Introductory Analysis I Fall 2014 Homework #5 Solutions

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1 Introductory Analysis I Fall 2014 Homework #5 Solutions 6. Let M be a metric space, let C D M. Now we can think of C as a subset of the metric space M or as a subspace of the metric space D (D being a subspace with the metric inherited from M). Using either of the definitions of connectedness given in class, for example: C is a connected subset of the metric space M if it is NOT possible to find open sets U, V in M such that C U V, C U = C V, and C U V =, show that C is a connected subset of D if and only if it is a connected subset of M. In particular, a subset of a metric space is connected if and only if it is connected as a metric space. Note: This exercise will not be graded. Solution. It is sufficient to prove that a subset of a metric space is connected if and only if it is connected as a metric space; equivalently, a subset of a metric space is disconnected if and only if it is disconnected as a metric space. Assume first C is a disconnected subset of the metric space M. There exist then U, V open in M, C U V, C U = C V, C U V =. If we now set U 1 = C U, V 1 = C V, then U 1, V 1 are open in C,C = U 1 V 1, U 1 = C U = C V = V 1, U 1 V 1 = C U V =. Thus C is disconnected as a metric space. Conversely, assume that C is disconnected as a metric space. There exist then U 1, V 1 open in C such that C = U 1 V 1, U 1 = = V 1, U 1 V 1 =. Because U 1, V 1 are open in C, there exist U, V open in M such that U 1 = C U, V 1 = C V. Then C U V, C U = U 1 = V 1 = C V, C U V = U 1 V 1 =. Thus C is a disconnected subset of M. 7. Let C be a subset of a metric space M. Show that C is connected if and only if the only subsets of C that are both open and closed in C are C and. Solution. Assume first C is connected. Let A C be both open and closed in C Then B = C\A is also open and closed in C. Now C = A B, A B = and, since both A, B are open (and closed), they will disconnect C except if one of them is empty. Thus either A = or B =, in which case A = C. Conversely, assume that C is a set such that the only open and closed subsets are C and. If C were disconnected, there would exist nonempty, disjoint open sets U, V such that C = U V. Then U = C\V is also closed in C. Moreover, = U C; the reason that U C is because V = C\U. Thus C cannot be disconnected.

2 2 8. Let M be a metric space. Let A M, assume A. We will define a relation in A as follows: If x, y A, we will write x y iff there exists a connected subset C of A such that x, y C. (a) Prove that is an equivalence relation. Solution. We have to see the relation is reflexive, symmetric and transitive. For reflexivity we must prove that x x for all x M. This is immediate since singleton sets are always connected and x, x {x}. Symmetry is also immediate, if x y, then there is C connected with x, y C; then y, x C and y]simx. For transitivity assume x y, y z. There are then connected subsets C, D of A such that x, y C, y, z D. Since y C D, we see that C D is connected, since x, z C D we get x z. (b) Since is an equivalence relation, it partitions A into pairwise disjoint equivalence classes. Show that each one of these equivalence classes is a connected subset of A. Solution. Let E be one of these equivalence classes and, for a contradiction, assume it is disconnected. Assume E U V, E U V = and E U E V. Let x E U and let y E V. Then x y and there is a connected subset C such that x, y C. Now C E; in fact, if z C, then z x, z y, thus z is in the same equivalence class containing x; namely, E. The contradiction should be clear; the sets U, V disconnect C. (c) Show that the equivalence classes are maximal connected subsets of A; if C is a connected subset of A and C E for some equivalence class E, then C E. Conversely, show that if E is a maximal connected subset of A, then E is one of these equivalence classes. Solution. Let E be one of these equivalence classes and let C be a connected subset of A. If C E then E C is connected and everything in E C is equivalent under to everything in E, so E C E; i.e., C E. Conversely, if E is a maximal connected subset of A, then (because E is connected) x y for all x, y E and there exists an equivalence class F such that F E. But then E = F since F is connected and E is maximal connected. These equivalence classes are called the connected components of A. Thus, a connected component of a subset A of a metric space is a connected subset E of A such that if C is a connected subset of A, and C E, then C A. 9. Determine the connected components if the following subsets A of the given metric space M. (a) M = R 2 and A = R 2 \{(0, 0)}. (b) M = R and A = {0} {1/n : n N}. (c) M = R m, A = R m \S m 1.

3 3 (d) M a discrete metric space, A = M. Solution. I ll be very brief here. (a) A = R 2 \{(0, 0)} is connected, thus there is only one connected component, namely A. (b) The case M = R and A = {0} {1/n : n N} is perhaps the most interesting example of these four. Every point of A determines a connected component; that is, the connected components are {{ 1n } }, n N { {0} }. It is fairly easy to see that all the sets {1/n}, n = 1, 2, 3... are maximal connected sets because they are all both open and closed in A. But {0} is not open in A, but any set containing 0 and some other point, say 1/n, cannot be connected since {1/n} would be a closed and open subset of this set that is neither empty nor equal to the whole set. (c) In the case of M = R m, A = R M \S m 1, the connected components are E = {x R m : x < 1} and F = {x R m : x > 1}. (d) M a discrete metric space, A = M. Here the family of connected components is { {x}, x A }. 10. Let M be a metric space and let A be a subset of M. Let C be a connected component of A. Prove: (a) C is closed in A. Solution. Since C is a maximal connected subset of A and the closure of a connected subset is connected, C must equal its closure in A, hence be closed in A. (b) If A has only a finite number of connected components, then C is open in A. Solution. Assume that A has only a finite number of connected components, say C 1,..., C n. They are all closed in A, thus the union of any finite number of them is also closed in A. Now C 1 is the complement in A of C 2 C m, hence the complement of a closed set, hence open. (c) Give an example of a metric space M having a connected component that is not open. Solution. See Exercise 9b. 11. Let U be an open subset of R n. Prove that all connected components of U are open. Solution. Assume C is a connected component of the open set U in R m. Let x C. Since x U and U is open, there is r > 0such that

4 4 B(x; r) = {y R m : y x < r} U. But B(x; r) is connected, B(x, r) C (it contains x), thus B(x; r) C. 12. Let U be an open subset of R. Using some of the material developed about connected sets, prove that U can be written as a countable union of pairwise disjoint intervals. Prove that this decomposition is essentially unique; given two decompositions, one is just a permutation of the other. Note: This exercise won t be graded. Solution. Let U be an open subset of R. As any set, U is a union of its connected components. Since U is open, these connected components are open by Exercise 11. As we saw in class, the only connected subsets of R are intervals, thus U is a union of pairwise disjoint open intervals. Every open interval contains rational numbers; selecting one rational number from every open interval defines a one-to-one map from the family of intervals to Q, proving that the cardinality of this family is less than or equal that of Q; i.e., the family is at most countable. This shows that U is a countable union of pairwise disjoint open intervals. For uniqueness, assume U = I j = H k, j J k K where J, K are either equal to N or to a section of N; i.e., to a set of the form {1, 2..., m} for some m N. The sets I j, H k are open intervals. Now let j J. Then I j = I j U = I j H k. k K If I j intersects 2 or more H k s, the H k s would disconnect I j, thus I j will intersect exactly on H k, say H k0. The equality above becomes I j = I j U = k K I j H k = I j H k0. Thus I j H k0. We thus have a 1-1 map ϕ from J to K where ϕ(j) is this k 0 ; that is ϕ(j) is the unique element of K such that I j H ϕ(j). Reversing the roles of J, K, one sees that ϕ : J K is a bijection, hence a permutation, and that I j = h ϕ(j). 13. (From this year s qualifier) Let A, B be subsets of R m. One defines Prove: A + B = {a + b : a A, b B}. (a) If one of A, B is open, then A + B is open.

5 5 Solution. Assume A is open. For b R m, the map x x + b is a homeomorphism, thus it maps open sets to open sets. It follows that A + b is open for all b R m ; in particular for all b B. Thus A + B = b B A + b, as a union of open sets, is open. (b) If A, B are compact, then A + B is compact. Solution. Here is the preferred proof of the people who got this problem right. Let (x n ) be a sequence in A + B. Then x n = a n + b n for a n A, b n B, n N. Since A is compact, the sequence (a n ) in A has a convergent subsequence (a nk ), converging to some a A. Now consider the sequence (b nk ) in B; since B is compact, it has a convergent subsequence (b nkj ), converging to some b B. Now (a nkj ) is a subsequence of (a nk ), thus also converges to a. Thus (a nkj + b nkj ) is a subsequence of (x n ) converging to the element a + b A + B. This proves A + B is sequentially compact, thus compact. But there is a more elegant solution. The set A B is a compact subset of R m ; the map f : R m R m R m defined by f(x, y) = x+y is continuous, thus A + B = f(a B) is compact. (c) If A, B are connected, then A + B is connected. Solution. Again I l provide two solutions. The second one is nicer. Here is the first one. We may assume A = B; otherwise it is trivial. For every a R m, a+b is connected because it is homeomorphic to B. Unfortunately all of the sets {a+b} a A have no points in common so we can t just declare their union, which is A+B, as being connected. More is needed. A trick, perhaps. The trick (used by at least one successful solver in our last qualifier) is to fix a 0 A (here is where we need A ). For every b B, the set A + b is also connected (it is homeomorphic to b) and its intersection with a 0 + B is not empty; in fact, (a 0, b) (A+b) (a 0 +B). Thus (A+b) (a 0 +B) is connected for all b B. Since (a 0 + B) (A + b) (a 0 + B) for all b B, all of these sets have a point (or more) in common (because B ), thus their union is connected. We claim that ((a 0 + B) (A + b)) = A + B. b B The inclusion ((a 0 + B) (A + b)) A + B is obvious; conversely, let x b B ((a 0 + B) (A + b)). b B For some b B, x (a 0 + B) (A + b), so either x = a 0 + b for some b B or x = a + b for some a A. In either case x A + B. And now for the elegant proof. The set A B is a connected subset

6 6 of R m ; the map f : R m R m R m defined by f(x, y) = x + y is continuous, thus A + B = f(a B) is connected. I will add (d) Prove: If one of A, B is closed and the other one compact, then A+B is closed. Give an example of two closed subsets A, B of R m such that A+B is not closed. (There are examples with m = 1; the easiest examples are with m = 2, I think) Solution. Here we may need an approach similar to the non-elegant proof that A+B is compact if both A, B are compact. Assume that A is compact and B is closed. To prove A + B closed we will show that A + B contains all of its limit points, so assume (x n ) is a sequence in A + B, converging to some point x R m. Then x n = a n + b n for a n A, b n B, n N, and we have to prove that x = a + b for some a A, b B. Since A is compact, the sequence (a n ) in A has a convergent subsequence (a nk ), converging to some a A. We have our a! Now consider the sequence (b nk ) in B; we have b nk = x nk a nk and since (x nk ) is a subsequence of (x n ) it still converges to x; thus (b nk ) converges to x a. Since B is closed x a = b B, proving x = a + b A + B. 14. Here is a nice one from the Berkeley Qualifier section. Prove that the interval [0, 1] cannot be written as a countable infinite union of pairwise disjoint closed intervals. Solution. I write these solutions before checking the answers in the handed in homework; if I find that one of the handed in solutions is nicer than mine I will add it here. Let us assume that [0, 1] = where each I n = [a n, b n ] for some a n, b n R, a n b n and I n I m = if n m. Visualizing this is quite hard and one should not make too many assumptions. For example that you can list the intervals by order of appearance. I will consider the set n=1 I n A = {a n : n N} {b n : n N} of endpoints and prove that it is perfect or contains a perfect subset. This will be a contradiction; since R is complete all perfect, non-empty subsets of R, are uncountable, but A is obviously countable. Because these endpoints could be all over the place, a sequence of endpoints is NOT necessarily a subsequence of (a n ) or (b n ); in fact, most likely it is not. This is something to keep in mind. I will use the generic letter c to denote endpoints. 1) A is closed. Assume (c k ) is a sequence of endpoints converging to some x R. Since [0, 1] is closed, x [0, 1] thus there exists I n = [a n, b n ]

7 7 such that x I n. If x = a n or x = b n, we are done. The alternative is a n < x < b n, but then we would have c k (a n, b n ) for k large enough; since c k is the endpoint of some interval I m this would imply I m I n (since the endpoint of I m is in the interior of I n and I n I m. The alternative can t hold; A is closed. 2) Every point of A, except possibly 0 or 1, is a cluster point of A. Let c A; we have to show that for every > 0 there is c A, 0 < c c <. Assume first c is a left endpoint of one of the intervals, say I m = [a m, b m ]. If a m = b m = 0, a possibility, then c is also a right endpoint of an interval and we deal with it next. If 0 = a m < b m, then 0 cannot be a cluster point of A. Assume thus 0 < a m = c. Let > 0 be given; there is a point x [0, 1], c < x < c. Now x has to be in some interval I j = [a j, b j ]; then b j A and c < b j < c. Assume next that c is a right endpoint of one of the intervals, say c = b m. The argument is identical (mutatis mutandis); if b m = 1, either 1 is not a cluster point, or c = a m = b m and we deal with c as we did with left endpoints > 1. Other wise for every > 0 there will be a j (c, c + ). The perfect subset of A is then either A itself (if both 0,1 are cluster points of A) or A with 0 or 1 removed, if either is not a cluster point. If a set A is closed, if x A, x / A, then A\{x} is still closed. A note on subsets of R Here is a simple subset of R. Select > 0. Order the rationals as a sequence, as one can because they are countable, say Q = {r 1, r 2, r 3,...}. Let U = n=1 ( r n 2 n+1, r n + ) 2 n+1. Let s look at a finite number of these intervals, say N and add all their lengths. We get (since the length of ( ) r n 2, r n+1 n + 2 is 2 n+1 2 = n+1 2 ) n N n=1 2 n = ( 1 2 N ) <. So as we keep adding intervals, we are really covering a length of at most ; much less probably since the intervals overlap. In fact, consider one of these intervals, say ( ) r n 2, r n+1 n + 2. This intervals contains an n+1 infinity of rational numbers, thus an infinity of rational numbers that have index > n in the given ordering, thus it contains an infinity of the intervals that make up U. But wait! There is more. Since U is open, we can write U = I m, m=1 where each I m = (a m, b m ) is an open interval and I m I k = if m k.

8 8 Try to visualize this. How are the endpoints a m, b m distributed? Notice that the open set U is a very small portion of R, yet it contains an interval about each rational number. The only relation of this with exercise 14 is to make you aware that infinity can have surprising properties. One should not assume hastily.