1 The topology of metric spaces

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1 Introductory Analysis I Fall 2014 Notes on Metric Spaces These notes are an alternative to the textbook, from and including Closed Sets and Open Sets (page 58) to and excluding Cantor Sets (page 95) 1 The topology of metric spaces Assume M is a metric space with distance function d. I will depart a bit from the book s notation; instead of using M r (p), I will denote by B(p; r) the open ball of center p M and radius r > 0; i.e, B(p; r) = {q M : d(q, p) < r}. The reason is that once we get to concrete metric spaces like R n, it may be silly to write (R n ) p (r). More importantly, I like my notation better. Incidentally a notation such as r > 0 means: r is a positive real number. The basic definition is Definition 1 A subset U of M is said to be open iff for every p U there is r > 0 such that B(p; r) U. In other words, the set U is open if and only if for every p U there is r > 0 such that if q U and d(q, p) < r, then q U. Examples: 1. Let M be an arbitrary metric space. Then the empty set and M are open. 2. Let M be an arbitrary metric space. Let p M and r > 0. Then B(p; r) is open. 3. Let M be a discrete metric space; i.e., d(p, q) = 1 whenever p q. Then every subset of M is open. 4. Let M = R m. The following sets are open: U 1 = {x = (x 1,..., x m ) R m : x 1 > 0}, U 2 = {x = (x 1,..., x m ) R m : x 1 > x 2 }, U 3 = {x = (x 1,..., x m ) R m : x x m < 1}, 5. Let M = R. A subset U of R is open if and only if for each x R there exist a, b R such that x (a, b) U. In particular, all open intervals are open. You should be able to prove that the sets declared open in these examples are indeed open.

2 1 THE TOPOLOGY OF METRIC SPACES 2 Theorem 1 Let M be a metric space and let T be the family of all open subsets of M. This family has the following properties: 1., M T. 2. If U α T for all α A; A some index set, then α A U α T. In words, T is closed under arbitrary unions. An equivalent way this property is sometimes formulated is: Assume U is a collection of open subsets of M (finite, infinite, countable or non-countable). Then {U : U U} is open. 3. If U, V T, then U V T. The intersection of two open sets is open. By induction, it follows that the intersection of any finite number of open sets is open. Proof. 1. That, M are open was already mentioned. 2. Let U α T for all α A; A some index set, let U = α A U α T. Let p U. Then there exists α A such that p U α. Since U α is open, there is r > 0 such that B(p; r) U α. Since U α U, B(p; r) U. 3. Let U, V be open. If p U V, then p U and p V. Since U is open, p U implies that there exists r 1 > 0 such that B(p; r 1 ) U. Similarly, because V is open, there exists r 2 > 0 such that B(p; r 2 ) V. Let r = min(r 1, r 2 ). Then r > 0 and clearly B(p; r) B(p; r 1 ) B(p; r 2 ) U V. Exercise 1 Let M = R. 1. Prove that in R singleton sets are NOT open. 2. For n N let U n = {x R : 1/n < x < 1/n}. Prove that U n is open for all n N, but n=1 U n is not open. This shows that, in general, the intersection of an infinite family of open sets may not be open. Note: If M is any set, a family of subsets of M satisfying properties 1, 2, and 3 of Theorem 1 is called a topology for M. Definition 2 Let M be a metric space. complement M\F is open. A subset F of M is closed iff its Examples:

3 1 THE TOPOLOGY OF METRIC SPACES 3 1. Let M be an arbitrary metric space. Then the empty set and M are closed. 2. Let M be an arbitrary metric space. Let p M and r 0. Then the closed ball of center p, radius r; that is, the set {q M : d(q, p) r} is closed. 3. A particular case of the previous result, the case r = 0, is that in every metric space singleton sets are closed. 4. Let M be a discrete metric space; i.e., d(p, q) = 1 whenever p q. Then every subset of M is closed. 5. Let M = R m. The following sets are closed: U 1 = {x = (x 1,..., x m ) R m : x 1 0}, U 2 = {x = (x 1,..., x m ) R m : x 1 x 2 }, U 3 = {x = (x 1,..., x m ) R m : x x m 11}, 6. Let M = R. All closed intervals are closed. As mentioned in the examples, in R open intervals are open and closed intervals are closed. Intervals of the form (a, b], [a, b) with a, b R, a < b are neither open nor closed. By taking complements, the following theorem is an immediate consequence of Theorem 1 Theorem 2 Let M be a metric space. 1., M T are closed. 2. If F α is a closed subset of M for all α A; A some index set, then α A U α is closed. In words, arbitrary intersections of closed subsets are closed. Equivalently: Assume F is a collection of closed subsets of M (finite, infinite, countable or non-countable). Then {F : F F} is closed. 3. If F, G, are closed then F G is closed. By induction, it follows that the Union of any finite number of closed sets is closed. Our next task will be to show that the basic metric properties can be described exclusively in terms of open sets. Another way of stating this is to say that one never has to mention the distance function again, except if one wants to. It will be convenient to first have another topological concept, that of neighborhood. Definition 3 Let M be a metric space. Let p M. A subset V of M is said to be a neighborhood of p in M (or just a neighborhood of p) iff there exists an open set U such that p U V.

4 1 THE TOPOLOGY OF METRIC SPACES 4 Exercise 2 V is a neighborhood of p if and only if there exists r > 0 such that B(p; r) V. We also have the following simple lemma Lemma 3 A subset U of a metric space is open if and only if it is a neighborhood of each of its points. Proof. Assume first U is open and let p U. Then p U U, showing U is a neighborhood of p. Conversely, assume U is a neighborhood of each of its points. Then for each p U, there is W p open, p W p U. Let W = p U W p. By property 2 of Theorem 1, W is open. Since W p U for all p U, we have W U. On the other hand, if p U, then p W p W, thus U W. It follows that U = W is open. Let us begin with sequences. We ll state it in the form of a theorem, but it is fairly immediate. Theorem 4 Let (p n ) be a sequence in the metric space M; let p M. The following statements are equivalent: 1. For each ϵ > 0 there exists N N such that n N implies d(p n, p) < ϵ. 2. For each ϵ > 0 there exists N R such that n > N implies d(p n, p) < ϵ. 3. For each open set U in M such that p U, there exists N R such that n > N implies p n U. 4. For each neighborhood V of p, there exists N R such that n > N implies p n V. 5. If V is a neighborhood of p, then {n N : p n / V } is finite. As mentioned before, most of the equivalences are obvious. In lieu of a proof I ll just make some comments about what could be the less obvious parts. Concerning the equivalence of 1 and 2, notice first that if the following statement For each there exists N N such that n N implies (fill the blanks any way you wish), is equivalent to For each there exists N N such that n > N implies In fact, if something holds for n N, it also holds for n > N. But the converse is also true; if we can find N N so that a property holds for n > N, we can also find N so it holds for n N ; namely N = N + 1. Maybe in words: if there is an element of N so the property holds when n is strictly larger than that element of N, then there is an element of N (namely the previous element plus 1) such that the property holds for n greater than or equal this element. Concerning now the equivalence of 1 and 3, it is a consequence of N R and the fact that if r R, there is n N, n > r. That 3 implies 1 is a consequence of the fact that B(p; r) is open; the converse is an easy consequence of the definition of open sets. A similar argument (sort of)

5 1 THE TOPOLOGY OF METRIC SPACES 5 shows the equivalence of 3 and 4. To see 5 is equivalent to any and all of the previous statements notice that if (p n ) is a sequence and B is a set, to say there exists N N such that p n B for all n N is equivalent to saying that the set of indices for which p n / B is at most N 1, so finite. Conversely, if the set of indices n for which p n / B is finite, let M be the largest index for which p n / B. Then p n B for all n M + 1. Convergence can be defined exclusively using the concept of open sets. Before we continue, it may be convenient to define a few more topological concepts. Definition 4 Let S be a subset of the metric space M. A point p M is a limit point of S iff there is a sequence (p n ) of points of S converging to p. It is an accumulation or cluster point of S iff there is a sequence (p n ) of points of S such that p n p for all n, converging to p Note: I use limit point as in our textbook. A lot of other textbooks (maybe all other textbooks?), if they define limit point at all they mean accumulation point. To see the difference between the two concepts notice first the obvious: A cluster point is a limit point. The second obvious thing is that every point of a set is a limit point of the set. If p S, let p n = p for all n N to get a sequence of points of S converging to p. Also fairly obvious is that if p is a limit point of S and p / S, then p has to be a cluster point of S. Can this happen? Yes, it can. Here are a few examples. A convenient notation is the following: Let S be a set. We denote by S the set of all cluster points of S; p S if and only if there exists (p n ), p n S, p n p for all n N such that lim n p n = p. Examples. 1. Let M be a discrete metric space. Then S = for all S M. 2. In R 2, consider the set S = {(x, y) R 2 } such that x < y. Claim S = {(x, y) R 2 : x y}. There are more elegant ways to prove this then how I am about to do it, but at this level brute force may be necessary. Let us see first that S {(x, y) R 2 : x y}. Let (x 0, y 0 ) S. there is then a sequence ((x n, y n )) in S converging to (x 0, y 0 ). I ll use a calculus result; if x n x 0, y n y 0 and x n < y n for all n, then x 0 y 0, to conclude (x 0, y 0 ) S. Notice that we did not have to assume (x n, y n ) (x 0, y 0 ) for all n N. Conversely, assume x 0 y 0. Then set x n = x 0 1 n, y n = y 0 for n N. Clearly (x n, y n ) S, (x n, y n ) (x 0, y 0 ) for all n N, and (x n, y n ) (x 0, y 0 ). It follows that (x 0, y 0 ) S. 3. In R consider the set S = (0, 1] {2}. It is then easy to see that S = [0, 1]. The point 2 is a limit point of S, but not an accumulation point of S. 4. In R consider the set { 1 n : n N}. Then S = {0}.

6 1 THE TOPOLOGY OF METRIC SPACES 6 5. In R consider the subset Z of all integers. Then Z =. The fact is that the useful concept is that of cluster point. Limit points are less interesting, except when they are cluster points. But... Here is a another quite simple result; let s call it a proposition for variety s sake. Proposition 5 Let M be a metric space, let p M and let S M. 1. The following statements are equivalent. (a) p is a limit point of S. (b) If V is a neighborhood of p, then V S. (Each neighborhood of p has non-empty intersection with S.) (c) If r > 0, then B(p; r) S 2. The following statements are equivalent. (a) p is a cluster point of S. (b) If V is a neighborhood of p, then V (S\{p}). (Each neighborhood of p contains a point of S different from p.) (c) If r > 0, then B(p; r) (S\{p}). (d) If V is a neighborhood of p, then V S is an infinite set. Proof. I ll only prove the equivalence of the statements about being a cluster point; the proof of the equivalence of the statements about being a limit point is similar, a bit easier because one doesn t have to worry about not being equal to p. And, of course, there is no fourth statement. a) b). Assume p is a cluster point of S and let V be a neighborhood of p. By definition of cluster point, there is (p n ), p n S, p n p for all n N, such that p n p as n. Since V is a neighborhood of p, there is N N such that p n V for n N. Then (for example) p N V (S\{p}). b) c). Since B(p; r) is a neighborhood of p for each r > 0, this is immediate. c) d). Assume that for each r > 0, B(p; r) (S\{p}) is not empty. Let V be a neighborhood of p. There is then r 0 > 0 such that B(p; r 0 ) V. Let r 1 = min(r 0, 1); then r 1 > 0, B(p; r 1 ) B(p r 0 ) V, and by assumption c), there is p 1 B(p; r 1 ) S such that p 1 p. Then d(p 1, p) > 0 (notice that this is the first time in quite a while that we actually mention the distance function), let r 2 = min(d(p 1, p), 1/2). Then r 2 > 0 and there exists p 2 B(p; r 2 ) S such that p 2 p. In addition, p 2 p 1 since d(p 2, p) < d(p 1, p). Assume now that for some n 1 we have found points p 1,..., p n V S such that p i p j if i j and such that 0 < d(p i, p) < 1/i for i = 1, 2,..., n. We have done this already for n = 2. Let 1 r n+1 = min{d(p 1, p),..., d(p n, p), n + 1 }. Then r n=1 > 0, hence there exists p n+1 B(p; r n+1 ) S such that p n+1 p.

7 1 THE TOPOLOGY OF METRIC SPACES 7 What we have done here is to show inductively the existence of a sequence (p n ) such that the set {p n : n N} is an infinite set contained in V S. So we are done. As a bonus we also have lim n p n = p and p n p for all n. That is, as a bonus we proved in passing: If p is a cluster point of S, there exists a sequence (p n ) of points of S such that p n p m if n m, p n p for all n, (p n ) converging to p. d) a). Let n N.The set B(p; 1/n) is a neighborhood of p, hence B(p; 1/n) is an infinite set, hence it must contain points other than p. Lots of them, an infinite number of them. Select one of these points, call it p n. Thus p n B(p; 1/n) S, p n p. The sequence (p n ) clearly converges to p, p n p for all n N, thus p is a cluster point of S. We can now give a convenient characterization of closed sets in terms of cluster or limit points. It is again a TFAE theorem. Theorem 6 Let S be a subset of the metric space M. The following statements are equivalent. 1. S is closed. 2. S contains all of its limit points. 3. S S (S contains all of its cluster points). Proof. 1) 2 Assume S is closed and let p M\S. Since M\S is open, a sequence of points converging to p must have all but a finite number of terms in M\S, thus it cannot be a sequence of points of S. It follows that p is not a limit point of S. Therefore, if p is a limit point of S, p S. 2) 3) Since cluster points are limit points, this is obvious. 3) 1) Assume S S. Let U = M\S, we need to show U is open. Let p U. Then p / S, hence p / S ; p is not a cluster point of S. By Theorem 5, there exists a neighborhood V of p such that V (S\{p}) =. Since p / S, this is equivalent to V S =. Thus V U. It is (or should be) clear that supersets of neighborhoods are neighborhoods, thus U is a neighborhood of p. Since p is an arbitrary point of p, this proves U open. Definition 5 Let S be a subset of the metric space S. The closure of S, denoted by S, is the intersection of all closed subsets of M containing S; in symbols, S = {F : F is a closed subset of M and S F } Some comments are in order. First of all, because M itself is closed and contains (of course!0 all its subsets, the family {F : F is a closed subset of M and S F } is never empty. The definition of S makes sense. Second, since S is the intersection of a family of sets all of which contain S, we have S S. Third, since the intersection of closed sets is closed, S is a closed set. It is also easy to see

8 1 THE TOPOLOGY OF METRIC SPACES 8 that S = S if and only if S is closed. In fact, if S = S, then S is closed because, as mentioned, S is closed. Conversely, if S is closed, then S is a member of the family of closed sets containing S, hence S S. The converse inclusion always holds, thus S = S. One problem with this definition, however, is that it is fairly useless to determine closures because to find S we need to know the family of closed sets containing S and, unfortunately, bars is one of these sets. To find S we need to know S. Fortunately, there are intrinsic definitions. We state them as a theorem. Theorem 7 Let S be a subset of the metric space M. Then S = S S = {p M : p is a limit point of S}. Proof. We see first that S S is closed. This is equivalent to proving that M\(S S ) is open. If p M\(S S ), then p / S, p / S and we are exactly in the situation of the proof of 3) 1) of Theorem 6. The same proof shows M\(S S ) is open. Since S S is closed, we have S S S. Conversely, let p S S and let F be a closed set containing S. Since it is obvious that S F and F F since F is closed, we see that S S F for all closed sets F S. Thus S S is in the intersection of all these closed sets, hence S S S. Concerning the characterization in terms of limit points, notice that {p M : pis a limit point of S} = S S. Since cluster points are limit points, the inclusion is obvious. But so is the converse inclusion since a limit point not in S is a cluster point of S. As a simple but useful corollary we get: Corollary 8 Let S be a subset of the metric space M. Then p S if and only if V S for all neighborhoods V of p. Equivalently, p S if and only if V B(p; r) for all r > 0. It might be a moment to do some exercises. It is important to do simple exercises like those that follow to get a feeling for all these objects. Exercise 3 Let M = R m and let r > 0, x R m. Prove that B(x; r) = {y R m : y x r}. Exercise 4 Let M = R m. Prove that Q m = R m. A subset of a metric space whose closure equals the whole space is said to be dense in the space. Thus Q m is dense in R m. In particular Q is dense in R. Exercise 5 In R 2, let S = {(x, sin(1/x)) : x R, x 0}. Prove that S = S {(0, y) : 1 y 1}. Exercise 6 It seems reasonable to assume that the result of Exercise 3 is valid in every metric space. But that would be wrong. Assume M is a metric space with the discrete metric, and assume that M has at least two points. Let p M. Determine B(p; 1) and show that B(p; 1) {q M : d(q, p) 1}.

9 1 THE TOPOLOGY OF METRIC SPACES 9 Exercise 7 Assume (p n ) is a sequence in a metric space and let S be the range of the sequence, i.e., S = {o n : n N}. Prove: If q is a cluster point of S, then there is a subsequence of )p n ) converging to q. Is the converse true? That is, if q is the limit of a subsequence of (p n ), is p a cluster point of S? Here are a few other properties of the closure: Proposition 9 Let M be a metric space. 1. If S T M, then S T. 2. If S is a subset of T, then S = S. 3. If S, T are subsets of M, then S T = S T. Proof. The first statement is obvious, any definition one wants to use. The second statement is also obvious; the closure of a closed set is the set, and S is closed. For the third statement, notice first that Ā B is a closed set containing A B, thus it contains A B. Conversely, A B is a closed set containing A B, since A A B, A A B, hence Ā A B. Similarly, B A B, thus Ā B A B. As a corollary we get: If S 1,..., S m M (a metric space), then m i=1 S i = m i=1 S i. Exercise 8 Provide an example of a metric space M and a countable family {S n } n N of subsets of M such that n N U n n N S n. Here are two final topological concepts (for now). Definition 6 Let S be a subset of the metric space M. The interior of S is the set S of all points p of S such that there exists r > 0 with B(p; r) S. That is: p S if and only if there exists r > 0, B(p, r) S. Equivalently: p S if and only if S is a neighborhood of p. It is also easy to see that S is the largest open subset of S; that is, if V is open and V S, then V S. Also equivalently S = {V : V is open and V S}. If we compare this last equivalent way of defining the interior and compare with the definition of closure, we get at once thanks to our friend DeMorgan, S = M\(M\S). We have talked of points very much inside a set, and points outside of it. We need also to talk of points that are halfway between being in and out.

10 2 SUBSPACES 10 Definition 7 Let M be a metric space, and S M. The boundary of S is the set S = S (M\M). Several equivalent definition are: S = S\S 0 ; S is the set of all points p such that each neighborhood of p has non-empty intersection with S and M\S. Equivalently, p S if and only if for each r > 0, B(p; r) S = B(p; r) (M\S). 2 Subspaces As mentioned in class, if M is a metric space, if N M, then N becomes a metric space by simply restricting the distance function of M to points of N. We then say N is a (metric) subspace of M. All the topological concepts make sense in N, of course, and we should know how they relate to the similar concept in M. So suppose S is a subset of N and it is closed in the metric space N. Is it closed in M? Conversely? Since we now have two spaces, I will write B M (p; r) for the open ball of center p, radius r > 0 in M, and B N (p; r) for the related object in N (which only makes sense if p N). A first trivial observation is that if p N, r > 0, then B N (p, r) = B M (p, r) N. This has as consequence that a subset U of N is open in N if and only if U = W N for some W open in M. In fact, assume first W is open in M. Let p W N. Because p W, there is r > 0 such that B M (p; r) W, implying that B N (p; r) = B M (p; r) N W N. This proves W N is open in N. Conversely, let U be open in N. For each p U there is r p > 0 such that B N (p; r p ) U. The set W = p U B M (p; r p ) is open in M and W N = U. A couple of immediate consequences are that if U N and U is open in M, then it is open in N. If N is open in M, then being open in N is equivalent to being open in M for subsets of N. One gets similar/exoected results for closed sets. Suppose F N. then F is closed in N (by definition) if and only if N\F is open in N, which happens if and only if there is W open in M such that N\F = W N. Observing that N\(W N) = (M\W ) N, we see that if F is closed in N, then F = G N, G closed in M. If F = G N, with G closed in M, then N\F = N\(G N), showing that N\F is open in N. Thus F is closed in N if and only if there is G closed in M such that F = G N. In particular, if N is closed in M then a subset F of N is closed in N if and only if it is closed in M. Exercise 9 Let M be a metric space and let N be a subspace of M. Let S N. closure closure Prove: The interior of S in N is the intersection of the interior boundary boundary

11 3 EQUIVALENCE OF METRICS 11 of S in M intersection N. 3 Equivalence of Metrics We saw in a homework that two distance functions for the same set are equivalent iff convergence for one is equivalent to convergence for the other one. A nicer definition is: Definition 8 Assume d 1, d 2 are distance functions for a set M. They are equivalent, and we write d 1 d 2, iff they generate the same tropology. In other words, a set is open for d 1 if and only if it is open for d 2. Let us write for p M, r > 0, B 1 (p; r) = {q M : d 1 (p, q) < r}, B 2 (p; r) = {q M : d 2 (p, q) < r}. Assume d 1 d 2. Let r > 0, p M. Since B 1 (p; r) is open with respect to d 1 and p B 1 (p; r), the fact that it is also open with respect to d 2 implies that there exists r > 0 such that B 2 (p, r ) B 1 (p, r). The same must hold true if we switch the roles of d 1, d 2. We see thus that if d 1 d 2, for every r > 0, p M, there exists r > 0 such that B 2 (p, r ) B 1 (p, r), and for every r > 0 there is r > 0 such that B 1 (p, r ) B 2 (p, r ). Conversely, suppose this property holds. If U is open with respect to d 1, let p U. Then there is r > 0 such that B 1 (p, r) U; by the property there exists r > 0 such that B 2 (p, r ) B 1 (p, r), hence B 2 (p, r ) U. Since p was arbitrary in U, we see that U is open with respect to d 2. Since the property is symmetric in d 1, d 2 ; it also follows that being open with respect to d 2 is the same as being open with respect to d 1. Concerning sequences, if two metrics define the same open sets, since convergence can be defined exclusively in terms of open sets, convergence will be the same for both metrics. On the other hand, if two metrics have the same notion of convergence then given any subset of the space, the notion of limit point will be the same for both metrics. It follows that the concept of being a closed set is the same for both metrics, hence also the notion of being the complement of a closed set; i.e., being an open set. We thus have Theorem 10 Let M be a set and let d 1, d 2 be distance functions in M. The following statements are equivalent. 1. A subset of M is open with respect to d 1 if and only if it is open with respect to d For each p M, r > 0 there exists r > 0 such that B 2 (p; r ) B 1 (p; r), and for each p M, r > 0 there exists r > 0 such that B 1 (p; r ) B 2 (p; r). 3. A sequence in M converges with resepct to d 1 if and only if it converges with resepct to d 2.

12 4 AN EXCURSION INTO SET THEORY 12 4 An Excursion into Set Theory Let M, N be sets (not necessarily metric spaces) and let f : M N. If A M, the image of A under f is the set f(a) = {y B : x M, y = f(x)}; briefly, f(a) = {f(x) : x A}. The statement y f(a) is equivalent to y = f(x) for some x A. In particular, f(m) is the range of f. If B is a subset of N, one can define the pre-image of B under f. The author of our textbook denotes it by f pre (B) and gives some excellent reasons why this notation is preferable to the usual one, which is f 1 (B), except that it isn t really so much the usual notation as the universal notation, used everywhere I know of, except our textbook. So I will stick with the conventional notation. The pre-image of B under f is the set defined by f 1 (B) = {x M : f(x) B}. Notice that x f 1 (B) does NOT imply x = f 1 (y) for some y B; f 1 as a function on B might not even be defined. (Actually, if f 1 is defined, it does mean that, but that requires a proof.) x f 1 (B) means precisely the same as f(x) B. Here are some examples. Examples. 1. Let f : R R be defined by f(x) = 1 for all x R. Then f(a) = {1} if A R and A ; { f 1, if 1 / B, (B) = R, if 1 B. 2. Let f : R R be given by f(x) = sin x for all x R. Then f(r) = [ 1, 1]; f 1 {0} = {kπ : k Z}. 3. If A M and A, then f(a). In fact, x A implies f(x) f(a). However, if B N one can have f 1 (B) = even if B is not empty. f 1 (B) = if and only if B f(m) =. Let us prove some basic facts about these operations. A bit of common sense helps. Proposition 11 Let f : M N. 1. If B λ N for λ Λ (Λ an index set; finite or infinite), then ( ) f 1 B λ = Λf 1 (B λ ) (1) λ Λ λ ( ) f 1 B λ = Λf 1 (B λ ) (2) λ Λ λ

13 4 AN EXCURSION INTO SET THEORY If B, C are subsets of N, then f 1 (C\B) = f 1 (C)\f 1 (B). (3) 3. If B N, then f 1 (N\B) = M\f 1 (B). (4) Proof. These properties prove themselves; if one simply does the obvious (and uses the definitions as provided). To prove (1) notice that x f ( 1 λ Λ B ) λ if and only if f(x) λ Λ B λ, which happens if and only if there is λ Λ such that f(x) B λ ; this is equivalent to saying that there is λ Λ such that x f 1 (B λ ), which occurs if and only if x λ Λf 1 (B λ ). The proof of (2) is almost identical; replace there is or there exists by for all (or for each ): x f ( 1 λ Λ B ) λ if and only if f(x) λ Λ B λ, which happens if and only if for each λ Λ it holds that f(x) B λ ; this is equivalent to saying that for each λ Λ, x f 1 (B λ ), which occurs if and only if x λ Λf 1 (B λ ). The proof of (3) is just as immediate: x f 1 (C\B) f(x) C\B f(x) C and f(x) / B x f 1 (C) and x / f 1 (B) x f 1 (C)\f 1 (B). Property (14) is a consequence of (3); take C = N and notice that f 1 (N) = M. Are similar properties true for images? That is, if A λ M for λ Λ, is it true that f applied to the union of the A λ s is the union of the f(a λ ); in other words, is the image of a union the union of the images? Is the image of an intersection the intersection of the images? The answer is yes for the union, no for the intersection. Proofs are easy. Concerning intersections, suppose A, B M. If f is not one-to-one, there could be a A\B, b B\A, such that f(a) = f(b). Then y = f(a) is in f(a) f(b), but there is no reason to assume that it also is in f(a B). In general, f(a B) f(a) f(b), but f(a B) f(a) f(b) in general. Equality holds if f is one-to-one. Here are a few more properties, as an exercise. If you understood the definitions, you should have no trouble with this exercise. If you have any trouble, review the definitions carefully. Exercise 10 Assume M, N are sets and f : M N. 1. Let A M. Prove: A f 1 (f(a)). Show, by a counterexample, that in general equality does not hold; that is, provide an example in which A f 1 (f(a)). After doing this, give a condition on f such that equality will hold whenever f satisfies that condition. 2. Let B N. Prove: f ( f 1 (B) ) B. Show, by a counterexample, that in general equality does not hold, but give a condition on f such that equality will hold whenever f satisfies that condition.

14 5 CONTINUITY 14 Finally, assume f : M N is one-to-one and onto so that f 1 : N M exists. Then it would seem we have two interpretations of f 1 (B) for B N: As the pre-image of B under f or as the direct image of B under f 1. Fortunately both interpretations are the same in this case. Suppose x f 1 (B). If we interpret f 1 (B) as the pre-image of B,this means f(x) B. If we interpret f 1 (B) as the image of B under f 1, this means there exists y B such that x = f 1 (y), hence f(x) = y B. And, of curse, the only possible value of y is f(x) in this case. 5 Continuity In this section we will assume M, N are metric spaces. I will use the symbol d for the distance function in M as well as for the distance function in N; this should not cause any confusion; one should know from the context which one is being referred to. Anyway, after a while, there probably won t be any explicit mention of distance functions. I will however denote the open ball of radius r center q by B M (q, r) if we are in M, by B N (q, r) in N. Definition 9 Let f : M N and let p M. We say f is continuous at p iff for every ϵ > 0 there is δ > 0 such that d(f(q), f(p)) < ϵ whenever q M and d(q, p) < δ. Definition 10 Let f : M N. We say f is continuous (or, continuous on M)iff f is continuous at each p M. We already went over a lot of this in class. A characterization of continuity in terms of sequences is given by the following result. Theorem 12 Let f : M N and let p M. Then f is continuous at p if and only if the following property holds: Whenever (p n ) is a sequence in M converging to p, the sequence (f(p n )) is a sequence in N converging to f(p). Proof. Assume first f is continuous at p. Let (p n ) be a sequence in M converging to p. To prove (f(p n )) converges to f(p), let ϵ > 0 be given. By the continuity of f at p, there exists δ > 0 such that d(f(q), f(p)) < ϵ whenever q M and d(q, p) < δ. Because δ > 0 and lim n p n = p, there is N N such that n N implies d(p n, p) < δ, hence d(f(p n ), f(p)) < ϵ. Conversely, assume the property holds. We proceed by contradiction. Assume f is not continuous at p. There exists then ϵ > 0 for which no δ > 0 works, thus for every δ > 0 there exists p δ M, d(p δ, p) < δ and f(p δ ), f(p)) ϵ. Then (p 1/n ) n N is a sequence in M converging to p; since d(f(p 1/n, f(p)) ϵ for all n; f(p 1/n ) does not converge to f(p), contradicting our hypothesis. Moving a bit away from explicit mention of the distance functions, we have the following simple lemma. I prove it because perhaps we all need to become a bit more familiar with this way of proving things. Lemma 13 Let f : M N, let p M. The following are equivalent.

15 5 CONTINUITY f is continuous at p. 2. For each neighborhood W of f(p), there exists a neighborhood V of p such that f(v ) W. 3. If W is a neighborhood of f(p), then f 1 (W ) is a neighborhood of p. Proof. 1) 2) Assume f is continuous at p and assume W is a neighborhood of f(p). There is then ϵ > 0 such that B N (f(p); ϵ) W. By continuity, there is δ > 0 such that q M, d(q, p) < δ implies d(f(q), f(p)) < ϵ, thus f(q) B N (f(p); ϵ) W. Take V = B M (p; δ); then W is a neighborhood of p in M and if z f(v ), then z = f(q), some q V = B M (p, δ), thus z = f(q), d(q, p) < δ, hence d(z, f(p)) < ϵ and z W. That is, f(v ) W. 2) 3) Assume 2) Let W be a neighborhood of f(p) in N. By 2), there exists a neighborhood V of p, with f(v ) W. Applying f 1 and using (10), part 1, V f 1 (f(v )) f 1 (W ). Since supersets of neighborhoods are neighborhoods, we are done. 3) 1) Assume 3) Let ϵ > 0 be given. Then B N (f(p); ϵ) is a neighborhood of f(p), hence f 1 (B N (f(p); ϵ) is a neighborhood of p, hence contains an open set containing p, which in turn contains an open ball centered at p. That is, there is δ > 0 such that B M (p; δ) f 1 (B N (f(p); ϵ). It follows that if q M and d(q, p) < δ, then d(f(q), f(p)) < ϵ. An important but simple result is the following. Theorem 14 Let M, N, P be metric spaces; let f : M N, g : N P. Let p M. If f is continuous at p and g is continuous at f(p), then the composition g f is continuous at p Proof. Let W be a neighborhood of g f(p) = g(f(p)) in P. Since g is continuous, g 1 (W ) is a neighborhood of f(p), thus (because f is continuous), f 1 ( g 1 (W ) ) is a neighborhood of p. But f 1 ( g 1 (W ) ) = (g f) 1 (W ) and, since W was an arbitrary neighborhood of g f(p), we proved that the pre-image of every neighborhood of g f(p) under g f is a neighborhood of p. Continuity of g f at p follows. An obvious corollary is Corollary 15 Let M, N, P be metric spaces; let f : M N, g : N P. If both f and g are contibuous, then g f is continuous. Here are a few exercises, to keep practising. Exercise 11 Let M, N be metric spaces and let p M be an isolated point of M; that is p M but p / M. Equivalently, there is r > 0 such that B M (p, r) = {p}. Prove: If f : M N, then f is continuous at p.

16 6 ALGEBRA OF CONTINUOUS FUNCTIONS 16 Exercise 12 Let M, N be metric spaces, f : M N, A M. The restriction of f to A is the function f A : A N defined by f A (p) = f(p) for p A. It makes sense to ask about the continuity of f A since A is a metric space with the metric of M. Prove: If p A and f is continuous at p,then f A is continuous at p. Show (by a counterexample) that the converse is false; one can have p A M, f A continuous at p, but f is not continuous at p. Exercise 13 Let M, N be metric spaces, f : M N, A M. Let p A. Assume there exists U open in M such that p U A. Prove: If f A is continuous at p, then f is continuous at p. Exercise 14 Let M, N be metric spaces, f : M N. Let A, B M be such that M = A B and assume p A B. Prove: f is continuous at p if and only if both f A, f B are continuous at p. We are already characterizing continuous functions in terms of topological concepts. For functions continuous at all points we have some nicer characterizations. The following theorem will be used with some frequency. Theorem 16 Let M, N be metric spaces and let f : M N. The following statements are equivalent. 1. f is continuous. 2. f 1 (U) is open in M whenever U is open in N; in words: The inverse image of open sets is open. 3. f 1 (F ) is closed in M whenever F is closed in N; in words: The inverse image of closed sets is closed. Proof. 1) 2) Assume f is continuous and let U be open in N. Let p f 1 (U). Then f(p) U. Since U is open, U is a neighborhood of f(p) thus, by Lemma 13, f 1 (U) is a neighborhood of p. Since p U was arbitrary, itfollows that f 1 (U) is a neighborhood of each on of its points hence, by Lemma 3, it is open. 2) 3) Assume the inverse image under f of all open sets is open and assume F is a closed subset of N. Then F = N\U where U is open; thus f 1 (F ) = M\f 1 (U), proving F is closed since f 1 (U) is open. 3) 2) TO prove that 3) implies 2) all one needs todo is to exchange the words open and closed in the proof that 2) implies 3). 2) 1) Assume the inverse image of all open sets is open. Let p M. Let W be a neighborhood of f(p). There is then U open in N, f(p) U W. Then p f 1 (U) f 1 (W ). Since f 1 (U) is open, f 1 (W ) is a neighborhood of p. By Lemma 13, f is continuous at p. 6 Algebra of continuous functions Assume f : M R m. Then f(p) is an m-vector for each p M and we can define functions f 1, f 2,..., f m : M R by f k (p) is the k=th component of p. In other words, f(p) = (f 1 (p),..., f m (p)). We write f = (f 1,..., f m ). We have

17 6 ALGEBRA OF CONTINUOUS FUNCTIONS 17 Proposition 17 Let M be a metric space and let f = (f 1,..., f m ) : M R m. Then f is continuous at p M if and only if each f k is continuous at p; k = 1,..., m. It follows that f is continuous if and only if each f k is continuous. Proof. For once, the proof might be easier using the distance functions. Assume first f is continuous at p. Let k {1,..., m}; we want to prove f k is continuous at p. Let ϵ > 0 be given. Because f is continuous, there exists δ > 0 such that if q M and d(q, p) < δ, then f(q) f(p) < ϵ. Since x k x = n j=1 x j 2 for all x = (x 1,..., x m ) R m, q M, d(q, p) < δ implies f k (q) f k (p) f(q) f(p) < ϵ. Conversely, assume each f k is continuous at p. To prove f is continuous at p, let ϵ > 0 be given. Since then ϵ/ n > 0, there exist δ 1,..., δ m > 0 such that q M; d(q, p) < δ k implies f k (q) f k )(p) < ϵ/ m. Let δ = min 1 k m δ k. Then δ > 0 and d(q, p) < δ implies d(q, p) < δ k for k = 1,..., n, hence f(q) f(p) = m (f k (q) f k (p)) 2 < m ϵ 2 m = ϵ. k=1 k=1 Generally speaking, a function from a metric space (or any set, actually) to R m is just a convenient way to deal with m functions at once. Once again generally speaking (there could be exceptions), if you have f : M R m (M not necessarily a metric space, just a set) you can write f = (f 1,..., f m ) where f k : M R for k = 1,..., m. Chances are that given a property such an f can have it is either defined by, or a theorem states it, that f : M R m has that property if and only if each f k : M R has the property. We just saw it is so for continuity. I will now define a number of functions. The symbols I use will probably be reused later on with different interpretations. What I mean is that these symbols are just one-use symbols. Here are the functions: µ : R m R m R m, defined by µ(x, y) = x + y if x, y R m. ν : R R m R m, defined by µ(c, x) = cx if c R and x R m. λ : R m R m to R, defined by λ(x, y) = x y = m i=1 x iy i if x = (x 1,..., x m ), y = (y 1,..., y m ) R m. abs : R m R, defined by abs(x) = x x2 m if x = (x 1,..., x m ) R m. All of these functions are continuous. For a proof we need to know a bit about product of metric spaces. If (M, d 1 ), (N, d 2 ) are metric spaces, then M N can be made into a metric space in several equivalent ways. One can define the metric d for M N by d((p, q), (p, q )) = d(p, p ) + d(q, q ). Or one could define d((p, q), (p, q )) = d 1 (p, p ) 2 + d 2 (q, q ) 2. Or d((p, q), (p, q )) =

18 6 ALGEBRA OF CONTINUOUS FUNCTIONS 18 max(d 1 (p, p ), d 2 (q, q )). We should all have enough experience by now to realize a) All three of these are metrics, and b)they are equivalent. So use the one you like most; the main thing is that if we denote the ball of center p radius r > 0 in M by B M (p; r), the one in N by B N (q, r). then U is open in M N if and only if for every p, q) U, there exist r 1 > 0, r 2 > 0 such that B M (p, r 1 ) B N (q; r 2 ) U. One can always assume r 1 = r 2, if necessary. Or if one wants to. Notice that B M (p, r) B N (q; r) is not the same as B M N ((p, q), r) if we use the first two metrics we defined above for M N, but is equal to B M N ((p, q), r) for the third of these metrics. When M = R m and N = R n, then one sees that if we identify R m R n with R m+n by ((x 1,..., x m ), (y 1,..., y n )) (x 1,..., x m, y 1,..., y n ), then they also get identified as metric spaces. In fact, if we use as product metric the second definition, namely d((p, q), (p, q )) = d 1 (p, p ) 2 + d 2 (q, q ) 2, and d 1, d 2 are the Euclidean metrics for R m, R n, respectively, then d is the Euclidean metric for R n+m. One more thing. The standard way to prove by the definition f : M N P is continuous, where (M, d 1 ), (N, d 2 ), and (P, d) are metric spaces is as fllows. One begins with Let (p 0, q 0 ) M N. Let ϵ > 0 be given. (or similar). Now one somehow has to find δ > 0; so one will have a statement such as Let δ =. The right hand side of this equality will most likely be a function of ϵ, p 0, q 0. If it isn t 100% obvious that this d is positive, one has to verify that it is; one should be able to say δ > 0. Next one assumes p M, q N satisfy d 1 (p, p 0 ) < δ and d 2 (q, q 0 ) < δ. One hasto show that this implies d(f(p, q), f(p 0, q 0 )) < ϵ. With all this said, the continuity of the functions mentioned above becomes quite easy to prove. I will use the Euclidean distance(s) throughout. So x is the absolute value of x if x R; x = m i=1 x2 i if x R m, and ( x, y) = m i=1 x2 i + m i=1 y2 i if (x, y) R R m = R 2m. Continuity of µ Let (x 0, y 0 ) R m R m. We have to prove µ is continuous at (x 0, y 0 ). Let ϵ > 0. Let δ = ϵ/2. Then δ > 0 and x x 0 < δ, y y 0 < δ implies µ(x, y) µ(x 0, y 0 ) = (x+y) (x 0 +y 0 ) = (x x 0 )+(y y 0 ) x x 0 + y y 0 < 2δ = ϵ. Continuity follows. Continuity of ν Let (c 0, x 0 ) R R m. We have to prove ν is continuous at (c 0, x 0 ). Let ϵ > 0. Let ( ) ϵ δ = min 1, 2( c 0 + 1), ϵ. 2( x 0 + 1) Then δ > 0. If c c 0 < δ, x x 0 < δ then because δ 1, we have c c 0 < 1, x x 0 < 1, hence c = c c 0 +c 0 c c 0 + c 0 < 1+ c 0, x = x x 0 +x 0 x x 0 + x 0 < 1+ x 0.

19 6 ALGEBRA OF CONTINUOUS FUNCTIONS 19 Thus cx c 0 x 0 = c(x x 0 ) + (c c 0 )x 0 c x x 0 + c c 0 x 0 (1 + c 0 ) x x 0 + (1 + x 0 ) x x 0 < (1 + c 0 )δ + ( x 0 )δ < ϵ 2 + ϵ 2 = ϵ. Continuity of λ Let (x 0, y 0 ) R m R m. We have to prove λ is continuous at (x 0, y 0 ). The proof uses the same idea as the proof of the continuity of ν, plus the Cauchy-Schwarz inequality Let ϵ > 0. Let ( δ = min 1, x y x y. ϵ 2( x 0 + 1), ) ϵ. 2( y 0 + 1) Then x x 0 < δ implies x = x x 0 + x 0 < 1 + x 0. If now x x 0 < δ, y y 0 < δ, then λ(x, y) λ(x 0, y 0 ) = x y x 0 y 0 = x (y y 0 ) + (x x 0 ) y 0 x x x 0 + x x 0 y 0 < x δ + y 0 δ < ( x 0 + 1)δ + y 0 δ ϵ 2 + ϵ 2 = ϵ. Notice that if m = 1, then λ is just the usual product. Continuity of abs Absolutely trivial. As a corollary we obtain the following theorem. Theorem 18 Let M be a metric space. 1. Let f, g : M R m be continuous. Then the functions f + g, f g from M to R m are continuous. (f ± g : M R m is defined by f ± g(x) = f(x) ± g(x).) 2. Let f, g : M R m be continuous. Then f g : R m ]tor is continuous, where f g(x) = f(x) g(x). 3. Let f : M R m be continuous. Then f : R m ]tor is continuous, where f (x) = f(x). 4. Let f : M R m be continuous and let ϕ : R m R. Then ϕf : R m R m is continuous, where ϕf(x) = ϕ(x)f(x). Proof. If f, g : M R m are continuous, the map p (f(p), g(p) : M R m R m is continuous (Proposition 17). Now f + g is just this map followed by µ. If we follow it by λ, we get that f g is continuous. If f : M R m, ϕ : M R are continuous, then p (ϕ(p)mf(p)) : M R R m is continuous; following by ν we get the continuity of ϕf. Specializing to ϕ(p) = 1 for all p M we get g is continuous if g is continuous, hence we get f g = f + ( g) is continuous. Following a continuous f with abs proves f continuous.

20 7 COMPLETENESS AND SOME REALITY CHECKS 20 7 Completeness and Some Reality Checks An important property that a metric space may or may not have is completeness. To discuss what is meant by completeness, we need to introduce the notion of a Cauchy sequence Definition 11 Lat M be a metric space. A sequence (p n ) in M is said to be ]extbfcauchy sequence iff for each ϵ > 0 there exists N N such that d(p n, p k ) < ϵ whenever n, k N. In a way, the following theorem presents the main example of a Cauchy sequence. Theorem 19 Assume (p n ) is a sequence of points in the metric space M. If (p n ) converges, it is Cauchy. Proof. Assume (p n ) converges and let p be the limit. Let ϵ > 0 be given. Then ϵ/2 > 0 and there exists N such that n N implies d)p n, p) < ϵ/2. If now n, k N, then d(p n, p k ) d(p n, p) + d(p, p k ) < ϵ 2 + ϵ 2 = ϵ. Is the converse true? The answer is: sometimes. Not in general. For example, consider the following sequence (x n ) of rational numbers, defined defined by x n = n k=1 1, n = 1, 2, 3... k2 If we consider Q as a metric space with the metric inherited from R, (x n ) is a Cauchy sequence. To prove this without introducing gadgets like the integral test, I will prove first that n+p k=n 1 k 2 1 n 1 for all n N, n 2, p N {0}. The proof is simple, it relies on the inequality and equality 1 k 2 = 1 (k 1)k = 1 k 1 1 k. The proof of (5) can now be done as follows. Notice, we assume n 2 to be sure that k 2 and 1/k 1 is defined. n+p k=n n+p 1 k 2 k=n 1 (k 1)k = 1 n 1 1 n + 1 n 1 n n + p 1 1 n + p. The last sum telescopes, all terms but the first and last cancel so that we have n+p 1 k 2 1 n 1 1 n + p 1 n 1 k=n (5)

21 7 COMPLETENESS AND SOME REALITY CHECKS 21 which is (5). To prove now it is a Cauchy sequence, let ϵ > 0 be given. Let N = 1 ϵ +1. If n, m N, without loss of generality we may assume m > n, write m = n p for some p 0. Then x m x n = n+1+p k=n+1 1 k 2 1 n 1 N < ϵ by (5) with n replaced by n + 1. (We can always assume m > n N instead of m, n N when dealing with Cauchy sequences because if m = n then d(p m, p n ) = 0 and there is nothing to prove; if m < n, we can switch the roles of m, n.) We see that (x n ) is a Cauchy sequence in Q. In 1644, the Italian mathematician Pietro Mengoli wondered about the limit of this sequence. He could not figure it out. It was only almost a century later, in 1735, that Euler proved lim n x n = π 2 /6. In 1798, Legendre proved π 2 irrational. It follows that the sequence (x n ) does not converge in the metric space Q. Definition 12 A metric space is said to be complete iff all of its Cauchy sequences converge. The example we just did shows that the metric space Q is not complete. Exercise 15 Prove that if M is a discrete metric space, M is complete. lack of completeness is not a terribly bad sin. One can show that every metric space can be embedded into a complete space; essentially, all one does is to add a limit for each Cauchy sequence. Doing for Q is another way of producing R. But let us see that R is complete. There are many proofs; here is one way of doing it. We will use the concepts of lim sup and lim inf (which always exist for every sequence of real numbers) and the fact that a sequence of real numbers converges if and only if it is bounded and its liminf equals its limsup. First we will see that Cauchy sequences are always bounded; in fact, let s have it as an exercise. Exercise 16 Let M be a metric space, let (p n ) be a Cauchy sequence in M. Then (p n ) is bounded in the following sense: There is q M and a non-negative real number M such that d(p n, q) M for all n N. Some comments may be in order. In the first place, if one q M works, every other q in M works. In fact, if q, q M and d(q, p n ) M for all n, then d(q, p n ) M + d(q, q ) for all n N. A good choice for q, M for the case of the exercise is q = p N, where N is such that d(p n, p m ) 1 if n, m N, and M = max(d(p 1, p N ),..., d(p N 1, p N ), 1). It is now easy to prove: Theorem 20 The metric space R is complete.

22 7 COMPLETENESS AND SOME REALITY CHECKS 22 Proof. Let (p n ) be a Cauchy sequence in R. To prove it converges, we will prove that lim inf n p n equals lim sup n p n. Let L = lim inf n p n, L = lim sup n p n. Since L L, to prove L = L it suffices to prove L L. In the first place, by Exercise 16, L, L R; neither is or. Let ϵ > 0 be given. Because the sequence is Cauchy, there is N such that n, m N imply x n x m < ϵ/2. Recall the definitions of lim sup and lim inf. First lim sup: lim sup x n = lim sup x k. n n It follows that there is N 2 such that L sup n k x k < ϵ/2 for n N 1. All we ll use from this is that n N 1 implies that L < sup k n x k + (ϵ/2) for n N 1. Let N 2 = max)n, N 1 ). Then n N implies n N 1, thus L ϵ 2 < sup k n x k, hence there exists k n such that L ϵ 2 < x k. But now k n N 2 N, thus x n x k < ϵ/2, hence n k x n > x k ϵ 2 > L ϵ. That is, we proved that if n N 2, then x n > L ϵ. Assume n N 2. Then k N 2 if k n so that x k L ϵ. It follows that inf k n x k L ϵ for all n N 2, thus L = lim inf x k L ϵ. n k n Since ϵ > 0 is arbitrary, this implies L L. Thew next set of exercises are important results from real analysis; they provide an alternative proof of the completeness of R. Thus, do not use completeness to prove any of them. Exercise 17 Let (x n ) be a sequence of real numbers and assume it is increasing; x n x n+1 for all n N. Prove: If the sequence is bounded above, it converges to a finite value. Otherwise it diverges to infinity. In fact, in either case lim x n = sup{x n : n N}. n Similarly, if (x n ) is a decreasing sequence of real numbers, then lim x n = inf{x n : n N}; n it follows that such a sequence converges if and only if it is bunded below; otherwise it diverges to. The proof is quite easy; letting σ = sup{x n : n N}, assume σ <. Given ϵ > 0 there is some N N with x N > σ ϵ; since the sequence increases x n > σ ϵ if n N. If σ = one argues similarly,but differently. This result, together with is one of the best ways of proving convergence of sequences of real numbers. Here are is a classical example.

23 7 COMPLETENESS AND SOME REALITY CHECKS 23 Example 1. Let (f n ) be the Fibonacci sequence; the sequence defined inductively by f 1 = 1, f 2 = 1, f n = f n 1 + f n 2 if n 3. We will prove that f n+1 lim = 1 n f n 2 (1 + 5). Let us set g n = f n+1 /f n. Then, of course, g n = (f n + f n 1 )/f n = 1 + (1/g n 1 ) and taking limits here we conclude that IF there is a limit l it must satisfy l = 1 + 1/l, hence l 2 l 1 = 0, thus l = (1 ± 5)/2 and, since it can t be 0, l = (1 + 5)/2. But why does a limit exist? Here are the first few terms of the sequence (g n ) (up to three decimal places). 1, 2, 1.5, 1.667, 1.6, 1.625, Writing out terms, one sees that the odd terms increase, the even terms decreases. Can we prove this? It might be a good idea to have a formula giving g n interms of g n 2 rather than g n 1. That, of course, is quite easy: g n = = 1 + g n = 2g n g n 2 g n = 2 1 g n Let us write, as is customary, φ = 1 2 (1 + 5). Notice that 2φ + 1 φ + 1 = φ. With this we can prove: If g n 2 < φ, then g n < φ, and conversely, g n 2 > φ, implies g n < φ. In fact, assume g n 2 < φ. Then g n 2 +1 < φ+1, 1/(g n 2 +1) > 1/(φ + 1), and g n = 2 1 g n < 2 1 φ + 1 = 2φ + 1 φ = 1 = φ. The proof that g n 2 > φ implies g n > φ is identical. Consider now the difference g n g n 2. We have g n g n 2 = 2g n g n g n 2 = g2 n 2 g n 2 1. g n Suppose g n 2 < φ. Since f(x) = x 2 x 1 < 0 for (1 5)/2 < x < (1 + 5)/2 = φ, in this case g n g n 2 > 0 and g n < φ. Notice that g 1 = 1 < φ, thus the sequence of odd terms is increasing. In case that s not clear, we have by induction: g 2k 1 < φ for all k, and because g 2k 1 < φ we also have that g 2k+1 g 2k 1 > 0. The sequence (g 2k 1 ) is increasing and bounded above by φ. Thus it converges. Set α = lim k g 2k 1. Similarly one sees that the sequence (g 2k ) decreases and is bounded below by φ. Let β = lim k g 2k. Finally, we need to see that α = β = φ. That is easy. From g 2k = 2g 2(k 1) + 1 g 2(k 1) + 1 we get taking limits α = (2α + 1)/(α + 1), hence α 2 α 1 = 0 and, since α must be 0, α = φ. Similarly β = φ. It now follows that lim n g n = φ.