Transfinite Numbers. 453 



This can be shown as follows : — 

 From (A), if y be transfinite 



0(0,0 = 0(7), 

 C(c»'+ 1 )=C( 7 +l)=C(7) ) 

 {C( 7 )}*=C( 7 ); 



and if 8 be transfinite, 



{C(S)}*=C(S). 

 Therefore, if both 7 and 8 be transfinite, 



G{co' / + 1 S\ = C(y) or C(S), whichever be the greater. 



Therefore 



C (a)y +1 8 + co y v) = C(y) or C(S), whichever be the greater. 



If 7 be finite and 8 transfinite the above expression equals 

 C(8) ; and if y be transfinite and 8 finite then the above ex- 

 pression equals 0(7) ; and if both 7 and 8 be finite then it is 

 equal to K . 



In any case, therefore, 



C0 / + l 8-\-Q) V V <C0 K) 



provided that 7 and 8 be less than co K . 



It follows from this that every ordinal in the above scheme 

 is less than co K . 



All the ordinals in the above scheme are different, for if 



8 2 + co<*v 2 (/3) 



a)' 2 



/ 2 +i 



Then, since a series of type (a) has a last segment of type 

 » y i, and a series of type (ft) has a last segment of type ft) y 2, a 

 last segment of a series of type ft) Yl is in ordinal correspon- 

 dence with a last segment of a series of type ft> /2 , but this is 

 not possible unless 7 X =7 2 , for every last segment of a series 

 of type g) y is itself of type co y . 



Therefore 71 = 72? and it obviously follows that v 1 = v 2 and 

 8i = 8 2 . 



Now as 8 may be any ordinal whatever less than co K , the 

 number of ordinals in each set (i. e. in each horizontal line) 

 is C(ft> K )=K K , and the series of sets is a series of type &> K , 

 since 7 may be any ordinal less than co K . Therefore there 

 are 8 K sets. 



Therefore the total number of ordinals in the scheme is 

 X K , but the number is also tf~ K ; .'. ^ = ^ K . 



We have thus proved that if the propositions (A) are true 



