False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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FALSE: an arbitrary intersection of open subsets of $R$ is open

Statement

False claim: for every family $U$ of open subsets of $R$ (Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen), the intersection $⋂ U$ is open.

The true statement is Arbitrary unions and finite intersections of open subsets of $R$ are open, and dually for closed sets, claim 2, which asserts this for finite families only. The claim above deletes the word "finite", and the refutation below shows that the word cannot be deleted.

Facts & Assumptions

Given: For each natural $k \geq 1$ the interval $U_{k} : = (- 1 / k, 1 / k)$ , where $1 / k$ abbreviates the inverse of the canonical natural $k \cdot 1_{R}$ , which is positive for $k \geq 1$ .

[A1]

The false claim: for every family $U$ of open subsets of $R$ , the set $⋂ U$ is open.

[L1]

$U$ is open when every $x \in U$ admits $ε > 0$ with $N_{ε} (x) \subseteq U$ , and each interval of the form $(a, b)$ is an open set (Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L2]

$N_{ε} (x) = {y : ∣ y - x ∣ < ε} = (x - ε, x + ε)$ (The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ ).

[L3]

Reciprocal Archimedean property: 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 $1 / n < ε$ , Every complete ordered field is Archimedean).

[L4]

Absolute value: $∣ z ∣ \geq 0$ , $∣ z ∣ = 0$ exactly when $z = 0$ , and for $c > 0$ one has $∣ z ∣ < c$ exactly when $- c < z < c$ (Basic properties of the absolute value).

[L5]

Canonical naturals are positive for $k \geq 1$ and their inverses are positive (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order); $0 < 1$ , so $2 : = 1 + 1 > 0$ and $0 < d \cdot 2^{- 1} < d$ for $d > 0$ (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field, Complete ordered field (least-upper-bound property)). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used below follow by adjoining the equality case, in which the two sides coincide.

Refutation

technique · direct

1.1

Each $U_{k}$ is an open subset of $R$ , being an interval of the form $(a, b)$ with $a = - 1 / k$ and $b = 1 / k$ , and $1 / k > 0$ by [L5].

L1L5

1.2

$0 \in U_{k}$ for every $k \geq 1$ , since $∣ 0 - 0 ∣ = 0 < 1 / k$ by [L4] and [L5].

L4L5

1.3

The singleton ${0}$ is not open: for every real $ε > 0$ the element $ε \cdot 2^{- 1}$ satisfies $0 < ε \cdot 2^{- 1} < ε$ by [L5], so it lies in $N_{ε} (0)$ by [L2] and [L4] and differs from $0$ ; hence no $N_{ε} (0)$ is contained in ${0}$ .

L1L2L4L5

2.1

$⋂_{k \geq 1} U_{k} = {0}$ : the inclusion $\supseteq$ is step 1.2, and for the other inclusion let $x \neq 0$ ; then $∣ x ∣ > 0$ by [L4], so [L3] supplies a natural $n \geq 1$ with $1 / n < ∣ x ∣$ , and $x \in U_{n}$ would mean $∣ x ∣ < 1 / n$ by [L4], which trichotomy forbids; hence $x \notin U_{n}$ and $x$ is not in the intersection.

step 1.2L3L4

3.1

The family ${U_{k} : k \geq 1}$ consists of open subsets of $R$ by step 1.1, and its intersection is ${0}$ by step 2.1, which is not open by step 1.3. So the claim [A1] fails for this family and is false.

step 1.1step 1.3step 2.1A1∎

Remarks

Exactly one word is deleted, and it is load bearing. Arbitrary unions and finite intersections of open subsets of $R$ are open, and dually for closed sets proves that a finite intersection of open sets is open, and the proof takes the minimum of finitely many positive radii. That minimum is positive because it is one of the radii. An infinite family need supply no such minimum, and the family here supplies none: the radii at the point $0$ are the numbers $1 / k$ , which have no positive lower bound, precisely by For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ .
The failure is Archimedean, not merely set-theoretic. In a non-Archimedean ordered field a positive infinitesimal lies in every $(- 1 / k, 1 / k)$ , so the intersection there is strictly larger than ${0}$ and the computation of the intersection above is false there. What makes the claim fail over $R$ is that the reciprocals of the naturals really do get below every positive real.
The named witness is $⋂_{k} (- 1 / k, 1 / k) = {0}$ is not open ↗, which records the same family as a counterexample; the refutation itself is carried out here.
The dual statement fails too, by complementation (Arbitrary unions and finite intersections of open subsets of $R$ are open, and dually for closed sets): an arbitrary union of closed sets need not be closed, and $R ∖ {0}$ , the union of the closed sets $R ∖ U_{k}$ , is the witness.

Depends on

Used by

⋂ₖ (-1/k, 1/k) = {0} is not open Counterexample

Dependency tree · two levels

24 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

Open set (Wikipedia) (standard reference, not scraped)
Archimedean property (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 2 (Thm 2.24 and the remark following it) (standard reference, not scraped)
J. K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)

FALSE: an arbitrary intersection of open subsets of RR is open