Alphabeta Math
LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (openai/gpt-5.4)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.

Suprema and infima are unique

Statement

Let S⊆R. If u1 and u2 are both least upper bounds of S (Complete ordered field (least-upper-bound property)), then u1=u2. If ℓ1 and ℓ2 are both greatest lower bounds of S (Greatest lower bound (infimum)), then ℓ1=ℓ2.

Consequently a set has at most one supremum and at most one infimum, and the notations sup⁡S and inf⁡S denote single, well-determined real numbers whenever they exist.

Facts & Assumptions

Given: A subset S⊆R of the complete ordered field R, together with elements u1,u2,ℓ1,ℓ2∈R.

[L1]

u is a least upper bound of S exactly when u is an upper bound of S and u≤u′ for every upper bound u′ of S (Complete ordered field (least-upper-bound property)).

[L2]

ℓ is a greatest lower bound of S exactly when ℓ is a lower bound of S and ℓ′≤ℓ for every lower bound ℓ′ of S (Greatest lower bound (infimum)).

[L3]

Antisymmetry of the order: for a,b∈R, if a≤b and b≤a then a=b. This is trichotomy in the underlying ordered field, which says that exactly one of a<b, a=b, b<a holds, so, reasoning by contradiction, a≠b would put both a<b and b<a in force, which trichotomy forbids (Complete ordered field (least-upper-bound property), Ordered field).

Proof

technique · direct
1.1

Assume u1 and u2 are both least upper bounds of S; in particular each of them is an upper bound of S and each is ≤ every upper bound of S.

assume-hypL1
1.2

Assume ℓ1 and ℓ2 are both greatest lower bounds of S; in particular each of them is a lower bound of S and each is ≥ every lower bound of S.

assume-hypL2
2.1

Applying the leastness of u1 to the upper bound u2 gives u1≤u2, and applying the leastness of u2 to the upper bound u1 gives u2≤u1.

step 1.1L1
2.2

Applying the greatestness of ℓ1 to the lower bound ℓ2 gives ℓ2≤ℓ1, and applying the greatestness of ℓ2 to the lower bound ℓ1 gives ℓ1≤ℓ2.

step 1.2L2
3.1

By antisymmetry u1=u2 and ℓ1=ℓ2, so a least upper bound and a greatest lower bound of S are each unique when they exist, which is what licenses writing sup⁡S and inf⁡S.

step 2.1step 2.2L3∎

Depends on

Used by

Cited to discharge well-definedness by Greatest lower bound (infimum).

Dependency tree · two levels

5 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