190 



Round Pk as centre we take an interval 7^ (a quadrangle, a cube, 

 etc. according to the number of dimensions of tlie space in which 

 E is given), so that E Ik is an inner limiting set, taking care that 

 the boundary of Ik contains no point of E, wliicli is possible on 

 account of E being enumerable. 



By Ik we understand the 0|)eii interval, by Ik we shall indicate 

 the closed one, by an accent, the complement of a set. Now 



E= EI, + Ei, iT,r-{-Ei, iinir + . . . 



From N°. 1 there follows now immediately that E is an inner 

 limiting set. 



4. Let E be enumerable and not an inner limiting set. In this 

 case according lo N°. 3 the set D of the points E in which E is 

 not an inner limiting set, is not empty. Let P be a point of Z> and 

 ƒ an interval with P as centre. EI is according to N°. 2 not an 

 inner limiting set, hence neither is EJ — P; according lo N°. 3, 

 El — P contains a point Q in which E[ — P is not an inner limiting 

 set, hence E is not an inner limiting set in Q, so that Q lies in D. 

 From this there follows that D is dense in itself and from that the 

 theorem which was to be proved. 



