Math 220A Complex Analysis Solutions to Homework #2 Prof: Lei Ni TA: Kevin McGown

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1 Math 220A Complex Analysis Solutions to Homework #2 Prof: Lei Ni TA: Kevin McGown Conway, Page 14, Problem 11. Parts of what follows are adapted from the text Modular Functions and Dirichlet Series in Number Theory by Tom Apostol. There are shorter proofs, but I am trying to avoid compactness arguments. Before giving the solution, we establish some notation. We will write x to denote the floor of a real number x; this is defined to be the greatest integer not exceeding x. We will write {x} to denote the fractional part of x; this is defined to be x x. In order to solve the problem, we will prove the following: Theorem 1 The set { exp(2πinθ) n Z + } is dense in the unit circle if and only if θ is irrational. Taking arguments and normalizing by 2π, the above theorem is clearly equivalent to the following. Theorem 2 The sequence {θ}, {2θ}, {3θ},... is dense in [0, 1] if and only if θ is irrational. Proof. If θ is rational, then it is easy to see that {{nθ} n Z + } is a finite set and hence it cannot be dense in the unit circle. (A finite subset of C is closed and thus cannot be dense in an infinite set.) Suppose θ is irrational. Without loss of generality, we will assume that θ [0, 1]. Indeed, one may replace θ by {θ} after making the observation that {nθ} = {n{θ}}. Claim: If n m, then {nθ} {mθ}. If {nθ} = {mθ}, then nθ nθ = mθ mθ which implies a contradiction. θ = nθ mθ n m Q, Let ε > 0 be given and choose α [0, 1]. We must find an integer m so that {mθ} α < ε. 1

2 Claim: There exists h, k Z + such that kθ h < ε. Subdivide [0, 1] into half-open intervals of equal length less than ε. Since {nθ} {mθ} for n m, one of these subintervals must contain two of these fractional parts. Thus there exists m, n Z + with {nθ} {mθ} < ε. That is, (n m)θ ( mθ nθ ) < ε. Setting k = n m and h = mθ nθ proves the claim. Suppose that kθ > h. (The argument is similar if kθ < h.) Choose the largest N Z + such that {kθ} < 1/N. Consider the sequence 0, {kθ}, {2kθ},..., {Nkθ} In light of the fact that {mkθ} = m{kθ} if and only if {kθ} < 1/m, we see that the sequence given above is increasing and equally spaced with common distance {kθ}. Thus we have subdivided the unit interval into N +1 subintervals with the partition 0 < {kθ} < {2kθ} < < {Nkθ} < 1. ( ) Since α lies in one of the subintervals defined by ( ), it suffices to prove the following: Claim: The partition ( ) subdivides the unit interval into subintervals of length less than ε. By earlier remarks, the length of the first N subintervals is {kθ} < ε. We need only look at the last subinterval. By the definition of N we have {kθ} > 1/(N + 1) and hence which implies N{kθ} > N N + 1 = 1 1 N + 1, 0 < 1 {Nkθ} < 1 N + 1. We have shown that the distance between {Nkθ} and 1 is less than 1/(N + 1). Finally we observe that 1 N + 1 < {kθ} < ε. This establishes the claim. 2

3 Conway, Page 17, Problem 4. Let (X, d) be a metric space and suppose {D j j J} is a collection of connected subsets of X such that for all j, k J we have D j D k. Show that the set D = D j is connected. Proof. Let A D be nonempty, open, and closed. We will show that A = D. Observe that A = A D j ; then since A, we know that A D k for some k. But now we see that A D k is a nonempty, open, closed subset of the connected set D k and hence A D k = D k ; it follows that D k A. Let l J be arbitrary. By hypothesis, there exists x 0 D k D l. Thus x 0 D k A and x 0 D l. Hence A D l is a nonempty, open, closed subset of D l ; as before we have A D l = D l and hence D l A. Since l was arbitrary, we have D A. 3

4 Conway, Page 17, Problem 5. Let (X, d) be a metric space. For a subset F X and a, b F we define an ε-chain in F from a to b to be a finite sequence of points z 0, z 1,..., z n F such that z 0 = a, z n = b, and d(z k 1, z k ) < ε for each k = 1,..., n. Show that if F X is closed and connected then for every pair of points a, b F there is an ε-chain in F from a to b. Is the hypothesis that F is closed needed? If F is a set which satisfies this property then F is not necessarily connected, even if F is closed. Give an example. Proof. Fix a F and let ε > 0 be given. Consider the set T = {b F there is an ε-chain in F from a to b}. By definition T F and clearly T since a T. If we can show that T is both open and closed (in the subspace topology of F ), then it will follow by the connectedness of F that T = F, which is what we want to show. Claim: T is open Let c T. Then B ε (c) F T. Indeed, if b F with d(b, c) < ε and z 0,..., z n is an ε-chain in F from a to c, then we define z n+1 = b to obtain an ε-chain from a to b; hence b T. Claim: T is closed We show that T contains its limit points. Let c n T and suppose that c n converges to c F. Choose N such that d(c N, c) < ε. Since c N T, there is an ε-chain z 0,..., z n from a to c N. Setting z n+1 = c, we obtain an ε-chain from a to c. This establishes the claim and therefore the result. Note that the fact that F is closed was not used. Finally, the set {(x, 1/x) R 2 x > 0} {(x, 1/x) R 2 x < 0} is an example of a closed, disconnected set satisfying the condition. 4

5 Conway, Page 20, Problem 8. Let (X, d) be a metric space. Suppose that {x n } is Cauchy and that the subsequence {x nk } converges to x. Show that {x n } converges to x. Proof. Let ε > 0. Since {x n } is Cauchy we may choose N 1 such that n, m N 1 implies d(x n, x m ) < ε/2. Since x nk x, we may choose N 2 such that k N 2 implies d(x nk, x) < ε/2. Set N = max{n 1, N 2 } and note that, in particular, n N N. For n N we have d(x n, x) d(x n, x nn ) + d(x nn, x) < ε/2 + ε/2 = ε. Since ε > 0 was arbitrary, x n x. 5