Homework 3. d(x, y) = x i y i. d(x, y) = sup x i y i. d(x, y) = max{d 1 (x 1, y 1 ), d 2 (x 2, y 2 )}.

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1 Homework 3 Page 61 of Rosenlicht: 1, 2, 4, 5, 6, 15, 16(a),(c), 17, Verify that the following are metric spaces (a) all n-tuples of real numbers with d(x, y) = n x i y i i=1 (b) all bounded infinite sequences x = (x 1, x 2,...) of elements of R with d(x, y) = sup x i y i. (c) (E 1 E 2, d) where (E j, d j ) are metric spaces and d(x, y) = max{d 1 (x 1, y 1 ), d 2 (x 2, y 2 )}. Solution: Part (a) was an example from class, so no solution will be provided. (b): Let x and y be bounded infinite sequences in R. It is clear that we have since x i y i 0 for all i. 0 d(x, y) If x = y, then we clearly have that d(x, y) = 0. Suppose that d(x, y) = 0, then we have 0 = sup x i y i and so x i y i = 0 for all i, and so x i = y i for all i, or x = y. So we have that d(x, y) = 0 if and only if x = y. We also have d(x, y) = sup x i y i = sup (x i y i ) = sup y i x i = d(y, x). Finally, for the triangle inequality for absolute value, note that x i y i x i z i + z i y i sup x i z i + sup z i y i = d(x, z) + d(z, y). But, this implies that d(x, y) = sup x i y i d(x, z) + d(z, y) so the triangle inequality holds for the metric d.

2 (c) The non-negativity of the metric d is easy to see so we omit that. Also, if (x 1, x 2 ) = x = y = (y 1, y 2 ) then we trivially have d(x, y) = 0. Further, if d(x, y) = 0 then we have that d j (x j, y j ) = 0 for j = 1, 2. But this means that x j = y j for j = 1, 2, or x = y. It is also easy to see that d(x, y) = d(y, x). Suppose, with out loss of generality, that max{d 1 (x 1, y 1 ), d 2 (x 2, y 2 )} = d 1 (x 1, y 1 ). Then note, d(x, y) = max{d 1 (x 1, y 1 ), d 2 (x 2, y 2 )} = d 1 (x 1, y 1 ) d 1 (x 1, z 1 ) + d 1 (z 1, y 1 ) max{d 1 (x 1, z 1 ), d 2 (x 2, z 2 )} + max{d 1 (z 1, y 1 ), d 2 (z 2, y 2 )} = d(x, z) + d(z, y). Here we used the triangle inequality for the metric space (E 1, d 1 ). Thus d satisfies the triangle inequality and so we have that (E 1 E 2, d) is a metric space. 2. Show that (R 2, d) is a metric space, where { y + y d((x, y), (x, y )) = + x x : x x y y : x = x. Solution: This is essentially definition checking as well. No solution will be provided. 4. Show that the set of R 2 given by {(x 1, x 2 ) R 2 : x 1 > x 2 } is open. Solution: Let E = {(x 1, x 2 ) R 2 : x 1 > x 2 } and note that E is the half plane that has boundary given by the line x 2 = x 1. Suppose that x E, so we have that x 1 > x 2. We want to determine a radius r so that { (y 1, y 2 ) R 2 : x 1 y x 2 y 2 2 < r } = B r (x) E which would prove that E is open. Drawing a picture of the situation, we see that the optimal choice of r > 0 would be given by the distance of the point (x 1, x 2 ) from the line (x 1, x 1 ) in the plane R 2. A basic linear algebra computation gives that this distance is given by x 1 x 2 2. Choose r < x 1 x 2 2, and for concreteness take, r = x 1 x 2 2. Geometrically 2 it is obvious then that we have that B r (x) E. 5. Prove that any bounded open subset of R is the union of disjoint open intervals. Solution: Let A denote this set and let x A. Since A is open, there is an open interval containing x entirely in A. Define a x = inf{y R : (y, x) A}

3 and b x = sup{z R : (x, z) A}. both these sets are bounded since A is bounded, and non-empty since A is open and there is an interval about the point x contained in A. So, by the least upper bound property, the sup and inf exist. We now note that (a x, b x ) A. This follows from the definitions of the numbers a x and b x. Repeat this procedure for every point in A, and we have constructed a (possibly infinite) collection of open intervals whose union is A, i.e. A = x A (a x, b x ). Now we claim that this collection can be reduced to a disjoint collection of open intervals. To show this it suffices to demonstrate that for any two points x 1, x 2 A, either: 1. x 1 and x 2 lie in the same interval, or 2. x 1 and x 2 lie in different intervals. We may assume that x 1 (a 1, b 1 ) and x 2 (a 2, b 2 ). Suppose x 2 (a 1, b 1 ). Then since b 1 is the least upper bound of all z such that (x 2, z) A, b 1 b 2. Similarly, a 2 a 1. Thus, x 2 (a 2, b 2 ), and since b 2 is the least upper bound of all z such that (x 1, z) A, b 2 b 1, and similarly, a 1 a 2. So a 1 = a 2, and b 1 = b 2. On the other hand suppose x 2 / (a 1, b 1 ). Then (a 1, b 1 ) (a 2, b 2 ) =. Suppose not, and let w be an element of (a 1, b 1 ) (a 2, b 2 ) = (c, d). By the same argument as before a 1 = c, and b 1 = d, and also a 2 = c, and b 2 = d. Hence a 1 = a 2 and b 1 = b 2, but then x 2 (a 1, b 1 ), which is a contradiction, and so (a 1, b 1 ) and (a 2, b 2 ) must be disjoint. 6. Show that the subset of R 2 given by {(x 1, x 2 ) R 2 : x 1 x 2 = 1, x 1 > 0} is closed. Solution: The set E = {(x 1, x 2 ) R 2 : x 1 x 2 = 1, x 1 > 0} is the branch of the hyperbola that lies in the first quadrant. If we consider the set E c, then we will see that this set is open using the definitions. Note that E c = R 2 \ E = Y R 2 + \ E, where Y is the second, third and fourth quadrants in the plane. It is clear that if we take a point in Y, then there exists a ball that is wholey contained in the set E c, by choosing the radius of the ball smaller than the minimum of the coordinates. So it only remains to show the same statement for points in R 2 + \ E. Let p R 2 + \ E. So we have p = (x 0, y 0 ) and x 0 y 0 1 with x 0, y 0 > 0. Suppose that x 0 y 0 > 1, the case of x 0 y 0 < 1 can be handled similarly. Let r 2 = inf{(x x 0 ) 2 +(y y 0 ) 2 : xy = 1}. Note that r exists since the set is non-empty ((1, 1) is in there) and bounded from below, since r 2 > x y 2 0. Then we have that B r 2 (p) R2 + \ E. 15. Let S be a subset of the metric space E. A point p S is called an interior point of S if there is an open ball in E with center p which is contained in S. Prove the set of all interior points of S is an open subset of E that contains all other open subsets of E that are contained in S.

4 Solution: Let S = {p S : p is an interior point of S}. We first show that S is open. Let p S, so we know that p is an interior point of S. Thus, there exists a open ball B r (p) S. We claim now that B r (p) S and to see this it suffices to show that each q B r (p) is an interior point of S. For this q B r (p), since B r (p) is open, there is an open ball B s (q) B r (p) S. Thus, q is an interior point, and S is open as claimed. We now need to prove the following statement: Let G S be an open set, then G S. Let p G, and since G is open, then there exists a ball B r (p) such that B r (p) G S. This implies that p is an interior point S, and so p S. Combining things, we have that G S as claimed. 16. Let S be a subset of a metric space E. Define the closure of S, denoted by S, to be the intersection of all closed subsets of E that contain S. Show that (a) S S and S is closed if and only if S = S. (c) A point p E is in S if and only if any ball in E centered at p contains points of S, which happens if and only if p is not an interior point of S c. Solution: (a) Note that from the definition we have that S = {F : S F, F is a closed subset of E}. Let p S c, and then we have by De Morgan s Laws that S c = {F : S F, F is a closed subset of E} c = {F c : S F, F is a closed subset of E}. So there exists a closed set F with S F, such that p F c. Since p F c, we have that p S c. Thus, we have that S c S c, or equivalently S S. We now prove that S is closed if and only if S = S. Suppose that S is closed, then S S, and so we have that S S from the definition of the closure. But, by above we proved that S S, and so S = S. Now suppose that S = S. We want to prove that S is closed, and this is the same as S c being open. Since S = S, we have that S c = S c, and by the above we have S c = S c = ( {F : S F, F is a closed subset of E}) c = {F c : S F, F is a closed subset of E}. But, since each F is closed, each F c is open, and since this is a union of open sets, it is open. Thus, S c is open as desired, and so S is closed. (c) Suppose that p S. Proceed by contradiction, and suppose that there exists a ball centered at p that does not contain points of S. Then B r (p) S =, so we have that S B r (p) c, and B r (p) c is closed since B r (p) is open. Since p S, we have that

5 p B r (p) c, which is a contradiction. So for any ball centered at p, we must have that B r (p) S. Conversely, suppose now that for all r > 0 we have that B r (p) S. Then we claim that p S. Let F be a closed subset such that S F. We claim that p F. Suppose this wasn t the case, then p F c, and since F c is open, we have that a ball B r (p) F c. Note that B r (p) S = since B r (p) F c S c. But, this contradicts the hypothesis that B r (p) S for all radii r > 0. Thus, we have that p F for all F closed such that S F, and so p S. We finally demonstrate the claim that B r (p) S for all r > 0 if and only if p is not an interior point of S c. First, suppose that p is not an interior point of S c. Then there exists no r > 0 such that B r (p) S c. Thus, we must have that B r (p) S for all r > 0. Since if there existed a r > 0 such that B r (p) S =, then we would have that B r (p) S c, which would imply p is an interior point of S c. Conversely, suppose that B r (p) S for all r > 0. We claim that p is not an interior point of S c. If p were an interior point of S c then we would have an r > 0 such that B r (p) S c. But this is a contradiction to the condition that B r (p) S. 17. Let E be a metric space. The boundary of S is defined to be S S c, denoted by S. Show that (a) E is the disjoint union of S, (S c ), and S. (b) S is closed if and only if S contains its boundary. (c) S is open if and only if S and its boundary are disjoint. Solution: (a): We clearly have that S (S c ) S E. It is easy to see that each of these sets are disjoint from each other, and this then gives us that S (S c ) S = E since we have that S ( S) c = E, and ( S) c = S c S cc = (S c ) S. Suppose that S (S c ). Then there is a p S and p (S c ), and so p is an interior point of both S and S c. Thus, there are balls such that B r (p) S and B s (p) S c. So p S S c, but this is impossible. So we have S (S c ) =. If S S, then there are p S and p S S c. Since p is an interior point, there is a ball centered at p such that B r (p) S. Also, since p S S c, any ball centered at p must contain points of S c. This leads to a contradiction since for the ball used for the interior point we have that B r (p) S, and so B r (p) S c =. This, we have that S S =. The case of (S c ) S = is obtained by swapping the roles of the set and its complement.

6 (c) We use part (a) to deduce this. Note that we have S = E S = (S S ) (S (S c ) ) (S S). (1) Note that S (S c ) =, and (S S ) = S so S = E S = S (S S). (2) If S S =, then we have that S = S, and since S is open, we have that S is open. On the other hand, if S is open then we have that S = S, and since the decomposition given in (1) is disjoint, we have from (2) that S (S S) = and so S S =. (b) One can use part (c) and the fact that S is closed if and only if S c is open. Applying part (c) with S c we see that S c is open if and only if S c at its boundary are disjoint. Observe now that (S c ) = S c (S c ) c = S S c = S. So S c is open if and only if S c S =. But, this is logically equivalent to S is closed if and only if S S. 22. Let V be a vector space over R with a norm denoted x for x V. Consider the function d(x, y) = x y for x, y V. Show that (V, d) is a metric space. Solution: Let V be a real normed vector space. Let d : V V R + be the function d(x, y) = x y for x, y V. Let x, y, z V. Since x y V and v 0 for all v V, then d(x, y) 0. Likewise, for all v V, v = 0 if and only if v = 0, so d(x, y) = 0 if and only if x y = 0 if and only if x = y. Since x y = ( 1)(y x) and av = a v for all v V, a R, then d(x, y) = x y = 1 y x = d(y, x). Since x z, z y V and v + u u + v for all u, v V, then d(x, y) = x y = (x z) + (z y) x z + z y = d(x, z) + d(z, y). Since d(x, y) is positive definite, symmetric, and satisfies the triangle inequality, it is a metric on V.