Math 2.0

Order, lattices, ordered algebraic structures Proof

Proof of the uniqueness of the supremum

Builds on SupremumSet bounded abovePartial order

Proof

  1. Let (P,≤)(P, \le) be a partially ordered set, let S⊆PS \subseteq P, and let bb and b′b' be suprema of SS.
  2. By the first condition of the definition, bb is an upper bound of SS.
  3. By the second condition for b′b', applied to the upper bound bb, we get b′≤bb' \le b.
  4. In the same way, b′b' is an upper bound of SS and bb is a supremum, so b≤b′b \le b'.
  5. The relation ≤\le is a partial order, so it is antisymmetric. From b′≤bb' \le b and b≤b′b \le b' we get b=b′b = b'.

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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Each page uses the pages above it. The colour of the bar on a box shows how strong the evidence is, and the box also gives it as text. The list has the same pages.
  • no evidence level
  1. Element definition
  2. Proposition concept
  3. Set bounded above definition
  4. Set definition
  5. Implication definition
  6. Predicate definition
  7. Supremum definition
  8. Ordered pair definition
  9. Universal quantifier definition
  10. Cartesian product definition
  11. Subset definition
  12. Binary relation definition
  13. Antisymmetric relation definition
  14. Reflexive relation definition
  15. Transitive relation definition
  16. Preorder definition
  17. Partial order definition
  18. Proof of the uniqueness of the supremum proof

Where this leads

1 page builds on this one.

Further reading

  1. Web https://proofwiki.org/wiki/Supremum_is_Unique
About this page
TagQRBWM2V1RK
URIm2://QRBWM2V1RK
Kindproof
WritingW0 stub
MSC06A06
ProvesUniqueness of the supremum
UsesSupremum, Set bounded above, Partial order
Changed4 November 2026

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