110 Chapter^2is not empty. Hence the family of closed sets { G~ : a E I} has the finite
intersection property. However, from
LJ Ga= A
a EIit follows, by taking complements with respect to A, that
n G~=0
a EIThis shows that if A is not compact, then not every infinite family of closed
sets with the finite intersection property has a nonempty intersection.
The concept of total boundedness, which is introduced by the following
definition, is clearly related to that of compactness.
Definition 2.37 Let (S, d) be a metric space. A set M C S is called
totally bounded iff for every E > 0 the set M can be covered by finitely
many open c;-neighborhoods Ne(x) with x E M.
Theorem 2.18 Any compact subset of a metric space is totally bounded.
Proof Let A be a compact subset of a metric space ( S, d). Clearly, LJ Ne(x)
is an open covering of A for every E > 0. By the compactness of A there
exists a finite subcover of A by €-neighborhoods.
Theorem 2.19 A totally bounded subset of a metric space (S, d) 1s
bounded.
Proof If M C S is totally bounded, we have
nMC LJ Ne(xk)
k=lLet p1 and pz be any two points of M. Then we have
d(p1 , pz) < 2c + max d( x j , x k)
where j, k, E {l, 2, ... , n}, and we conclude that
6.(M) = supd(p1,p2):::; 2E +maxd(xj,Xk)Remarks I. From Theorems 2.18 and 2.19 it follows that any compact
subset of a metric space is bounded. This gives an alternative proof of one
of the conclusions of Theorem 2.14.
II. The converse of Theorem 2.19 is not true in general, i.e., a bounded
subset of a metric space is not necessarily totally bounded.
Example Consider the so-called real sequential Hilbert space l 2 , namely,
the set of all real sequences x = {xn}~ such that 2:~ x; converges, with