162 Solutions and Hints for Selected Exercises SECTION 3.2 Exercise 3.2.5. The given condition implies that if both x1 and x2 are close to x0, then they are close to each other and, hence, f (x1) and f (x2) are close to each other. This suggests the use of the Cauchy ε criterion for limit to solve the problem. Given any ε > 0, choose δ = . If x1,x2 ∈ D\{x0} 2(k + 1) with |x1 − x0| < δ and |x2 − x0| < δ , then ε | f (x1) − f (x2)|≤ k|x1 − x2|≤ k(|x1 − x0| + |x2 − x0|) < k(δ + δ )= 2k < ε. 2(k + 1) Therefore, limx →x0 f (x) exists. SECTION 3.3 Exercise 3.3.6. (a) Observe that f (a)= g(a)= h(a) and, hence, ( |g(x) − g(a)|, if x ∈ Q ∩ [0,1]; | f (x) − f (a)| = |h(x) − h(a)|, if x ∈ Qc ∩ [0,1]. It follows that | f (x) − f (a)|≤|g(x) − g(a)| + |h(x) − h(a)| for all x ∈ [0,1]. Therefore, limx →a f (x)= f (a) and, so, f is continuous at a. (b) Apply part (a). Exercise 3.3.13. At any irrational number a ∈ (0,1], we have f (a)= 0. If x is near a and x is irrational, it is obvious that f (x)= 0 is near f (a). In the case when x is near a and x is rational, f (x)= 1/q where p,q ∈ N. We will see in part (a) that for any ε > 0, there is only a fnite number of x ∈ (0,1] such that f (x) ≥ ε. So f (x) is close to f (a) for all x ∈ (0,1] except for a fnite number of x ∈ Q. Since a is irrational, we can choose a suffciently small neighborhood of a to void such x. (a) For any ε > 0, p 1 p 1 Aε = {x ∈ (0,1] : f (x) ≥ ε} = x = ∈ Q : f (x)= ≥ ε = x = ∈ Q : q ≤ . q q q ε p Clearly, the number of q ∈ N such that q ≤ 1 ε is fnite. Since 0 < ≤ 1, we have p ≤ q. Therefore, q Aε is fnite. (b) Fix any irrational number a ∈ (0,1]. Then f (a)= 0. Given any ε > 0, by part (a), the set Aε is fnite, so we can write Aε = {x ∈ (0,1] : f (x) ≥ ε} = {x1,x2,...,xn}, for some n ∈ N, where xi ∈ Q for all i = 1,...,n. Since a is irrational, we can choose δ > 0 such that xi ∈/ (a−δ ,a+δ ) for all i = 1,...,n (more precisely, we can choose δ = min{|a−xi| : i = 1,...,n}). Then | f (x) − f (a)| = f (x) < ε whenever |x − a| < δ .
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