Abel's theorem and the allied theory, including the theory by H.F. Baker

By H.F. Baker

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But since every non-empty subset of N has both a first and a last element, one sees that every subset of N + {m}which is not empty has this property as well. Therefore MeT, which means that the theorem is true for M. Theorem 45. If a set M is well-ordered and also inversely well-ordered, it is inductive finite. Proof. Let us assume the existence of elements y of M such that the set of all x = y was not inductive finite. Among these y there is then a least one, say m. There is a predecessor mi of m.

Proof. We obtain all sets M by taking in (1) all propositional functions which, by the operations of the predicate calculus, can be built from atomic statements yez and only contain x as a free variable. We may replace x by XQ, letting the bound variables be denoted by Xi, x2, .... Further, 0(x) may be written in prenex normal form, while its matrix is written in conjunctive normal form. Then we will get an enumeration of all 0(x) by enumerating all finite sequences consisting first of some pairs of integers corresponding to the quantifiers of the prefix, the first number being the index of the x which occurs as quantifier, the last number being 0 or 1 according as the quantifier is universal or existential.

We have seen that the ordinal numbers are well-ordered by the relation < . It is then natural to ask how the cardinal numbers behave. Because of the comparability of the ordinals it is immediately clear that the cardinal numbers are comparable; indeed, if M and N are any two sets and they are in some way well-ordered, then either M is similar to, and thus equivalent to, some initial part of N or inversely. Thus we have either M = N or N = M. Now let T be a set of sets. I assert that the cardinal numbers represented by the elements A,B,C,....

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