False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)verified 2026-08-04 (gpt-5.6-sol-codex-subscription)

How statement and proof provenance work

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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FALSE: a totally bounded metric space is compact

Statement

False claim: every totally bounded metric space (Finite $ε$ -net and totally bounded metric space) is compact (Open cover, subcover, compact metric space, and compact subset of a metric space).

Where the claim comes from, and what is actually true. A compact metric space is totally bounded, and it is also complete (A compact metric space is complete and totally bounded, and neither implication uses any choice principle); the converse needs both of those conditions, not one of them, and, as stated in this library, it also assumes the Axiom of Countable Choice (A complete, totally bounded metric space is compact, proved from countable choice used exactly once, The Axiom of Countable Choice ( ${A C}_{ω}$ )). The claim above drops completeness, and dropping it is fatal.

The refutation takes the open interval $(0, 1)$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length) as a metric subspace of $R$ with its usual metric $∣ x - y ∣$ (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, Isometry, isometric embedding, and the subspace metric on a subset).

Facts & Assumptions

Given: The interval $(0, 1) = {x \in R : 0 < x < 1}$ as a metric subspace of $(R, d_{R})$ , $d_{R} (x, y) = ∣ x - y ∣$ .

[A1]

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

[L1]

A space is totally bounded when for every real $ε > 0$ it has a finite $ε$ -net, a finite subset $F$ with the balls $B (y, ε)$ , $y \in F$ , covering the space (Finite $ε$ -net and totally bounded metric space, Open ball, closed ball and sphere in a metric space).

[L2]

A subset $A$ of a metric space is compact exactly when every family of open subsets of the ambient space whose union contains $A$ has finitely many members whose union contains $A$ ; and the sets open in the subspace $A$ are the traces on $A$ of the open subsets of the ambient space (A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it, Open cover, subcover, compact metric space, and compact subset of 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).

[L3]

In $(R, d_{R})$ the ball $B (x, r)$ is the interval $(x - r, x + r)$ , and the subspace metric on $(0, 1)$ is the restriction of $d_{R}$ (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, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Isometry, isometric embedding, and the subspace metric on a subset, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L4]

Every nonempty subset of $N$ has a least element (The well-ordering principle).

[L5]

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

[L6]

For every real $η > 0$ there is a natural $m \geq 1$ with $1 / m < η$ ; reciprocals of positives are positive and reverse the order (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, Inverses of positives are positive, and reciprocation reverses order).

Refutation

technique · direct

1.1

Let $ε > 0$ be real and take a natural $m \geq 1$ with $1 / m < ε$ ; the points $j / (m + 1)$ for $1 \leq j \leq m$ lie in $(0, 1)$ , since $0 < j / (m + 1) < 1$ , and they form a finite subset $F$ of $(0, 1)$ .

L3L6

2.1

$F$ is a finite $ε$ -net for $(0, 1)$ : given $x \in (0, 1)$ , the set of naturals $j \geq 1$ with $x < (j + 1) / (m + 1)$ is nonempty, containing $m$ because $x < 1 = (m + 1) / (m + 1)$ , so it has a least element $j$ , and $j \leq m$ .

L4L6step 1.1

3.1

For that $j$ one has $x < (j + 1) / (m + 1)$ and also $(j - 1) / (m + 1) < x$ : for $j = 1$ because $x > 0$ , and for $j \geq 2$ because minimality gives $x \geq j / (m + 1) > (j - 1) / (m + 1)$ . Hence $∣ x - j / (m + 1) ∣ < 1 / (m + 1) < 1 / m < ε$ , so $x$ lies in the subspace ball of radius $ε$ about $j / (m + 1)$ .

L3L6step 2.1

4.1

As $ε > 0$ was arbitrary, $(0, 1)$ with the restricted metric is totally bounded.

L1step 1.1step 3.1

5.1

For each $k \in N$ put $U_{k} : = (1 / (k + 2), 1)$ , an open subset of $R$ contained in $(0, 1)$ ; the family $(U_{k})_{k \in N}$ has union $(0, 1)$ , because any $x \in (0, 1)$ admits a natural $m \geq 1$ with $1 / m < x$ and then $1 / (m + 2) < 1 / m < x < 1$ .

L3L6step 4.1

6.1

No finitely many of the $U_{k}$ have union containing $(0, 1)$ : given $U_{k_{0}}, \dots, U_{k_{p}}$ , put $t : = \min {1 / (k_{i} + 2) : i \leq p}$ , a positive real; each $U_{k_{i}}$ is contained in $(t, 1)$ because $1 / (k_{i} + 2) \geq t$ , so the union of the finite subfamily is contained in $(t, 1)$ , while the real $x : = \min {t, 1 / 2} / 2$ satisfies $0 < x < 1$ and $x \leq t$ , so $x \in (0, 1)$ and $x$ lies in no $U_{k_{i}}$ .

L5L6step 5.1

7.1

Hence $(0, 1)$ is not a compact subset of $R$ , that is the metric subspace $(0, 1)$ is a totally bounded metric space that is not compact, and the claim [A1] is false.

A1L2step 4.1step 5.1step 6.1∎

Remarks

What the witness lacks is completeness. A compact metric space is complete (A compact metric space is complete and totally bounded, and neither implication uses any choice principle, Complete metric space: every Cauchy sequence converges in the space), and $(0, 1)$ is not: the terms $1 / (k + 2)$ form a Cauchy sequence in $(0, 1)$ whose only candidate limit in $R$ is $0$ , which is not a point of the space. Adding completeness to total boundedness does restore compactness (A complete, totally bounded metric space is compact, proved from countable choice used exactly once), at the cost of the Axiom of Countable Choice.

The same interval also witnesses that boundedness is far from compactness, and it is the standard example behind the failure of the extreme value theorem and of Heine-Cantor off a compact domain (On $(0, 1)$ the identity is bounded with no greatest value and $x \mapsto 1 / x$ is continuous and unbounded, so the extreme value theorem needs compactness and not merely boundedness of the domain ↗, $x \mapsto 1 / x$ is continuous on $(0, 1)$ and not uniformly continuous, so Heine-Cantor needs compactness of the domain ↗).

Depends on

Used by

The open interval (0,1) is totally bounded and not compact, the cover by the intervals (1/(k+2), 1) having no finite subcover Counterexample

Dependency tree · two levels

63 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)
Compact space (Wikipedia) (standard reference, not scraped)