# Sequences and Convergence in Metric Spaces

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1 Sequences and Convergence in Metric Spaces Definition: A sequence in a set X (a sequence of elements of X) is a function s : N X. We usually denote s(n) by s n, called the n-th term of s, and write {s n } for the sequence, or {s 1, s 2,...}. See the nice introductory paragraphs about sequences on page 23 of de la Fuente. By analogy with R n, we use the notation R to denote the set of sequences of real numbers, and we use the notation X to denote the set of sequences in a set X. (But is not an element of N, a natural number, so this notation is indeed simply an analogy.) Example: Let V be a vector space. The set V of all sequences in V is a vector space under the natural component-wise definitions of vector addition and scalar multiplication: {x 1, x 2,...} + {y 1, y 2,...} := {x 1 + y 1, x 2 + y 2,...} and α{x 1, x 2,...} := {αx 1, αx 2,...}. Earlier, when we defined R n as a vector space, we defined vector addition and scalar multiplication in R n component-wise, from the addition and multiplication of the real-number components of the n-tuples in R n, just as we ve done in the example above for V. Unlike R n, however, the vector spaces R and V are not finite-dimensional. For example, no finite subset forms a basis of the set R of sequences of real numbers (i.e., sequences in R). Because R is a vector space, we could potentially define a norm on it. But the norms we defined on R n don t generalize in a straightforward way to R. For example, you should be able to easily provide an example of a sequence for which neither max{ x 1, x 2,...} nor n=1 x n is a real number. This is one symptom of the fact that the set of all sequences in a space generally doesn t have nice properties. But we re often interested only in sequences that do have nice properties for example, the set of all bounded sequences. Definition: A sequence {x n } of real numbers is bounded if there is a number M R for which every term x n satisfies x n M. More generally, a sequence {x n } in a normed vector space is bounded if there is a number M R for which every term x n satisfies x n M. Remark: We use the notation l for the set of all bounded real sequences, equipped with the norm {x n } := sup{ x 1, x 2,...}. Note that this is a subset of R. Note too that we needed to change max to sup in the definition of {x n } in order that the norm be well-defined: the sequence x n = 1 (1/n), for example, is in l (it s bounded), and sup{ x n n N} = 1, but max{ x n n N} is not defined.

2 Exercise: Verify that l is a vector subspace of R and that {x n } is indeed a norm. Therefore l is a normed vector space. We can easily convert our definition of bounded sequences in a normed vector space into a definition of bounded sets and bounded functions. And by replacing the norm in the definition with the distance function in a metric space, we can extend these definitions from normed vector spaces to general metric spaces. Definition: A subset S of a metric space (X, d) is bounded if x X, M R : x S : d(x, x) M. A function f : D (X, d) is bounded if its image f(d) is a bounded set. Since a sequence in a metric space (X, d) is a function from N into X, the definition of a bounded function that we ve just given yields the result that a sequence {x n } in a metric space (X, d) is bounded if and only if x X, M R : n N : d(x n, x) M, so that the definition we gave earlier for a bounded sequence of real numbers is simply a special case of this more general definition. Remark: Every element of C[0, 1] is a bounded function. Question: Is C[0, 1] a bounded set, for example under the max norm f? Exercise: Let S be the set of all real sequences that have only a finite number of non-zero terms i.e., S = {{x n } R x n 0 for a finite set A N}. Determine whether S is a vector subspace of l. If it is, provide a proof; if it isn t, show why not. Another important set of sequences is the set of convergent sequences, which we study next. 2

3 Convergence of Sequences Definition: A sequence {x n } of real numbers converges to x R if ɛ > 0 : n N : n > n x n x < ɛ. Example 1: The sequence x n = 1 n converges to 0. Example 2: Example 3: The sequence x n = ( 1) n does not converge. The sequence { 1, if n is a square, i.e. if n {1, 4, 9, 16,...} x n = 0, otherwise does not converge, despite the fact that it has ever longer and longer strings of terms that are zero. If we replace x n x in this definition with the distance notation d(x n, x), then the definition applies to sequences in any metric space: Definition: Let (X, d) be a metric space. A sequence {x n } in X converges to x X if ɛ > 0 : n N : n > n d(x n, x) < ɛ. We say that x is the limit of {x n }, and we write lim{x n } = x, x n x, and {x n } x. Example: A convergent sequence in a metric space is bounded; therefore the set of convergent real sequences is a subset of l. You should be able to verify that the set is actually a vector subspace of l. This is not quite as trivial as it might at first appear: you have to show that the set of convergent sequences is closed under vector addition and scalar multiplication that if two sequences {x n } and {y n } both converge, then the sequences {x n }+{y n } and α{x n } both converge. Example: For each n N, let a n = n and define F n+1 n : [0, 1] R by 1 x, if x < a n F n (x) = a n 1, if x a n. Then F n F in C([0, 1]) with the max-norm, where F (x) = x. What is the value of F n F? Note that F n is the cdf of the uniform distribution on the interval [0, a n ], and {F n } converges to the cdf of the uniform distribution on [0, 1]. 3

4 Definition: Let (X, d) be a metric space, let x X, and let r R ++. (1) The open ball about x of radius r is the set B( x, r) := {x X d(x, x) < r}. (2) The closed ball about x of radius r is the set B( x, r) := {x X d(x, x) r}. Remark: Let (X, d) be a metric space. A subset S X is bounded if and only if it is contained in an open ball and equivalently, if and only if it is contained in a closed ball. Remark: Let (X, d) be a metric space. A sequence {x n } in X converges to x if and only if for every ɛ > 0, x n is eventually in B( x, ɛ) i.e., n N : n > n x n B( x, ɛ). Here s a proof that the sum of two convergent sequences in a normed vector space is a convergent sequence: Proposition: If {x n } and {y n } are sequences in a normed vector space (V, ) and if {x n } x and {y n } ȳ, then {x n } + {y n } x + ȳ. Proof: Let ɛ > 0. Because {x n } x and {y n } ȳ, there are n x and n y such that n > n x x n x < ɛ/2 and n > n y y n ȳ < ɛ/2. Let n = max{ n x, n y }; then n > n x n x + y n ȳ < ɛ/2 + ɛ/2 = ɛ. But we have (x n + y n ) ( x + ȳ) = (x n x) + (y n ȳ) x n x + y n ȳ, by the Triangle Inequality. Therefore n > n (x n + y n ) ( x + ȳ) < ɛ, and since x n + y n is the n th term of the sequence {x n } + {y n }, this concludes the proof. 4

5 The Least Upper Bound Property and the Completeness Axiom for R Definitions: Let S be a subset of R. An upper bound of S is a number b such that x b for every x S. A least upper bound of S is a b such that b b for every b that s an upper bound of S. Remark: In R, a set can have no more than one least upper bound, so it makes sense to talk about the least upper bound of S. It is also called the supremum of S, denoted sup S or lub S. Lower bound, greatest lower bound, glb and inf are defined analogously. Definition: A partially ordered set X has the LUB Property if every nonempty set that has an upper bound has a least upper bound. The Completeness Axiom: R has the LUB property any nonempty set of real numbers that has an upper bound has a least upper bound. Most sequences, of course, don t converge. Even if we restrict attention to bounded sequences, there is no reason to expect that a bounded sequence converges. Here s a condition that is sufficient to ensure that a sequence converges, and it tells us what the limit of the sequence is. The Monotone Convergence Theorem: Every bounded monotone sequence in R converges to an element of R. Proof: Let {x n } be a monotone increasing sequence of real numbers. Since it s bounded, it has a least upper bound b. We will show that {x n } b. Suppose {x n } doesn t converge to b. Then for some ɛ > 0, infinitely many terms of the sequence satisfy x n b ɛ i.e., x n b ɛ (x n cannot be greater than b if b is an upper bound). It follows that x n b ɛ for all n N: since {x n } is increasing, if x m > b ɛ for some m, then x m > b ɛ for all larger n, contradicting that x n b ɛ for infinitely many n. Thus we have x n b ɛ for all n N; i.e., b ɛ is an upper bound of {x 1, x 2,...}, and therefore b is not a least upper bound of {x 1, x 2,...}, a contradiction. Therefore {x n } does converge to b. If {x n } is a monotone decreasing sequence, the above proof shows that the increasing sequence { x n } converges, and therefore {x n } converges. 5

6 Subsequences and Cluster Points Definition: Let f : X Y be a function and let A be a subset of X. The restriction of f to A, denoted f A, is the function f A : A Y defined by x A : f A (x) = f(x). Definition: Let {x n } be a sequence in X i.e., x : N X. A subsequence of {x n }, denoted {x nk }, is the restriction of the function x( ) to an infinite subset of N. Here is an alternative, equivalent definition: Definition: Let {x n } be a sequence in X. A subsequence of {x n } is the sequence {x nk } for a strictly increasing sequence {n k } in N. Remark: A subsequence {x nk } of a sequence {x n } in X is also a sequence in X. Example 1: Let x n = n, and n k = k 2. Then x n = {1, 2, 3, 4,...} and {x nk } = {1, 4, 9, 16,...}. Example 2: Let x n = ( 1) n, i.e., {x n } = { 1, 1, 1, 1,...}. If n k = 2k, then {x nk } = {1, 1, 1, 1,...}. If n k = 2k 1, then {x nk } = { 1, 1, 1, 1,...}. If n k = k 2, then {x nk } = { 1, 1, 1, 1,...}. Example 3: Let x n = ( 1) n n 1 n, i.e., {x n} = {0, 1 2, 2 3, 3 4, 4 5, 5 6,...}. If n k = 2k, then {x nk } = { 1 2, 3 4, 5 6,...}. If n k = 2k 1, then {x nk } = {0, 2 3, 4 5, 6 7,...}. If n k = k 2, then {x nk } = {0, 3 4, 8 9, 15 16, 24 25,...}. Definition: Let {x n } be a sequence in a metric space (X, d). A cluster point of {x n } is an element x X such that every open ball around x contains an infinite number of terms of {x n } i.e., such that for every ɛ > 0, the set {n N x n B( x, ɛ} is an infinite set. We also say that for every ɛ > 0, x n is frequently in B( x, ɛ}. Remark: If {x n } converges to x, then x is its only cluster point. Therefore, if {x n } has more than one cluster point, it doesn t converge. Remark: If {x n } converges to x, then every subsequence of {x n } converges to x. Therefore, if {x n } has a subsequence that doesn t converge, then {x n } doesn t converge. 6

7 Remark: x is a cluster point of {x n } if and only if there is a subsequence that converges to x. Examples: In Example 1, {x n } does not converge and in fact has no cluster points, so it has no convergent subsequences. In Example 2, {x n } has exactly two cluster points, 1 and -1, so the sequence doesn t converge; we exhibited a subsequence that converges to 1, a subsequence that converges to -1, and a subsequence that doesn t converge. In Example 3, {x n } has the same two cluster points, 1 and -1, and for each one we exhibited a subsequence that converges to it. Terminology: We ve introduced the terminology {x n } is eventually in B( x, ɛ} and {x n } is frequently in B( x, ɛ}. More generally, for any property P that a sequence might have, we say that {x n } eventually has Property P if n N : n > n x n has Property P, and that {x n } frequently has Property P if {x n } has Property P for all n in an infinite subset of N i.e., if some subsequence of {x n } has Property P. Clearly, a sequence eventually has a property P if and only if the sequence does not frequently have Property P (the negation of P ); and a sequence frequently has Property P if and only if it does not eventually have Property P. 7

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