Topology of Plane Sets of Points 117
Proof 1. Necessity. Suppose that S is complete, and let En have center
at an and radius rn,. Since
Em c Enc EN
whenever m > n > N, the centers an and am belong to EN, so that
d(am,an) < 2rN. Since 'T"N--+ 0, for each 6 > 0 there.is an integer N such
that r N <^1 / 2 6, and hence d( am, ~n) < 6 provided that m > n > N. This
means that the sequence {an} of the centers of the spheres is a Cauchy
sequence. Since S 'is complete, lim·an = a exists and belongs to S.
Now each sphere En contains all the points an with the possible ex-
ception of the points a~, a 2 , ••• , an-I· Hence En contains a for each n,
since En. is a closed set. Therefore, a en;: 1 En: i.e., the intersection of
the closed nested spheres is not ·empty.· In fact, the intersection consists
of the single point a.. · ·
- Sufficiency. Suppose that the condition in the statement of the
theorem is satisfied yet S is not complete. Under this assumption there
is some Cauchy sequence { x n} in S that does not converge to a point of
S, and we.can find a strictly increasing sequence {kn} of positive integers
such that
1
d(xm,Xkn) < 2 n+l
form> kn. Now consider the sequence {En} of closed spheres with center
at xkn and radii 'T"n l/2n. This sequence is nested because w E En+I
implies that
so that
and
1
d(w,Xkn) ~ d(w,xkn+i) + d(xkn+ 1 ,Xkn) <
2
n
which means that w E En. Hence En+I C En for every n.
Let
00
x En En
n=l
this intersection being nonvoid by assumption. Since 'T"n --+ 0 as n --+ oo,
it follows that the sequence { Xkn} of the centers of the spheres tends to x.
but this is a subsequence of {xn}. Hence, by Theorem 2.26, Xn--+ x, which
is a contradiction.
In a similar way one can prove the following proposition:
