Page 328 - Advanced Linear Algebra

312 Advanced Linear Algebra

Theorem 12.6 Let 4 be a metric space.
)
1 Any complete subspace of 4 is closed.
2 If 4 ) is complete, then a subspace of 4 : is complete if and only if it is
closed.
Proof. To prove 1 , assume that is a complete subspace of 4 . Let % ² ³ be a
)
:
sequence in for which % ² ³ ¦ %  4 . Then % ² ³ is a Cauchy sequence in :
:
:
:
and since is complete, % ² ³ must converge to an element of . Since limits of
:
sequences are unique, we have %: . Hence, is closed.
)
To prove part 2 , first assume that is complete. Then part 1 shows that is
)
:
:
closed. Conversely, suppose that is closed and let % ² ³ be a Cauchy sequence
:
:
in . Since ² % ³ is also a Cauchy sequence in the complete space 4 , it must
converge to some %4 . But since is closed, we have ²% ³¦%: . Hence,
:

: is complete.
Now let us consider some examples of complete and incomplete metric spaces.
)
(
Example 12.9 It is well known that the metric space is complete. However, a
(
s
proof of this fact would lead us outside the scope of this book. Similarly, the
³
complex numbers are complete.
d
Example 12.10 The Euclidean space s and the unitary space d are complete.
Let us prove this for s . Suppose that ²% ³ is a Cauchy sequence in s , where

% ~ ²% ÁÃÁ% Á ³

Á
Thus,

²% Á % ³ ~ ²% Á c % Á ³ ¦ as Á ¦ B

~
and so, for each coordinate position ,

²% Á c % Á ³  ²% Á % ³ ¦

which shows that the sequence ²% ³ ~ Á ÁÃ2 of th coordinates is a Cauchy
Á
sequence in . Since is complete, we must have
s
s
²% ³ ¦ & as ¦ B
Á
If & ~ ²& ÁÃÁ& ³ , then

²% Á &³ ~ ²% Á c & ³ ¦ as ¦ B

~
and so ²% ³ ¦ &  s . This proves that s is complete.