False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)verified 2026-08-04 (gpt-5.6-sol-codex-subscription)

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FALSE: a closed and bounded subset of a metric space is compact

Statement

False claim: in every metric space $(X,d)$ (Metric space: $d(x,y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric), a subset that is closed in $X$ (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) and bounded (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space) is a compact subset of $X$ (Open cover, subcover, compact metric space, and compact subset of a metric space).

Where the claim comes from, and what is actually true. One half of the Heine-Borel property does hold in every metric space: a compact subset is closed and bounded (A compact subset of a metric space is closed and bounded). The converse holds in $\mathbb{R}^n$ with the Euclidean metric (Heine-Borel in $\mathbb{R}^n$ : with the Euclidean metric a subset of $\mathbb{R}^n$ 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), and the claim above is that reading of Heine-Borel transplanted to an arbitrary metric space, where it fails. What survives in general is that a compact space is complete and totally bounded (A compact metric space is complete and totally bounded, and neither implication uses any choice principle), and it is total boundedness, not boundedness, that the witness below lacks.

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

Facts & Assumptions

Given: The set $\mathbb{N}$ of natural numbers (The natural numbers $\mathbb{N}$ (von Neumann)) and the function $d : \mathbb{N} \times \mathbb{N} \to \mathbb{R}$ with $d(m,n) = 0$ for $m = n$ and $d(m,n) = 1$ for $m \ne n$ .

[A1]

The false claim: in every metric space a closed bounded subset is compact.

[L1]

A metric on a set is a real-valued function satisfying (M1) $d(x,y) = 0$ exactly when $x = y$ , (M2) $d(x,y) = d(y,x)$ and (M3) $d(x,z) \le d(x,y) + d(y,z)$ (Metric space: $d(x,y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L2]

$B(x,r) = \{y : d(x,y) < r\}$ ; a set is open when each of its points has a ball around it inside it; a set is closed when its complement is open; and a subset is bounded when it is empty or lies in a ball (Open ball, closed ball and sphere in a metric space, 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, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

[L3]

A subset $A$ of a metric space is compact exactly when the metric subspace $(A,d_A)$ is a compact metric space; and a compact metric space has, for every family of open subsets with union the space, a finite subfamily with union the space (Open cover, subcover, compact metric space, and compact subset of a metric space).

[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 $N \ge 1$ with $M < \iota(N)$ , where $\iota$ is the canonical natural of $\mathbb{R}$ (Every complete ordered field is Archimedean, The canonical natural $\iota(n) = n \cdot 1_F$ of a field).

Refutation

technique · direct

1.1

$d$ is a metric on $\mathbb{N}$ : (M1) holds because $d(m,n) = 0$ was defined to mean $m = n$ ; (M2) because the defining condition is symmetric in $m$ and $n$ ; and (M3) because the left side is $0$ or $1$ , and when it is $1$ one has $x \ne z$ , so $y$ differs from at least one of $x$ and $z$ and the right side is at least $1$ .

2.1

In $(\mathbb{N},d)$ one has $B(n,1) = \{n\}$ , since $d(n,m) < 1$ forces $d(n,m) = 0$ and hence $m = n$ ; consequently every subset of $\mathbb{N}$ is open, each of its points $n$ having $B(n,1)$ inside it, and every subset is closed as well.

L2step 1.1

3.1

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

L2step 2.1

3.2

The family $(B(n,1))_{n \in \mathbb{N}}$ consists of open subsets of $\mathbb{N}$ and has union $\mathbb{N}$ , because $n \in B(n,1)$ for every $n$ .

L2step 2.1

4.1

No finite subfamily has union $\mathbb{N}$ : such a subfamily is $B(n_0,1), \dots, B(n_k,1)$ for some $k \in \mathbb{N}$ and naturals $n_0, \dots, n_k$ , with union $\{n_0, \dots, n_k\}$ by step 2.1; the reals $\iota(n_0), \dots, \iota(n_k)$ have a maximum $M$ , and a natural $N \ge 1$ with $M < \iota(N)$ then satisfies $\iota(N) \ne \iota(n_i)$ and hence $N \ne n_i$ for every $i \le k$ , so $N$ lies in $\mathbb{N}$ and in no member of the subfamily.

L4L5step 2.1step 3.2

5.1

Hence $(\mathbb{N},d)$ is not a compact metric space, so $\mathbb{N}$ is a closed and bounded subset of the metric space $(\mathbb{N},d)$ that is not compact, and the claim [A1] is false.

A1L3step 3.1step 3.2step 4.1∎

Remarks

What the witness fails is total boundedness, not boundedness. The space $(\mathbb{N},d)$ has diameter $1$ , so it is as bounded as a nonempty space can be; but a finite $1/2$ -net would have to contain every point, and $\mathbb{N}$ is not finite ( $\mathbb{N}$ with the discrete metric is bounded and is not totally bounded ↗, FALSE: a bounded metric space is totally bounded). Since a compact space is totally bounded (A compact metric space is complete and totally bounded, and neither implication uses any choice principle), that alone already settles non-compactness; the explicit cover of step 3.2 is given because it makes the failure visible without any theory.

The witness is complete, so completeness is not the missing ingredient either. In $(\mathbb{N},d)$ a Cauchy sequence is eventually constant, hence convergent, so this is a complete, bounded, closed space that is not compact. The pair that is equivalent to compactness, once the Axiom of Countable Choice and the Axiom of Dependent Choice are assumed, is completeness together with total boundedness (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).

A second, analytically natural witness is the closed unit ball of the bounded real-valued functions on $\mathbb{N}$ under the supremum metric, where the indicator functions of the singletons are pairwise at distance $1$ (In the bounded real-valued functions on $\mathbb{N}$ with the supremum metric, the closed unit ball is closed and bounded and is not compact: the indicator functions of the singletons are pairwise at distance $1$ ↗).

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 135 results over 23 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

Heine-Borel theorem (Wikipedia) (standard reference, not scraped)
Discrete space (Wikipedia) (standard reference, not scraped)
Compact space (Wikipedia) (standard reference, not scraped)