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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13
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  • 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.
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Set has all small limits, realized as compatible tuples in a set-indexed product

Statement

Every small diagram D:J→Set has a limit. It is the set

L={(xj)j∈Ob⁡J∈∏jD(j):D(u)(xj)=xk for every u:j→k}

with its coordinate projections.

Facts & Assumptions

Given: A small category J and a diagram D:J→Set.

[F1]

A small category has sets of objects and morphisms, and completeness means existence of limits for all small diagrams (Finite, small, and large limits and colimits; complete and cocomplete categories).

Proof

technique · construction
1.1F1F2F3

By [F1] and [F3], the displayed product and its subset L are sets. For each j, let pj:L→D(j) be the coordinate function. The defining equalities give D(u)pj=pk, so (L,p) is a cone.

1.2F3

If Ob⁡J is empty, the product is the singleton containing the empty function and all compatibility conditions are vacuous. Thus the construction still gives the terminal set.

1.3F2given

Let (X,ξj) be any cone. Define h:X→∏jD(j) by h(x)j=ξj(x). The cone equations imply h(x)∈L, so h corestricts to a function hˉ:X→L satisfying pjhˉ=ξj.

2.1step 1.2step 1.3

If g:X→L has the same composites, then for every x and j, g(x)j=pjg(x)=ξj(x)=pjhˉ(x). Equality of functions gives g=hˉ. This remains true for the empty index, where there is one function to the singleton.

3.1F1F4step 1.1step 1.3step 2.1∎

By [F4], steps 1.1, 1.3, and 2.1 prove that (L,p) is a limit, with step 1.2 covering the empty boundary. Since D was an arbitrary small diagram, Set is complete.

Depends on

Used by

Dependency tree · two levels

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Sources