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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)
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Zorn's lemma implies the Axiom of Choice

Statement

Assume Zorn's lemma: every nonempty poset in which every chain has an upper bound has a maximal element. Then every family of nonempty sets has a choice function (Choice function); that is, the Axiom of Choice holds.

Facts & Assumptions

Given: A family F all of whose members are nonempty, and Zorn's lemma.

[A1]

Every member of F is nonempty.

[A2]

By the assumed Zorn principle, a nonempty poset in which every chain has an upper bound has a maximal element.

[L2]

A choice function for a family G is a function h with domain G such that h(S)∈S for every S∈G (Choice function).

[L3]

m is maximal when there is no element strictly above it (Maximal element and greatest element).

[L4]

A partial order is a relation that is reflexive, antisymmetric and transitive, and a poset is a set carrying one (Partial order and partially ordered set).

[L5]

A chain is a subset of a poset in which any two members are comparable (Chain in a poset).

Proof

technique · contradiction
1.1

Suppose F has no choice function.

assume-contra
1.2

Let P be the set of partial choice functions: pairs (G,h) with G⊆F and h a choice function for G, ordered by (G,h)≤(G′,h′) when G⊆G′ and h′ restricted to G equals h.

L2construct
2.1

P is nonempty, since the empty function is a choice function for the empty subfamily, so (∅,∅)∈P.

step 1.2L2
2.2

Every chain in P has an upper bound: given a chain, take the union of the domains and the union of the functions. Any two partial choice functions in the chain are comparable by [L5], so the smaller is a restriction of the larger; their values therefore agree on overlapping domains. Thus the union is a function and is a choice function for the union of the domains.

step 1.2L2L5
2.3

The relation just defined is a partial order on P: it is reflexive, since G⊆G and h restricted to G is h; antisymmetric, since (G,h)≤(G′,h′) and (G′,h′)≤(G,h) give G⊆G′ and G′⊆G, hence G=G′, and then h=h′ restricted to G=G′, which is h′ itself; and transitive, since G⊆G′⊆G′′ gives G⊆G′′, while restricting h′′ to G is the same as first restricting it to G′, which gives h′, and then restricting h′ to G, which gives h. So (P,≤) is a poset.

step 1.2L4
3.1

By the assumed Zorn principle P has a maximal element (G,h).

step 2.1step 2.2step 2.3A2
4.1

If G=F then h is a choice function for F, contrary to the assumption; so there exists S∈F with S∉G.

step 3.1step 1.1L2choose
5.1

The set S is nonempty, so there exists an element of S; fix one and call it a.

step 4.1A1choose
6.1

Then (G∪{S},h∪{(S,a)}) lies in P and is strictly above (G,h), since S∉G.

step 4.1step 5.1step 1.2L2
7.1

This contradicts maximality of (G,h), so the assumption fails and F has a choice function.

step 6.1step 3.1L3discharge-contradiction∎

Remarks

Depends on

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Dependency tree · two levels

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Sources