MT804 Analysis Homework II

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1 MT804 Analysis Homework II Eudoxus October 6, 2008 p , p part a only) p Exercise Use the Intermediate Value Theorem to prove that every polynomial of with real coefficients and odd degree has a zero in R. Proof: Let px) = a n x n + a n 1 x n a 1 x + a 0 be a polynomial with n odd and all a i R, with a n 0. Dividing px) by a n, we may assume a n = 1. Let qx) = a n 1 x n a 1 x + a 0, so that px) = x n 1 + qx) ). x n Since qx) has degree < n, the term in parentheses goes to 1 as x. Choose b > 0 such that 1 + qx) > 0 for x b. Then x n pb) = b n 1 + qb) ) > 0, b n while p b) = b n 1 + q b) ) < 0, b) n since n is odd. By the I.V.T., there is a zero of px) in the interval [ b, b]. Exercise Let I 0 = [a 0, b 0 ], a 0 < b 0, and let f CI) be continuous with fa 0 ) < 0 < fb 0 ). Use the bisection method to prove that there exists z [a 0, b 0 ] with fz) = 0. Then use this method find n such that the midpoint z n approximates 2 to ten decimal places. 1

2 Proof: Let z 0 be the midpoint of [a 0, b 0 ]. If fz 0 ) = 0, we re done. Otherwise, define a new interval [a 1, b 1 ] by: { [a 0, z 0 ] if fz 0 ) > 0 [a 1, b 1 ] = [z 0, b 0 ] if fz 0 ) < 0 so that we again have f < 0 at the left endpoint and f > 0 at the right endpoint. Let z 1 be the midpoint of [a 1, b 1 ]. Repeating this process, we may eventually get a zero of f at one of the midpoints z n, in which case we have our zero and the process stops. Assume that fz n ) 0 for all n. We therefore have a nested sequence of closed intervals [a 0, b 0 ] [a 1, b 1 ] [a 2, b 2 ] each half the length of the previous one. So b n a n = b 0 a 0 ) 2 n 0. By the Nested Intervals Theorem, there is a number z [a n, b n ]. n=0 If z is any point in this intersection and ɛ > 0 then choose n that b n a n < ɛ. Since z, z are both in [a n, b n ] we have z z < ɛ. Since ɛ was arbitrary, this means z = z. We have shown that [a n, b n ] = {z}. n=0 It remains to prove that fz) = 0. Suppose fz) 0. Then the number ɛ = fz) is positive. Since f is uniformly continuous on I, there exists δ > 0 such that for all x, x I with x x < δ we have fx) fx ) < ɛ = fz). Choose n large enough that b n a n < δ. If fz) > 0, take x = z, x = a n. Since z [a n, b n ], we have z a n < δ so so fa n ) > 0, a contradiction. fz) fa n ) = fz) fa n ) < ɛ = fz), 2

3 If fz) < 0, take z = b n, x = z. Again we have b n z < δ, so fb n ) fz) = fb n ) fz) < ɛ = fz), so fb n ) < 0, another contradiction. It follows that fz) = 0. Now let s approximate 2. Take fx) = x 2 2 on the interval [0, 2]. Since each of the midpoints will be averages of rational numbers, hence rational, and 2 is irrational, we get an infinite sequence of intervals as above. We want or or b 0 a 0 ) 2 n < 10 10, n > 10 2 n > 10 10, log 10 log 2 = So the rational number z 34 approximates 2 to ten decimal places. Exercise 4.5.3a Show that any two non-empty open intervals are homeomorphic. Proof: Since homeomorphisms are invertible, it suffices to show that any nonempty open interval I is homeomorphic to 1, 1). First suppose I = a, b) is bounded. We construct an affine function fx) = αx + β so that fa) = 0 and fb) = 1. This gives two equations to be solved for α and β and we get the function fx) = 2x a b b a which is a homeomorphism f : a, b) 1, 1). Indeed, the inverse of f is f 1 x) = 1 [b a)x + a + b]. 2 Now suppose I is unbounded. If I = R then the function fx) = x/ 1 + x 2 is a homeomorphism f : R 1, 1), as shown in the text, p The remaining two cases are I =, b) and I = a, ). Each of the first kind I =, b) is homeomorphic to b, ) via the map x x and b, ) is of the second kind. So it suffices to find a homeomorphism from a, ) to 1, 1). Here s a nice one: fx) = x a 1 x a + 1, f 1 x) = 3 1 a)x + a x

4 Note that the discontinuity of f is at a 1, which is not in a, ). Likewise, the discontinuity of f 1 x) is at x = 1 which is not in 1, 1). Exercise Prove that sin1/x) is not uniformly continuous on 0, 1]. Proof: By the Continuous Extension Theorem Thm 4.6.2) it suffices to show that sin1/x) does not extend to a continuous function on [0, 1]. Let c R and suppose we define fx) = sin1/x) on 0, 1] and f0) = c. Let x n = 1/nπ and let x n = 1/2nπ + π/2). Then fx n ) = 0 and fx n) = 1 for all n. So, no matter what c is, we cannot have lim x 0 fx) = c. Hence fx) is discontinuous at x = 0 and there is no continuous extension of sin1/x) to 0, 1]. Extra Exercise 1 Let K be a compact subset of R. Prove that supk) and infk) exist and belong to K. Proof: By the Heine-Borel theorem, K is both closed and bounded. Since K is bounded, the real numbers s = supk) and l = infk) exist. Suppose s is not in K. Then s K c, which is open, since K is closed. Choose ɛ > 0 such that V 2ɛ s) K c. Then s ɛ is an upper bound for K, contradicting the fact that s is the least upper bound of K. A similar argument shows that l K, as well. Extra Exercise 2 Let f : R R be the function defined by 1 if x = 0 fx) = 1/n if x = m/n is rational in lowest terms and n > 0 0 if x Q c. Prove that f is continuous at every irrational number and discontinuous at every rational number. Proof: First, if x = m/n Q then fx) = 1/n > 0. Choose a sequence x k ) in Q c converging to x. Then fx k ) = 0 for all k, so lim k fx k ) = 0 fx). The case where x Q c is trickier. Since x is not an integer, there is N N such that x N, N + 1). Let ɛ > 0. Since fx) = 0, we must find δ > 0 such that fx ) < ɛ for all x V δ x). I claim that there are only finitely many points in N, N + 1) where this can fail. Indeed, suppose y N, N + 1) and fy) ɛ. Since fy) > 0, we must have y Q, say y = m/n in lowest terms with n > 0, and fy) = 1/n ɛ. This means n 1/ɛ, so there are only finitely many possibilities for the integer 4

5 n. Also, since y N, N + 1) we have so N < m n < N + 1, nn < m < nn + 1). with N fixed and only finitely many possibilities for n, we see there are only finitely many possibilities for m as well. Hence there are only finitely many y N, N + 1) for which fy) ɛ. Since fx) = 0, the number δ defined by δ := min{ x N, x N 1, x y : y N, N + 1), and fy) ɛ} is positive and for x x < δ we have fx ) < ɛ. 5