CounterexampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck 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.
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$\mathbb{N}$ 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 $\mathbb{N}$ (The natural numbers $\mathbb{N}$ (von Neumann)) with the discrete metric $d(m,n) = 0$ for $m = n$ and $d(m,n) = 1$ for $m \ne n$ (With the discrete metric $d(x,y) = 1$ for $x \ne y$ , 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 $\varepsilon$ -net and totally bounded metric space), because a finite $1/2$ -net would have to be the whole of $\mathbb{N}$ , 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 $\mathbb{N}$ are all checked. This item records the witness and says what makes it work.

Facts & Assumptions

Given: The set $\mathbb{N}$ 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 $\mathbb{N}$ ; $B(y,r) = \{y\}$ for $0 < r \le 1$ ; and $\mathbb{N} = B(0,2)$ , so the space is bounded (FALSE: a bounded metric space is totally bounded, With the discrete metric $d(x,y) = 1$ for $x \ne y$ , 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 $F \subseteq \mathbb{N}$ with the balls $B(y,1/2)$ , $y \in F$ , covering $\mathbb{N}$ ; a nonempty finite set can be listed (Finite $\varepsilon$ -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 $N \ge 1$ with $M < \iota(N)$ , $\iota$ being the canonical natural of $\mathbb{R}$ (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 $\iota(n) = n \cdot 1_F$ of a field).

Counterexample

technique · direct

1.1

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

2.1

If $F$ were a finite $1/2$ -net, then $\mathbb{N} = \bigcup_{y \in F} B(y,1/2) = \bigcup_{y \in F}\{y\} = F$ , so $\mathbb{N}$ would be finite and, being nonempty, listable as $\{n_0, \dots, n_k\}$ .

L1L2step 1.1

3.1

The reals $\iota(n_0), \dots, \iota(n_k)$ then have a maximum $M$ , while a natural $N \ge 1$ with $M < \iota(N)$ satisfies $\iota(N) \ne \iota(n_i)$ , hence $N \ne n_i$ , for every $i \le k$ ; so $N$ is a natural number outside $\mathbb{N}$ , which is impossible.

L3step 2.1

4.1

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

A1L2step 1.1step 3.1∎

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 $x \ne y$ , a space is compact iff it is totally bounded iff it is finite, and it is complete whatever its size).

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 96 results over 21 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Totally bounded space (Wikipedia) (standard reference, not scraped)
Discrete space (Wikipedia) (standard reference, not scraped)

N\mathbb{N}N with the discrete metric is bounded and is not totally bounded