452 Mr. A. E. Harward on the 



If /3= 7 + l, 



then 0(/3) = C(y) and o>0=<» y <», 



and CK)=C(« y )N , 



= C(y)K, as 7 <A 



= 0(7) = C(/9). 



I£ /3 denote a type of series with no last term, then the 

 series of ordinals less than co? (which is a series of type o>^) 

 is partitioned by the numbers 



i, &), a) 2 , ... ft*v..o> 1 ', ••• (y<£) 



into segments of which these numbers are the first terms. 

 Each segment is of type co s .($<ft). 



Therefore the cardinal number of the aggregate of terms 

 in each segment is equal to or less than C(ft) *. 



The series of segments is of type ft; therefore the cardinal 

 number of segments =C(/3). 



.-. C(a>*)<{0(/3)}». 



If {C(/3)P=C(/3), then it follows that C(©")=C(/3). 



Now if ft be not the first ordinal of a class, there is some 

 ordinal <y</3 such that C(ft) = 0(7); as the propositions A are 

 true for y, it follows that {C(ft)\ 2 = C(ft), and therefore in 

 this case the propositions A are true for ft also. 



But if ft be the first ordinal of a class, then we have to 

 prove that \C(ft)\ 2 —C(ft). This it is easy to do with the 

 aid of the assumption that the propositions A are true for all 

 ordinals less than ft. 



Let ft = u K the first ordinal of a class. 



We can classifv all the ordinals less than co K as follows : — 



(1) 



ordinals 



of the 



type 



V 



or 



coS + v, 



(2) 



)■> 



•>) 





cov 



1) 



(O^S + (OV, 



(3) 



55 



5? 





(0 2 V 



17 



G) d 8 + a/V, 



0) 



)"> 



?> 





(0 W V 



?5 



ft) w + 1 S + <*>uv, 



(y) 



55 



J5 





(o'v 



JJ 



a»-r+ 1 8 + g) y v. 



In the above scheme 7 and 8 stand for any ordinals less 

 than co,c = ft, and v stands for any finite ordinal (1, 2, ...). 

 It has to be proved that all these ordinals are less than eo K . 



* Because C(w*) = C(8) (6 transfinite) and =^ (6 finite). 



