458 Mr. A. E. Harward on the 



we can go on indefinitely performing a certain endless process. 

 We have no right to assume that we may regard that process 

 as completed. In fact it would require a non-enumerable 

 infinity of steps to complete it. 



In order to complete the proof on the lines indicated by 

 Mr. Jourdain, it is necessary that some rule or formula 

 f analogous to the formula 



for finite numbers) should be given by which the required 

 correlation can be established once for all. As I could not 

 succeed in constructing such a formula, I adopted a different 

 method of proof. 



I have recently discovered that by a slight modification of 

 Mr. Jourdnin's method a simple and rigorous proof can be 

 obtained. Instead of considering the double series a up take 

 the aggregate of couples * of ordinals less than © 1? and let 

 them be arranged in the following scheme : — 



(1, 2) (1,3X1, 4) (1,51 . 



.. (l,a>)(l,co + l) . 



.. (1,0) 



(2, 3)(2, 4)(2, 5) .. 



.. (2,a,)(2,«, + l) .. 



■ (2,/3) 



(3, 4) (3, 5) .. 



.. (3, <a)(3, <b + 1) . 



■■ (3,/3) 



(4,5) . 



.. (4, a>)(4, » + l) . 



■■ (4,/3) 



(©,© + 1) ... (©,£) ... 



(7,£J - 7</3 



The couples in the same horizontal line have the same lesser 

 component, and the couples in the same vertical column have 

 the same greater component- 

 Reading by horizontal lines the couples form a series of 

 type mf, but reading by vertical columns the couples form a 

 series of different type, and by means of the proposition 

 Kq = K it is easy to prove that the type of this latter series 

 is &)i. 



indefinitely, and there is no point short of eoj where we are compelled to 

 stop. But we may not infer that we can correlate all the ordinals less 

 than (o x with the terms of the progression. For the residue of the 

 progression is enumerable, and the residue of the other series is always 

 non-enumerable. In assuming that his process can be completed Mr, 

 Jourdain really assumes that the residue of his double series has the 

 same cardinal number as the residue of his single series. 



* (a /3) is the same couple as (/3 a) and (a a) is not a couple. 



