TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-04 (gpt-5.6-sol-codex-subscription)

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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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

Statement

Let $(X_{i}, T_{i})_{i \in I}$ be topological spaces, let $P : = \prod_{i \in I} X_{i}$ , and let $T^{Π}$ and $T^{□}$ be the product and the box topology on $P$ (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:

$T^{Π} \subseteq T^{□}$ : the box topology is finer than the product topology (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).
If $I$ is a natural number then $T^{Π} = T^{□}$ ; this is a theorem of ZF.
Assume the Axiom of Choice. Suppose every $X_{i}$ is nonempty and let $J : = {i \in I : X_{i} has an open subset U with \emptyset \neq U \neq X_{i}} .$ If $J$ is not finite then $T^{Π} ⊊ T^{□}$ : the inclusion of claim 1 is strict.

Claim 3 spends the Axiom of Choice twice (The Axiom of Choice), once to produce a point of $P$ and once to select an open set together with a point of it in each factor indexed by $J$ ; both uses are flagged at the steps that make them. The hypothesis is stated in terms of open subsets rather than as "infinitely many factors are non-trivial", because a factor may have more than one point and still have no open set other than $\emptyset$ and itself, as the indiscrete topology shows (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), and for such a factor the conclusion fails.

Facts & Assumptions

Given: Topological spaces $(X_{i}, T_{i})_{i \in I}$ , the product set $P = \prod_{i \in I} X_{i}$ with its two topologies, and the set $J$ of the statement.

[A1]

A basis for $T^{Π}$ is the family $R^{Π}$ of boxes $\prod_{i} U_{i}$ with every $U_{i}$ open and $U_{i} = X_{i}$ for all but finitely many $i$ ; a basis for $T^{□}$ is the family $R$ of all boxes $\prod_{i} U_{i}$ with every $U_{i}$ open (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).

[A2]

A basic product-open set is an intersection of finitely many sets $π_{i}^{- 1} [U]$ , so its exceptional index set ${i : U_{i} \neq X_{i}}$ is contained in a set listed as ${i_{0}, \dots, i_{n - 1}}$ 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, A family is a basis for a unique topology iff it covers the set and every point of an intersection of two members lies in a member inside that intersection; finite intersections of any subbasis form a basis).

[L1]

If $B_{1} \subseteq B_{2}$ are bases for topologies $T_{1}$ and $T_{2}$ on the same set, then $T_{1} \subseteq T_{2}$ , every member of $T_{1}$ being a union of members of $B_{1} \subseteq B_{2} \subseteq T_{2}$ (Basis and subbasis for a topology, and the topology generated by a family of sets, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[L2]

A subset of a finite set is finite; this is fact (i) discharged in The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies.

[L3]

If every member of a family of nonempty sets is nonempty then the family has a choice function, and the product of the family is nonempty; this is the Axiom of Choice (The Axiom of Choice, Choice function).

Proof

technique · direct

1.1

$R^{Π} \subseteq R$ , a box with all but finitely many factors equal to $X_{i}$ being in particular a box.

1.2

If $I$ is a natural number $n$ then $R \subseteq R^{Π}$ , since the exceptional index set of any box is a subset of $n$ and hence finite by [L2], so every box is basic product-open.

A1L2

1.3

Assume every $X_{i}$ is nonempty and $J$ is not finite. For $i \in J$ let $A_{i} : = {(U, u) : U \in T_{i}, u \in U, U \neq X_{i}}$ ; each $A_{i}$ is nonempty, since $J$ supplies an open $U$ with $\emptyset \neq U \neq X_{i}$ and $U$ nonempty supplies a $u \in U$ .

givenL3

2.1

By [L3] applied to $(A_{i})_{i \in J}$ there are $(U_{i}, u_{i}) \in A_{i}$ for every $i \in J$ ; and by [L3] applied to $(X_{i})_{i \in I}$ there is a point $p \in P$ .

step 1.3L3choose

2.2

Claim 1 follows from step 1.1 and [L1], and claim 2 from steps 1.1 and 1.2 with [L1] applied in both directions.

step 1.1step 1.2L1

3.1

Define $x \in P$ by $x_{i} : = u_{i}$ for $i \in J$ and $x_{i} : = p_{i}$ for $i \notin J$ , and put $B : = \prod_{i} V_{i}$ with $V_{i} : = U_{i}$ for $i \in J$ and $V_{i} : = X_{i}$ otherwise. Then $B \in R \subseteq T^{□}$ and $x \in B$ .

step 2.1A1

4.1

Suppose $B$ were in $T^{Π}$ . Then by [A1] there is a basic product-open $O = \prod_{i} O_{i}$ with $x \in O \subseteq B$ , and by [A2] its exceptional index set is contained in ${i_{0}, \dots, i_{n - 1}}$ for some natural $n$ .

step 3.1A1A2assume-hyp

5.1

The set $J$ is not contained in ${i_{0}, \dots, i_{n - 1}}$ : otherwise $J$ would be a subset of a finite set and hence finite by [L2], contrary to the hypothesis of step 1.3. So there is $j \in J$ with $O_{j} = X_{j}$ .

step 1.3step 4.1L2

6.1

For $j$ as in step 5.1 and any $t \in X_{j}$ , the point $y$ with $y_{j} : = t$ and $y_{i} : = x_{i}$ for $i \neq j$ lies in $O$ , since $x \in O$ and $O_{j} = X_{j}$ ; hence $y \in B$ and $t = y_{j} \in V_{j} = U_{j}$ . So $X_{j} \subseteq U_{j}$ , contradicting $U_{j} \neq X_{j}$ from step 2.1.

step 2.1step 3.1step 4.1step 5.1

7.1

Therefore $B \notin T^{Π}$ while $B \in T^{□}$ , so the inclusion of claim 1 is strict, which is claim 3; with step 2.2 all three claims are proved.

step 2.2step 3.1step 6.1∎

Remarks

Both hypotheses of claim 3 are needed. If some factor is empty then $P$ is empty and there is exactly one topology on it, so the two agree. If only finitely many factors have a nonempty proper open subset then every box is, after replacing the unrestricted factors by $X_{i}$ , already basic product-open, and again the two agree; the indiscrete topology on a set with many points is the standard case (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).
The strictness argument is not constructive, and it does not have to be. Claim 3 as stated quantifies over arbitrary factors, so a choice principle is unavoidable. In every concrete instance on the companion page the sets $U_{i}$ are written down by a formula and no choice is spent; the false statement on this page uses $R^{N}$ with $U_{k} = (- 1 / (k + 1), 1 / (k + 1))$ and is choice free.
Finer means more open sets, and here it means too many. The box topology has so many open sets that maps into it are hard to make continuous, which is exactly the failure the characteristic property of A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice avoids. That is why "product" without qualification means the product topology in this library.

Depends on

Used by

Dependency tree · two levels

23 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)
J. Munkres, Topology, 2nd ed., §19 (standard reference, not scraped)