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.