Math 320: Real Analysis MWF 1pm, Campion Hall 302 Homework 7 Solutions Please write neatly, and in complete sentences when possible.

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1 Math 320: Real Analysis MWF 1pm, Campion Hall 302 Homework 7 Solutions Please write neatly, and in complete sentences when possible. Do the following problems from the book: 4.2.1, 4.2.3, 4.2.6, 4.2.8, 4.2.9, Problem D. Let f : A B be a map of sets (picture a function of real numbers if you prefer). Let U i f(a) for each i I. (1) Show that f 1 (U i ) f 1 (U j ) if and only if U i U j. (2) Show that ( ) f 1 U i = f 1 (U i ). (3) Show that f 1 ( i I U i ) = i I f 1 (U i ). Problem E. Let W R be an open set, and f : W R. Show that f is continuous on W if and only if f 1 (U) is open whenever U R is open. Problem F. Show that f(x) = x is continuous on R. Solution 4.2.1(a). Given ɛ > 0, we need (2x + 4) 8 = 2 x 2 < ɛ. Let δ = ɛ/2, so that if 0 < x 2 < δ, then 2 x 2 < 2δ = ɛ, as desired. Solution 4.2.1(b). Given ɛ > 0, we need x 3 < ɛ. Let δ = 3 ɛ, so that if 0 < x < δ, then x 3 = x 3 < δ 3 = ɛ, as desired. Solution 4.2.1(c). Given ɛ > 0, we need x 3 8 = x 2 x 2 +2x+4 < ɛ. As long as δ < 1, for x 2 < 1 we will have x < 3, so that x 2 + 2x + 4 < = 19. Let δ = min{1, ɛ/19}, so that if 0 < x 2 < δ, then x 3 8 = x 2 x 2 + 2x + 4 < ɛ 19 = ɛ, 19 1

2 2 as desired. Solution 4.2.1(d). Since π = , for x π <.01 we will have 3 < π.01 < x < π +.01 < 4. For such x, we have [[x]] = 3, so that [[x]] 3 = 0. Thus, for any ɛ > 0, we choose δ =.01, and 0 < x π <.01 implies that [[x]] 3 < ɛ, as desired. Solution 4.2.3(a). Let x n = 1/n and y n = 1/n. Then we have lim x n = lim y n = 0, but x n /x n = 1 and y n /y n = 1, for all n. Thus x n lim n x n y n lim, n y n and we ve demonstrated the divergence criterion. Solution 4.2.3(b). Let x n = 1 1 and y n n = 1 + 2, for each n. n Note that x n is rational, whereas y n is irrational for each n. Then we have lim x n = lim y n = 1, but g(x n ) = 1 and g(y n ) = 0, for all n. Thus lim g(x n ) lim g(y n ), and we ve demonstrated the divergence criterion. Solution Let ɛ > 0. Since lim x c g(x) = 0, there exists a δ > 0 so that if 0 < x c < δ we have g(x) < ɛ/m. (This is like handing the definition of lim g(x) = 0 the value ɛ/m in place of ɛ.) In this case, for any 0 < x c < δ, we have g(x)f(x) = g(x) f(x) < ɛ M = ɛ, M as desired. Solution Denote the limits lim x c f(x) and lim x c g(x) by L and M, respectively (and assume they exist! this should be assumed in the statement of the problem). Since c is a limit point of A, there is a sequence {x n } in A converging to c, with x n c for all n. By the Sequential Criterion for Functional Limits (Theorem 4.2.3), lim n f(x n ) = L and lim n g(x n ) = M. As x n A, by assumption we have f(x n ) g(x n ) for all n. By the Order Limit Theorem (Theorem 2.3.4), we have lim f(x n ) lim g(x n ), so that L M, as desired. Solution Choose a sequence {x n } in A converging to c, as above. Then lim n f(x n ) = L = lim n h(x n ) by the Sequential Criterion for Functional Limits (Theorem 4.2.3). Note that f(x n ) g(x n ) h(x n ) for all n, so that by Exercise (on Homework 3),

3 3 we conclude lim n g(x n ) = L as well. lim x c g(x) = L, as desired. By Theorem again, Solution 4.3.1(a). Given ɛ > 0, let δ = ɛ 3. Then if x < δ, then 3 x < δ 1/3 = ɛ, as desired. (Note that a < b if and only if a 3 < b 3, so that a < b implies a 1/3 < b 1/3.) Solution 4.3.1(b). Let ɛ > 0, and suppose first that c > 0. Note that we need to demonstrate x 1/3 c 1/3 < ɛ. Since we have x c = x 1/3 c 1/3 x 2/3 + x 1/3 c 1/3 + c 2/3, x 1/3 c 1/3 = x c x 2/3 + x 1/3 c 1/3 + c 2/3. For x c < c we have c < x < 2c, and thus x 2/3 +x 1/3 c 1/3 +c 2/3 > c 2/3. Let δ = min{c, ɛc 2/3 }. For x c < δ, we have x 1/3 c 1/3 < ɛc 2/3 c 2/3 = ɛ, as desired. The case c < 0 can be dealt with by observing that 3 x = 3 x. Since f(x) = 3 x is continuous on (0, ), and f(x) = x is continuous on R, the composition of these two functions is continuous, so that 3 x is continuous on (, 0). Solution D(1). Suppose that f 1 (U i ) f 1 (U j ), and let y U i. Since U i f(a), there is an x A so that f(x) = y, so that x f 1 (U i ), which implies that x f 1 (U j ). Thus y = f(x) U j, so that U i U j. On the other hand, suppose that U i U j, and choose x f 1 (U i ). This implies that f(x) U i, so that f(x) U j and x f 1 (U j ), showing that f 1 (U i ) f 1 (U j ). Solution D(2). We have: x f ( 1 i I U i) if and only if f(x) i I U i, if and only if there is an i 0 I so that f(x) U i0, if and only if there is an i 0 I so that x f 1 (U i0 ), if and only if x i I f 1 (U i ). Thus x f ( 1 i I U ) i if and only if x i I f 1 (U i ), so that ( ) f 1 U i = f 1 (U i ).

4 4 Solution D(3). We have: x f ( 1 i I U i) if and only if f(x) i I U i, if and only if f(x) U i for each i I, if and only if x f 1 (U i for each i I, if and only if x i I f 1 (U i ). Thus x f ( 1 i I U ) i if and only if x i I f 1 (U i ), so that ( ) f 1 U i = f 1 (U i ). Solution E. Suppose that f is continuous on W and U R is an open set. If f 1 (U) is empty, f 1 (U) is open, so we assume that there is an element x f 1 (U), and f(x) U. Since U is open and f(x) U, there is an ɛ > 0 so that V ɛ (f(x)) U. Since f is continuous at x, there is a δ > 0 so that f(v δ (x)) V ɛ (f(x)). As V ɛ (f(x)) U, we have f(v δ (x)) U, which implies that V δ (x) f 1 (U). Thus f 1 (U) is open. Conversely, suppose that f 1 (U) is open for any open set U R, and let c W and ɛ > 0. Since V ɛ (f(c)) is open, we have that f 1 (V ɛ (f(c))) is open. Since f(c) V ɛ (f(c)), we have that c f 1 (V ɛ (f(c))), and by openness there is a δ > 0 so that V δ (c) f 1 (V ɛ (f(c))). This implies that f(v δ (c)) V ɛ (f(c)), and f is continuous at c. Solution F. First, we show that f is continuous at c = 0: Given ɛ > 0, let δ = ɛ. Then x = x 0 < δ implies that f(x) f(0) = x < ɛ. For the remaining cases, note that f(x) = x for x > 0, and f(x) = x for x < 0. That is, for any c 0 the given function f is equal to a linear polynomial in an open set containing c. Since polynomials are continuous on R, the lemma below proves that f is continuous at any c 0, and we are done. Lemma. Suppose that g and h are functions that agree on an open interval containing a, except possibly at x = a, and that lim x a g(x) exists. Then lim g(x) = lim h(x). x a x a Proof: Let L = lim x a g(x), and let I = (x 1, x 2 ) be the open interval containing a. Let δ 0 = min{ x 1 a, x 2 a }, so that V δ0 (a) I. Choose ɛ > 0, so that lim x a g(x) = L implies that there is a δ > 0

5 so that g(v δ (a) \ {a}) V ɛ (L). Choose δ = min{δ 0, δ}, so that for any x V δ (a) \ {a} we have g(x) = h(x). Thus h(v δ (a) \ {a}) = g(v δ (a) \ {a}), so that h(v δ (a) \ {a}) V ɛ (L). This demonstrates that lim x a h(x) = L, and the lemma is proved.