CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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$\bigcap_k (-1/k, 1/k) = \{0\}$ is not open

Statement refuted

Refuted claim: an arbitrary intersection of open subsets of $\mathbb{R}$ is open (FALSE: an arbitrary intersection of open subsets of $\mathbb{R}$ is open, Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen).

The witness is $U_k := (-1/k,\ 1/k)$ for $k \ge 1$ : each $U_k$ is a nonempty open interval, the family is nested, and

$\bigcap_{k \ge 1} \Big( -\frac{1}{k},\ \frac{1}{k} \Big) \;=\; \{0\},$

which is not open. The index runs over $k \ge 1$ because $1/0$ is undefined; read as a family indexed by $\mathbb{N}$ it is $j \mapsto (-1/(j+1),\ 1/(j+1))$ . The refutation is carried out in full in FALSE: an arbitrary intersection of open subsets of $\mathbb{R}$ is open and is recorded here as the named counterexample. The comparison worth keeping in view is the finite case: any finite subfamily of $\{U_k\}$ has intersection the smallest of its members, which is open.

Facts & Assumptions

Given: For each natural $k \ge 1$ the interval $U_k := (-1/k,\ 1/k)$ , where $1/k$ is the inverse of the canonical natural $k \cdot 1_{\mathbb{R}}$ .

[A1]

The refuted claim: for every family of open subsets of $\mathbb{R}$ , the intersection is open.

[L1]

Each $U_k$ is open, the intersection of the family is $\{0\}$ , and $\{0\}$ is not open (FALSE: an arbitrary intersection of open subsets of $\mathbb{R}$ is open).

[L2]

$U$ is open when every point of it has a neighbourhood inside it, and each interval $(a,b)$ is an open set; $N_\varepsilon(x) = \{\, y : |y-x| < \varepsilon \,\}$ (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen, Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length, The $\varepsilon$ -neighbourhood and the punctured $\varepsilon$ -neighbourhood of a point of $\mathbb{R}$ ).

[L3]

Reciprocal Archimedean property: for every real $\varepsilon > 0$ there is a natural $n \ge 1$ with $1/n < \varepsilon$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , Every complete ordered field is Archimedean).

[L4]

Canonical naturals are positive for $k \ge 1$ and their inverses are positive (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order); $|z| \ge 0$ with $|z| = 0$ only for $z = 0$ , and $|z| < c$ exactly when $-c < z < c$ for $c > 0$ (Basic properties of the absolute value); $0 < 1$ , 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.

Counterexample

technique · direct

1.1

Each $U_k$ is a nonempty open subset of $\mathbb{R}$ , being the interval $(-1/k, 1/k)$ with $1/k > 0$ , so the family $\{\, U_k : k \ge 1 \,\}$ is an instance of the claim [A1], which asserts that its intersection is open.

A1L1L2L4

1.2

$0 \in U_k$ for every $k \ge 1$ , since $|0| = 0 < 1/k$ by [L4].

L1L4

2.1

If $x \ne 0$ then $|x| > 0$ by [L4], so [L3] supplies a natural $n \ge 1$ with $1/n < |x|$ , and $x \in U_n$ would give $|x| < 1/n$ , which trichotomy forbids; hence $x \notin U_n$ . With step 1.2 this gives $\bigcap_{k \ge 1} U_k = \{0\}$ .

step 1.2L3L4

3.1

The singleton $\{0\}$ is not open, since for every real $\varepsilon > 0$ the point $\varepsilon \cdot 2^{-1}$ lies in $N_\varepsilon(0)$ and differs from $0$ by [L4]. So the family of step 1.1 consists of open sets and its intersection, computed in step 2.1, is not open: the claim [A1] is refuted.

step 1.1step 2.1A1L1L2L4∎

Remarks

Exactly one hypothesis of Arbitrary unions and finite intersections of open subsets of $\mathbb{R}$ are open, and dually for closed sets is missing. That theorem asserts openness for finite intersections, and its proof takes the minimum of finitely many positive radii. Here the family is infinite and the radii at the surviving point $0$ are the numbers $1/k$ , which have no positive lower bound (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ ). So the true theorem is not contradicted; its finiteness hypothesis cannot be dropped.
The Archimedean property is doing the work in step 2.1. In a non-Archimedean ordered field a positive infinitesimal lies in every $(-1/k, 1/k)$ , and the intersection is then strictly larger than $\{0\}$ . So this is a counterexample about $\mathbb{R}$ , supplied by For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , and not a formal consequence of openness alone.
The intersection here is a single point, and that is not forced. An infinite intersection of open sets can be open, for instance when all members are equal. The claim refuted is a universal one, so one witness settles it.

Depends on

Used by

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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)

⋂k(−1/k,1/k)={0}\bigcap_k (-1/k, 1/k) = \{0\}⋂k​(−1/k,1/k)={0} is not open