Transfinite Numbers. 455 



In the same way it can be proved that 



obviously if j3 < a C (a^) = C (a) . 



13. We can subdivide the classes of ordinals into sub- 

 classes o£ which the first terms are ordinals K €$, where K e x 

 denotes the first ordinal such that 



©f< K e 1 if P< K e 1} 



and K 6|3 denotes the first ordinal, such that 



if 7^/3 and 5< K e /S . 



The cardinal number of such sub-classes in any class is 

 obviously equal to the cardinal number of ordinals in that 

 class. 



The following propositions follow immediately from the 

 proposition that 



If in any class we select any aggregate of ordinals, of 

 cardinal number not greater than the cardinal associated 

 with the class, then there are ordinals of the class after all 

 the ordinals of that aggregate. If the ordinals of a class be 

 divided into segments in any way so that there is no last 

 segment, then the cardinal number of the segments is equal 

 to the cardinal number of the aggregate of the ordinals of 

 the class, i. e. the cardinal associated with the next class. 



I think it probable that 



and =K y+ i(7>/3), 



but I cannot prove it at present. 



United Service Club, 

 Calcutta. 



Note A. 



*The proof which Mr. Jourdain gives of Schroder and 

 Bernstein's theorem (Phil. Mag. Jan. 190-4, pp. 71-73) 

 appears to me to be defective. The proof involves the 

 theorem that if a = a -rl) then R=& + XJb. The argument 



* A note on these points was submitted with the above article to 

 Mr. Jourdain, but the note was defective and incorrectly expressed, and 

 the reply with which Mr. Jourdain has favoured me. does not appear to 

 me to meet the real objection to his arguments which I failed to make 

 clear in my former note. 



