MATH 31BH Homework 1 Solutions

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1 MATH 3BH Homework Solutions January 0, 04 Problem.5. (a) (x, y)-plane in R 3 is closed and not open. To see that this plane is not open, notice that any ball around the origin (0, 0, 0) will contain points whose third component is not 0, and hence not in the (x, y)-plane. To show that the plane is closed, we will prove that its complement is open. Indeed, suppose that (a, b, c) is a point in the complement of the (x, y)-plane. Then c is not zero. Consider the ball B = B c ((a, b, c)). The closest point to (a, b, c) in the (x, y)-plane is its orthogonal projection: (a, b, 0). This is a distance of c > c away from (a, b, c). Hence the ball B contains no points in the (x, y)-plane. So B is contained completely in the complement of the (x, y)-plane, and hence the complement of the plane is open. (b) R C is closed and not open. R is not open in C because, for example, a ball of radius ɛ around 0 will contain the purely imaginary point ɛ i. We now show that C R is open. Let z belong to C R. Then z = x + iy, where y 0. Consider the ball B = B y (z). Now the closest point on the real line to z is the real number x. This is a distance of y away from z. It follows that no point in B is purely real. Hence B is contained in C R, and so C R is open. We conclude that R is closed in C. (c) The line x = 5 is closed and not open in the (x, y)-plane. x = 5 is not open in the (x, y)-plane. Notice, in particular, a ball of radius ɛ around (5, 0) will contain the point (5 + ɛ, 0), which is not on the line x = 5. To show that x = 5 is closed we show that its complement is open. Let (s, t) belong to the complement of the line x = 5. That is to say, s 5. Consider the ball B = B s 5 ((s, t)). Now the closest point on the line x = 5 to (s, t) is (5, t). This is a distance of s 5 away from (s, t). It

2 follows that no point in B is on the line x = 5. Hence B is contained in the complement of the line, and so the complement is open. (d) (0, ) C is neither open nor closed. To see that (0, ) is not open, we note that any ball around will contain non-real points. To see that (0, ) is not closed, we will show that its complement is not open. Notice that the real number is in C (0, ). Now any ball of radius ɛ around will contain the real number ɛ, and hence will not be contained in C (0, ). So C (0, ) is not open. (e) R n R n is both open and closed. Let x R n. Then B (x) is completely contained in R n. Since x was an arbitrary point, it follows that R n is open in R n. Now the complement of R n is the empty set. Since there are no points in, the statement for all points in there exists a ball centered around the point contained completely in is vacuously true. (That is to say, there are no points around that have to satisfy the required criterion.) So is open, and hence R n is closed. (f) The unit sphere in R 3 is closed and not open. To see that the unit sphere is not open, Notice that any ball of radius ɛ around the point (, 0, 0) will contain the point ( + ɛ, 0, 0), which is at a distance of more than from the origin. Now to show that the unit sphere is closed in R 3, we show that its complement is open. Let (a, b, c) be a point in the complement of the unit sphere. In other words, (a, b, c) is at a distance not equal to away from the origin. Let s call this distance + d (i.e., (a, b, c) is exactly d units away from the closest point on the boundary). d may be positive or negative, but it cannot be 0. Consider the ball B = B d ((a, b, c)). Since the closest point on the sphere to (a, b, c) is d > d units away from (a, b, c), B must avoid the sphere completely. Therefore B is contained in the complement of the unit sphere, and hence the complement of the unit sphere is open. Problem The sequence x n = Proof. Let ɛ > 0 be given. Let N(ɛ) = ɛ. ( ) n n +, n converges to the it (, 0).

3 Now suppose that n > N(ɛ). Then we have ( ) ( ) n n +, n (, 0) n = n +, n ( = ) n +, n ( = ) n + + n n + + n n + n = n < N(ɛ) = ɛ. We conclude that n x n = (, 0). Problem.5.4 (a) (x,y) (,) x x + y = 3. Proof. By corollary.5.30 since this function is rational and x + y 0 at the point (, ), the function is continuous at (, ). Therefore (x,y) (,) x x + y = () + = 3. (b) (x,y) (0,0) x y x does not exist. + y Proof. We examine the it along the paths y = k x. As x approaches 0 on these paths, so 3

4 does y. We have x y (x,y) (0,0);y=k x x + y = k x x 0 x + k x = x 0 k x x + k x k = x 0 x + k = k. Since this it depends on the choice of k, it follows that the original it does not exist. (c) (x,y) (0,0) xy does not exist. x + y Proof. We consider the it along the paths y = kx. As x approaches 0 on these paths, so does y. We have (x,y) (0,0);y=kx xy x + y = x 0 k x x + k x x k = x 0 x + k = k + k. This it depends on the choice of k, so the original it does not exist. (d) (x,y) (,) x + y 3 3 = 6. Proof. The function in the expression is a polynomial. Therefore, by corollary we have (x,y) (,) x + y 3 3 = () + () 3 3 = 6. Problem.5.0 For this problem we identify Mat(n, m) with R nm so that the notion of its of matrices makes sense. (a) Consider the sequence of matrices A k for k N, where [ ] a a A = a a 4

5 for some real a. When a < the sequence converges to the 0 matrix. When a =, the sequence converges to the matrix [ When a > or a =, the sequence does not converge. Proof. By induction, we can show that [ A k = k a k k a k ] k a k k a k ]. [ ] = k a k. Therefore, by Theorem.5.6.(), we need only investigate the convergence of the real sequence k a k. Notice that k a k = (a)k. Now the geometric sequence (a) k will converge if and only if a < or a =. In the case that a <, the sequence converges to 0 (and hence so does the sequence A k ). In the case where a =, this sequence is the constant sequence consisting of the real number. Hence it converges to, and so the sequence Ak converges to the matrix [ We note that when a =, the sequence k a k will oscillate between and, and in the case where a > the sequence k a k is unbounded in magnitude. (b) We may extend the above methods to any n n matrix A whose entries are all equal to some real number a. Inductively we can see that n k a k n k a k A k =... n k a k n k a k We notice that each entry is again a geometric sequence that converges to 0 if a < n, converges to n if a = n, and diverges otherwise. EC (a) The possible topologies on X = {a, b} are T = {, {a}, {a, b}} T = {, {b}, {a, b}} 5 ].

6 T = {, {a}, {b}, {a, b}} (b) Define T to be the collection of subsets of R which are either empty or whose complements are finite. Then T is a topology on R. Proof. Certainly and R belong to T. The former is a member because it is empty, and the latter because its complement (the empty set) has finite cardinality. Now suppose that, for some index set I, {O i } i I T. That is to say, each O i is either empty or has a finite complement. If all the O i are empty, then i I O i is empty as well, and hence belongs to T. Otherwise, at least one of the O i is nonempty and hence has finite complement. Let us call this set O. In this case, we see that ( i I O i ) = i I O i. Here, O denotes the complement of a set O. Since i I O i O and O is finite, it follows that i I O i is finite also. Therefore i I O i T. Finally, let O and O be members of T. If either of these sets is empty, then O O is empty, too, and thus belongs to T. If both O and O are nonempty, then they must have finite complements. In this case, (O O ) = O O. O O is the union of two finite sets, and therefore is finite itself. We conclude that O O T. By induction, it follows that any finite intersection of members of T is a member of T. These three facts show that T is a topology on R. EC (a) R is a metric space with the distance function d(x, y) = x y. Proof. Let x, y, z R. First suppose that x y = 0. Then we also have x y = 0, and so x = y. Conversely, if x = y then x y = 0 and hence x y = 0. Next, note that x y = (x y) = y x. Finally, by the triangle inequality for R n, we have x y x z + z y. Thus d(x, y) satisfies the three properties of a metric, and so (R, d) is a metric space. 6

7 Notice that the open balls in this metric space are intervals, and therefore open sets are just unions of open intervals. (b) Any metric space (X, d) induces a topological space (X, T ), where T = {U X : for all x U there exists a small ball B r (x) U}. Proof. We show that T satisfies the three properties of a topology. First, we see that T vacuously. Moreover, X T because any ball in X will be contained in X. Next, let {U i } i I be a collection of subsets in T. We need to show that for any point in i I U i we can find a ball around that point contained completely inside the union. Let x i I U i be given. Then x U j for some j I. By the fact that U j T, there exists some real number r such that B r (x) U j i I U i. Thus i I U i T. Finally, let U and U be sets in T. Let x U U. Then x U and x U. It follows that we may find real numbers r and r such that B r (x) U and B r (x) U. Let r = min{r, r }. Then B r (x) B r (x) and B r (x) B r (x). Therefore B r (x) U U. By induction, we see that any finite intersection of members of T is also a member of T. Therefore T is a topology on X. (c) Consider an infinite set X and let Then (X, d) is a metric space. d = { if x y 0 if x = y. Proof. Let x, y, z X. By definition of d we already have that d(x, y) = 0 if and only if x = y. Moreover, x = y if and only if y = x. Hence we have d(x, y) = d(y, x). Finally, we consider two cases. If x = y, then d(x, y) = 0 and it must be the case that d(x, y) d(x, z) + d(z, y) regardless of the value of z. If x y, then d(x, y) =. But it is also the case that z may only equal one of x or y. Therefore at least one of d(x, z) and d(z, y) is guaranteed to be. So in this case we also have d(x, y) d(x, z) + d(z, y). We conclude that (X, d) is a metric space. 7

8 Notice that there are two kinds of open balls in this metric space. For radii less than or equal to, the open balls are just points of X. For any open ball with radius strictly greater than, the ball is the entirety of X. Since an open set is just a union of open balls, and since any point in X is an open ball, it follows that any subset of X is an open set in (X, d). (d) Consider X = R and let Then (X, d) is a metric space. d((x, y ), (x, y )) = max( x x, y y ). Proof. Let (x, y ), (x, y ), (x 3, y 3 ) R. Suppose that d((x, x ), (y, y )) = 0. Then max( x x, y y ) = 0. Since these are both non-negative numbers, it must be that x x = y y = 0. In other words, x = x and y = y. So (x, y ) = (x, y ). Conversely, suppose that (x, y ) = (x, y ). Then x x = 0 and y y = 0, so d((x, y ), (x, y )) = 0. Next, we notice that x x = x x and y y = y y. As a result, d((x, y ), (x, y )) = d((x, y ), (x, y )). Finally, we must show that max( x x, y y ) max( x x 3, y y 3 ) + max( x 3 x, y 3 y ). Certainly by real triangle inequality we have max( x x, y y ) max( x x 3 + x 3 x, y y 3 + y 3 y ). However, we can get an even larger sum than the two arguments in the maximum on the right by choosing the largest of each summand and adding those together. That is, max( x x 3 + x 3 x, y y 3 + y 3 y ) max( x x 3, y y 3 ) + max( x 3 x, y 3 y ). By transitivity, this proves the triangle inequality (property (iii)) of d. Therefore (X, d) is a metric space. Notice that the open balls B r (x) of the induced topological space are open squares with side length r and center x. Thus the open sets are unions of squares. 8