Solutions for Math 217 Assignment #3

Solutions for Math 217 Assignment #3 (1) Which of the following sets in R n are open? Which are closed? Which are neither open nor closed? (a) {(x, y) R 2 : x 2 y 2 = 1}. (b) {(x, y, z) R 3 : 0 < x + y + z < 1}. (c) {(x, y) R 2 : x + y Q}. (d) {(x, y) R 2 : x + y Q}. You must justify your answer. Solution. (a) S = {(x, y) R 2 : x 2 y 2 = 1} is closed and not open. It is not open because p = (1, 0) S and for all r > 0, (1, r/2) B r (p) and (1, r/2) S. It is closed because its complement S c = {x 2 y 2 1} is open with the proof as follows. Let p = (x 0, y 0 ) S, N = max( x 0 + 1, y 0 + 1) and a = 1 (x 2 0 y 2 0). Then for q = (x 1, y 1 ) satisfying p q < r < 1, (x 2 1 y 2 1) (x 2 0 y 2 0) = (x 1 x 0 )(x 1 + x 0 ) (y 1 y 0 )(y 1 + y 0 ) Consequently, (x 1 x 0 )(x 1 + x 0 ) + (y 1 y 0 )(y 1 + y 0 ) < 2Nr + 2Nr = 4Nr. 1 (x 2 1 y 2 1) 1 (x 2 0 y 2 0) (x 2 1 y 2 1) (x 2 0 y 2 0) > a 4Nr 0 for all q = (x 1, y 1 ) B r (p) satisfying r = min(1, a/(4n)). It follows that B r (p) S and S c is open. (b) S = {(x, y, z) R 3 : 0 < x + y + z < 1} is open and not closed. It is not closed because (0, 0, r) S for all 0 < r < 1 and hence p = (0, 0, 0) is a cluster point of S but p S. It is open by the following argument. Let p = (x 0, y 0, z 0 ) S and a = min(x 0 + y 0 + z 0, 1 (x 0 + y 0 + z 0 )). For q = (x 1, y 1, z 1 ) satisfying p q < r, (x 1 + y 1 + z 1 ) (x 0 + y 0 + z 0 ) x 1 x 0 + y 1 y 0 + z 1 z 0 Therefore, < r + r + r = 3r. x 1 +y 1 +z 1 (x 0 +y 0 +z 0 ) (x 1 +y 1 +z 1 ) (x 0 +y 0 +z 0 ) > a 3r > 0 1

2 and 1 (x 1 + y 1 + z 1 ) 1 (x 0 + y 0 + z 0 ) (x 1 + y 1 + z 1 ) (x 0 + y 0 + z 0 ) > a 3r > 0 for all q B r (p) with r satisfying r < a/3. Therefore, B r (p) S and S is open. (c) (d) S = {(x, y) R 2 : x + y Q} is neither open nor closed. The same holds for S c = {(x, y) R 2 : x + y Q}. Since Q is dense in R, Q Q S is dense in R 2. So S is dense in R 2. If S is closed, then S = R 2 and hence S is not closed. And since R\Q is dense in R, Q (R\Q) S c is dense in R 2. Hence S c is dense in R 2. It is not closed by the same argument as before. (2) For two sets S 1 and S 2 in R n, show that (a) S 1 + S 2 is open if both S 1 and S 2 are open; (b) S 1 + S 2 is closed if both S 1 and S 2 are closed; (c) S 1 + S 2 is bounded if both S 1 and S 2 are bounded. Are the converses of these statements true? Prove or disprove their converses. Proof. (a) Actually, we only need one of S i is open. That is, it is true that S 1 +S 2 is open if either S 1 or S 2 is open. WLOG, we assume that S 1 is open. First, we show that S 1 +{q} is open for all q R n. Since S 1 is open, for every point p S 1, B r (p) S 1 for some r > 0. Hence B r (p + q) = B r (p) + {q} S 1 + {q} and S 1 + {q} is open. Now we have S 1 + S 2 = q S 2 (S 1 + {q}) is a union of open sets. Hence S 1 + S 2 is open. (b) This statement is false by the counter example below. Let { S 1 = n + 1 } { n + 2 : n Z+ and S 2 = n + 1 } n + 2 : n Z+. Both S 1 and S 2 are closed and 0 is a cluster point of S 1 + S 2 but 0 S 1 + S 2. (c) Since S i is bounded, there exists R i > 0 such that S i B Ri (0) for i = 1, 2. That is, p i < R i for all p i S i. Therefore,

p 1 + p 2 p 1 + p 2 < R 1 + R 2 and hence S 1 + S 2 B R1 +R 2 (0) and it is bounded. The converses of (a) and (b) are both false. Consider S 1 = {1} and S 2 = R\{1} in R. Then R = S 1 + S 2 is both open and closed. But S 1 is not open and S 2 is not closed. The converse of (c) is true by the following argument. Since S 1 + S 2 is bounded, S 1 + S 2 B R (0) for some R > 0. Fixing q S 2 and we have p p + q + q < R + q for all p S 1. So S 1 is bounded. The same argument shows that S 2 is bounded. 3 (3) For a point p R n and a set S R n, we define the distance from p to S to be ρ(p, S) = inf p q. q S By convention, we set ρ(p, ) =. Show that p is a cluster point of S if and only if ρ(p, S\{p}) = 0. Proof. Suppose that p is a cluster point of S. Then B r (p) S\{p} = for all r > 0. Therefore, there exists a point p r S and p r p such that p p r < r for all r > 0. Therefore, ρ(p, S) = inf p q = 0. q S On the other hand, suppose that ρ(p, S) = 0. Then r is not a lower bound of the set { p q : q S, p q} for all r > 0. That is, there exists a point p r S and p r p such that p p r < r. Therefore, B r (p) S\{p} = for all r > 0 and hence p is a cluster point of S. (4) Find all the cluster points of the set { ( )} 1 S = (x, y) : x 0, y = sin x You must justify your answer. R 2.

4 Solution. First, every point p = (x 0, y 0 ) S is a cluster point by the following argument. For x 0 0 and x 1 0, ( ) ( ) 1 1 sin sin x 0 x 1 ( ) ( = 2 sin x1 x 0 1 cos + 1 ) 2x 0 x 1 2x 0 2x 1 < x 1 x 0 x 0 x 1. Therefore, for all r > 0, ( ) ( ) p q = x 0 x 1 + 1 1 sin sin x 0 x 1 < x 0 x 1 + x 1 x 0 x 0 x 1 r 2 + x 1 x 0 x 0 (x 0 /2) < r 2 + r 2 = r for q = (x 1, y 1 ) S satisfying ( r x 1 x 0 < min 2, x ) 0 2, x2 0r. 4 Therefore, every point p S is a cluster point of S. We claim that every point of {(0, y) : 1 y 1} is also a cluster point of S. Let p = (0, y 0 ) with 1 y 0 1. Let a = sin 1 (y 0 ) and ( p n = 1 2nπ + a, y 0 for n Z +. Clearly, p n S. Since p p n = ) 1 2nπ + a p is a cluster point of S. Finally, we claim that there are no other cluster points of S than the ones given above. Suppose that p = (0, y 0 ) is a cluster point of S with y 0 > 1. Then ( ) p q 1 y 0 sin y 0 1 x 1 for all q = (x 1, y 1 ) S. Contradiction.

Suppose that p = (x 0, y 0 ) is a cluster point of S for some x 0 0 and y 0 sin(1/x 0 ). Let ( ( ) ) a = min 1, x 0, 1 y 0 sin. Then for q = (x 1, y 1 ) S, ( ) ( ) ( ) y 0 y 1 1 y 0 sin 1 1 x 0 sin sin x 0 x 1 a x 1 x 0 x 0 x 1. So p q max( x 0 x 1, y 0 y 1 ) ( ) max x 0 x 1, a x 1 x 0 x 0 x 1. If x 0 x 1 > a 3 /3, then If x 0 x 1 a 3 /3, then and hence Therefore, p q > a3 3. x 0 x 1 x 0 x 0 x 1 a a3 3 2a 3 x 1 x 0 x 0 x 1 a 3 3 x 0 (2a/3) = a 2. ( a 3 p q min 3, a ) = a3 2 3. Contradiction. In conclusion, the cluster points of S are S {(0, y) : 1 y 1}. 5 (5) Let S be a convex set in R n. If p and q are two cluster points of S, then tp+(1 t)q is also a cluster points of S for all 0 t 1.

6 Proof. This is trivial for t = 0 or t = 1. Suppose that 0 < t < 1. Since p and q are cluster points of S, there exist points p r p S and q r q S such that p p r < r and q q r < r for all r > 0. Fix r > 0 and p r. Let a = p p r. Since p r p, a > 0. Let ( ) ta b = min r,. 1 t Then (tp + (1 t)q) (tp r + (1 t)q b ) t p p r + (1 t) q q b < r. We have to show that tp+(1 t)q tp r +(1 t)q b. This follows from (tp + (1 t)q) (tp r + (1 t)q b ) t p p r (1 t) q q b > ta (1 t)b 0. Therefore, tp + (1 t)q is a cluster point of S.

Solutions for Math 217 Assignment #4 7 (1) Find the boundaries, interiors and closures of the following sets: (a) S = {(x, y) : 1 < x + y 2} R 2 ; (b) S = {(x, y) : x + y Q} R 2 ; (c) S = {(x, y) : x + y Z} R 2. Justify your answers. Solution. (a) The interior of S is int(s) = {(x, y) : 1 < x + y < 2} by the following argument. For every point p = (x 0, y 0 ) with 1 < x 0 + y 0 < 2, ( x 1 + y 1 ) ( x 0 + y 0 ) x 1 x 0 + y 1 y 0 < 2r 1 < x 1 + y 1 < 2 for q = (x 1, y 1 ) B r (p) if 2r min( x 0 + y 0 1, 2 x 0 y 0 ). Also a point p { x + y = 2} cannot be an interior point of S since for p = (x 0, y 0 ) satisfying x 0 + y 0 = 2, we have either x 0 + y 0 + r > 2 or x 0 + y 0 r < 2 for all r > 0. Therefore, int(s) = {(x, y) : 1 < x + y < 2}. Similarly, we can show that int(s c ) = {(x, y) : x + y < 1 or x + y > 2}. Therefore, S = {(x, y) : x + y = 1 or x + y = 2} and S = S S = {(x, y) : 1 x + y 2}. (b) By assignment #3 1(c) (d), both S and S c are dense in R 2. So S = S c = R 2, int(s) = (S c ) c =, int(s c ) = (S) c = and S = R 2. (c) We will show that S is open. For p = (x 0, y 0 ) with x 0 + y 0 Z, let a = min n Z x 0 + y 0 n. Clearly, a > 0. Then x 1 + y 1 n x 0 + y 0 n x 0 x 1 y 0 y 1 > a 2r 0 for all n Z and (x 1, y 1 ) B r (p) with r a/2. Therefore, B r (p) S and S is open. Hence int(s) = S. Also int(s c ) = since for every point p = (x 0, y 0 ) satisfying x 0 + y 0 Z, x 0 + y 0 + r Z and hence B r (p) S c for all 0 < r < 1. Therefore, S = {x + y Z} and S = R 2. (2) Let S 1 and S 2 be sets in R n. Show that (a) int(s 1 ) + int(s 2 ) int(s 1 + S 2 ); (b) S 1 + S 2 S 1 + S 2.

8 Is it true that int(s 1 ) + int(s 2 ) = int(s 1 + S 2 )? How about S 1 + S 2 = S 1 + S 2? Prove or disprove these statements. Proof. (a) Since int(s 1 ) S 1 and int(s 2 ) S 2, int(s 1 ) + int(s 2 ) S 1 + S 2. We have proved that int(s 1 ) + int(s 2 ) is open (see assignment #3 2(a)). Therefore, int(s 1 ) + int(s 2 ) int(s 1 + S 2 ). (b) Let p S 1 and q S 2. We want to prove that p + q S 1 + S 2. Since p S 1 and q S 2, there exist p n S 1 and q n S 2 such that p p n < 1 n and q q n < 1 n for all n Z +. Therefore, (p + q) (p n + q n ) < 2 n for all n Z + with p n + q n S 1 + S 2. Consequently, p + q S 1 + S 2. The converses of both (a) and (b) are false. For the converse of (a), let S 1 = {1} (2, 3) and S 2 = (0, 1). Then int(s 1 ) = (2, 3), int(s 2 ) = (0, 1) and int(s 1 + S 2 ) = (1, 2) (2, 4). For the converse of (b), use the same counter example for assignment #3 2(b). (3) Let S be a convex set in R n. Show that both its interior int(s) and its closure S are convex. Proof. Let p and q be two points in int(s). We want to show that tp + (1 t)q int(s) for all 0 < t < 1. Since p int(s), B r (p) S for some r > 0. Then for x B tr (tp + (1 t)q), and hence Then x (tp + (1 t)q) < tr ( 1 t x 1 t ) q p t < r 1 t x 1 t q B r (p) S. t ( 1 x = t t x 1 t ) q t + (1 t)q S since S is convex. This shows that B tr (tp + (1 t)q) S and hence tp + (1 t)q int(s).

Let p and q be two points in S. We want to show that tp + (1 t)q S for all 0 < t < 1. Since p, q S, there exist p n, q n S such that 9 p n p < 1 n and q n q < 1 n for all n Z +. It follows that (tp + (1 t)q) (tp n + (1 t)q n ) t p p n + (1 t) q q n < 1 n. And since tp n + (1 t)q n S, tp + (1 t)q S. (4) Show that S 1 + S 2 = S 1 + S 2 for two bounded sets S 1 and S 2 in R n. Proof. Actually, we only need one of S i to be bounded. WLOG, we assume that S 1 is bounded. We have proved that S 1 + S 2 S 1 + S 2 and hence it suffices to show that S 1 + S 2 S 1 + S 2. Suppose that x S 1 + S 2. Then there exist p n S 1 and q n S 2 such that (p n + q n ) x < 1 n for all n Z +. Since S 1 is bounded, there exists a point p and an infinite subsequence such that {n l : l Z + } Z + p nl p < 1 l for all l Z +. Hence p S 1. Let q = x p. Then q nl q (p nl + q nl ) x + p p nl 1 n l + 1 l. Therefore, q S 2 and hence x = p + q S 1 + S 2.

10 (5) Find the closure of the set S = { m n : m, n Z + } in R. Justify your answer. Solution. We claim that S = R. Clearly, x S if and only if x S. So it suffices to show that R + S. Let r n = n + 1 n for n Z +. Then r n S and 0 < r n = n + 1 1 n = < 1 n + 1 + n 2 n. For x R +, let m = x/r n. That is, m is the integer such that mr n x < (m + 1)r n. Then x mr n < r n < 1 2 n. On the other hand, mr n = m( n + 1 n) = m 2 (n + 1) m 2 n and hence mr n S. Therefore, x S and R + S. We are done.