The Real And Complex Number Systems

In Exercises 4.29 through 4.33, we assume that f : S T is a function from one metric
space S, d S to another T, d T .
4.29 Prove that f is continuous on S if, and only if,
f 1 intB intf 1 B for every subset B of T.
Proof: () Suppose that f is continuous on S, and let B be a subset of T. Since
intB B, wehavef 1 intB f 1 B. Note that f 1 intB is open since a pull back of
an open set under a continuous function is open. Hence, we have
intf 1 intB f 1 intB intf 1 B.
That is, f 1 intB intf 1 B for every subset B of T.
() Suppose that f 1 intB intf 1 B for every subset B of T. Given an open subset
U T, i.e., intU U, sowehave
f 1 U f 1 intU intf 1 U.
In addition, intf 1 U f 1 U by the fact, for any set A, intA is a subset of A. So,f is
continuous on S.
4.30 Prove that f is continuous on S if, and only if,
fclA clfA for every subset A of S.
Proof: () Suppose that f is continuous on S, and let A be a subset of S. Since
fA clfA, then A f 1 fA f 1 clfA. Note that f 1 clfA is closed
since a pull back of a closed set under a continuous function is closed. Hence, we have
clA clf 1 clfA f 1 clfA
which implies that
fclA ff 1 clfA clfA.
() Suppose that fclA clfA for every subset A of S. Given a closed subset
C T, and consider f 1 C as follows. Define f 1 C A, then
fclf 1 C fclA
clfA clff 1 C
clC C since C is closed.
So,wehaveby(fclA C)
clA f 1 fclA f 1 C A
which implies that A f 1 C is closed set. So, f is continuous on S.
4.31 Prove that f is continuous on S if, and only if, f is continuous on every compact
subset of S. Hint. If x n p in S, the set p, x 1 , x 2 ,... is compact.
Proof: () Suppose that f is continuous on S, then it is clear that f is continuous on
every compact subset of S.
() Suppose that f is continuous on every compact subset of S, Givenp S, we
consider two cases.
(1) p is an isolated point of S, then f is automatically continuous at p.
(2) p is not an isolated point of S, thatis,p is an accumulation point p of S, then there
exists a sequence x n S with x n p. Note that the set p, x 1 , x 2 ,... is compact, so we
know that f is continuous at p. Sincep is arbitrary, we know that f is continuous on S.
Remark: If x n p in S, the set p, x 1 , x 2 ,... is compact. The fact is immediately