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PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13
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A poset category is complete exactly when every small family has an infimum, and cocomplete exactly when every small family has a supremum

Statement

A poset regarded as a category is complete if and only if every set-indexed family has an infimum, including the empty family. It is cocomplete if and only if every set-indexed family has a supremum, including the empty family. Hence it is both complete and cocomplete exactly when it is a complete lattice.

Facts & Assumptions

Given: A poset P regarded as a category.

[F1]

In the associated category, x→y exists exactly when x≤y, and there is at most one such arrow (A preorder is a category with at most one morphism between any two objects, and its functors are exactly monotone maps).

[F2]
[L1]

Products plus equalizers characterize completeness, and coproducts plus coequalizers characterize cocompleteness (A category is complete exactly when it has all small products and equalizers, and cocomplete exactly when it has all small coproducts and coequalizers).

Proof

technique · translate universal properties into inequalities
1.1

A cone from x to a discrete family (pi) is precisely the collection of inequalities x≤pi. By [F1] and [F2], a product is therefore a lower bound above every lower bound, namely inf⁡ipi. For the empty family this is a greatest element.

F1F2
1.2

Reversing inequalities, a coproduct is sup⁡ipi, with the empty coproduct a least element. Coequalizers are identities for the same at-most-one-arrow reason. The dual half of [L1] proves both directions of the cocompleteness equivalence.

F1F2L1
2.1

Parallel arrows in a poset category are equal whenever they exist. The identity of their common domain is consequently an equalizer, since every factor is unique by [F1]. Hence [L1] and step 1.1 prove both directions of the completeness equivalence.

F1L1step 1.1
3.1

Having all set-indexed infima and suprema, including empty ones, is exactly the complete-lattice condition, which proves the last assertion.

step 2.1step 1.2∎

Depends on

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