Hence, (f(x) f(x 0 )) 2 + (g(x) g(x 0 )) 2 < ɛ

Matthew Straughn Math 402 Homework 5 Homework 5 (p. 429) 13.3.5, 13.3.6 (p. 432) 13.4.1, 13.4.2, 13.4.7*, 13.4.9 (p. 448-449) 14.2.1, 14.2.2 Exercise 13.3.5. Let (X, d X ) be a metric space, and let f : X R and g : X R be uniformly continuous functions. Show that the direct sum f g : X R 2 defined by f g(x) := (f(x), g(x)) is uniformly continuous. Proof. Since f, g are uniformly continuous we know that for all x 0 X given ɛ > 0 there exists δ > 0 such that if d X (x, x 0 ) < δ, then d R (f(x), f(x 0 )) < ɛ, and d R (g(x), g(x 0 )) < ɛ. So given ɛ > 0 take ɛ = ɛ/ 2, and δ = δ. If d X (x, x 0 ) < δ = δ, then from the uniform continuity of f, g we get that: d R (f(x), f(x 0 )) = f(x) f(x 0 ) < ɛ = ɛ/ 2 d R (g(x), g(x 0 )) = g(x) g(x 0 ) < ɛ = ɛ/ 2. Thus, (f(x) f(x 0 )) 2 + (g(x) g(x 0 )) 2 < ɛ 2 Hence, (f(x) f(x 0 )) 2 + (g(x) g(x 0 )) 2 < ɛ But, (f(x) f(x 0 )) 2 + (g(x) g(x 0 )) 2 = d R 2((f(x), g(x)), (f(x 0 ), g(x 0 ))) < ɛ. Therefore f g is uniformly continuous. Exercise 13.3.6. Show that the additive function (x, y) x + y and the subtraction function (x, y) x y are uniformly continuous from R 2 to R, but the multiplication function (x, y) xy is not. Conclude that if f : X R and g : X R are uniformly continuous functions on a metric space (X, d), then f + g : X R and f g : X R are also uniformly continuous. Give an example to show that fg : X R need not be uniformly continuous. What is the situation for max(f, g), min(f, g), f/g, and cf for a real number c? Proof. Note that to deduce the second part we can: First let a denote the addition function, and s denote the subtraction function, then note that f +g = a (f g) and f g = s (f g). Now using exercises 13.3.4 and 13.3.5 we immediately get that f +g, f g are both uniformly continuous. We will, however, just prove the generalization (second part) which automatically gives us the first. Since f, g are uniformly continuous we know that given ɛ > 0 there is δ > 0 such that if d X (x, x 0 ) < δ then d R (f(x), f(x 0 )) < ɛ and d R (g(x), g(x 0 )) < ɛ for all x X. Given ɛ > 0 let ɛ = ɛ/2 and take δ = δ. If d X (x, x 0 ) < δ = δ then from the uniform continuity of f, g we know that d R (f(x), f(x 0 )) = f(x) f(x 0 ) < ɛ = ɛ/2 and d R (g(x), g(x 0 )) = g(x) g(x 0 ) < ɛ = ɛ/2 1 and

But, (f + g)(x) (f + g)(x 0 ) = f(x) + g(x) (f(x 0 ) + g(x 0 )) = f(x) f(x 0 ) + g(x) g(x 0 ) f(x) f(x 0 ) + g(x) g(x 0 ) < ɛ/2 + ɛ/2 = ɛ Thus, f + g is uniformly continuous. A similar argument shows that f g is uniformly continuous. To show that fg is not uniformly continuous we take f : (0, ) R, g : (0, ) R defined by f(x) = g(x) = x. So fg = x 2 which we know is not uniformly continuous. max(f, g), min(f, g) are both uniformly continuous. Proof. Suppose f, g are both uniformly continuous. That is, given ɛ > 0 there exists δ > 0 such that if d X (x, x 0 ) < δ then f(x) f(x 0 ) < ɛ and g(x) g(x 0 ) < ɛ for all x X. (which we can do if we just take the smaller of the two deltas that we would get from the original definition). Now we want to show that given ɛ > 0 there exists δ > 0 such that if d X (x, x 0 ) < δ then max(f(x), g(x)) max(f(x 0 ), g(x 0 )) < ɛ for all x X. So let us break this into four cases. Case 1: max(f(x), g(x)) = f(x), max(f(x 0 ), g(x 0 )) = f(x 0 ). Given ɛ > 0 take ɛ = ɛ and δ = δ then we immediately get that max(f(x), g(x)) max(f(x 0 ), g(x 0 )) < ɛ. Case 2: max(f(x), g(x)) = g(x), max(f(x 0 ), g(x 0 )) = g(x 0 ). This case is exactly the same as Case 1. Case 3: max(f(x), g(x)) = f(x), max(f(x 0 ), g(x 0 )) = g(x 0 ). This case is a bit more complicated. So we have f(x) > g(x) and g(x 0 ) > f(x 0 ). And we can assume that f(x) g(x 0 ) (if f(x) = g(x 0 ) then we are clearly done). Without loss of generality we can assume that x < x 0. Consider h = f g on the interval [x, x 0 ]. Then h(x) > 0 and h(x 0 ) < 0. And since both f, g are continuous we know that h is continuous and hence by the Intermediate Value Theorem we know that there exists z [x, x 0 ] such that h(z) = 0. That is f(z) = g(z). Now given ɛ > 0 take ɛ = ɛ/2 and take δ = δ and we get: 2

As desired. max(f(x), g(x)) max(f(x 0 ), g(x 0 )) = f(x) g(x 0 ) = f(x) g(x 0 ) h(z) since h(z) = 0 = f(x) f(z) + g(z) g(x 0 ) f(x) f(z) + g(z) g(x 0 ) ɛ/2 + ɛ/2 = ɛ Case 4: max(f(x), g(x)) = g(x), and max(f(x 0 ), g(x 0 )) = f(x 0 ). This is exactly the same as case 3. Therefore max(f, g) is uniformly continuous whenever f, g are uniformly continuous. f/g is not uniformly continuous. Consider f = x, g = 1/x where f, g : (1, ) R. Then f/g = x 2 which is not uniformly continuous. Clearly cf is uniformly continuous. Exercise 13.4.1. Let (X, d disc ) be a metric space with the discrete metric. Let E be a subset of X which contains at least two elements. Show that E is disconnected. Proof. Let x, y E and consider V = {x}, W = E\{x}. So we have V W = E, V W =, and W as y W, now we just need V, W to be open in E. But this is clear since V = {x} = B disc (x, 1/2) and W = E\{x} (since {x} is closed, and so the complement is open) which are both open in E. Hence E is disconnected. Exercise 13.4.2. Let f : X Y be a function from a connected metric space (X, d) to a metric space (Y, d disc ) with the discrete metric. Show that f is continuous if and only if it is constant. Proof. ( ) Suppose that f is continuous, and also suppose that f is not constant. That is, there are at least two values in Y, say x, y Y. From the previous exercise we know that (Y, d disc ) is disconnected. However, this contradicts Theorem 13.4.6. Thus f must be constant. ( ) Clearly if f is constant it is also continuous. Exercise 13.4.7. Let (X, d) be a metric space, and let E be a subset of X. Show that every path-connected set is connected. 3

Proof. Let E be a path-connected set. That is, for every x, y E there is a continuous function γ : [0, 1] E such that γ(0) = x and γ(1) = y. Now suppose that E is not connected. That is, there exists V, W non-empty, disjoint, and open in E such that V W = E. We will now show that [0, 1] is disconnected (which gives us a contradiction). Since γ is continuous and V, W are open in E, we know that γ 1 (V ), γ 1 (W ) are open in [0, 1]. Since γ(0) = x and γ(1) = y we have that both γ 1 (V ), γ 1 (W ) are non-empty. Suppose z γ 1 (V ) γ 1 (W ). That is, γ(z) V and γ(z) W, but V, W are disjoint, a contradiction. Hence γ 1 (V ) γ 1 (W ) =. Let z [0, 1]. Then γ(z) E, but E = V W, and V, W are disjoint. So either γ(z) V z γ 1 (V ) or γ(z) W z γ 1 (W ). Thus [0, 1] = γ 1 (V ) γ 1 (W ). Thus we have that [0, 1] is disconnected, a contradiction (Theorem 13.4.5.). Hence our original assumption was wrong, and therefore every path-connected set is connected. Exercise 13.4.9. Let (X, d) be a metric space. Let us define a relation x y on X by declaring x y if and only if there exists a connected subset of X which contains both x and y. Show that this is an equivalence relation. Also, show that the equivalence classes of this relation are all closed and connected. Proof. First let us show that is an equivalence relation. x x: Take E = {x} X. E is certainly connected, and both x and x are contained in E. Thus x x. x y y x: Clearly true by definition of. x y, y z x z: If x y then there is some connected set E X such that x, y E. And if y z then there is some connected set F X such that y, z F. Since E F we can use the result from exercise 13.4.6 to conclude that E F is connected. But x, z E F. Hence x z. Thus is an equivalence relation on X. Now let us show that the equivalence classes are connected: Let [x] be an equivalence class. Now suppose (for sake of contradiction) that [x] is disconnected. That is, there are two non-empty, disjoint, open sets V, W [x] such that V W = [x]. If [x] has only one element then it is clear that [x] is connected. So suppose that [x] has at least two elements, x, y [x]. And without loss of generality, let x V, and y W. x y, so by definition there is some connected set E X with x, y E. We also know that E [x] since if x E connected, then for all y E there is a connected set, E, that contains x and y. So E [x]. But, E = E [x] = E (W V ) = (E W ) (E V ) 4

And since E V E [x], and V is open in [x] we can use Proposition 12.3.4 to conclude that E V is open in E. The same argument gives us that E W is open in E. But, E = (E W ) (E V ), (E W ) (E V ) = E V W = and x E V, and y E W, so these sets are non-empty, and from above they are open in E. Thus, by definition, E is disconnected, a contradiction. Thus [x] is connected. Now let us show that [x] is closed. First note that [x] is the largest connected set that contains x by the following argument: If there were a larger connected set containing x, say A, then there would be some element y A\[x]. But A is connected, so we have to have y x which gives us y [x]. From Exercise 13.4.8 we know that [x] is a connected set. But x is always in [x] and hence is always in [x] thus [x] [x] [x]. Thus [x] = [x], and therefore [x] is closed. Exercise 14.2.1. Let f : R R be a function. For any a R, let f a : R R be the shifted function f a (x) := f(x a). (a) Show that f is continuous if and only if, whenever (a n ) n=0 is a sequence of real numbers which converges to zero, the shifted functions f an converge pointwise to f. (b) Show that f is uniformly continuous if and only if, whenever (a n ) n=0 is a sequence of real numbers which converges to zero, the shifted functions f an converge uniformly to f. Proof. ( ) Let x 0 R. Suppose f is continuous at x 0. That is, given ɛ > 0 there exists δ > 0 such that if x x 0 < δ then f(x) f(x 0 ) < ɛ. Assume a n 0 as n and let x n = x 0 a n. So given δ > 0 there exists N > 0 such that x n x 0 < δ for all n > N. Given ɛ > 0 take ɛ = ɛ, and take N = N. So by the continuity of f we get that x n x 0 < δ for all n > N = N Thus, f(x n ) f(x 0 ) < ɛ = ɛ. But, f(x n ) = f(x 0 a n ) = f an (x 0 ) for all n > N So we have f an (x 0 ) f(x 0 ) < ɛ for all n > N. Thus f an converges pointwise to f. ( ) Suppose given a n 0 that f an converges pointwise to f. Given x 0 take a sequence x n which converges to x 0. That is, a n = x 0 x n 0 Since f an converges pointwise to f, given ɛ > 0 there is some N > 0 such that f an (x 0 ) f(x 0 ) < ɛ for all n > N. But, f an (x 0 ) f(x 0 ) = f(x 0 a n ) f(x 0 ) = f(x n ) f(x 0 ). Hence, f(x n ) f(x 0 ) < ɛ for all n > N. That is, f(x n ) converges to f(x 0 ). Thus, f is continuous. For part (b) the argument is the same, except that now δ, N are independent of the base point x 0. 5

Exercise 14.2.2. (a) Let (f (n) ) n=1 be a sequence of functions from one metric space (X, d X ) to another metric space (Y, d Y ), and let f : X Y be another function from X to Y. Show that if f (n) converges uniformly to f, then f (n) converges pointwise to f. (b) For each integer n 1, let f (n) : ( 1, 1) R be the function f (n) (x) := x n. Prove that f (n) converges pointwise to the zero function, but does not converge uniformly to any function f : R R. (c) Let g : ( 1, 1) R be the function g(x) := x/(1 x). With the notation as in (b), show that the partial sums N n=1 f (n) converges pointwise as N to g, but does not conform uniformly to g on the open interval ( 1, 1). What would happen if we replaced the open interval ( 1, 1) with the closed interval [ 1, 1]? Proof. Part (a): This is obvious from the definitions. Part (b): lim n f (n) (x) = lim n x n = 0 for all x ( 1, 1) (from previous work). Thus f (n) converges pointwise to 0. To show that f (n) does not conver uniformly to any function it is enough to show that it does not converge uniformly to 0 (because of the result in part (a)). So we need to show that there exists ɛ > 0 such that for all N > 0 there is an n > N and x ( 1, 1) such that x n ɛ. Consider, 0 < ɛ < 1, n = N + 1, x = ɛ 1/n ( 1, 1) Then we get x n = (ɛ 1/n ) n = ɛ So x n ɛ. And hence f (n) does not conver uniformly to any function. Part (c): See Joshua s work. 6