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By Thoralf Skolem

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Example text

But in this case it is obvious that there is no other maximal element than u itself, which proves the theorem. We might therefore just as well define a finite set as a set with property that there is a maximal subset in every set of subsets. We have seen that this notion coincides with the notion inductive finite, and we may notice that we have proved this without any use of the axiom of choice. A further definition of finiteness is the following: A set M is called Dedekind finite, if there is no one-to-one correspondence between M and any proper subset MT of M.

But then y0U {z} is a maximal element in M. Hence, since u is inductive finite, the theorem is true for u. The inverse is also true, namely: Theorem 40. If every set of subsets ofu then u is inductive finite. contains a maximal element, Proof. In particular there is a maximal element in every set x of subsets such that Oex and (yex) & (z-eu) —»(y U {z}eu). But in this case it is obvious that there is no other maximal element than u itself, which proves the theorem. We might therefore just as well define a finite set as a set with property that there is a maximal subset in every set of subsets.

We let u - Sxo be mapped identically onto itself while every g(y), where yex 0 , shall be the image of g(y-i) for the corresponding y _ j . This provides a mapping of Sxo onto the proper part Sxo - (g(0)}. Indeed every zeSxo must be a g(y) for some yexo, because otherwise we could remove all elements y containing the element z from XQ and still have a subset x with the properties 1) and 2). Theorem 44. If an inductive finite set is well-ordered, it is also inversely well-ordered by the same ordering.

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