Mid Term-1 : Practice problems

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1 Mid Term-1 : Practice problems These problems are meant only to provide practice; they do not necessarily reflect the difficulty level of the problems in the exam. The actual exam problems are likely to be easier because of the time constrains. 1. Let F be a collection of some subsets of R k, and let S = A F A and T = A F A. For each of the following statements, either give a proof or provide a counterexample. (a) If p is a limit point of T, then p is a limit point of each A F. Solution: TRUE. If p is a limit point of T, then every ball B r (p) intersects with T at a point other than p. But this means that B r (p) A has a point other than p for all A F and so p is a limit of every A F. (b) If p is a limit point of S, then p is a limit point of at least one set A F. Solution: FALSE. Consider the family of open sets A n = (1/n, 1). Then S = (0, 1) and so 0 is a limit point of S. But clearly 0 is not the limit point for any A n. 2. Consider the function d : R 2 R 2 R defined by the formula where x = (x 1, x 2 ) and y = (y 1, y 2 ). d(x, y) = max( x 1 y 1, x 2 y 2 ), (a) Show that d defines a metric on R 2. This is called the Manhatten metric. Solution: Positive definiteness. Clearly d(x, y) 0 for all x and y. Suppose d(x, y) = 0. Then x 1 y 1 = 0 and x 2 y 2 = 0 or x 1 = y 1 and x 2 = y 2. So d(x, y) = 0 = x = y. Symmetry. Clearly d(x, y) = d(y, x) since is symmetric. Triangle inequality. It is enough to show the triangle inequality with u = (u 1, u 2 ), v = (v 1, v 2 ) and 0 = (0, 0) ie. d(u, v) d(u, 0) + d(v, 0). (1) 1

2 This is because if x, y, z are three points, then d(x, y) = d(x z, y z), d(x, z) = d(x z, 0) and d(y, z) = d(y z, 0). So then the above inequality applied to u = x z and v = y z, we obtain d(x, y) d(x, z) + d(y, z). Now we prove inequality (1). Note that d(u, 0) = max( u 1, u 2 ) and d(v, 0) = max( v 1, v 2 ). In particular for i = 1, 2, u i d(u, 0), and we have a similar inequality for coordinates of v. By the usual triangle inequality for the absolute value, u 1 v 1 u 1 + v 1 d(u, 0) + d(v.0), and similarly u 2 v 2 u 2 + v 2 d(u, 0) + d(v.0). Taking the max of both the above inequalities we obtain the inequality (1). (b) Draw the ball of radius one centered at the origin in the above metric. What geometric shape do you obtain? Solution: That is, we want the set B 1 (0) := {(x 1, x 2 ) x 1, x 2 < 1}. This is clearly the open square with edges x 1 = 1, x 1 = 1, x 2 = 1 and x 2 = 1. (c) Are closed and bounded sets compact with this new metric? If so, give a proof. If not, give an example of a closed and bounded set (in this new metric) which is not compact. Solution: YES. First, we claim that a set U R 2 is open with respect the metric d if and only if it is open with respect to the Euclidean metric d E. To see this, note that a ball B d r (p) in the metric d is a square of side length 2r and sides parallel to the two axes. Now suppose U is open in the Euclidean metric, and let p U. Then there exists an r > 0 such that the ball B r (p) U. But then one can also draw a square (of a possibly smaller side length r < r) with center p and sides parallel to the axes contained completely in B r (p). But then B d r (p) B r (p) U, and so p is also an interior point with respect to the metric d. This is true for any p U, and shows that U is open in the metric d also. The converse is also similar. It follows that F is closed with respect to d if and only if it closed with respect to the Euclidean metric d E. Now let K be a closed and bounded subset with respect to the metric d. And let {U α } be an open cover where U α is an open set with respect to the metric d. Then by the above arguments, K is also a closed and bounded with respect to the Euclidean metric, and hence is compact with respect to the Euclidean metric. By the above each U α is also an open set with respect to the metric d E. So there exists α 1,, α n such that K n k=1 U α k. 2

3 So we have managed to extract a finite sub-cover, and hence K is compact even with respect to the metric d. 3. Let {a n } be a sequence of real numbers such that a n 2, and a n+2 a n a2 n+1 a 2 n for all n 1. Show that a n converges. Hint. Show that the sequence is Cauchy. You might have to use the fact that 1 4 k = 4 3. k=0 Solution: Since a n < 2, we conclude that a n+2 a n a2 n+1 a 2 n = 1 8 a n+1 a n a n+1 + a n 1 4 a n+1 a n. Inductively we can see that a n+2 a n+1 < 1 4 n a 2 a 1 From this, it follows by triangle inequality that for any n < m, a m a n a m a m 1 + a m 1 a m 2 + a n+1 a n ( 1 4 m m ) 4 n 1 a 2 a 1 4 (n 1)( 1 4 m n ) 4 m n a 2 a 1 4 (n 1) a 2 a 1 = 4 (n 1) a 2 a Since a 2 a 1 a 2 + a 1 < 4, we see that k=0 1 4 k a m a n 4 (n 3). 3 Given ε > 0, if we choose N big enough so that 4 (N 3) < 3ε, then for any n, m > N we will have a m a n < ε, and so the sequence is Cauchy. Since R is complete, the sequence converges. 4. Let 2 N denote the collection of subsets of N. That is, 2 N = {A A N}. 3

4 Show that 2 N is uncountable. Hint. Proceed by contradiction and use anx argument similar to Cantor diagonalization. Solution: Suppose 2 N, the set of subsets of N, is countable. Let us the list all the subsets of N as {A 1, A 2, }. Consider the subset A N defined by A = {k N k / A n }. We claim that A / {A 1, A 2, }. But this would be a contradiction since we are assuming that the list exhausts all possible subsets of N. To see the claim, consider A n. If n A n, then n / A, and if n / A n, then n A. So the set A is not equal to A n for any n N. An alternate viewpoint. An alternative viewpoint is to think of subsets of N as an infinite string of 0 and 1 in the following way - For a subset A N we associate the string {a n } n=1, where { 1, n A a n = 0, n / A. For instance the empty set will correspond to a string of all zeroes and the set N will correspond to a string of all ones. The subset of even numbers will correspond to a string with zeroes in odd places and ones in even places, and so on. The proof that the set of all infinite binary strings is uncountable is exactly our Cantor diagonalization argument. To go over it once more - Suppose {A 1, A 2,, A k, } is an exhaustive list of all such strings with A k = {a kn } n=1. Consider the string { 1, a nn = 0 a n = 0, a nn = 1. Then A differs from A n in the n th spot, and so A is not in the list {A 1, } but A itself is an infinite string of zeroes and ones. Contradiction! 5. Let A and B be disjoint closed subset of a metric space (X, d). Show that there exist disjoint open sets U and V such that A U and B V. Hint. First show that there is an open set U containing A such that U and B are disjoint. Solution: Let p A. Since B is closed, B c is open. Now A and B are disjoint, and so p B c. Then there exists an r p > 0 such that B 2rp (r) B c. Now let U = p A B rp (p). The clearly A U. Similarly for each q B, there exists a r q such that B 2rq (q) A c, and take V = q B B rq (q). Then B V. Claim. U V = φ. Proof Suppose x U V. Then there exists a p A and q B such that d(x, p) < r p and d(x, q) < r q. By triangle inequality d(p, q) d(p, x) + d(q, x) < r p + r q. 4

5 Now suppose r p r q, then d(p, q) < 2r q, and so p B 2rq (q) A. But by choice this intersection is empty since B 2rq (q) A c. So a contradiction. The other case r p r q is similar. 6. Let (X, d) be a complete metric space. (a) Suppose {E n } is a decreasing (ie. E n+1 E n for all n) sequence of non-empty, closed and bounded sets with lim n 0 diam(e n) = 0, where for any subset E, the diameter is defined as diam(e) = sup x,y E d(x, y). Show that n=1 E n consists of exactly one point. Solution: Note that the hypothesis of bounded is superfluous since the diameters go to zero, and so the sets have to be bounded! Claim-1. The intersection E n is non-empty. Proof. Pick a point x n E n. Since diam(e n ) 0, for any ε > 0, there exists an N such that diam(e N ) < ε. But then if n, m > N, since E n, E m E N, by the definition of diameter, and our choice of points, we must have that d(x n, x m ) < ε. This shows that the sequence {x n } is Cauchy, and hence must converge (X is complete) to some p X. We claim that p lies in the intersection. To see this, note that for any N, x n E N if n > N. And so p is a limit point of a sequence of points {x n } n=n in E N, and hence must belong to E N since E N is closed. This shows that p E N for all N, and so p n=1 E n. Claim-2. p is the uniques such point. Proof. If not, then there is some q which also lies in the intersection. Fix an ε > 0, and choose N such that diam(e N ) < ε. Then since p, q n=1 E n, they both belong to E N in particular. But then d(p, q) < diam(e N ) < ε. So for any ε > 0, d(p, q) < ε, and hence d(p, q) = 0 which means that p = q. (b) If {G k } is a sequence of dense open subsets of R k, show that k=1 G k is non-empty. Hint. Find a shrinking sequence of neighborhoods E n such that E n G n and apply part (a). Solution: Let E 1 G 1 be open such that E 1 G 1. One way to acheive this is to take any x 1 G 1. Then since G 1 is open, there is some r > 0 such that B r (x 1 ) G 1. Then one can take E 1 = B r/2 (x 1 ). Now since G 2 is dense, E 1 G 2 is non-empty. It is also open, so by the same argument as above, there exists an open set E 2 E 1 G 2 such that E 2 E 1 G 2 We can take diam(e 2 ) < diam(e 1 )/2. We continue this process. So in the end we will have a sequence of open sets E n such that E 1 E 2 E n G n. diam(e n ) < 2 (n 1) diam(e 1 ). 5

6 Since {E n } is a decreasing sequence of closed sets with diameter converging to zero, by part(a) E n is non-empty. But E n G n, and so the latter is also non-empty. (c) From the above part show directly that if R k = k=1 F k for closed subsets F k, then at least one of F k has a non-empty interior. Hint. Argue by contradiction by considering G k = F c k. Solution: We proceed by contradiction. Note that a set F has an empty interior if and only F c is dense (prove it!). So suppose all F k s have an empty interior. Then G k = F c k satisfy G k is open and dense. G k = R k F k = φ. But this contradicts part(c). Hence at least one of F k has a non-empty interior. 6