CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passverified 2026-08-04 (gpt-5.6-sol-codex-subscription)

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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.

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$N$ with the discrete metric is bounded and is not totally bounded

Statement refuted

Refuted claim: every bounded metric space is totally bounded (FALSE: a bounded metric space is totally bounded).

The witness is $N$ (The natural numbers $N$ (von Neumann)) with the discrete metric $d (m, n) = 0$ for $m = n$ and $d (m, n) = 1$ for $m \neq n$ (With the discrete metric $d (x, y) = 1$ for $x \neq y$ , a space is compact iff it is totally bounded iff it is finite, and it is complete whatever its size). It is bounded (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space), lying inside the ball $B (0, 2)$ ; it is not totally bounded (Finite $ε$ -net and totally bounded metric space), because a finite $1 / 2$ -net would have to be the whole of $N$ , which is not finite.

The full verification is carried out in FALSE: a bounded metric space is totally bounded, where the metric axioms, the identity $B (y, 1 / 2) = {y}$ and the impossibility of listing $N$ are all checked. This item records the witness and says what makes it work.

Facts & Assumptions

Given: The set $N$ with the discrete metric $d$ , and the false claim that every bounded metric space is totally bounded.

[A1]

The refuted claim: every bounded metric space is totally bounded.

[L1]

$d$ is a metric on $N$ ; $B (y, r) = {y}$ for $0 < r \leq 1$ ; and $N = B (0, 2)$ , so the space is bounded (FALSE: a bounded metric space is totally bounded, With the discrete metric $d (x, y) = 1$ for $x \neq y$ , a space is compact iff it is totally bounded iff it is finite, and it is complete whatever its size, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric, Open ball, closed ball and sphere in a metric space, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

[L2]

A finite $1 / 2$ -net is a finite $F \subseteq N$ with the balls $B (y, 1 / 2)$ , $y \in F$ , covering $N$ ; a nonempty finite set can be listed (Finite $ε$ -net and totally bounded metric space, Open cover, subcover, compact metric space, and compact subset of a metric space, Finite, countably infinite, countable, uncountable).

[L3]

A nonempty finite set of reals has a maximum, one of its members, and for every real $M$ there is a natural $N \geq 1$ with $M < ι (N)$ , $ι$ being the canonical natural of $R$ (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set, Every complete ordered field is Archimedean, The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

Counterexample

technique · direct

1.1L1

$(N, d)$ is a bounded metric space, since $d (0, n) \leq 1 < 2$ for every $n$ gives $N = B (0, 2)$ .

2.1L1L2step 1.1

If $F$ were a finite $1 / 2$ -net, then $N = ⋃_{y \in F} B (y, 1 / 2) = ⋃_{y \in F} {y} = F$ , so $N$ would be finite and, being nonempty, listable as ${n_{0}, \dots, n_{k}}$ .

3.1L3step 2.1

The reals $ι (n_{0}), \dots, ι (n_{k})$ then have a maximum $M$ , while a natural $N \geq 1$ with $M < ι (N)$ satisfies $ι (N) \neq ι (n_{i})$ , hence $N \neq n_{i}$ , for every $i \leq k$ ; so $N$ is a natural number outside $N$ , which is impossible.

4.1A1L2step 1.1step 3.1∎

So no finite $1 / 2$ -net exists and $(N, d)$ is not totally bounded, while being bounded by step 1.1; the claim [A1] is refuted.

Remarks

The implication that does hold is the converse. A totally bounded metric space is bounded (A totally bounded metric space is bounded, every subspace of a totally bounded space is totally bounded, and the closure of a totally bounded subset is totally bounded), so this witness also shows that claim 1 of that lemma does not reverse.

The same space refutes more. It is closed in itself and bounded and not compact, which is the witness recorded in FALSE: a closed and bounded subset of a metric space is compact, and it is complete, so no one of boundedness, closedness and completeness, nor all three together, implies compactness (With the discrete metric $d (x, y) = 1$ for $x \neq y$ , a space is compact iff it is totally bounded iff it is finite, and it is complete whatever its size).

$N$ with the discrete metric is bounded and is not totally bounded

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

Sources

NN with the discrete metric is bounded and is not totally bounded

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

Sources

$N$ with the discrete metric is bounded and is not totally bounded