The Real And Complex Number Systems

continuous at x, then
0
lim n
gx n
g lim n
x n by continuity of f at x
gx
x.
So, the function g may be continuous at 0. In fact, g is continuous at 0 which prove as
follows. Given 0, choose , as|x| , wehave
|gx g0| |gx| |x| . So,g is continuous at 0. Hence, from the preceding,
we know that g is continuous only at x 0.
3. Write
1ifx 0,
hx 0ifx R Q 0, 1,
1/n if x m/n, g. c. d. m, n 1.
Consider a 0, 1 and given 0, there exists the largest positive integer N such that
N 1/. LetT x : hx , then
0, 1 x : hx 1 x : hx 1/2...x : hx 1/N if 1,
T
if 1.
Note that T is at most a finite set, and then we can choose a 0 such that
a , a acontains no points of T and a , a 0, 1. So,if
x a , a a, wehavehx . It menas that
lim xa
hx 0.
Hence, we know that h is continuous at x 0, 1 R Q. For two points x 1, and
y 0, it is clear that h is not continuous at x 1, and not continuous at y 1bythe
method mentioned in the exercise of part 1 and part 2. Hence, we have proved that h is
continuous only at the irrational points in 0, 1.
Remark: 1. Sometimes we call f Dirichlet function.
2. Here is another proof about g, we write it down to make the reader get more.
Proof: Write
0ifx R Q 0, 1,
gx
x if x Q 0, 1.
Given a 0, 1, andifg is continuous at a, then given 0 a, there exists a 0
such that as x a , a 0, 1, wehave
|gx ga| .
If a R Q, choose 0 so that a Q. Then a a , a which
implies |ga ga| |ga | a a. But it is impossible.
If a Q, choose 0 so that a R Q. a a , a which
implies |ga ga| |a| a a. But it is impossible.
If a 0, given 0 and choose , thenas0 x , wehave
|gx g0| |gx| |x| x . It means that g is continuous at 0.
4.17 For each x 0, 1, letfx x if x is rational, and let fx 1 x if x is