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

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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

Statement refuted

Refuted claim: a continuous real-valued function on a nonempty bounded metric space is bounded and attains a greatest value.

The true statement replaces bounded by compact (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value), and the difference is not cosmetic. The witness is the interval $(0, 1)$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length) as a metric subspace of $R$ (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), which is bounded (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space) and not compact (The open interval $(0, 1)$ is totally bounded and not compact, the cover by the intervals $(1 / (k + 2), 1)$ having no finite subcover), together with two continuous functions on it (Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form):

the identity $f (x) = x$ , which is bounded, has supremum $1$ , and attains no greatest value;
the map $g (x) = 1 / x$ , which is continuous and unbounded.

So on a merely bounded domain a continuous function may fail to attain its supremum, and may fail to be bounded at all.

Facts & Assumptions

Given: The interval $(0, 1)$ with the metric $∣ x - y ∣$ restricted to it, and the functions $f (x) = x$ and $g (x) = 1 / x$ on it.

[A1]

The refuted claim: a continuous real-valued function on a nonempty bounded metric space is bounded and attains a greatest value.

[L1]

$(0, 1)$ is a nonempty bounded metric subspace of $(R, d_{R})$ , $d_{R} (x, y) = ∣ x - y ∣$ , and it is not compact (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, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, The open interval $(0, 1)$ is totally bounded and not compact, the cover by the intervals $(1 / (k + 2), 1)$ having no finite subcover, Open cover, subcover, compact metric space, and compact subset of a metric space, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L2]

$h$ is continuous at $c$ when for every real $ε > 0$ there is a real $δ > 0$ with $∣ h (x) - h (c) ∣ < ε$ whenever $∣ x - c ∣ < δ$ (Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form).

[L3]

$u = \sup S$ for a nonempty $S$ bounded above exactly when $u$ is an upper bound and for every real $ε > 0$ some $s \in S$ has $u - ε < s$ ; a maximum is a member of the set that bounds it above (Epsilon characterisation of the supremum, Maximum and minimum of a set, Lower bound, bounded below, bounded set, Complete ordered field (least-upper-bound property)).

[L4]

For every real $M$ there is a natural $N \geq 1$ with $M < ι (N)$ , for every real $η > 0$ a natural $N \geq 1$ with $1 / N < η$ , and reciprocals of positives are positive and reverse the order (Every complete ordered field is Archimedean, For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Inverses of positives are positive, and reciprocation reverses order).

[L5]

A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

Counterexample

technique · direct

1.1

The identity $f$ is continuous on $(0, 1)$ , $δ : = ε$ serving at every point, and its image is $(0, 1)$ , which is bounded above by $1$ and below by $0$ .

L1L2

2.1

$\sup f [(0, 1)] = 1$ : the value $1$ is an upper bound, and for a real $ε > 0$ the point $x : = \max {1 - ε / 2, 1 / 2}$ lies in $(0, 1)$ and satisfies $x > 1 - ε$ .

L3step 1.1

3.1

$f$ attains no greatest value: for $x \in (0, 1)$ the point $(x + 1) / 2$ lies in $(0, 1)$ and satisfies $(x + 1) / 2 > x$ , so no member of the image bounds the image above.

L3step 2.1

4.1

$g (x) = 1 / x$ is defined on $(0, 1)$ , every $x$ there being positive, and it is continuous at each $c \in (0, 1)$ : given a real $ε > 0$ , the choice $δ : = \min {c / 2, ε c^{2} / 2}$ gives, for $∣ x - c ∣ < δ$ , first $x > c / 2 > 0$ and then $∣ 1 / x - 1 / c ∣ = ∣ c - x ∣ / (x c) < (2 / c^{2}) ∣ x - c ∣ < ε$ .

L2L4step 3.1

5.1

$g$ is unbounded on $(0, 1)$ : given a real $M$ , take a natural $N \geq 1$ with $M < ι (N)$ ; the point $x : = 1 / ι (N + 1)$ lies in $(0, 1)$ and $g (x) = ι (N + 1) > ι (N) > M$ .

L4step 4.1

6.1

So on the nonempty bounded non-compact space $(0, 1)$ the continuous function $f$ is bounded and attains no greatest value, and the continuous function $g$ is not even bounded; the claim [A1] is refuted, and the compactness hypothesis of the extreme value theorem cannot be weakened to boundedness.

A1L1L5step 1.1step 3.1step 5.1∎

Remarks

Which hypothesis each failure isolates. The identity shows that the attainment half of A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value fails on a bounded non-compact domain even for a bounded function; the map $1 / x$ shows that the boundedness half fails too. Compactness is what supplies both, through the image being closed and bounded (The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset, A compact subset of a metric space is closed and bounded).

Closing the interval repairs the first example and rules out the second. On $[0, 1]$ , which is compact (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), the identity attains the value $1$ . The map $1 / x$ has no continuous extension to $[0, 1]$ at all: such an extension would be bounded by [L5], whereas its restriction to $(0, 1)$ is unbounded by step 5.1. So the failure of the second example is a failure of the domain, not of the theorem.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

49 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

Extreme value theorem (Wikipedia) (standard reference, not scraped)
Compact space (Wikipedia) (standard reference, not scraped)

On (0,1)(0,1) the identity is bounded with no greatest value and x↦1/xx↦1/x is continuous and unbounded, so the extreme value theorem needs compactness and not merely boundedness of the domain