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

Statement refuted

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

The witness is the interval $(0,1)$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length) as a metric subspace of $(\mathbb{R}, d_{\mathbb{R}})$ , $d_{\mathbb{R}}(x,y) = |x-y|$ (The absolute value makes $\mathbb{R}$ 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). It is totally bounded (Finite $\varepsilon$ -net and totally bounded metric space), the points $j/(m+1)$ for $1 \le j \le m$ forming a finite net of mesh $1/(m+1)$ ; and it is not compact (Open cover, subcover, compact metric space, and compact subset of a metric space), the family

$U_k \;:=\; \Big(\tfrac{1}{k+2},\ 1\Big), \qquad k \in \mathbb{N},$

having union $(0,1)$ and no finite subfamily with that union. The index is written $k+2$ because $\mathbb{N}$ contains $0$ , so that $1/(k+2) \le 1/2 < 1$ for every $k$ and each $U_k$ is a nonempty subinterval of $(0,1)$ .

The full verification is carried out in FALSE: a totally bounded metric space is compact; this item records the witness and says what makes it work.

Facts & Assumptions

Given: The interval $(0,1)$ with the metric $|x-y|$ restricted to it, and the sets $U_k = (1/(k+2), 1)$ for $k \in \mathbb{N}$ .

[A1]

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

[L1]

$(0,1)$ with the restricted metric is totally bounded, and the family $(U_k)_{k \in \mathbb{N}}$ consists of open subsets of $\mathbb{R}$ contained in $(0,1)$ with union $(0,1)$ , no finitely many of which have union $(0,1)$ (FALSE: a totally bounded metric space is compact, Finite $\varepsilon$ -net and totally bounded metric space, Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length, 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, Open ball, closed ball and sphere in a metric space, The absolute value makes $\mathbb{R}$ 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, Metric space: $d(x,y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[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 $A$ (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).

[L3]

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).

[L4]

For every real $\eta > 0$ there is a natural $m \ge 1$ with $1/m < \eta$ , and reciprocals of positives are positive and reverse the order (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , Every complete ordered field is Archimedean, Inverses of positives are positive, and reciprocation reverses order).

Counterexample

technique · direct

1.1

$(0,1)$ with the restricted metric is a totally bounded metric space.

2.1

The family $(U_k)_{k \in \mathbb{N}}$ consists of open subsets of $\mathbb{R}$ and has union $(0,1)$ : given $x \in (0,1)$ , a natural $m \ge 1$ with $1/m < x$ gives $1/(m+2) < 1/m < x < 1$ , so $x \in U_m$ .

L1L4step 1.1

3.1

Given finitely many members $U_{k_0}, \dots, U_{k_p}$ , put $t := \min\{1/(k_i+2) : i \le p\}$ , a positive real; each $U_{k_i}$ is contained in $(t,1)$ because $1/(k_i+2) \ge t$ , so the union of the finite subfamily lies in $(t,1)$ , while $x := \min\{t,1/2\}/2$ satisfies $0 < x < 1$ and $x \le t$ , so $x \in (0,1)$ lies in no $U_{k_i}$ .

L3L4step 2.1

4.1

Hence no finitely many of the $U_k$ have union containing $(0,1)$ , so $(0,1)$ is not a compact subset of $\mathbb{R}$ , that is not a compact metric space, while being totally bounded by step 1.1; the claim [A1] is refuted.

A1L2step 1.1step 2.1step 3.1∎

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 it whose only candidate limit in $\mathbb{R}$ is $0$ , a point the space omits. Adding completeness to total boundedness does restore compactness, at the cost of the Axiom of Countable Choice (A complete, totally bounded metric space is compact, proved from countable choice used exactly once).

The same cover fails to have a Lebesgue number, which is the other thing compactness would have supplied (The cover of $(0,1)$ by the intervals $(1/(k+2), 1)$ has no Lebesgue number, so the Lebesgue number lemma needs compactness).

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 94 results over 20 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)
Compact space (Wikipedia) (standard reference, not scraped)

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