MATHEMATICS: R. L. MOORE 
367 
Pi, P 2 , P3, . . . such that every point of M is a limit point of N. For 
each ftj P n can be joined to P n + 1 by a simple continuous arc P n Pn + 1 lying 
entirely in M. In view of the fact that if M contains two arcs with only one 
point in common, that point must be an end-point of each of them, it is clear 
that, for every n, P1P2 + P2P3 +■..-.+ P n -i n P is either a simple closed 
curve or a simple continuous arc. It is clear that if, for some n, it is a simple 
closed curve, then this curve is identical with M. Suppose on the other hand, 
that, for every n, it is a simple continuous arc A n B n , the notation being so 
assigned that, in the order from A n to B n on the arc A n B n , every A { precedes 
every Bj (1 ^ i ^ n, 1 ^ n), Let TV* denote the point-set constituted 
by the sum of all the arcs A1B1, A 2 B 2 , A s Bz, ... If P h P 2 , P 3 , . . . is a 
set of points such that for every n, P n follows B n on some arc of the set A\Bu 
A 2 B 2 , . . . then the set Pi, P 2 , P3, . . . does not have more than one 
limit point. For suppose there exists such a set with two distinct limit points 
Oi and 0 2 . Then if d and C 2 are two distinct circles with center at Oi and 
such that 0 2 is without each of them, there clearly exists an infinite set of arcs 
G (Fig. 3) such that 1) each arc of G is, for some n, i and j (1 ^ i ^ n, 
1 ^ n) a sub-interval of the interval BiB 3 - of the arc A n B n , 2) no two arcs 
of G have a point in common, 3) each arc of G lies entirely between G and 
C 2 except that one of its endpoints is on Ci and the other one is on C 2 . But, 
in the course of the above proof that M is connected im kleinen, it was shown 
that M does not contain such a set of arcs. 
FIG. 
