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

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$Z$ and ${n + 1 / n : n \geq 2}$ are disjoint closed subsets of $R$ at distance $0$ , so the set-to-set distance is not a metric

Statement refuted

Refuted claim: the set-to-set distance $d (A, B) = \inf {d (a, b) : a \in A, b \in B}$ of Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space is a metric on the nonempty subsets of a metric space; specifically, that it satisfies the separation axiom (M1) of Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric, so that $d (A, B) = 0$ forces $A = B$ .

Work in $R$ with its usual metric (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) and write $m$ for the canonical copy of an integer $m$ inside $R$ (The integers as equivalence classes of pairs of naturals, The integers embed in the rationals, The unique embedding of ℚ into an ordered field). Put

$A : = {m : m \in Z}, B : = {n + \frac{1}{n} : n \in N, n \geq 2} .$

Then $A$ and $B$ are nonempty, disjoint, closed in $R$ , and

$d (A, B) = 0 while A \neq B .$

So (M1) fails for the set-to-set distance, and it fails on a pair of closed sets: closedness is not the missing hypothesis. Moreover $d (b, A) > 0$ for every individual $b \in B$ , so the infimum over pairs is not attained anywhere.

Facts & Assumptions

Given: The real line with $d_{R} (u, v) = ∣ u - v ∣$ ; the sets $A$ and $B$ above; and, for a subset $E \subseteq R$ , the property of being $c$ -separated for a real $c > 0$ , meaning $∣ e - e^{'} ∣ \geq c$ whenever $e, e^{'} \in E$ with $e \neq e^{'}$ .

[L1]

The embeddings $Z \to Q \to R$ are injective and order preserving (The integers embed in the rationals, The unique embedding of ℚ into an ordered field, The integers as equivalence classes of pairs of naturals), and $Z$ is totally ordered (The integers form a totally ordered ring); every integer $\geq 0$ is the image of a unique natural under $N \to Z$ , which is injective and order preserving (The naturals embed in the integers); and a natural $n \neq 0$ satisfies $n \geq 1$ (Discreteness: $σ (n)$ is the immediate successor, The natural numbers $N$ (von Neumann)).

[L2]

Canonical naturals in $R$ : $n \cdot 1_{R} > 0$ for $n \geq 1$ and $n \mapsto n \cdot 1_{R}$ is strictly increasing (Canonical naturals are positive and strictly increasing); reciprocals of positives are positive and reverse the order (Inverses of positives are positive, and reciprocation reverses order); and $r > 1$ gives $0 < 1 / r < 1$ (Reciprocals and order: $1 / r$ against $1$ ).

[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: $∣ t ∣ \geq 0$ , $∣ t ∣ = t$ for $t \geq 0$ , $∣ - t ∣ = ∣ t ∣$ , and $∣ t ∣ < c$ is equivalent to $- c < t < c$ (Basic properties of the absolute value, Absolute value in an ordered field).

[L5]

Infima: $d (x, E)$ and $d (E, F)$ exist for nonempty sets, being infima of nonempty sets bounded below by $0$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Nonnegativity of a metric is a consequence of the other axioms, not an axiom, Every nonempty set bounded below has an infimum, Greatest lower bound (infimum)); and $ℓ = \inf S$ for a lower bound $ℓ$ of a nonempty $S$ bounded below exactly when for every real $ε > 0$ some $s \in S$ has $s < ℓ + ε$ (Epsilon characterisation of the infimum).

[L6]

The closure of a nonempty $E$ is ${x : d (x, E) = 0}$ , and $E$ is closed exactly when $E = \overline{E}$ (The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset, 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, Isometry, isometric embedding, and the subspace metric on a subset).

[L7]

Order and field arithmetic in $R$ : halving a positive real, adding inequalities, scaling by a positive, the minimum of a two-element set, and trichotomy (Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, The multiplicative identity is positive, Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set, Field, Ordered field, Complete ordered field (least-upper-bound property)).

Counterexample

technique · direct

1.1

No canonical integer lies strictly between $0$ and $1$ : if $0 < k < 1$ in $R$ for an integer $k$ , then $0 < k < 1$ already in $Z$ , because the embedding is order preserving and injective and the order of $Z$ is total; then $k > 0$ makes $k$ the image of a natural $n \neq 0$ , so $n \geq 1$ and hence $k \geq 1$ , contradicting $k < 1$ .

L1L7

1.2

$B$ is $\frac{1}{2}$ -separated: for naturals $p > q \geq 2$ one has $p - q \geq 1$ and $0 < 1 / p < 1 / q \leq 1 / 2$ , so $(p + 1 / p) - (q + 1 / q) = (p - q) - (1 / q - 1 / p) \geq 1 - 1 / q \geq 1 - 1 / 2 = 1 / 2$ ; hence distinct elements of $B$ differ by at least $1 / 2$ in absolute value.

L2L4L7

1.3

A $c$ -separated set $E \subseteq R$ is closed: let $x \notin E$ . The ball $B (x, c / 2)$ contains at most one point of $E$ , since two distinct points of it would be within $c / 2 + c / 2 = c$ of each other, contradicting $c$ -separation with a strict inequality. If it contains none, then $B (x, c / 2) \cap E = \emptyset$ . If it contains exactly one point $e$ , then $x \neq e$ gives $t : = ∣ x - e ∣ > 0$ , and the radius $r : = \min {c / 2, t} > 0$ has $B (x, r) \cap E = \emptyset$ , because any point of $E$ in $B (x, r) \subseteq B (x, c / 2)$ would have to be $e$ , while $∣ x - e ∣ = t \geq r$ puts $e$ outside $B (x, r)$ . So the complement of $E$ is open and $E$ is closed.

L4L6L7

2.1

$A$ is $1$ -separated: for integers $m \neq m^{'}$ the difference $k : = m - m^{'}$ is a nonzero integer, so by step 1.1 and trichotomy either $k \geq 1$ or $- k \geq 1$ , and in both cases $∣ m - m^{'} ∣ = ∣ k ∣ \geq 1$ .

step 1.1L1L4L7

2.2

$A$ and $B$ are disjoint and both nonempty: $0 \in A$ and $2 + 1 / 2 \in B$ ; and if $n + 1 / n = m$ for a natural $n \geq 2$ and an integer $m$ , then the integer $m - n$ satisfies $m - n = 1 / n$ with $0 < 1 / n < 1$ by [L2], contradicting step 1.1.

step 1.1L1L2L7

3.1

$A$ and $B$ are closed subsets of $R$ , by steps 2.1, 1.2 and 1.3 with $c = 1$ and $c = 1 / 2$ respectively.

step 1.2step 1.3step 2.1

3.2

$d (A, B) = 0$ : the set ${∣ a - b ∣ : a \in A, b \in B}$ is nonempty and bounded below by $0$ , and for each natural $n \geq 2$ it contains $∣ n - (n + 1 / n) ∣ = 1 / n$ ; given a real $ε > 0$ , [L3] supplies a natural $n \geq 1$ with $1 / n < ε$ , and then $n + 1 \geq 2$ with $1 / (n + 1) < 1 / n < ε = 0 + ε$ , so the infimum is $0$ by the epsilon characterisation.

step 2.2L2L3L4L5

4.1

Yet every single point of $B$ is at positive distance from $A$ : for $b \in B$ we have $b \notin A = \overline{A}$ by steps 2.2 and 3.1, so $d (b, A) \neq 0$ by the description of the closure, and $d (b, A) \geq 0$ , hence $d (b, A) > 0$ .

step 2.2step 3.1L5L6L7

5.1

So $A$ and $B$ are nonempty disjoint closed subsets of $R$ with $d (A, B) = 0$ and $A \neq B$ : the separation axiom (M1) fails for the set-to-set distance, and it fails even on closed sets and even though each individual point-to-set distance is strictly positive.

step 2.2step 3.2step 4.1∎

Remarks

What survives. The set-to-set distance is symmetric and vanishes on $A = A$ , and on singletons it reduces to the metric. It satisfies no useful triangle inequality either: in $R$ the sets $A = {0}$ , $B = {0, 10}$ , $C = {10}$ have $d (A, B) = d (B, C) = 0$ while $d (A, C) = 10$ , so $d (A, C) \leq d (A, B) + d (B, C)$ fails. The construction that does give a metric on a family of sets is the Hausdorff distance, which is taken up on a later page and not defined here.
Where the intuition breaks. For a nonempty $A$ , the function $x \mapsto d (x, A)$ vanishes exactly on $\overline{A}$ (The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset), so a point at distance $0$ from a closed set does lie in it. The set-to-set distance takes an infimum over a second variable as well, and an infimum of a family of positive numbers can be $0$ ; step 4.1 is exactly the record of that.
The two sets approach each other only along their tails. The point $n + 1 / n$ of $B$ sits at distance exactly $1 / n$ from the integer $n$ ; those distances are all positive, and by step 3.2 their infimum is $0$ . That is the whole mechanism: the distance is driven to $0$ by pairs with $n$ arbitrarily large, never by any single pair.

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Sources

Hausdorff distance (Wikipedia) (standard reference, not scraped)
Hausdorff distance (Wikipedia) (standard reference, not scraped)
Closed set (Wikipedia) (standard reference, not scraped)

ZZ and {n+1/n:n≥2}{n+1/n:n≥2} are disjoint closed subsets of RR at distance 00, so the set-to-set distance is not a metric