False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

FALSE: an arbitrary intersection of open sets is open in every topological space

Statement

False claim: in every topological space $(X, T)$ (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), if ${U_{i} : i \in I}$ is any family of open sets then $⋂_{i \in I} U_{i}$ is open.

The topology axioms grant closure under arbitrary unions and under intersections of finitely many open sets, and the asymmetry is not a weakness of the axioms chosen: strengthening (T3) to arbitrary intersections would exclude the spaces this subject exists to study. Two witnesses are given below, one in a space with no metric in sight and one in $R$ with its usual topology, so that the failure cannot be blamed on exotic examples.

Facts & Assumptions

Given: An infinite set $X$ carrying the cofinite topology and a point $p \in X$ , with $I : = X ∖ {p}$ ; and $R$ with its usual topology, together with the family $U_{k} : = B (0, 1 / (k + 1))$ for $k \in N$ , where $1 / (k + 1)$ abbreviates the inverse of the canonical natural $(k + 1) \cdot 1_{R}$ .

[A1]

A topology is closed under arbitrary unions and binary intersections; a set is open exactly when it belongs to the topology (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[L1]

In the cofinite topology the open sets are $\emptyset$ together with the sets of finite complement; a subset of a finite set is finite and a union of two finite sets is finite (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies); "infinite" means "not finite" (Finite, countably infinite, countable, uncountable).

[L2]

$R$ with $d_{R} (x, y) = ∣ x - y ∣$ is a metric space, $B (x, r) = (x - r, x + r)$ , and the metric topology of $d_{R}$ is the usual topology of $R$ (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, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric via Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Open ball, closed ball and sphere in a metric space).

[L3]

Every ball is an open set of the metric topology, and $U \subseteq R$ is open in it exactly when every point of $U$ has a ball around it inside $U$ (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

[L4]

For every real $ε > 0$ there is a natural $n \geq 1$ with $1 / n < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ); for $n \geq 1$ the canonical natural $n \cdot 1_{R}$ is positive (Canonical naturals are positive and strictly increasing) and its inverse $1 / n$ is positive (Inverses of positives are positive, and reciprocation reverses order).

[L5]

$∣ u ∣ \geq 0$ , and $∣ u ∣ = 0$ only for $u = 0$ ; for $c > 0$ one has $∣ u ∣ < c$ if and only if $- c < u < c$ (Absolute value in an ordered field, Basic properties of the absolute value).

[L6]

Every nonzero natural number is a successor, so $n \geq 1$ gives $n = m + 1$ for some $m \in N$ (Every nonzero natural number is a successor).

Refutation

technique · direct

1.1

Since $X$ is infinite, $X ∖ {p}$ is infinite: were it finite, $X = {p} \cup (X ∖ {p})$ would be a union of two finite sets and hence finite. In particular $I \neq \emptyset$ .

givenL1

1.2

For each $x \in I$ the set $X ∖ {x}$ is open in the cofinite topology, its complement ${x}$ being finite; and $⋂_{x \in I} (X ∖ {x}) = X ∖ I = {p}$ .

givenL1

1.3

For every $k \in N$ the natural $k + 1$ satisfies $k + 1 \geq 1$ , so $1 / (k + 1)$ is a positive real and $U_{k} = B (0, 1 / (k + 1))$ is a legitimate ball; each $U_{k}$ is open in the usual topology of $R$ .

givenL2L3L4

1.4

$0 \in U_{k}$ for every $k$ , since $∣ 0 - 0 ∣ = 0 < 1 / (k + 1)$ .

givenL4L5

1.5

Let $x \in R$ with $x \neq 0$ ; then $∣ x ∣ > 0$ by [L5], so [L4] gives a natural $n \geq 1$ with $1 / n < ∣ x ∣$ , and [L6] writes $n = m + 1$ with $m \in N$ ; hence $∣ x - 0 ∣ = ∣ x ∣ > 1 / (m + 1)$ , so $x \notin B (0, 1 / (m + 1)) = U_{m}$ .

L4L5L6

1.6

${0}$ is not open in the usual topology of $R$ : a ball $B (0, r) = (- r, r)$ with $r > 0$ contains the point $1 / n$ for a natural $n \geq 1$ with $1 / n < r$ supplied by [L4], and $1 / n > 0$ , so $1 / n \in B (0, r)$ and $1 / n \neq 0$ ; hence no ball around $0$ lies inside ${0}$ .

L2L3L4L5

2.1

${p}$ is not open in the cofinite topology: it is nonempty, and its complement $X ∖ {p}$ is infinite by step 1.1, so it is neither $\emptyset$ nor a set of finite complement.

step 1.1L1

2.2

By steps 1.4 and 1.5, $⋂_{k \in N} U_{k} = {0}$ .

step 1.4step 1.5

3.1

By steps 1.2 and 2.1 the family ${X ∖ {x} : x \in I}$ consists of open subsets of the cofinite space $X$ , is nonempty, and has intersection ${p}$ , which is not open; so the claim fails already in a space defined without any reference to $R$ .

step 1.2step 2.1A1

4.1

By steps 2.2 and 1.6 the sets $U_{k}$ are open in $R$ , their intersection is ${0}$ , and ${0}$ is not open; with step 3.1 the false claim is refuted twice over, once in a non-metrizable setting and once in a metrizable one.

step 1.3step 3.1step 2.2step 1.6A1∎

Remarks

What is true instead. Intersections of finitely many open sets are open, which is axiom (T3) iterated (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison); and arbitrary intersections of closed sets are closed, which is its dual (C2). The intersection of an arbitrary family of open sets is in general only a set whose interior may be smaller than itself, and the interior operator exists precisely to name what survives.
The ℝ witness is the shape that recurs. A decreasing family of balls of radii shrinking to zero has the centre as its intersection, and a singleton is open only in a space where the point is isolated. The index shift is the usual one for this library: the radii are $1 / (k + 1)$ for $k \in N$ , not $1 / k$ , since $N$ contains $0$ .
The corresponding failure inside $R$ alone is already published (FALSE: an arbitrary intersection of open subsets of $R$ is open), stated there in the order-native vocabulary of the topology of $R$ . The present item is the statement about topological spaces in general, which that page explicitly declined to make.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

60 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

Topological space (Wikipedia) (standard reference, not scraped)
Open set (Wikipedia) (standard reference, not scraped)
Cofiniteness (Wikipedia) (standard reference, not scraped)