The Real And Complex Number Systems

Proof: Let D denote the set of dicontinuities of f. Consider
D m x a, b : fx fx m 1 , then D
m1 D m . Note that #D m , so
we have D is countable. That is, the set of dicontinuities of f is countable.
(c) Prove that f has points of continuity in every open subintervals of a, b.
Proof: By (b), f has points of continuity in every open subintervals of a, b, since
every open subinterval is uncountable.
Remark: (1) Here is another proof about (b). Denote Q x 1 ,...,x n ,..., and let x be
a point at which f is not continuous. Then we have fx fx 0. (If x is the end
point, we consider fx fx 0orfx fx 0) So, we have an open interval I x
such that I x fa, b fx. The interval I x contains infinite many rational numbers,
we choose the smallest index, say m mx. Then the number of the set of discontinuities
of f on a, b is a subset of N. Hence, the number of the set of discontinuities of f on a, b
is countable.
(2) There is a similar exercise; we write it as a reference. Let f be a real valued function
defined on 0, 1. Suppose that there is a positive number M having the following
condition: for every choice of a finite number of points x 1 ,..,x n in 0, 1, wehave
n
M i1
x i M. Prove that S : x 0, 1 : fx 0 is countable.
Proof: Consider S n x 0, 1 : |fx| 1/n, then it is clear that every S n is
countable. Since S
n1 S n , we know that S is countable.
4.64 Give an example of a function f, defined and strictly increasing on a set S in R,
such that f 1 is not continuous on fS.
Solution: Let
x if x 0, 1,
fx
1ifx 2.
Then it is clear that f is strictly increasing on 0, 1, sof has the incerse function
x if x 0, 1,
f 1 x
2ifx 1.
which is not continuous on fS 0, 1.
Remark: Compare with Exercise 4.65.
4.65 Let f be strictly increasing on a subset S of R. Assume that the image fS has one
of the following properties: (a) fS is open; (b) fS is connected; (c) fS is closed. Prove
that f must be continuous on S.
Proof: (a) Given a S, then fa fS. Given 0, wewantofinda 0 such
that as x Ba; S, wehave|fx fa| . SincefS is open, then there exists
Bfa, fS, where .
Claim that there exists a 0 such that fBa; S Bfa, . Choose
y 1 fa /2 and y 2 fa /2, then y 1 fx 1 and y 2 fx 2 ,wehave
x 1 a x 2 since f is strictly increasing on S. Hence, for x x 1 , x 2 S, wehave
fx 1 fx fx 2 since f is strictly increasing on S. So,fx Bfa, .Let
mina x 1 , x 2 a, then Ba; S a , a S x 1 , x 2 S which implis
that fBa; S Bfa, .( Bfa, )
Hence we have prove the claim, and the claim tells us that f is continuous at a. Sincea
is arbitrary, we know that f is continuous on S.