163 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 {sn} that converges to b. Since f (sn) =0 for all n∈N, the sequence {f (sn)} 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 Rby f (x) = 1 q , if x = p q , p,q∈N,where pandqhave 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 3.3.7. Consider f (x) =((x−a1)(x−a2)· · ·(x−ak), if x ∈Q; 0, if x ∈Qc. SECTION 3.4 Exercise 3.4.6. Let α=min{f (x) : x ∈[a,b]}andβ =max{f (x) : x ∈[a,b]}. Then f (x1)+f (x2)+· · ·+f (xn) n ≤ nβ n =β. Similarly, α≤ f (x1)+f (x2)+· · ·+f (xn) n . Then the conclusion follows from the Intermediate Value Theorem. Exercise 3.4.7. (a) Observe that | f (1/n)| ≤1/nfor all n∈N. (b) Apply the Extreme Value Theorem for the function g(x) = f (x) x on the interval [a,b]. Exercise 3.4.8. First consider the case where f is monotone decreasing on[0,1]. By Exercise 3.4.5, f has a fixed point in[0,1], which means that there exists x0 ∈[0,1] such that f (x0) =x0. Since f is monotone decreasing, f has a unique fixed point. Indeed, suppose that there exists x1 ∈[0,1] such that f (x1) =x1. If x1 <x0, thenx1 =f (x1)≥f (x0) =x0, which yields a contradiction. It is similar for the case where x1 >x0. Therefore, x0 is the unique point in[0,1] such that f (x0) =x0.
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