Here the power set notation gives . Suppose a nonempty set has no -minimal member. Form its intersection . This is a set: for any one , use the axiom schema of separation inside to define the elements common to all members of . Selecting a single witness to nonemptiness is not an application of the axiom of choice.
For an arbitrary , lack of minimality supplies some with . Since , every subset of is a subset of , so
This conclusion holds for each , without choosing all the corresponding simultaneously. Intersecting over gives .
For completeness, the contradiction in Cantor's theorem is completely explicit here. By the axiom schema of separation put . Then , and consequently
which is impossible. Thus no such exists:
This power-set-predecessor relation proof uses neither a descending sequence nor a comparison of possibly non-well-orderable cardinalities. It proves well-foundedness directly from the minimal-member definition.
Order the collection of every chain in a partial order, including the empty chain, by inclusion. Suppose that is an order-preserving function and an injection. Let be the Hartogs ordinal: by Hartogs theorem, no injection from to exists.
Use transfinite recursion to define, for ,
We verify both that is defined and that the values are strictly increasing. Assume inductively that the earlier values are strictly increasing. Then is a chain in a partial order. For any , , so the fact that is order-preserving gives . Furthermore , whereas by distinctness of the earlier values. Thus , and injectivity gives . We have proved
The empty initial chain starts the induction, and the same argument applies at every limit ordinal.
If a globally defined recursion rule is desired, apply when the earlier range is a chain and otherwise use the fixed value . The induction shows that the fallback case never occurs. Thus no choices of new points are being made: the function uniquely determines every stage.
The resulting function is an injection from to , contrary to its defining Hartogs ordinal property. Hence the chain-poset nonembedding lemma gives

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