つづき

選択公理からの証明
整列定理は選択公理から次のように導かれる。[ 9 ]
Let the set we are trying to well-order be
A, and let f be a choice function for the family of non-empty subsets of A.
For every ordinal α, define by transfinite recursion an element aα that is in A by setting
aα = f(A∖{aξ∣ξ<α})
if this complement A∖{aξ∣ξ<α}
is nonempty, or leaves aα undefined if the complement is empty.
That is, aα is chosen from the set of elements of
A that have not yet been assigned a place in the ordering (or undefined if the entirety of
A has been successfully enumerated).
Then the order < on A defined by aα<aβ if and only if
α<β (in the usual well-order of the ordinals) is a well-order of
A as desired, of order type
{α∣aα is defined}
(in the von Neumann representation).

選択公理の証明
選択公理は、整列定理から次のように証明できる。
To make a choice function for a collection of non-empty sets,
E, take the union of the sets in E and call it X. There exists a well-ordering of X; let R be such an ordering. The function that to each set
S of E associates the smallest element of S, as ordered by (the restriction to S of) R, is a choice function for the collection E.
An essential point of this proof is that it involves only a single arbitrary choice, that of
R; applying the well-ordering theorem to each member
S of E separately would not work, since the theorem only asserts the existence of a well-ordering, and choosing for each
S a well-ordering would require just as many choices as simply choosing an element from each S.
(引用終り)
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