18.100C Real Analysis: Practice Midterm 1 Solutions

Practice Midterm 1 Solutions, 18.100C 1 Suppose F is a field, and x, y ∈ F satisfy x2 = y 2 . We need to show that either x = y or x = −y. For the sake of sparing the proliferation of parenthesis, we will adopt the standard conventions about multiplication and addition, i.e. x + y · z is x + (y · z) rather than (x + y) · z. Now consider the quantity (x + (−y)) · (x + y). Using the distributive property, this is equal to (x + (−y))x + (x + (−y))y. Using distributiv­ ity again, we get (x2 + (−y) · x) + (x · y + (−y) · y). Using associativity of addition and commutativity of multiplication, we can rearrange this to x2 + ((x · y + x · (−y)) + (−y) · y). Now, we have x · y + x · (−y) = x · (y + (−y)) = x · 0 = 0 Where the first step follows from distributivity, the second from the defini­ tion of −y, and the third from noting that that x · 0 = x · (0 + 0) = x · 0 + x · 0 and adding −(x · 0) to both sides. Similarly, y · (−y) + y 2 = 0, and adding −y 2 on the right to both sides gives (−y) · y = −y 2 . Going back, we have (x − y) · (x + y) = x2 + ((x · y + x · (−y)) + (−y) · y) = x2 + ((0) + (−y 2 )) = x2 + (−y 2 ) But by assumption x2 = y 2 , so this is equal to y 2 + (−y 2 ) = 0. In other words, (x − y) · (x + y) = 0. If x = y, then we are done. So suppose x = y. Then x − y = 0, since if x − y = 0 we have y = 0 + y = (x − y) + y = x + (y − y) = x + 0 = x 1 using, respectively, definition of 0, substitution, associativity of addition, definition of −y, and definition of 0. So there must exist a multiplicative inverse (x − y)−1 . Then we have 0 = (x − y)−1 · 0 = (x − y)−1 ((x − y)(x + y) = ((x − y)−1 (x − y))(x + y) = 1 · (x + y) = x + y The first equality is a fact we already proved, the second is just substitu­ tion, the third is associativity of multiplication, the fourth is the definition of multiplicative inverse, and the fifth is the definition of 1. Now we get x = x + 0 = x + (y − y) = (x + y) − y = 0 − y = −y Using the definition of 0, definition of −y, associativity of addition, substi­ 6 y, then x = −y, which is tution, and definition of 0. In other words, if x = what we wanted to prove. 2 Suppose E is a finite dense subset of X. Then E contains no limit points. To see this, suppose x ∈ X is a limit point of E. Then by Theorem 2.20 of Rudin, every neighbourhood of x has to contain infinitely many points of E. But E only has finitely many points, so this is obviously impossible. Since E has no limit points, it is vacuously true that it contains all its limit points, so E is closed and E = E. But E is dense in X, so E = X. This means that X = E, so X itself must be finite since E is. 3 We will show that Z is not compact in the p-adic topology by constructing an infinite set with no limit points. We begin with an elementary Lemma 1: Let m ∈ N and k the largest natural number with pk |m. Let l be natural number with m < pl . Then pk is the largest power of p dividing m + pl 2 Proof: We have pk ≤ m < pl , so k < l and pk |pl , hence pk divides m + pl . Since k < l, k + 1 ≤ l, so pk+1 |pl . But then pk+1 cannot divide m + pl , since then it would divide (m + pl ) − pl = m, a contradiction. i=n 2i Now for n ∈ N, we introduce the finite sum sn = i=0 p , where p is the prime with respect to which we define the p-adic metric on Z. Let S = {sn |n ∈ N} be the set of all these numbers. S is obviously infinite, so if we can show that S has no limit points we are done. To do so, we need another elementary Lemma 2: 2sn < p2n+2 Proof: Recall that by the formula for the sum of a geometric series, sn = (p2n+2 − 1)/(p2 − 1). Since p ≥ 2 we have p2 > 3, and so p2n+4 −3p2n+2 +2 = p2n+2 (p2 − 3) + 2 > 0. Using these facts, the lemma follows from simple al­ gebra: 0 < p2n+4 −3p2n+2 +2 =⇒ 2p2n+2 −2 < p2n+2 (p2 −1) =⇒ 2· p2n+2 − 1 < p2n+2 p2 − 1 =⇒ 2sn < p2n+2 With these two lemmas, we can show that S has no limit points. Let x ∈ Z. Pick some n sufficiently large that |x| < sn , which is always possible since sn > n. Let k ∈ N be the largest power of p dividing sn − x, i.e. d(x, sn ) = p−k . sn > |x| means that sn − x ∈ N is positive, and sn − x < 2sn < p2n+2 , where the last step follows from Lemma 2. By Lemma 1, the largest power of p dividing (sn − x) + p2n+2 is k. But (sn − x) + p2n+2 = (sn + p2n+2 ) − x = sn+1 − x, and so d(x, sn+1 ) = p−k . Applying this argument again, d(x, sn+2 ) = p−k , and indeed by induction d(x, sm ) = pk for all m ≥ n. Now let E > 0 be a real number with E < p−k . Then if sm ∈ S ∩ NE (x), we must have m < n. So this neighbourhood contains only finitely many points of S, and hence x is not a limit point of S. Since x was arbitrary, this means that S has no limit points and Z is not compact. 3 4 We have K ⊂ X compact, E ⊂ X closed, and K ∩ E = ∅. We need to show that there exists D > 0 such that d(x, y) ≥ D for x ∈ X, y ∈ Y . Suppose not; we will derive a contradiction. Then for every D > 0, we can find x ∈ K, y ∈ E such that d(x, y) < D. For each n ∈ N, pick some xn ∈ K, and yn ∈ E, such that d(xn , yn ) < 1/n. Let S = {xn |n ∈ N} be the set of all the xn ’s. We consider two cases. Case 1: S is finite. This means there is a finite collection of points {z1 , z2 . . . zM } such that for all i ∈ N, there exists j ≤ M such that xi = zj . Now define the set Nj ⊂ N, 1 ≤ j ≤ M , by Nj = {i ∈ N|xi = zj }. By assumption, N is the union of the Nj ’s. But since there are only finite many Nj ’s, at least one of them must be infinite, since otherwise N would be a finite union of finite sets, and hence finite. So pick a j such that Nj is infinite. I claim that zj is a limit point of E. Fix any E > 0; we will find y ∈ NE (zj )∩E. Take some N0 ∈ N with 1/N0 < E. Since Nj is infinite and there are obvi­ ously only finite many positive integers less than N0 , there must exist some n ∈ Nj with n > N0 . Then 1/n < 1/N0 , and by definition zj = xn , and d(zj , yn ) = d(xn , yn ) < 1/n < E This means that zj is a limit point of E. Since E is closed, zj ∈ E. On the other hand, zj = xn and xn ∈ K by construction. Thus zj ∈ K ∩ E, but K ∩ E = ∅, contradiction. Case 2: S is infinite. S is an infinite subset of a compact set K, hence must have a limit point x. x is then obviously a limit point of K, and since compact sets are closed we must have x ∈ K. We will show that it is also a limit point of E. Let E > 0; as before, we want to find y ∈ E with d(x, y) < E. Since x is a limit point of E, the neighbourhood NE/2 (x) must contain in­ finitely many points of S. Pick N ∈ N with 1/N < E/2. Since there are only finitely many xn with n ≤ N , and S ∩ NE/2 (x) is infinite, there must be an xn ∈ S ∩ NE/2 (x) with n > N . Then using the triangle inequality we have d(x, yn ) ≤ d(x, xn ) + d(xn , yn ) < E/2 + 1/n < E/2 + E/2 = E 4 Since E was arbitrary, this means that x is a limit point of E. But E is closed, so x ∈ E, and we already knew that x ∈ K, contradiction. 5