Midterm Exam 1 with Solutions - Fundamental Concepts of Analysis | MATH 425A, Exams of Mathematics

Download Midterm Exam 1 with Solutions - Fundamental Concepts of Analysis | MATH 425A and more Exams Mathematics in PDF only on Docsity! MIDTERM 1 SOLUTIONS MATH 425A, OCTOBER 26, 2009 Problem 1. (a) Define: the sequence {xn} converges to x (in a metric space (M,d)). (b) Prove: a subset K of a metric space (M,d) is closed iff it has the property that whenever a sequence {xn} in K converges to a point x ∈M , it must follow that in fact x ∈ K. Use the definitions I gave in class. Solution. We say that {xn} converges to x in (M,d) if for all ε > 0 there exists N = N(ε) such that n ≥ N =⇒ d(xn, x) < ε. Definitions in mathematics are absolutely crucial. You must be precise. A statement like “There exists an N such that d(xn, x) < ε for n ≥ N , for every ε” just won’t cut it; it isn’t logical, and it’s not even good English. I asked you to use MY definitions. On an exam, that’s not really a request, but an order; if you don’t use my definitions, you should expect points to be taken off. I didn’t, this time, because using the book’s definition got people in some real trouble. Recall my definition of closed: K is closed iff its complement Kc is open. Now argue as follows: it’s a double implication. First, assume K is closed, and {xn} is a sequence in K with xn → x. If x /∈ K then x is an interior point of Kc, hence there exists ε > 0 such that Nε(x) ⊂ Kc. Since xn → x, for sufficiently large n we will have xn ∈ Nε(x); but this is impossible, since xn ∈ K. Conversely, suppose K has the given property; we must prove Kc is open. To this effect, let x ∈ Kc, and suppose it is false that Nε(x) ⊂ Kc for some ε > 0. Then for all ε > 0 we will have Nε(x) ∩K 6= ∅. In particular, taking ε = 1/n for n = 1, 2, . . . , we find xn ∈ K ∩N1/n(x); that is, d(xn, x) < 1/n (and thus xn → x) and xn ∈ K. By the property, we have x ∈ K, contradicting the assumption that x ∈ Kc. All right, what if you used the book’s definition? Then you’re trying to prove the equivalence of K contains all its limit points and the property xn ∈ K, xn → x in M =⇒ x ∈ K. Suppose the first condition holds, and xn ∈ K converges to x in M . Just because xn → x doesn’t mean that x is a limit point of {xn} (or, more properly speaking, of the range of the sequence {xn}). The sequence could be constant, for example. Instead, we must argue on whether the range of the sequence is infinite or finite. If it is infinite, then every neighborhood of x contains the terms of the sequence {xn} from some point on, and hence infinitely many points of its range; and in that case, x is a limit point of K, therefore x ∈ K. If the range of {xn} is finite, it’s even easier: some subsequence {xni} must be constant, and therefore constantly equal to x; since xni ∈ K, therefore x ∈ K. This is why I don’t like the book’s use of “limit point”. It has to do with limits of sequences, but requires quibbles about whether terms in the sequence repeat or not. ”Limit point” is a property of SETS, not sequences. I’ll leave the converse implication to you.  Problem 2. Prove: The intersection of an arbitrary family K of com- pact subsets of a metric space (M,d) is also compact. Solution. Compact sets are closed (a theorem in class); thus the inter- section of (K) is an intersection of closed sets, hence a closed subset of each K ∈ (K); but a closed subset of a compact set is closed, also by a theorem proved in class. SOLUTIONS If you want to prove this from first principles, you do not start off with open covers of the sets Kα in the class K. Instead, you must start with an open cover of the intersection K of the class. You then construct open covers of the members of the class. You end up reproving two of the theorems I quoted (compact sets are closed, and closed subsets of compact sets are compact). Yuk. By the way, the intersection of K may very well be empty (contrary to many assertions which I read). I never even said the SETS of K were nonempty. But the empty set is a perfectly good compact set.  Problem 3. Let M be the set in R given by M = {0} ∪ {1/n : n = 1, 2, 3, . . . }. (a) Prove that M is compact (with respect to the usual metric). (b) Prove that M is classically compact, i.e. sequentially compact. Proof. I was expecting people to do this from first principles. I’ll give such a proof first. (a) Let (U) be an open cover of M . Since 0 ∈ M , therre must be U ∈ U with 0 ∈ U . In particular, there is ε > 0 such that Nε(0) ⊂ U . Hence there is N > 0 such that 1/N < ε. Hence 1/n ∈ U for all n ≥ N ; on the other hand, there are sets U1, . . . , UN−1 in U so that 1/i ∈ Ui, so that {U, U1, . . . , UN−1} constitutes a finitte subcover of U . (b) Let {xn} be a sequence in M . If the range of the sequence is infinite, then there must be a subsequence which converges to 0; otherwise, if the range is FINITE, there must be a subsequence which is constant—therefore one of the elements m of M—and that is a subsequence which converges to an element of M . But I was surprised to find that most students used theorems to prove these. That’s fine too: (a) M is a closed bounded subset of R, hence by the Heine-Borel theorem is compact. It’s closed because it has only one limit point, 0, which it contains; and it’s bounded because it’s a subset of [0, 1]. (b) Sequential compactness is equivalent to compactness, and M is compact; therefore it’s sequentially compact. Actually, this uses only the easy half of the equivalence: that compact implies sequentially compact. (No digression in Lebesgue number of covers, separability, etc.) One student even PROVED that implication. (That was overkill.)  Problem 4. Complete this sketch of the proof of the following theorem: If A and B are nonempty subsets of a metric space M such that A is compact and B is closed (in M), with A ∩B = ∅, then inf{d(a, b) : a ∈ A and b ∈ B} > 0. Proof: if the inf were 0, you could find sequences {an} in A and {bn} in B so that d(an, bn)→ 0. (Why?) A is sequentially compact (why?), hence there is a subsequence of {an} which converges to some a ∈ A (why?). The corresponding subsequence of {bn} also converges, and to the same point (why?). But then the limit belongs to both A and B (why?), which is a contradiction. Proof. • If the inf is 0, then for any ε > 0 there must exist a ∈ A, b ∈ B with d(x, y) < ε. Otherwise, d(x, y) ≥ ε i.e. ε is a lower bound for the set which is greater than the greatest lower bound, 0. • A is sequentially compact because it’s compact. • The definition of sequentially compact. • If d(an, bn)→ 0 and an → a, then d(bn, a) ≤ d(bn, an) + d(an, a). Given ε > 0 then for n sufficiently large, each of the last two terms is < ε/2, hence bn → a also. • Since A is closed, an ∈ A, and an → a, we must have a ∈ A. Similarly, B is closed, bn ∈ B, and bn → a, therefore a ∈ B. Hence a ∈ A ∩B, contradicting A ∩B = ∅.  Problem 5. Let A be the upper branch of the hyperbola y = 1/x in R2, and B be the x-axis.