446 Mr. A. E. Harward on the 



a ~>ft\> a \>ft2 • - ' & c « Since there cannot be a progression 

 of ordinals in inverse order of magnitude, it follows that we 

 must after some finite number o£ steps come to a stage 

 where a n =/3 n +i or /3 w = a w , in either case we get a part of the 

 aggregate A which is similar both to A and B. Therefore A 

 is similar to B, and 8 = 0. 



We are now in a position to assert that if & and u be any 

 two different cardinals then either H>b orb>H; i. e. the 

 cardinals are a simply-ordered class *. 



6. It is easy to prove that the finite cardinals form a 

 progression, and that N which is defined as the cardinal 

 number of the aggregate of the terms of a progression, is the 

 smallest transfinite cardinal, for any initial segment of a 

 progression is finite. 



Since any transfinite well-ordered series with a last term 

 can be converted into a series with no last term by merely 

 transferring a finite number of the terms from the end to the 

 beginning ; it follows that any infinite aggregate can be 

 arranged in a well-ordered series with no last term, and since 

 the type of such a series is unaltered by substituting for each 

 term a finite series of n terms, it follows that rcft=tl 5 if a 

 be transfinite. 



If b<a, a + b<a + a=a; 



but a+b>a; .-. a+b=a(b<a). 



By diagonal enumeration of a progression of progressions 

 it is proved that tf 2 = tf . 



It is easy to prove now that tt c a = a if a be any transfinite 

 cardinal. 



For an aggregate of cardinal number 3 can be arranged 

 in a well-ordered series with no last term, i. e. a series of 

 type &)/5 where /3 is some ordinal. 



By substituting for each term of this series a progression 

 of terms we convert the series into one of type co 2 /3 and 

 the cardinal number of terms in the new series is K o a. But 

 in this series each component of type co 2 can be rearranged 

 in a series of type co ; the series is then of type co/3. 



K a=a. 



7. We are now in a position to classify the ordinals. I use 



* By following Cantor's method of procedure it can be proved, with- 

 out using Schroder & Bernstein's theorem, that the cardinals form a well- 

 ordered series. S. & B.'s theorem can then be interred as a corollary. 

 But the course of the argument is much simplified if S. & B.'s theorem 

 be first proved. 



