Alphabeta Math
LemmaStatement: Literature-sourcedProof: Literature-sourcedSession-authored (Fable 5 assisted)precheck 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 SRS \subseteq \mathbb{R}. If u1u_1 and u2u_2 are both least upper bounds of SS (Complete ordered field (least-upper-bound property)), then u1=u2u_1 = u_2. If 1\ell_1 and 2\ell_2 are both greatest lower bounds of SS (Greatest lower bound (infimum)), then 1=2\ell_1 = \ell_2.

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

Facts & Assumptions

Given: A subset SRS \subseteq \mathbb{R} of the complete ordered field R\mathbb{R}, together with elements u1,u2,1,2Ru_1, u_2, \ell_1, \ell_2 \in \mathbb{R}.

[L1]

uu is a least upper bound of SS exactly when uu is an upper bound of SS and uuu \le u' for every upper bound uu' of SS (Complete ordered field (least-upper-bound property)).

[L2]

\ell is a greatest lower bound of SS exactly when \ell is a lower bound of SS and \ell' \le \ell for every lower bound \ell' of SS (Greatest lower bound (infimum)).

[L3]

Antisymmetry of the order: for a,bRa, b \in \mathbb{R}, if aba \le b and bab \le a then a=ba = b. This is trichotomy in the underlying ordered field, which says that exactly one of a<ba < b, a=ba = b, b<ab < a holds, so, reasoning by contradiction, aba \ne b would put both a<ba < b and b<ab < a in force, which trichotomy forbids (Complete ordered field (least-upper-bound property), Ordered field).

Proof

technique · direct
1.1

Assume u1u_1 and u2u_2 are both least upper bounds of SS; in particular each of them is an upper bound of SS and each is \le every upper bound of SS.

assume-hypL1
1.2

Assume 1\ell_1 and 2\ell_2 are both greatest lower bounds of SS; in particular each of them is a lower bound of SS and each is \ge every lower bound of SS.

assume-hypL2
2.1

Applying the leastness of u1u_1 to the upper bound u2u_2 gives u1u2u_1 \le u_2, and applying the leastness of u2u_2 to the upper bound u1u_1 gives u2u1u_2 \le u_1.

step 1.1L1
2.2

Applying the greatestness of 1\ell_1 to the lower bound 2\ell_2 gives 21\ell_2 \le \ell_1, and applying the greatestness of 2\ell_2 to the lower bound 1\ell_1 gives 12\ell_1 \le \ell_2.

step 1.2L2
3.1

By antisymmetry u1=u2u_1 = u_2 and 1=2\ell_1 = \ell_2, so a least upper bound and a greatest lower bound of SS are each unique when they exist, which is what licenses writing supS\sup S and infS\inf S.

step 2.1step 2.2L3

Depends on

Used by

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

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 5 results over 5 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