The Real And Complex Number Systems

|a m a n| |a m a m1 a m1 a m2 ...a n1 a n |
|b m1 | ...|b n|
|b 1 | 1
2
m2
... 1
2
n1
, *
then a n is a Cauchy sequence since 1 2 k converges. Hence, we know that a n is a
convergent sequence.
Remark: In this exercise, we use the very important theorem, every Cauchy sequence
in the Euclidean space R k is convergent.
4.7 In a metric space S, d, assume that x n x and y n y. Prove that
dx n , y n dx, y.
Proof: Since x n x and y n y, given 0, there exists a positive integer N such
that as n N, wehave
dx n , x /2 and dy n , y /2.
Hence, as n N, wehave
|dx n , y n dx, y| |dx n , x dy n , y|
dx n , x dy n , y /2 /2
.
So, it means that dx n , y n dx, y.
4.8 Prove that in a compact meric space S, d, every sequence in S has a subsequence
which converges in S. This property also implies that S is compact but you are not required
to prove this. (For a proof see either Reference 4.2 or 4.3.)
Proof: Given a sequence x n S, and let T x 1 , x 2 ,.... If the range of T is finite,
there is nothing to prove. So, we assume that the range of T is infinite. Since S is compact,
and T S, wehaveT has a accumulation point x in S. So, there exists a point y n in T such
that By n , x 1 n . It implies that y n x. Hence, we have proved that every sequence in S
has a subsequence which converges in S.
Remark: If every sequence in S has a subsequence which converges in S, then S is
compact. We give a proof as follows.
Proof: In order to show S is compact, it suffices to show that every infinite subset of S
has an accumulation point in S. Given any infinite subset T of S, and thus we choose
x n T (of course in S). By hypothesis, x n has a subsequence x kn which converges
in S, say its limit x. From definition of limit of a sequence, we know that x is an
accumulation of T. So,S is compact.
4.9 Let A beasubsetofametricspaceS. IfA is complete, prove that A is closed. Prove
that the converse also holds if S is complete.
Proof: Let x be an accumulation point of A, then there exists a sequence x n such that
x n x. Sincex n is convergent, we know that x n is a Cauchy sequence. And A is
complete, we have x n converges to a point y A. By uniqueness, we know x y A.
So, A contains its all accumulation points. That is, A is closed.
Suppose that S is complete and A is closed in S. Given any Cauchy sequence
x n A, we want to show x n is converges to a point in A. Trivially, x n is also a
Cauchy sequence in S. SinceS is complete, we know that x n is convergent to a point x in
S. By definition of limit of a sequence, it is easy to know that x is an adherent point of A.
So, x A since A is closed. That is, every Cauchy sequence in A is convergent. So, A is