1549901369-Elements_of_Real_Analysis__Denlinger_

5.1 Continuity of a Function at a P oint 229

Proof. Let xo E R For contradiction, suppose f is continuous at x 0. Since
the rational numbers are dense^1 in JR, ::3 sequence { Xn} of rational numbers such
that Xn --4 x 0. Since the irrational numbers are also dense in JR, ::3 sequence {yn}
of irrational numbers su ch t hat Yn --4 xo. But f is continuous at xo. Thus, by
the sequential criterion,

f (xn) --4 f (xo), and
f(Yn) --4 f(xo).
Now, \:/n E N, f(xn) = 1 and f (Yn) = 0. Thus, f (xo) = 1 and f (xo) = 0.
Contradiction. Therefore, f is not continuous at xo. Since xo is a perfectly gen-
eral real number, we have proved that f is discontinuous at every real number.
D

Example 5.1.12 Thomae's Function (A function that is continuous at ev-
ery irrational number and discontinuous at every rational number) The follow-
ing function, known as Thomae's function, is extremely interesting:
1 m

• if x = - I-0, where m E Z, n E N and
  n n
  T(x) =

m and n h ave no common prime factors
1 if x = 0
0 if x is irrational.
Thomae's function T(x ) is continuous at every irra tional number, but is
discontinuous at every rational number.

y

1/2 • •

1 /3 • • • •

1/4 • • • •

• • •
  ··~
  ,. • • • • •
  •• ••• ,. • •• •• • ·~

l I .!. 2 3 2 5. l 2 2 2 x

4 3 2 3 4 4 3 2 3 4

Partial Graph ofThomae's Function

Figure 5.1

1. See Theorems 2.3.6 and 3.2.21.