Theorem 6.1: Let S and T two disjoint sets. Then

Proof: Let n = #(S ) and m = #(T ). Then by Definition 4.5, there exist one to one correspondences

and

We define a function

by

if s is an element of S and

if t e T. Since S and T are disjoint, the function is well defined. It is one to one because the elements of S go to smaller numbers than the elements of T, and it is onto because the elements of the domain are either less than or equal to n, in which case they are images of elements of S or if they are larger than n and less than or equal to n + m, they are some successor less than the m th successor of n and are thus images of elements of T.