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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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.
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The identity on $(0,1)$ is bounded with no greatest value, and on $[0,\infty)$ it is continuous and unbounded

Statement refuted

Refuted claim: for the conclusions of the extreme value theorem it is enough that the domain be bounded, or that it be closed; that is, a continuous real function on a bounded domain attains a greatest value, and a continuous real function on a closed domain is bounded (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point, Lower bound, bounded below, bounded set, Maximum and minimum of a set).

Both halves are false, and one function refutes both:

on $E_1 := (0,1)$ , which is bounded and not closed, the identity is continuous and bounded, $\sup$ of its image exists and equals $1$ , and no point of $E_1$ attains it;
on $E_2 := [0,\infty)$ , which is closed and not bounded, the identity is continuous and unbounded.

Neither $E_1$ nor $E_2$ is compact (A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded), so Extreme value theorem: a continuous real function on a nonempty compact subset of $\mathbb{R}$ attains a greatest and a least value does not apply to either, and both are instances of Rudin 4.20, the sharp converse: on a noncompact $E \subseteq \mathbb{R}$ there is an unbounded continuous function and a bounded continuous function with no greatest value, and if $E$ is bounded there is a continuous function on $E$ that is not uniformly continuous: the first is its bounded-not-closed case, the second its unbounded case. Together they show that neither half of "closed and bounded" can be dropped.

This item is the worked witness for FALSE: a continuous real function on a bounded domain attains a greatest value, which refutes the first half alone.

Facts & Assumptions

Given: The sets $E_1 := (0,1)$ and $E_2 := [0,\infty)$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length) and the identity $\mathrm{id}(x) = x$ on each of them.

[L1]

The identity is continuous on every subset of $\mathbb{R}$ (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point).

[L2]

A greatest value of $h$ on $E$ is a point $p \in E$ with $h(x) \le h(p)$ for every $x \in E$ ; a set is bounded when it lies between two reals (Maximum and minimum of a set, Lower bound, bounded below, bounded set).

[L3]

Suprema: a nonempty set bounded above has a least upper bound, and for $u = \sup S$ every real $\varepsilon > 0$ admits $s \in S$ with $u - \varepsilon < s$ (Complete ordered field (least-upper-bound property), Epsilon characterisation of the supremum).

[L4]

Archimedean property: for every real $t$ there is a natural $n \ge 1$ with $t < n$ (Every complete ordered field is Archimedean).

[L5]

Ordered-field arithmetic: for $0 < x < 1$ one has $x < (x+1)/2 < 1$ ; the maximum of a two-element set of reals exists and is one of them; and $n \ge 1 > 0$ for a natural $n \ge 1$ (Ordered field, Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L6]

$(0,1)$ is bounded and not closed, $[0,\infty)$ is closed and not bounded, and neither is compact (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length, Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen, Lower bound, bounded below, bounded set, A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded, Open cover, subcover, compact subset of $\mathbb{R}$ (every open cover has a finite subcover), and sequentially compact subset).

Counterexample

technique · direct

1.1

The identity is continuous on $E_1$ and on $E_2$ by [L1], and $E_1$ is bounded while $E_2$ is closed, by [L6].

L1L6

1.2

On $E_1$ the identity is bounded. Every $x \in (0,1)$ satisfies $0 \le x \le 1$ , so the image $\mathrm{id}[E_1] = E_1$ lies between $0$ and $1$ and is bounded by [L2].

L5L2

1.3

On $E_2$ the identity is unbounded. Let a real $M$ be given. By [L4] there is a natural $n \ge 1$ with $M < n$ , and $n \ge 0$ so $n \in [0,\infty) = E_2$ with $\mathrm{id}(n) = n > M$ . So no real bounds $\mathrm{id}[E_2]$ above, and by [L2] the identity is unbounded on $E_2$ .

L2L4L5

2.1

On $E_1$ there is no greatest value. Let $p \in E_1$ , so $0 < p < 1$ . By [L5] the point $p' := (p+1)/2$ satisfies $p < p' < 1$ and $p' > 0$ , so $p' \in E_1$ and $\mathrm{id}(p') = p' > p = \mathrm{id}(p)$ . Hence no $p \in E_1$ satisfies $\mathrm{id}(x) \le \mathrm{id}(p)$ for every $x \in E_1$ , and by [L2] the identity attains no greatest value on $E_1$ .

step 1.2L2L5

3.1

The supremum exists and equals $1$ . By step 1.2 and [L3] the nonempty set $\mathrm{id}[E_1] = (0,1)$ has a least upper bound $u$ , and $u \le 1$ since $1$ bounds it above. For a real $\varepsilon > 0$ the point $x := \max\{1/2,\ 1 - \varepsilon/2\}$ lies in $(0,1)$ by [L5] and satisfies $x > 1 - \varepsilon$ , so no real below $1$ bounds $(0,1)$ above; hence $u = 1$ . By step 2.1 no point of $E_1$ has value $1$ , so the supremum is not attained.

step 1.2step 2.1L3L5

4.1

So on the bounded set $E_1$ a continuous function attains no greatest value, and on the closed set $E_2$ a continuous function is unbounded: both halves of the refuted claim are false, and by [L6] neither domain is compact, so no conflict with Extreme value theorem: a continuous real function on a nonempty compact subset of $\mathbb{R}$ attains a greatest and a least value arises.

step 1.1step 2.1step 3.1step 1.3L6∎

Remarks

The two failures are of different kinds. On $(0,1)$ the function is bounded and the supremum exists as a real number; what is missing is a point at which it is attained, and adding the endpoint $1$ restores it. On $[0,\infty)$ there is no supremum at all, and no endpoint can be added. That is the distinction Rudin 4.20, the sharp converse: on a noncompact $E \subseteq \mathbb{R}$ there is an unbounded continuous function and a bounded continuous function with no greatest value, and if $E$ is bounded there is a continuous function on $E$ that is not uniformly continuous draws between its bounded-not-closed and unbounded cases.
The domains are exactly the two minimal ways to fail compactness. By A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded a subset of $\mathbb{R}$ fails compactness by failing closedness or by failing boundedness; $E_1$ fails only the first, $E_2$ only the second, and each already kills the theorem.
The same domains kill uniform continuity too, but with different witnesses: on $(0,1)$ it is $1/x$ ( $x \mapsto 1/x$ is continuous on $(0,1)$ and not uniformly continuous there, the pairs $1/(k+2)$ and $1/(k+3)$ defeating every $\delta$ ) and on an unbounded closed set it is $x^{2}$ ( $x \mapsto x^2$ is continuous on $\mathbb{R}$ and not uniformly continuous, the pairs $k+1$ and $k+1+1/(k+1)$ defeating every $\delta$ ). The identity itself is uniformly continuous on both, so a single witness cannot serve every conclusion at once.

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Sources

Extreme value theorem (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 4 (standard reference, not scraped)
J. Lebl, Basic Analysis I, §3.3 (standard reference, not scraped)
University of Edinburgh, The Extreme Value Theorem (standard reference, not scraped)

The identity on (0,1)(0,1)(0,1) is bounded with no greatest value, and on [0,∞)[0,\infty)[0,∞) it is continuous and unbounded