CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (openai/gpt-5.4)verified 2026-07-26 (claude-opus-5)

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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.

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An unbounded set has no supremum: the naturals inside $R$

Statement refuted

Refuted claim: every subset of $R$ has a supremum in $R$ (FALSE: every subset of $R$ has a supremum).

The witness here is the canonical copy of the natural numbers inside $R$ , $A = {n \cdot 1 : n \geq 1},$ where $n \cdot 1$ denotes the canonical natural $\underset{n}{\underset{⏟}{1 + \dots + 1}}$ of the field (Canonical naturals are positive and strictly increasing). The set $A$ is nonempty, so the nonemptiness hypothesis of the least-upper-bound property (Complete ordered field (least-upper-bound property)) is satisfied; what fails is boundedness above, and it fails as badly as possible, since $A$ has no upper bound whatsoever. That is precisely the Archimedean property of $R$ (Every complete ordered field is Archimedean), so this failure is a theorem about $R$ , not an accident of the set chosen.

Facts & Assumptions

Given: The complete ordered field $R$ and the set $A : = {n \cdot 1 : n \geq 1}$ of its canonical naturals.

[L1]

Canonical naturals: $1 \cdot 1 = 1$ and $n \cdot 1 > 0$ for every $n \geq 1$ (Canonical naturals are positive and strictly increasing).

[L2]

Archimedean property: $R$ is a complete ordered field, hence Archimedean, so for every $x \in R$ there is a natural $n \geq 1$ with $x < n \cdot 1$ (Every complete ordered field is Archimedean, Archimedean ordered field).

[L3]

Upper bound, bounded above, supremum: $u$ is an upper bound of $X$ when $x \leq u$ for every $x \in X$ ; $X$ is bounded above when it has an upper bound; and a supremum of $X$ is an upper bound of $X$ that is $\leq$ every upper bound of $X$ , so in particular every supremum is an upper bound (Complete ordered field (least-upper-bound property), Lower bound, bounded below, bounded set).

[L4]

The refuted claim: every $S \subseteq R$ has a supremum in $R$ (FALSE: every subset of $R$ has a supremum).

[L5]

Order: trichotomy holds, so $a < b$ and $b \leq a$ cannot both be true (Ordered field, Complete ordered field (least-upper-bound property)).

Counterexample

technique · direct

1.1

$A$ is a nonempty subset of $R$ : taking $n = 1$ gives $1 \cdot 1 = 1 \in A$ .

1.2

Let $x \in R$ be arbitrary.

assume-hyp

2.1

$x$ is not an upper bound of $A$ : the Archimedean property supplies a natural $n \geq 1$ with $x < n \cdot 1$ , and $n \cdot 1$ is an element of $A$ , so the requirement $n \cdot 1 \leq x$ for an upper bound fails by trichotomy.

step 1.2L2L3L5

3.1

Since $x$ was an arbitrary real number, no real number is an upper bound of $A$ ; hence $A$ is not bounded above.

step 2.1step 1.2L3

4.1

A supremum of $A$ would in particular be an upper bound of $A$ , and there is none, so $A$ has no supremum in $R$ even though $A$ is nonempty; the claim that every subset of $R$ has a supremum is refuted, and the boundedness hypothesis of the least-upper-bound property cannot be dropped.

step 3.1step 1.1L3L4∎

Remarks

The failure is of a specific shape: the set of upper bounds of $A$ is empty, so there is nothing among which to be least. The companion witness The empty set is bounded and has no supremum fails for the opposite reason, with a set of upper bounds so large that it has no least element either. Both are needed, since each alone would suggest that a single hypothesis carries all the weight.
In a non-Archimedean ordered field the same set is bounded above (Not every ordered field is Archimedean), so this counterexample really is using completeness by way of Every complete ordered field is Archimedean and is not a formal consequence of the ordered-field axioms.
Adjoining $+ \infty$ repairs the statement, at the price of leaving the field; this library does not adopt that convention silently (Conventions: $\sup \emptyset$ , unbounded sets, and the extended reals).

Depends on

Used by

The empty set is bounded and has no supremum Counterexample

Dependency tree · two levels

10 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

Archimedean property (Wikipedia) (standard reference, not scraped)
Least-upper-bound property (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 1 (standard reference, not scraped)
John K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)
MIT 18.100A, Complete Lecture Notes (standard reference, not scraped)

An unbounded set has no supremum: the naturals inside RR