MATHEMATICS: R. L. MOORE 
365 
exist (fig. 1), within y, points E', G' and K' and arcs E'K' and K'G' such that 
1) E' and G' are on the arc EFG in the order EE'FG'G and K' is on the arc 
FK, 2) the interval E'G' of the arc EFG lies wholly within y, 3) E'K' has in 
common with EFG only the point £' and, in common with FK, only the point 
K' , 4) K'G' has in common with EFG only the point G' and, in common with 
FK, only the point K', 5) the segment FK' of the arc FK lies wholly within 
the closed curve / formed by the arcs E'K' and K'G' and the interval E'FG' of 
the arc EFG. The arc E'K' together with the interval E'F of EFG and the 
interval FK' of Pi£ form a closed curve /i. The arc K'G' together with the 
interval FG' of EFG and the interval FK' of FK form a closed curve P 2 . Let 
7, 7i and 7 2 denote the interiors of /, J\ and P 2 respectively. Then 7 is the 
sum of I\, 1 2 and the segment FX' of the arc FK. Let P, Pi and P 2 denote 
points on the segments FK' , E'F and FG' respectively. The point P is a 
FIG. 1 
boundary point of R and the region 7 contains P. Hence there must be points 
of R in 7 and therefore in 7i or in 7 2 . The point Pi also is on the boundary 
of R. It follows that there exist, within the circle y, points of R lying either 
without / (and therefore without both Ji and J 2 ) or within Ji. Similarly either 
7 2 or the exterior of / contains points of R that are within y. Thus either both 
7i and 7 2 contain points of R or only one of them contains points of R in 
which latter cases there must exist points of R that are within y and without 
/. But if a connected point-set lies entirely within a and contains either 1) 
a point of Ii and a point of 7 2 or 2) a point of h or 7 2 and a point without /; 
then it must clearly contain at least one point of EFG or of FK. Therefore it 
can not be a subset of R. Thus the supposition that M contains two arcs 
EFG anql FK with only the point F in common leads to a contradiction. 
That M, the boundary of R, is connected is a consequence of a theorem of 
HausdorfFs. 6 I will proceed to show that it is connected im kleinen. Sup- 
