163 Therefore, f is continuous at a. Now fx any rational number b = q p ∈ (0,1]. Then f (b)= q 1. 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 defned on R by 1 p , if x = , p,q ∈ N,where p and q have no common factors; q q f (x)= 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 ( (x − a1)(x − a2)···(x − ak), if x ∈ Q; f (x)= 0, if x ∈ Qc . 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 (x1)+ f (x2)+ ··· + f (xn) nβ ≤ = β . n n Similarly, f (x1)+ f (x2)+ ··· + f (xn) α ≤ . 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. f (x) (b) Apply the Extreme Value Theorem for the function g(x)= on the interval [a,b]. x Exercise 3.4.8. First consider the case where f is monotone decreasing on [0,1]. By Exercise 3.4.5, f has a fxed 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 fxed point. Indeed, suppose that there exists x1 ∈ [0,1] such that f (x1)= x1. If x1 < x0, then x1 = 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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