ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck 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.

$\sup [0, 1] = 1 = \max [0, 1]$ and $\inf [0, 1] = 0 = \min [0, 1]$

Example

The closed unit interval $J = [0, 1] = {x \in R : 0 \leq x \leq 1}$ has $\sup J = 1$ and $\inf J = 0$ , exactly as the open interval $\sup (0, 1) = 1$ and $\inf (0, 1) = 0$ , with neither attained does, but here both values belong to $J$ : they are the maximum and the minimum (Maximum and minimum of a set).

That is the whole content of this example. The two intervals have the same supremum and the same infimum, and differ only in attainment, so the numerical value of a supremum says nothing about whether the set reaches it. The computation is also easier than for the open interval, because a set with a maximum has that maximum as its least upper bound for free, with no epsilon argument and no appeal to completeness (The supremum is attained exactly when a maximum exists).

Facts & Assumptions

Given: The complete ordered field $R$ and the closed interval $J : = {x \in R : 0 \leq x \leq 1}$ .

[L1]

Attainment: for a nonempty $X \subseteq R$ , if $X$ has a maximum then $\sup X$ exists and $\sup X = \max X$ ; and if $\sup X$ exists and lies in $X$ then $\max X$ exists and equals $\sup X$ (The supremum is attained exactly when a maximum exists).

[L2]

Maximum and minimum: $m = \max X$ means $m \in X$ and $x \leq m$ for every $x \in X$ ; $m = \min X$ means $m \in X$ and $m \leq x$ for every $x \in X$ ; each is unique when it exists (Maximum and minimum of a set).

[L3]

Bounds and the infimum: $u$ is an upper bound of $X$ when $x \leq u$ for every $x \in X$ (Complete ordered field (least-upper-bound property)), and $ℓ$ is a lower bound of $X$ when $ℓ \leq x$ for every $x \in X$ (Lower bound, bounded below, bounded set); and $ℓ = \inf X$ means $ℓ$ is a lower bound of $X$ with $ℓ^{'} \leq ℓ$ for every lower bound $ℓ^{'}$ of $X$ , unique when it exists (Greatest lower bound (infimum)).

[L4]

Order: $0 < 1$ ; the order is reflexive, so $a \leq a$ ; and $a < b$ implies $a \leq b$ (The multiplicative identity is positive, Complete ordered field (least-upper-bound property), Ordered field).

Verification

technique · direct

1.1

Both $0$ and $1$ lie in $J$ , so $J \neq \emptyset$ : from $0 < 1$ we get $0 \leq 1$ , and reflexivity gives $0 \leq 0$ and $1 \leq 1$ , so each of $0$ and $1$ satisfies the membership condition $0 \leq x \leq 1$ .

1.2

Every $x \in J$ satisfies $x \leq 1$ and $0 \leq x$ , directly by the membership condition; so $1$ is an upper bound of $J$ and $0$ is a lower bound of $J$ .

2.1

$1 = \max J$ : the number $1$ lies in $J$ and dominates every element of $J$ .

step 1.1step 1.2L2

2.2

$0 = \min J$ : the number $0$ lies in $J$ and is dominated by every element of $J$ .

step 1.1step 1.2L2

3.1

Since $J$ is nonempty and has the maximum $1$ , the attainment criterion gives that $\sup J$ exists and $\sup J = \max J = 1$ .

step 2.1step 1.1L1

3.2

$\inf J = 0$ : the number $0$ is a lower bound of $J$ , and any lower bound $ℓ^{'}$ of $J$ satisfies $ℓ^{'} \leq 0$ because $0$ is itself an element of $J$ ; so $0$ is the greatest lower bound of $J$ , and it equals $\min J$ .

step 2.2step 1.2L3

4.1

Therefore $\sup J = 1 \in J$ and $\inf J = 0 \in J$ : both bounds are attained, and they coincide with the maximum and the minimum of $J$ .

step 3.1step 3.2L1L2∎

Remarks

Claim 1 of The supremum is attained exactly when a maximum exists needs no completeness: a set with a maximum has a least upper bound because the maximum already is one. The least-upper-bound property of $R$ (Complete ordered field (least-upper-bound property)) is what handles sets with no maximum, such as the open interval of $\sup (0, 1) = 1$ and $\inf (0, 1) = 0$ , with neither attained.
The infimum was computed here directly from Greatest lower bound (infimum) rather than by reflecting through the origin. Both routes are available; the direct one is shorter when the bound is attained, since leastness of the upper bound and greatestness of the lower bound are then immediate from membership.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

18 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

Maximum and minimum (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]=1=max⁡[0,1]sup[0,1]=1=max[0,1] and inf⁡[0,1]=0=min⁡[0,1]inf[0,1]=0=min[0,1]