Order, lattices, ordered algebraic structures Proof
Proof of the uniqueness of the supremum
Builds on SupremumSet bounded abovePartial order
Proof
- Let be a partially ordered set, let , and let and be suprema of .
- By the first condition of the definition, is an upper bound of .
- By the second condition for , applied to the upper bound , we get .
- In the same way, is an upper bound of and is a supremum, so .
- The relation is a partial order, so it is antisymmetric. From and we get .
Idea
Each supremum is an upper bound, and each supremum lies below every upper bound. So two suprema lie below each other, and antisymmetry makes them equal.
Reading path
To understand this page from the ground up, read these 18 pages in order. The foundations come first.
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- no evidence level
- Element definition
- Proposition concept
- Set bounded above definition
- Set definition
- Implication definition
- Predicate definition
- Supremum definition
- Ordered pair definition
- Universal quantifier definition
- Cartesian product definition
- Subset definition
- Binary relation definition
- Antisymmetric relation definition
- Reflexive relation definition
- Transitive relation definition
- Preorder definition
- Partial order definition
- Proof of the uniqueness of the supremum proof
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Further reading
About this page
| Tag | QRBWM2V1RK |
|---|---|
| URI | m2://QRBWM2V1RK |
| Kind | proof |
| Writing | W0 stub |
| MSC | 06A06 |
| Proves | Uniqueness of the supremum |
| Uses | Supremum, Set bounded above, Partial order |
| Changed | 4 November 2026 |
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