CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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On $N$ with $d (m, n) = 1 + 1 / (m + n)$ for $m \neq n$ the sets ${n, n + 1, \dots}$ are nested, closed, bounded and complete with empty intersection

Statement refuted

Refuted claim: in Cantor's intersection theorem (In a complete metric space nested nonempty closed sets whose diameters tend to $0$ meet in exactly one point, and this property characterises completeness) the hypothesis $diam (F_{k}) \to 0$ may be dropped; in a complete metric space a nested sequence of nonempty closed bounded sets has nonempty intersection.

Let $X : = N$ (The natural numbers $N$ (von Neumann)) and define

$d (m, n) : = {\begin{cases} 0 & if m = n, \\ 1 + \frac{1}{m + n} & if m \neq n, \end{cases}$

where $m + n \geq 1$ whenever $m \neq n$ , so the reciprocal is defined. Put $F_{n} : = {k \in N : k \geq n}$ . Then $d$ is a metric, $(X, d)$ is complete, every $F_{n}$ is nonempty, closed, bounded and itself complete, the $F_{n}$ are nested, and

$⋂_{n \in N} F_{n} = \emptyset .$

What fails is only the diameter condition: $diam (F_{n}) = 1 + 1 / (2 n + 1) \geq 1$ for every $n$ .

Facts & Assumptions

Given: $X = N$ with the function $d$ above; the sets $F_{n} = {k \in N : k \geq n}$ ; naturals $m, n, p$ ; a real $ε > 0$ .

[L1]

$N$ contains $0$ , and for distinct naturals $m \neq n$ one has $m + n \geq 1$ , so $1 / (m + n)$ is a positive real, at most $1$ (The natural numbers $N$ (von Neumann), Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L2]

The metric axioms (M1), (M2), (M3), and nonnegativity (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric, Nonnegativity of a metric is a consequence of the other axioms, not an axiom).

[L3]

Open sets, balls, Cauchyness and convergence, tested with real $ε > 0$ (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, Open ball, closed ball and sphere in a metric space, Cauchy sequence in a metric space, Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ , The rationals embed densely in the reals).

[L4]

Bounded subset and diameter: $A$ is bounded when it lies in a ball, and $diam (A) = \sup {d (a, b) : a, b \in A}$ for nonempty bounded $A$ ; the supremum is the least upper bound (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Complete ordered field (least-upper-bound property), Maximum and minimum of a set).

[L5]

Reciprocals reverse order on the positives, and for every real $η > 0$ there is a natural $N \geq 1$ with $1 / N < η$ (Inverses of positives are positive, and reciprocation reverses order, 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).

[L6]

A closed subset of a complete metric space is complete (Closed subspaces of complete metric spaces are complete; the converse under countable choice, Complete metric space: every Cauchy sequence converges in the space).

Counterexample

technique · direct

1.1

$d$ is well defined and symmetric, and $d (m, n) = 0$ exactly when $m = n$ , since for $m \neq n$ the value $1 + 1 / (m + n)$ is at least $1$ . In fact $1 < d (m, n) \leq 2$ for all $m \neq n$ , because $0 < 1 / (m + n) \leq 1$ .

L1L2

2.1

$d$ satisfies the triangle inequality $d (m, p) \leq d (m, n) + d (n, p)$ . If $m = p$ the left side is $0$ ; if $n = m$ or $n = p$ one side of the right-hand sum is $0$ and the other equals the left side; and if $m$ , $n$ , $p$ are pairwise distinct then $d (m, n) + d (n, p) > 1 + 1 = 2 \geq d (m, p)$ . So $d$ is a metric on $N$ .

step 1.1L1L2

2.2

Every subset of $N$ is open, hence also closed: for $m \in N$ the ball $B (m, 1)$ is ${m}$ , since $d (m, n) > 1$ for $n \neq m$ , so every subset is a union of open balls; complements of subsets are subsets.

step 1.1L3

2.3

$(X, d)$ is complete: a Cauchy sequence $(x_{k})$ tested at $ε = 1$ has an index $K$ with $d (x_{k}, x_{l}) < 1$ for all $k, l \geq K$ , which by step 1.1 forces $x_{k} = x_{l}$ ; so the sequence is constant from $K$ on and converges to $x_{K} \in N$ .

step 1.1L3

2.4

Each $F_{n}$ is bounded: $d (0, k) \leq 2 < 3$ for every $k$ , so $F_{n} \subseteq X = B (0, 3)$ .

step 1.1L3L4

3.1

Each $F_{n}$ is nonempty (it contains $n$ ), closed by step 2.2, and complete by [L6]; and $F_{n + 1} \subseteq F_{n}$ , so the family is nested.

step 2.2step 2.3L6

3.2

$diam (F_{n}) = 1 + 1 / (2 n + 1)$ . Indeed the distances realised inside $F_{n}$ are $0$ and the values $1 + 1 / (j + k)$ for distinct $j, k \geq n$ ; among such pairs $j + k$ is least when ${j, k} = {n, n + 1}$ , giving $j + k = 2 n + 1$ , and $1 / (j + k)$ is largest there. So $1 + 1 / (2 n + 1)$ is an upper bound of those distances and is itself one of them, hence it is the least upper bound.

step 1.1step 2.4L4L5

4.1

In particular $diam (F_{n}) > 1$ for every $n$ , so the sequence $(diam (F_{n}))$ does not converge to $0$ : at $ε = 1$ no index makes the diameters smaller than $1$ .

step 3.2L3L5

4.2

The intersection is empty: for every $k \in N$ we have $k \notin F_{k + 1}$ , since $k < k + 1$ ; so no natural lies in all the $F_{n}$ .

step 3.1L1

5.1

So in the complete metric space $(X, d)$ the sets $F_{n}$ are nonempty, closed, bounded, complete and nested, and their intersection is empty; the only hypothesis of In a complete metric space nested nonempty closed sets whose diameters tend to $0$ meet in exactly one point, and this property characterises completeness that they fail is $diam (F_{n}) \to 0$ , so that hypothesis cannot be dropped.

step 2.3step 3.1step 2.4step 4.1step 4.2∎

Remarks

Indexing. $N$ contains $0$ here, so $d$ is written with $m + n$ and not with $m n$ or $1 / m$ : at $m = 0$ those would be undefined or degenerate. The clause $m \neq n$ is what guarantees $m + n \geq 1$ , and it is the reason the formula is stated by cases rather than as a single expression.
The metric is a small perturbation of the discrete metric, taking values in ${0} \cup (1, 2]$ , and its topology is discrete. Every subset is closed, so closedness is free and carries no information; what the example exploits is that a set can be closed, bounded and complete while its points stay a definite distance apart, so a nested family can drain away to nothing.
Contrast with the real line. On $R$ the same sets $[n, \infty)$ are nested, nonempty and closed with empty intersection, but they are not bounded, so they are not a Cantor chain either. The present example is sharper: it keeps boundedness and loses only the vanishing of the diameters.
Both conclusions of the theorem fail here, not just one. There is neither a common point nor a unique one, which is what one expects: the uniqueness half of In a complete metric space nested nonempty closed sets whose diameters tend to $0$ meet in exactly one point, and this property characterises completeness is exactly what the vanishing diameters buy.

Depends on

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Sources

Cantor's intersection theorem (Wikipedia) (standard reference, not scraped)
Complete metric space (Wikipedia) (standard reference, not scraped)

On NN with d(m,n)=1+1/(m+n)d(m,n)=1+1/(m+n) for m≠nm=n the sets {n,n+1,… }{n,n+1,…} are nested, closed, bounded and complete with empty intersection