False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-09 (gpt-5.6-terra-codex-subscription)

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FALSE: a continuous real function on a bounded domain attains a greatest value

Statement

False claim: if $E \subseteq R$ is nonempty and bounded (Lower bound, bounded below, bounded set) and $f : E \to R$ is continuous on $E$ (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point), then $f$ attains a greatest value on $E$ : there is $p \in E$ with $f (x) \leq f (p)$ for every $x \in E$ (Maximum and minimum of a set).

Why it is tempting. The extreme value theorem is often remembered as "a continuous function on a bounded interval attains its bounds", and on $[a, b]$ that is true. The hypothesis that actually does the work is compactness, which for a subset of $R$ is closed and bounded (A subset of $R$ is compact if and only if it is closed and bounded); dropping closedness loses the theorem even though the function may stay bounded.

What is true. Extreme value theorem: a continuous real function on a nonempty compact subset of $R$ attains a greatest and a least value gives the conclusion on a nonempty compact domain, and the hypothesis cannot be weakened: for every noncompact $E$ there is a bounded continuous function on $E$ whose supremum is not attained, which is Rudin 4.20, the sharp converse: on a noncompact $E \subseteq R$ 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 witness below is that theorem's simplest instance.

Facts & Assumptions

Given: The domain $E : = (0, 1) = {x \in R : 0 < x < 1}$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length) and the function $f : E \to R$ , $f (x) : = x$ .

[L1]

$E$ is nonempty and bounded: $1 / 2 \in E$ , and $0 \leq x \leq 1$ for every $x \in E$ (Lower bound, bounded below, bounded set, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Ordered field).

[L2]

The identity is continuous on every subset of $R$ (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 $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point).

[L3]

A greatest value of $f$ on $E$ is a point $p \in E$ with $f (x) \leq f (p)$ for every $x \in E$ ; equivalently a maximum of $f [E]$ lying in $f [E]$ (Maximum and minimum of a set).

[L4]

Ordered-field arithmetic in $R$ : for $0 < x < 1$ one has $x < (x + 1) / 2 < 1$ ; and $0 < 1$ (Ordered field).

[L5]

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

Refutation

technique · direct

1.1

$E$ is nonempty and bounded by [L1], and $f$ is continuous on $E$ by [L2], so the hypotheses of the claim are satisfied.

L1L2

2.1

Let $p \in E$ be arbitrary, so $0 < p < 1$ . Put $p^{'} : = (p + 1) / 2$ ; by [L4] we have $p < p^{'} < 1$ and $0 < p^{'}$ , so $p^{'} \in E$ and $f (p^{'}) = p^{'} > p = f (p)$ . Hence no $p \in E$ satisfies $f (x) \leq f (p)$ for every $x \in E$ , and by [L3] the function $f$ attains no greatest value on $E$ .

step 1.1L3L4

3.1

The claim is therefore false. Note also what the failure is not: $f [E] = E$ is nonempty and bounded above by $1$ , so $\sup f [E]$ exists by [L5] and equals $1$ , since $1$ bounds $f [E]$ and for every real $ε > 0$ the point $x : = \max {1 / 2, 1 - ε / 2}$ lies in $E$ with $x > 1 - ε$ . What fails is only that $1 \notin f [E]$ .

step 2.1L1L4L5∎

Remarks

The domain is bounded and not closed, and that is exactly the gap. The set $(0, 1)$ is not closed and hence is not compact by A subset of $R$ 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 $R$ attains a greatest and a least value does not apply. Adding the two endpoints repairs everything: on $[0, 1]$ the same $f$ attains the value $1$ at the point $1$ .
Boundedness of the function is not the issue either. The witness above is bounded, so the failure is not a blow-up; it is the loss of the point at which the supremum would be attained. A function on the same domain that is unbounded, such as $x \mapsto 1 / x$ , fails the conclusion for the cruder reason that no upper bound exists at all, and both failures are catalogued together in The identity on $(0, 1)$ is bounded with no greatest value, and on $[0, \infty)$ it is continuous and unbounded ↗ on the companion page.
The least value fails in the same way, by symmetry: $f$ has no least value on $(0, 1)$ either, and $\inf f [E] = 0 \notin f [E]$ . The statement is written for the greatest value only because that is the form the false claim usually takes.

Depends on

Used by

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

Dependency tree · two levels

43 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)
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)