The Real And Complex Number Systems

(b) Give an example of a metric space in which B a; r is not the closure of the open
ball Ba; r.
Solution: Consider discrete metric space M, then we have let x M
The closure of Ba;1 a
and
B a;1 M.
Hence, if we let a is a proper subset of M, then B a;1 is not the closure of the open ball
Ba;1.
3.32 In a metric space M, if subsets satisfy A S A , where A is the closure of A,
thenAissaidtobedense in S. For example, the set Q of rational numbers is dense in R. If
A is dense in S and if S is dense in T, prove that A is dense in T.
Proof: Since A is dense in S and S is dense in T, wehaveA S and S T. Then
A T. Thatis,A is dense in T.
3.33 Refer to exercise 3.32. A metric space M is said to be separable if there is a
countable subset A which is dense in M. For example, R 1 is separable becasue the set Q of
rational numbrs is a countable dense subset. Prove that every Euclidean space R k is
separable.
Proof: Since Q k is a countable subset of R k ,andQ k
R k , then we know that R k is
separable.
3.34 Refer to exercise 3.33. Prove that the Lindelof covering theorem (Theorem 3.28)
is valid in any separable metric space.
Proof: Let M, d be a separable metric space. Then there exists a countable subset
S x n : n N M which is dense in M. Given a set A M, and an open covering F
of A. Write P Bx n , r m : x n S, r m Q.
Claim that if x M, andG is an open set in M which contains x. Then
x Bx n , r m G for some Bx n , r m P.
Since x G, there exists Bx, r x G for some r x 0. Note that x clS since S is
dense in M. Then, Bx, r x /2 S . So, if we choose x n Bx, r x /2 S and r m Q
with r x /2 r m r x /3, then we have
x Bx n , r m
and
Bx n , r m Bx, r x
since if y Bx n , r m , then
dy, x dy, x n dx n , x
r m r x
2
r x
3 r x
2
r x
So, we have prvoed the claim x Bx n , r m Bx, r x G or some Bx n , r m P.
Use the claim to show the statement as follows. Write A GF G, and let x A, then
there is an open set G in F such that x G. By the claim, there is Bx n , r m : B nm in P
such that x B nm G. There are, of course, infinitely many such B nm corresponding to
each G, but we choose only one of these, for example, the one of smallest index, say
q qx. Then we have x B qx G.