HOMEWORK ASSIGNMENT 5

Transcription

1 HOMEWORK ASSIGNMENT 5 DUE 1 MARCH, ) Let f(x) = 1 if x is rational and f(x) = 0 if x is irrational. Show that f is not continuous at any real number. Solution Fix any x R. We will show that f is not continuous at x. We work by contradiction. Assume f is continuous at x. Then there exists δ > 0 such that for all y R with y x < δ, f(y) f(x) < 1. This means that for any y (x δ, x + δ), f(y) f(x) < 1. If x is rational, there exists y (x δ, x + δ) which is irrational, and f(y) f(x) = 0 1 = 1. If x is irrational, then there exists y (x δ, x + δ) which is rational, and f(y) f(x) = 1 0 = 1. In either case, we obtain a contradiction. Alternative proof: Since Q = R \ Q = R, any x is the limit of a sequence (x n ) of rational numbers and is the limit of a sequence of irrational numbers (y n ). Then if f is continuous at x, since lim n x n = x, f(x) = lim n f(x n ) = lim n 1 = 1. But since lim n y n = x, f(x) = lim n f(y n ) = lim n 0 = 0, a contradiction. 1

2 2 DUE 1 MARCH, ) Suppose f : A R is a function such that f(a) = {f(x) : x A} is a finite set. Let x int(a). Show that f is continuous at x if and only if there exists δ > 0 such that (x δ, x + δ) A and for all y (x δ, x + δ), f(y) = f(x). Solution First suppose there exists δ > 0 such that (x δ, x + δ) A and for all y (x δ, x+δ), f(y) = f(x). Then for any ε > 0, for any y (x δ, x+δ), f(y) f(x) = 0 < ε. This shows that f is continuous at x. Next, suppose f is continuous at x. Since x int(a), there exists r > 0 such that (x r, x + r) A. If f(a) has only one element, say f(a) = {c} then f(y) = c for all y A. In this case we may take δ = r. Consider the case that f(a) contains more than one element. Note that since f(a) is finite, there exists a positive number d such that for any distinct y 1, y 2 f(a), y 1 y 2 d. Indeed, we may take d = min{ y 1 y 2 : y 1, y 2 f(a), y 1 y 2 }. This is a finite, non-empty set, so it has a minimum (this is why the fact that f(a) is finite is necessary). Since f is continuous at x, there exists δ > 0 such that for any y A with y x < δ, f(y) f(x) < d. Let δ = {r, δ }. Then by our choice of r, (x δ, x + δ) (x r, x + r) A. Moreover, if y (x δ, x + δ) A, y x < δ δ, so f(y) f(x) < d. But as we noted above, if f(y), f(x) are different, then f(y) f(x) d, so it must be that f(y) = f(x).

3 HOMEWORK ASSIGNMENT 5 3 3) If f : R R and g : R R are continuous functions such that f(x) = g(x) for every rational number x, show that f(x) = g(x) for all x R. Solution Suppose f, g are continuous and f(x) = g(x) for every rational number x. Fix any real number x and fix a sequence (x n ) of rational numbers such that lim n x n = x. We may do this, since Q = R. Since x n is rational, f(x n ) = g(x n ). Then by continuity and the fact that lim n x n = x, we deduce that f(x) = lim n f(x n ) = lim n g(x n ) = g(x).

4 4 DUE 1 MARCH, ) Suppose f : A R is a function. Suppose that x int(a), f(x) > µ for some real number µ, and f is continuous at x. Show that there exists a number δ > 0 such that (x δ, x + δ) A and for all y (x δ, x + δ), f(y) > µ. Solution Let ε = f(x) µ. Note that since f(x) > µ, ε is positive. Since x int(a), there exists a number r > 0 such that (x r, x + r) A. Since f is continuous at x, there exists δ > 0 such that for all y A with y x < δ, f(y) f(x) < ε. Let δ = min{r, δ }. Then (x δ, x + δ) (x r, x + r) A and for any y (x δ, x + δ) A, y x < δ δ, f(y) f(x) < ε. Then f(x) f(y) f(x) f(y) = f(y) f(x) < ε = f(x) µ. Therefore f(x) f(y) < f(x) µ. Subtracting f(x) from both sides and multiplying by 1 (which flips the inequality), we obtain f(y) > µ.

5 HOMEWORK ASSIGNMENT 5 5 5) Suppose f : R R and g : R R are continuous functions. For each x R, let h(x) = max{f(x), g(x)}. Show that h is continuous. Solution Fix x R. We must show that h is continuous at x. Fix ε > 0. There exists δ f > 0 such that for all y R with x y < δ f, f(x) f(y) < ε. There exists δ g > 0 such that for all y R, g(x) g(y) < ε. Let δ = min{δ f, δ g }. Then for any y R with x y < δ δ f, δ g, f(x) ε < f(y) < f(x) + ε and g(x) ε < g(y) < g(x) + ε. Then Moreover, h(y) = max{f(y), g(y)} < max{f(x) + ε, g(x) + ε} = h(x) + ε. h(y) = max{f(y), g(y)} > max{f(x) ε, g(x) ε} = h(x) ε. Since h(x) ε < h(y) < h(x) + ε, h(y) h(x) < ε.

6 6 DUE 1 MARCH, ) Show that if I is an interval and f : I R is continuous, then f(i) = {f(x) : x I} is also an interval. Show that if I is closed and bounded, so is f(i). Note: We say the set A is an interval if for any x, y A with x < y, [x, y] A. Solution To show that f(i) is an interval, we need to fix y 1 < y 2 and show that for any c (y 1, y 2 ), c f(i). To that end, suppose y 1, y 2 f(i) are fixed with y 1 < y 2, and c (y 1, y 2 ) is fixed. Since y 1 f(i), there exists x 1 I with f(x 1 ) = y 1. Since y 2 f(i), there exists x 2 I such that f(x 2 ) = y 2. Define g : I R by g(x) = f(x) c. Note that g is continuous on I. We consider two cases: If x 1 < x 2, we apply the intermediate value theorem with a = x 1 and b = x 2 to the function g, noting that and g(a) = g(x 1 ) = f(x 1 ) c = y 1 c < 0 g(b) = g(x 2 ) = f(x 2 ) c = y 2 c > 0. By the intermediate value theorem, there exists x 3 [x 1, x 2 ] such that g(x 3 ) = 0. Note that since I is an interval and x 1, x 2 I, x 3 [x 1, x 2 ] I. Since 0 = g(x 3 ) = f(x 3 ) c, f(x 3 ) = c. Therefore in this case, c = f(x 3 ) f(i). The other case is that x 2 < x 1. We apply the intermediate value theorem with a = x 2 and b = x 1 to the function g, noting that and g(a) = g(x 2 ) = f(x 2 ) y = y 2 c > 0 g(b) = g(x 1 ) = f(x 1 ) c < 0. Then there exists x 3 [x 2, x 1 ] such that g(x 3 ) = 0. Again, c = f(x 3 ) f(i). This shows that f(i) is an interval. If I is closed and bounded, it is compact. Therefore f(i) has a maximum and a minimum (any continuous function on a compact domain has a maximum and a minimum). Therefore since f(i) is a bounded interval with a minimum and a maximum, f(i) = [min f(i), max f(i)]. We see this is closed and bounded.