Introduction to Mathematical Analysis I - Second Edition

153 Clearly, the number of q ∈ N such that q ≤ 1 ε is finite. Since 0 < p q ≤ 1, we have p ≤ q . Therefore, A ε is finite. (b) Fix any irrational number a ∈ ( 0 , 1 ] . Then f ( a ) = 0. Given any ε > 0, by part (a), the set A ε is finite, so we can write A ε = { x ∈ ( 0 , 1 ] : f ( x ) ≥ ε } = { x 1 , x 2 , . . . , x n } , for some n ∈ N , where x i ∈ Q for all i = 1 , . . . , n . Since a is irrational, we can choose δ > 0 such that x i / ∈ ( a − δ , a + δ ) for all i = 1 , . . . , n (more precisely, we can choose δ = min {| a − x i | : i = 1 , . . . , n } ). Then | f ( x ) − f ( a ) | = f ( x ) < ε whenever | x − a | < δ . Therefore, f is continuous at a . Now fix any rational number b = p q ∈ ( 0 , 1 ] . Then f ( b ) = 1 q . Choose a sequence of irrational numbers { s n } that converges to b . Since f ( s n ) = 0 for all n ∈ N , the sequence { f ( s n ) } does not converge to f ( b ) . Therefore, f is not continuous at b . In this problem, we consider the domain of f to be the interval ( 0 , 1 ] , but the conclusion remain valid for other intervals. In particular, we can show that the function defined on R by f ( x ) =   1 q , if x = p q , p , q ∈ N , where p and q have no common factors; 1 , if x = 0; 0 , if x is irrational , is continuous at every irrational point, and discontinuous at every rational point. Exercise 3.3.10 . Consider f ( x ) = ( ( x − a 1 )( x − a 2 ) · · · ( x − a k ) , if x ∈ Q ; 0 , if x ∈ Q c . SECTION 3.4 Exercise 3.4.6 . Let α = min { f ( x ) : x ∈ [ a , b ] } and β = max { f ( x ) : x ∈ [ a , b ] } . Then f ( x 1 )+ f ( x 2 )+ · · · + f ( x n ) n ≤ n β n = β . Similarly, α ≤ f ( x 1 )+ f ( x 2 )+ · · · + f ( x n ) n . Then the conclusion follows from the Intermediate Value Theorem. Exercise 3.4.7 . (a) Observe that | f ( 1 / n ) | ≤ 1 / n for all n ∈ N .