Cardinality, The Cardinals do not Form a Set

The Cardinals do not Form a Set

Let S be a set and let W be an ordinal that maps 1-1 into S. Let T be the image of W. Now W implements a well ordering on the members of T, and leaves the rest of S alone. Conversely, if T is a well ordered subset of S, an isomorphism maps T back to a unique ordinal W.

Let U be the set of all possible relations on S cross S. (Note that U cannot be built without the power set axiom.) Restrict U to those relations that implement a well ordering on a subset of S, and have no ordered pairs from the rest of S. Let D be the domain of a function f that carries the well ordered subset to the corresponding ordinal. Invoke the replacement axiom and the range of f becomes a set. Yet the range is precisely the corresponding ordinal that fits inside S. If every ordinal embeds in S then there is a set containing all the ordinals, which is impossible. Therefore there is some ordinal that does not map into S.

Let W be the least ordinal that does not map into S. If W can be mapped onto a lower ordinal, compose the two functions to carry W into S. Therefore W is a cardinal. For any set S, there is a cardinal, i.e. a bigger set, that cannot embed in S.

If S is the set of all cardinals, there is some cardinal W that is not a subset of S. Since both are ordinals, S is a proper subset of W. Now S is transitive, and if it contains W it contains all the members of W, which is impossible. Hence there are cardinals other than those found in S.