set is open in the usual topology of $R$ exactly when each of its points has a bounded open interval around it inside the set (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]

$ι (k + 1) \geq 1 > 0$ for every $k \in N$ , and $ι$ is strictly increasing, hence injective (Canonical naturals are positive and strictly increasing, The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[L2]

If $0 < u \leq v$ then $0 < 1 / v \leq 1 / u$ (Inverses of positives are positive, and reciprocation reverses order).

[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]

If $U$ belongs to a topology and $x \in U$ , then $U$ is open; and a topology is a family of subsets of the underlying set (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

Refutation

technique · direct

1.1

For every $k \in N$ : $1 / (k + 1) > 0$ by [L1] and [L2], so $0 \in (- 1 / (k + 1), 1 / (k + 1))$ by [A2]; hence $z \in B$ .

A2L1L2

1.2

For every $k \in N$ : $1 / (k + 1) \leq 1$ , since $1 \leq ι (k + 1)$ by [L1] and [L2] applied with $u = 1$ and $v = ι (k + 1)$ .

L1L2

1.3

Each factor $(- 1 / (k + 1), 1 / (k + 1))$ is open in the usual topology of $R$ , being a bounded open interval, so $B$ is a box with open factors and hence open in the box topology.

A1A2

1.4

Suppose $B$ were open in the product topology. Then by [A1] there is a basic product-open $O = \prod_{k} O_{k}$ with $z \in O \subseteq B$ , 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 $O_{j} = R$ : if $n = 0$ the list is empty and $j : = 0$ serves; if $n \geq 1$ then by [L3] the set ${ι (j_{0}), \dots, ι (j_{n - 1})}$ has a maximum, attained at some index $m_{0} < n$ , and $j : = j_{m_{0}} + 1$ satisfies $ι (j) > ι (j_{m})$ for every $m < n$ by [L1], hence $j \neq j_{m}$ for every $m < n$ .

step 1.4L1L3

3.1

Let $y \in P$ be the point with $y_{j} : = 1$ and $y_{k} : = z_{k} = 0$ for $k \neq j$ . Then $y \in O$ , since $y_{j} = 1 \in R = O_{j}$ and $y_{k} = z_{k} \in O_{k}$ for $k \neq j$ .

step 1.4step 2.1

4.1

$y \notin B$ : by step 1.2 one has $1 / (j + 1) \leq 1 = y_{j}$ , so $y_{j} \notin (- 1 / (j + 1), 1 / (j + 1))$ by [A2].

step 1.2step 3.1A2

5.1

Steps 3.1 and 4.1 contradict $O \subseteq B$ from step 1.4, so $B$ is not open in the product topology; by step 1.3 it is open in the box topology, so the two topologies on $P$ are different and the claim is false.

step 1.3step 1.4step 3.1step 4.1∎

Remarks

What is true is the inclusion, in one direction only. The box topology is always finer than the product topology, and the two agree whenever the index set is finite (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, claims 1 and 2); it is only the converse inclusion for infinite index sets that fails, and the box above is the cheapest witness of the failure.
The failure is not an artefact of $R$ . 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 shows that any infinite family of nonempty factors, infinitely many of which have an open subset that is neither empty nor everything, produces the same separation. The real line is used here only because its open intervals are written down without effort.
The practical consequence is the failure of the characteristic property. A map into a box-topologised product can have every component continuous and still fail to be continuous; the diagonal of $R^{N}$ does exactly that, and it is worked on the companion page as The diagonal $x \mapsto (x, x, \dots)$ from $R$ into $R^{N}$ is continuous for the product topology and not for the box topology ↗. That is why the product topology, and not the box topology, is what $\prod_{i} X_{i}$ carries by default.

Depends on

Used by

The diagonal x ↦ (x,x,…) from ℝ into ℝ^ℕ is continuous for the product topology and not for the box topology Counterexample

Dependency tree · two levels

56 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

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