CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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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 $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length) as a metric subspace of $\mathbb{R}$ (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), 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 $\varepsilon$ - $\delta$ 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 $(\mathbb{R},d_{\mathbb{R}})$ , $d_{\mathbb{R}}(x,y) = |x-y|$ , and it is not compact (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, Intervals of $\mathbb{R}$ : 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 $\varepsilon > 0$ there is a real $\delta > 0$ with $|h(x) - h(c)| < \varepsilon$ whenever $|x - c| < \delta$ (Continuity of a map between metric spaces, at a point and globally, in the $\varepsilon$ - $\delta$ form).

[L3]

$u = \sup S$ for a nonempty $S$ bounded above exactly when $u$ is an upper bound and for every real $\varepsilon > 0$ some $s \in S$ has $u - \varepsilon < s$ ; 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 $N \ge 1$ with $M < \iota(N)$ , for every real $\eta > 0$ a natural $N \ge 1$ with $1/N < \eta$ , and reciprocals of positives are positive and reverse the order (Every complete ordered field is Archimedean, For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , 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 $(0,1)$ , $\delta := \varepsilon$ 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 $\varepsilon > 0$ the point $x := \max\{1-\varepsilon/2,\ 1/2\}$ lies in $(0,1)$ and satisfies $x > 1 - \varepsilon$ .

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 $\varepsilon > 0$ , the choice $\delta := \min\{c/2,\ \varepsilon c^2/2\}$ gives, for $|x-c| < \delta$ , first $x > c/2 > 0$ and then $|1/x - 1/c| = |c-x|/(xc) < (2/c^2)|x - c| < \varepsilon$ .

L2L4step 3.1

5.1

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

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 $\mathbb{R}^n$ : 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), 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

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Sources

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

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