Exercise 2311 Let us prove a stronger version of Theorem 237

Exercise 2.3.11: Let us prove a stronger version of Theorem 2.3.7. Suppose that {xn} is a sequences such that every subsequence {xn,} has a subsequence {xnn} that converges to x. a) First show that {xn} is bounded. b) Now show that {xn} converges to x.

Solution

(a):- Let (X,d) be a metric space. A sequence in X is a function from N to X. If : N X is a sequence we write (xn), (xn)nN or (x0,x1,...) for , where xn = (n) is the nth term of the sequence = (x0,x1,...). Sequences in R are called real sequences.

Definition 5.1. A sequence (xn) in X converges to x if for every neighborhood U of x there exists N N such that xn U for all n N. In this case we write xn x or limn xn = x or limn xn = x. A sequence (xn) which does not converge is called divergent.

Proposition 5.2. The following are equivalent. (i) limn xn = x. (ii) For every > 0, there is N N such that xn B(x,) for all n N. (iii) For every > 0, there is N N such that d(xn,x) < for all n N

Proposition 5.3. (i) (Uniqueness of the limit) A sequence cannot have more than one limit. (ii) (Boundedness) If (xn) converges to x, then there is r > 0 such that xn Br(x). Proof of (ii). Assume that (xn) converges to x. By (iii) of the previous proposition with = 1, there is N so that xn B1(x) for all n N. Now take r be any number bigger or equal to d(x1,x),... ,d(xN1,x) and 1. Then xn Br(x) for all n, as claimed.

(b):- Let (xnk ) be a subsequence of (xn). Take a neighborhood U of x. Since lim xn = x, there is N N such that xn U for all n N. From the definition of a subsequence, nk k for all k N. So, nk N for all k N. Thus xnk U for all k N. Hence (xnk ) converges to x as claimed.Let (xnk ) be a subsequence of (xn). Take a neighborhood U of x. Since lim xn = x, there is N N such that xn U for all n N. From the definition of a subsequence, nk k for all k N. So, nk N for all k N. Thus xnk U for all k N. Hence (xnk ) converges to x as claimed.