The Real And Complex Number Systems

this is because a can be an isolated point. However, if a is an accumulation point, we then
have
f is continuous at a if, and only if, lim xa
fx fa.
4.16 Let f, g, andh be defined on 0, 1 as follows:
fx gx hx 0, whenever x is irrational;
fx 1andgx x, whenever x is rational;
hx 1/n, ifx is the rational number m/n (in lowest terms);
h0 1.
Prove that f is not continuous anywhere in 0, 1, that g is continuous only at x 0, and
that h is continuous only at the irrational points in 0, 1.
Proof: 1. Write
0ifx R Q 0, 1,
fx
1ifx Q 0, 1.
Given any x R Q 0, 1, andy Q 0, 1, and thus choose x n Q 0, 1
such that x n x, andy n R Q 0, 1 such that y n y. Iff is continuous at x,
then
1 lim n
fx n
f lim n
x n by continuity of f at x
fx
0
which is absurb. And if f is continuous at y, then
0 lim n
fy n
f lim n
y n by continuity of f at y
fy
1
which is absurb. Hence, f is not continuous on 0, 1.
2. Write
0ifx R Q 0, 1,
gx
x if x Q 0, 1.
Given any x R Q 0, 1, and choose x n Q 0, 1 such that x n x. Then
x
lim n
x n
lim n
gx n
limg lim n
x n by continuity of g at x
gx
0
which is absurb since x is irrational. So, f is not continous on R Q 0, 1.
Given any x Q 0, 1, and choose x n R Q 0, 1 such that x n x. Ifg is