ExampleConstruction: Literature-sourcedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)verified 2026-07-26 (claude-opus-5)

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.

$\sup (0, 1) = 1$ and $\inf (0, 1) = 0$ , with neither attained

Example

The open unit interval $I = (0, 1) = {x \in R : 0 < x < 1}$ has $\sup I = 1$ and $\inf I = 0$ , and neither value belongs to $I$ . This is the smallest interesting computation in the subject: both bounds exist by the least-upper-bound property (Complete ordered field (least-upper-bound property)) and its dual (Every nonempty set bounded below has an infimum), and both are missed by the set, so $I$ has neither a maximum (The supremum is attained exactly when a maximum exists) nor a minimum. The maximum is ruled out by the attainment criterion; the minimum is ruled out directly, since a minimum would be a lower bound of $I$ lying in $I$ and therefore at most $\inf I = 0$ (Maximum and minimum of a set, Greatest lower bound (infimum)).

Nothing here is asserted by inspection. That $1$ bounds $I$ above is checked from the definition; that no smaller number does is checked with the epsilon characterisation (Epsilon characterisation of the supremum), by exhibiting, for each $ε > 0$ , an explicit element of $I$ lying strictly above $1 - ε$ . The witness is $\max {1 / 2, 1 - ε / 2}$ : the second entry does the approximating, and the first keeps the witness inside $I$ when $ε$ is large. The infimum is handled symmetrically with Epsilon characterisation of the infimum and the witness $\min {1 / 2, ε / 2}$ .

Facts & Assumptions

Given: The complete ordered field $R$ , the open interval $I : = {x \in R : 0 < x < 1}$ , the abbreviation $2 : = 1 + 1$ , and the notation $c / 2 : = c \cdot 2^{- 1}$ for $c \in R$ .

[L1]

Epsilon characterisation of the supremum: for a nonempty $X \subseteq R$ bounded above and an upper bound $w$ of $X$ , one has $w = \sup X$ if and only if for every $ε > 0$ there is $x \in X$ with $w - ε < x$ (Epsilon characterisation of the supremum).

[L2]

Epsilon characterisation of the infimum: for a nonempty $X \subseteq R$ bounded below and a lower bound $m$ of $X$ , one has $m = \inf X$ if and only if for every $ε > 0$ there is $x \in X$ with $x < m + ε$ ; such an $\inf X$ exists for every nonempty $X$ bounded below (Epsilon characterisation of the infimum, Every nonempty set bounded below has an infimum, Greatest lower bound (infimum)).

[L3]

Order: trichotomy holds, so exactly one of $a < b$ , $a = b$ , $b < a$ is true, $a < a$ is impossible, the negation of $a \leq b$ is $b < a$ , and $a < b$ implies $a \leq b$ ; the order is transitive; and adding a constant preserves it, that is, $a < b$ implies $a + c < b + c$ (claim 1 of Order is preserved by adding a constant and by adding inequalities) (Complete ordered field (least-upper-bound property), Ordered field, Order is preserved by adding a constant and by adding inequalities).

[L4]

Halving: $0 < 1$ , so $2 = 1 + 1 > 1 > 0$ ; hence $2 \neq 0$ and $2^{- 1}$ exists, and $2 > 1$ gives $0 < 2^{- 1} < 1$ ; consequently $0 < c / 2 < c$ for every $c > 0$ (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Reciprocals and order: $1 / r$ against $1$ , When $a b < b$ for positive $a, b$ , Field).

[L5]

Every set ${a, b}$ of two reals has a maximum and a minimum, so the notations $\max {a, b}$ and $\min {a, b}$ are legitimate: each is an element of ${a, b}$ , the maximum dominates both entries and the minimum is dominated by both (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L6]

Upper bound, lower bound, maximum, minimum: $u$ bounds $X$ above when $x \leq u$ for all $x \in X$ , and $ℓ$ bounds $X$ below when $ℓ \leq x$ for all $x \in X$ ; $m = \max X$ means $m$ is an upper bound of $X$ with $m \in X$ , and $m = \min X$ means $m$ is a lower bound of $X$ with $m \in X$ (Complete ordered field (least-upper-bound property), Lower bound, bounded below, bounded set, Maximum and minimum of a set).

[L7]

Attainment of the supremum: for nonempty $X$ whose supremum exists, $\sup X \in X$ holds exactly when $X$ has a maximum, and then $\sup X = \max X$ (The supremum is attained exactly when a maximum exists).

[L8]

The infimum is the greatest lower bound: $\inf X$ is a lower bound of $X$ , and $ℓ \leq \inf X$ for every lower bound $ℓ$ of $X$ (Greatest lower bound (infimum)).

Verification

technique · direct

1.1

The element $2^{- 1}$ lies in $I$ , so $I \neq \emptyset$ : by [L4] one has $0 < 2^{- 1} < 1$ , which is exactly the membership condition.

L4L6

1.2

The number $1$ is an upper bound of $I$ and the number $0$ is a lower bound of $I$ : every $x \in I$ satisfies $0 < x < 1$ , hence $0 \leq x$ and $x \leq 1$ ; in particular $I$ is bounded above and bounded below.

L3L6

1.3

Neither $1$ nor $0$ lies in $I$ : membership requires $x < 1$ and $0 < x$ , and both $1 < 1$ and $0 < 0$ are impossible by irreflexivity.

L3L6

1.4

Let $ε > 0$ be arbitrary and put $s_{ε} : = \max {2^{- 1}, 1 - ε / 2}$ and $t_{ε} : = \min {2^{- 1}, ε / 2}$ , both of which exist.

assume-hypL5choose

2.1

$s_{ε} \in I$ : on the one hand $s_{ε} \geq 2^{- 1} > 0$ , so $s_{ε} > 0$ ; on the other hand $s_{ε}$ is one of the two entries $2^{- 1}$ and $1 - ε / 2$ , and each of them is $< 1$ , the first by [L4] and the second because $ε / 2 > 0$ gives $1 - ε / 2 < 1$ ; hence $s_{ε} < 1$ .

step 1.4L4L3

2.2

$1 - ε < s_{ε}$ : from $ε / 2 < ε$ one gets $1 - ε < 1 - ε / 2$ , and $1 - ε / 2 \leq s_{ε}$ because $s_{ε}$ dominates that entry.

step 1.4L4L3L5

2.3

$t_{ε} \in I$ : $t_{ε}$ is one of the two entries $2^{- 1}$ and $ε / 2$ , both of which are positive by [L4], so $t_{ε} > 0$ ; and $t_{ε} \leq 2^{- 1} < 1$ , so $t_{ε} < 1$ .

step 1.4L4L3L5

2.4

$t_{ε} < 0 + ε$ : indeed $t_{ε} \leq ε / 2 < ε = 0 + ε$ , using $ε / 2 < ε$ from [L4] and transitivity.

step 1.4L4L3L5

3.1

$I$ is nonempty and bounded above by $1$ , and for every $ε > 0$ the element $s_{ε}$ of $I$ satisfies $1 - ε < s_{ε}$ ; the epsilon characterisation therefore gives $\sup I = 1$ .

step 1.1step 1.2step 2.1step 2.2L1

3.2

$I$ is nonempty and bounded below by $0$ , and for every $ε > 0$ the element $t_{ε}$ of $I$ satisfies $t_{ε} < 0 + ε$ ; the dual characterisation therefore gives $\inf I = 0$ .

step 1.1step 1.2step 2.3step 2.4L2

4.1

$I$ has no maximum: if $m$ were one, the attainment criterion would give $\sup I = m \in I$ , whereas $\sup I = 1 \notin I$ ; the two are incompatible, so no maximum exists.

step 3.1step 1.3L7

4.2

$I$ has no minimum: a minimum of $I$ would be a lower bound of $I$ lying in $I$ , hence at most the greatest lower bound $\inf I = 0$ ; but every element of $I$ satisfies $0 < x$ , and $x \leq 0$ together with $0 < x$ is impossible by trichotomy.

step 3.2L3L6L8

5.1

Hence $\sup I = 1$ and $\inf I = 0$ , while $1 \notin I$ and $0 \notin I$ , and $I$ has neither a maximum nor a minimum, so both bounds are approached by $I$ and reached by neither.

step 3.1step 3.2step 1.3step 4.1step 4.2∎

Remarks

The two entries of the witness $\max {2^{- 1}, 1 - ε / 2}$ play different roles, and dropping either breaks the argument. For small $ε$ the approximating entry $1 - ε / 2$ is the larger one and does the work; for $ε \geq 2$ that entry falls to $0$ or below and leaves $I$ , which is exactly what the guard $2^{- 1}$ prevents. Using $1 - ε$ itself as the witness fails for the same reason and also fails the strict inequality $1 - ε < s$ .
The interval $I$ is the standard witness that a supremum need not belong to its set (FALSE: the supremum of a set belongs to the set); that consequence is recorded separately as A supremum need not belong to its set: $\sup (0, 1) = 1 \notin (0, 1)$ .
Compare $\sup [0, 1] = 1 = \max [0, 1]$ and $\inf [0, 1] = 0 = \min [0, 1]$ : the closed interval $[0, 1]$ has the same supremum and the same infimum, and attains both. The value of a supremum therefore carries no information at all about whether it is attained.

Depends on

Used by

A supremum need not belong to its set: sup(0,1) = 1 ∉ (0,1) Counterexample
(0,1) + (2,3) = (2,4), with supremum 4 = sup(0,1) + sup(2,3) Example
sup[0,1] = 1 = max[0,1] and inf[0,1] = 0 = min[0,1] Example

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 30 results over 8 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Infimum and supremum (Wikipedia) (standard reference, not scraped)
Interval (mathematics) (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)

sup⁡(0,1)=1sup(0,1)=1 and inf⁡(0,1)=0inf(0,1)=0, with neither attained

Example

Facts & Assumptions

Verification

Remarks

Depends on

Used by

Dependency tree · next 3 levels

Sources

$\sup (0, 1) = 1$ and $\inf (0, 1) = 0$ , with neither attained