Appendix to Chapter III
321
The first is meaningless unless X is embedded in an "ambient" space:
X is always closed in X. The condition that X be "bounded" means that
(7.1)
sup d(a,x) < +00
xEX
for some a E X (and so for all a E X), but one needs more to ensure (BW) or
(BL). On the other hand it is clear that (BW) implies (1), as in the classical
case.
Moreover,
(BW) = } X is complete,
since if a Cauchy sequence (xn ) in X contains a convergent subsequence, it
is obvious that it converges to the same limit.
Since, for all r > 0, the open balls B (x, r) centred at the points of X cover
X, (BL) implies the following property:
(PC) For every r > 0, X is the union of a finite number of open
balls of radius r.
Neither does this property, precompactness, suffice to ensure compactness:
in C any bounded set satisfies it trivially (see, in Chap. V, nO 6, beginning
of the proof of BL). But if X is complete, then (PC) implies both (BW) and
(BL).
Proof of (BW): let (x(n)) be a sequence of points of X; by applying (PC)
for r = lone obtains an open ball BI of radius 1 in X which contains infinitely many terms of the given sequence, i.e. a subsequence of the given
sequence; then, using (PC) for r = 1/2, one finds a ball B2 of radius 1/2
which contains infinitely many terms of the first subsequence, so a subsequence extracted from it. Pursuing this "construction", one obtains balls Bn
of radius l/n and integers PI < P2 < ... such that X(Pk) E Bn for all k > n.
Now d [x (Pk) , x (Ph)] < 2/n for k, h > n; whence a Cauchy sequence, which
converges if X is complete.
Proof of (BL): this reduces to showing that if (Ui) is a covering of X by
open sets then there exists an r > 0 such that, for all x, the ball B(x, r) is
contained in one of the Ui ; but since (PC) implies BW as we have just seen,
one can argue as in Chap. V, n° 6: there is nothing to change in the proof.
The right definition then consists of declaring that a metric space is compact if it is precompact and complete. One can show without difficulty that
conversely (BW) or (BL) implies compactness, in other words that these three
properties are equivalent (see Dieudonne, Treatise on Analysis, Vol. 1, III.16).
It is almost obvious that all we have said in Chap. III and V about compact sets in C extends to the general case. First, a subset K of a metric space
X will be called compact if it is so as a metric space "in itself"; clearly this
forces K to be closed and bounded in X, but this condition is not sufficient if
one imposes no other hypothesis on X; for example, the unit ball Ilxll :::; 1 of
321
The first is meaningless unless X is embedded in an "ambient" space:
X is always closed in X. The condition that X be "bounded" means that
(7.1)
sup d(a,x) < +00
xEX
for some a E X (and so for all a E X), but one needs more to ensure (BW) or
(BL). On the other hand it is clear that (BW) implies (1), as in the classical
case.
Moreover,
(BW) = } X is complete,
since if a Cauchy sequence (xn ) in X contains a convergent subsequence, it
is obvious that it converges to the same limit.
Since, for all r > 0, the open balls B (x, r) centred at the points of X cover
X, (BL) implies the following property:
(PC) For every r > 0, X is the union of a finite number of open
balls of radius r.
Neither does this property, precompactness, suffice to ensure compactness:
in C any bounded set satisfies it trivially (see, in Chap. V, nO 6, beginning
of the proof of BL). But if X is complete, then (PC) implies both (BW) and
(BL).
Proof of (BW): let (x(n)) be a sequence of points of X; by applying (PC)
for r = lone obtains an open ball BI of radius 1 in X which contains infinitely many terms of the given sequence, i.e. a subsequence of the given
sequence; then, using (PC) for r = 1/2, one finds a ball B2 of radius 1/2
which contains infinitely many terms of the first subsequence, so a subsequence extracted from it. Pursuing this "construction", one obtains balls Bn
of radius l/n and integers PI < P2 < ... such that X(Pk) E Bn for all k > n.
Now d [x (Pk) , x (Ph)] < 2/n for k, h > n; whence a Cauchy sequence, which
converges if X is complete.
Proof of (BL): this reduces to showing that if (Ui) is a covering of X by
open sets then there exists an r > 0 such that, for all x, the ball B(x, r) is
contained in one of the Ui ; but since (PC) implies BW as we have just seen,
one can argue as in Chap. V, n° 6: there is nothing to change in the proof.
The right definition then consists of declaring that a metric space is compact if it is precompact and complete. One can show without difficulty that
conversely (BW) or (BL) implies compactness, in other words that these three
properties are equivalent (see Dieudonne, Treatise on Analysis, Vol. 1, III.16).
It is almost obvious that all we have said in Chap. III and V about compact sets in C extends to the general case. First, a subset K of a metric space
X will be called compact if it is so as a metric space "in itself"; clearly this
forces K to be closed and bounded in X, but this condition is not sufficient if
one imposes no other hypothesis on X; for example, the unit ball Ilxll :::; 1 of