The Real And Complex Number Systems

from the statement that every infinite subset p, x 1 , x 2 ,... of has an accumulation point in
p, x 1 , x 2 ,....
4.32 A function f : S T is called a closed mapping on S if the image fA is closed
in T for every closed subset A of S. Prove that f is continuous and closed on S if, and only
if,
fclA clfA for every subset A of S.
Proof: () Suppose that f is continuous and closed on S, and let A be a subset of S.
Since A clA, wehavefA fclA. So, we have
clfA clfclA fclA since f is closed. *
In addition, since fA clfA, wehaveA f 1 fA f 1 clfA. Note that
f 1 clfA is closed since f is continuous. So, we have
clA clf 1 clfA f 1 clfA
which implies that
fclA ff 1 clfA clfA. **
From (*) and (**), we know that fclA clfA for every subset A of S.
() Suppose that fclA clfA for every subset A of S. Gvien a closed subset C
of S, i.e., clC C, then we have
fC fclC clfC.
So, we have fC is closed. That is, f is closed. Given any closed subset B of T, i.e.,
clB B, we want to show that f 1 B is closed. Since f 1 B : A S, wehave
fclf 1 B fclA clfA clff 1 B clB B
which implies that
fclf 1 B B clf 1 B f 1 fclf 1 B f 1 B.
That is, we have clf 1 B f 1 B. So,f 1 B is closed. Hence, f is continuous on S.
4.33 Give an example of a continuous f and a Cauchy sequence x n in some metric
space S for which fx n is not a Cauchy sequence in T.
Solution: Let S 0, 1, x n 1/n for all n N, andf 1/x : S R. Then it is clear
that f is continous on S, andx n is a Cauchy sequence on S. In addition, Trivially,
fx n n is not a Cauchy sequence.
Remark: The reader may compare the exercise with the Exercise 4.54.
4.34 Prove that the interval 1, 1 in R 1 is homeomorphic to R 1 . This shows that
neither boundedness nor completeness is a topological property.
Proof: Since fx tan x : 1, 1 R is bijection and continuous, and its
2
converse function f 1 x arctanx : R 1, 1. Hence, we know that f is a Topologic
mapping. (Or say f is a homeomorphism). Hence, 1, 1 is homeomorphic to R 1 .
Remark: A function f is called a bijection if, and only if, f is 1-1 and onto.
4.35 Section 9.7 contains an example of a function f, continuous on 0, 1, with
f0, 1 0, 1 0, 1. Prove that no such f can be one-to-one on 0, 1.
Proof: By section 9.7, let f : 0, 1 0, 1 0, 1 be an onto and continuous function.
If f is 1-1, then so is its converse function f 1 . Note that since f is a 1-1 and continous
function defined on a compact set 0, 1, then its converse function f 1 is also a continous