False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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FALSE: a bounded metric space is totally bounded

Statement

False claim: every bounded metric space (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric) is totally bounded (Finite $ε$ -net and totally bounded metric space).

Where the claim comes from, and what is actually true. The implication holds in the other direction: a totally bounded metric space is bounded, which is claim 1 of 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. The claim above is its converse, and the converse fails. Boundedness asks for one ball containing the space; total boundedness asks for finitely many balls of every prescribed radius, and no amount of shrinking the diameter forces the second condition.

The refutation builds its witness: the set $N$ carrying the metric that assigns distance $1$ to distinct points.

Facts & Assumptions

Given: The set $N$ (The natural numbers $N$ (von Neumann)) and the function $d$ with $d (m, n) = 0$ for $m = n$ and $d (m, n) = 1$ for $m \neq n$ .

[A1]

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

[L1]

A metric satisfies (M1) $d (x, y) = 0$ exactly when $x = y$ , (M2) symmetry and (M3) the triangle inequality (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L2]

$B (x, r) = {y : d (x, y) < r}$ , and a space is bounded when it is empty or is contained in a ball (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).

[L3]

A finite $ε$ -net is a finite $F \subseteq X$ with $X = ⋃_{y \in F} B (y, ε)$ , and a space is totally bounded when it has one for every real $ε > 0$ ; a nonempty finite set can be listed as ${n_{0}, \dots, n_{k}}$ (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).

[L4]

A nonempty finite set of reals has a maximum, one of its members (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L5]

For every real $M$ there is a natural $N \geq 1$ with $M < ι (N)$ , where $ι$ is the canonical natural of $R$ (Every complete ordered field is Archimedean, The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

Refutation

technique · direct

1.1

$d$ is a metric on $N$ : (M1) holds by the definition of $d$ ; (M2) because the defining condition is symmetric; and (M3) because the left side is $0$ or $1$ , and if it is $1$ then $x \neq z$ , so $y$ differs from at least one of $x$ and $z$ and the right side is at least $1$ .

2.1

$(N, d)$ is bounded: $d (0, n) \leq 1 < 2$ for every $n$ , so $N = B (0, 2)$ .

L2step 1.1

2.2

In $(N, d)$ one has $B (y, 1 / 2) = {y}$ , since $d (y, n) < 1 / 2$ forces $d (y, n) = 0$ and hence $n = y$ .

L2step 1.1

3.1

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

L3step 2.2

4.1

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

L4L5step 3.1

5.1

So no finite $1 / 2$ -net exists, $(N, d)$ is a bounded metric space that is not totally bounded, and the claim [A1] is false.

A1L3step 2.1step 3.1step 4.1∎

Remarks

Diameter is not a measure of how spread out a space is at small scales. The witness has diameter $1$ , and yet every ball of radius $1 / 2$ contains a single point, so no finite family of them can cover an infinite space. Total boundedness is exactly the condition that rules this out, and it is what makes the pair "complete and totally bounded" equivalent to compactness (For a metric space, compact, countably compact, limit point compact, sequentially compact, and complete together with totally bounded are all equivalent, given countable choice and dependent choice).

The same witness refutes more. Because it is complete and closed in itself, it also shows that a closed bounded subset of a metric space need not be compact (FALSE: a closed and bounded subset of a metric space is compact). The counterexample page records it once, in $N$ with the discrete metric is bounded and is not totally bounded ↗.

In $R^{n}$ the claim is true, since a bounded subset lies in a box and a box is compact, hence totally bounded (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, A compact metric space is complete and totally bounded, and neither implication uses any choice principle). What fails is the general metric statement, and the witness is a space whose points are pairwise equidistant, which is exactly what the geometry of $R^{n}$ forbids for infinitely many points at once.

Depends on

Used by

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

Dependency tree · two levels

44 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

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