1549901369-Elements_of_Real_Analysis__Denlinger_

230 Chapter 5 • Continuous Functions

Proof. (a) Suppose xis an irrational number. Then T(x) = 0.

Let n EN. Since x is not rational, '1im E Z 3 x = m. For each n EN, let

n m

On = the distance from x to the nearest rational number of the form -. Then

N 0 n (x) contains no rational number with denominator n.

n

1

Let€> 0. By the Archimedean property, 3no EN 3 - < €. Choose o =

no

min{o 1 ,o 2 , · · · , On 0 }. Then N 0 (x ) contains no rational number with denominator

~no.

Consider arbitrary y E N 0 (x).

m

(i) Suppose y is rational, say y = - -=I- 0 (where m E Z, n E N, and

n

. 1
m and n have no common pnme factors). Then T(y) = -, where n >no, so
n
1 1
IT(y)I = - < - < €.
n no
(ii) If y is irrational, then T(y) = 0.

In either case, (i) or (ii),

y E N 0 (x) =? IT(y) - OI < €

=? IT(y) -T(x)I < €.

Therefore, Tis continuous at x, an arbitrary irrational number.

(b) Let x be a rational number. Then T( x) -=I-0. However, since the irra-
tionals are dense in JR, there exists a sequence {Zn} of irrational numbers such
that Zn ----t x. By the sequential criterion for continuity at x, if T were continu-
ous at x, then T(zn) ----t T(x). But Vn EN, T(zn) = 0. This would imply that
T(x) = 0, which would contradict T(x) -=I-0. Therefore, T cannot be continuous
at x. •

Theorem 5 .1.1 3 (Algebra of Continuous Functions) Suppose f and g
are continuous at a point x 0 and c E JR. Then,

(a) cf is continuous at xo;

(b) f + g is continuous at xo;

/

(c) f - g is continuous at x 0 ;

(d) f · g is continuous at xo;

(e)

1

• is continuous at xo, if g(xo) -=I-O;
  g