208 
MATHEMATICS: R. L. MOORE 
Lemma 1. If in a class L every set of points which contains the point P in its 
interior contains at least one point of the set M distinct from P, then the point P 
is a Ihniting element of the set M, 
Proof. If P were not a limiting element of M then it would be interior to 
L — M -\- P. But this set of elements contains no point of M distinct from 
P. 
Theorem 1. //, in a class 5*, M is a closed and compact set of points and /3 is 
a well-ordered sequence of point- sets such that M is covered hy the family composed 
of all the members of ^, then there exists a member g of the sequence /3 such that 
M is covered by the family composed of g together with all those members of ^ 
that precede g. 
Proof. Suppose there exists no such member g. Then for each member x 
of the sequence (3 let denote the set of all points P belonging to M such that 
P is not in the interior of x or of any member of (3 that precedes x. For every 
X, Mx contains at least one point. For every x, is closed. For suppose 
P is a limiting element of M^. If P were not in then it would necessarily 
be in the interior either of the point-set x or of some point-set of the sequence 
/5 that precedes x and therefore, by a lemma of Hedrick's,^ this particular point- 
set would contain, in its interior, at least one point of M^, which is contrary to 
the definition of M^. It follows that the family composed of all MJs for all 
members x of the sequence /3 is a monotonic family of closed point-sets. Hence 
there exists a point Pq which is in every M^. But Po is in the interior of some 
point-set Xq of the sequence jS. Hence Po is not in Mxq. Thus the supposi- 
tion that Theorem 1 is not true leads to a contradiction. 
Theorem 2. In order that the Heine-Borel-Lebesgue Theorem should hold true 
in a given class S it is necessary and sufficient that that class S should be also a 
class 5*. 
Proof. Suppose that in a given class 5* the closed and compact point-set 
M is covered by the infinite family G of point-sets. The members of the 
family G can^ be arranged in a well-ordered sequence /?. By Theorem 1, 
there exists at least one member g oi ^ such that M is covered by the family 
composed of g and all those members of /? that precede g, together with not 
more than a finite number of those elements of /3 that follow g. Let gi denote 
the first^ such g and let g2, ^3, g4, • • • gn denote a finite set of members fol- 
lowing gi such that M is covered by gi,g2,gz, . . • gn together with all 
those members of /3 that precede gi. There clearly cannot exist a member of 
jS immediately preceding ^i. I will show that gi is the first member of /3. 
Suppose it is not. Then if and all the succeeding members of the sequence 
j(5 are removed, there remains a well-ordered sequence (3i whose members 
are the remaining members of /3 arranged in exactly the same order as 
in the original sequence /3. Let us now construct a third well-ordered 
sequence ^2 having for its first n members the point-sets ^1, g2, gs, . . . gn 
arranged in the order indicated and having as its remaining members the 
members of /3i arranged in the same order as in ^i. The point-set ^2 has no 
