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Question: suppose an is a sequence in x that converges to...

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Suppose an) is a sequence in X that converges to x and z1,.... Zm is a finite collec tion of points in X. Define a new sequence (yn) in X by letting yk Zk for 1sksm and yk = Xk-m when k > m + 1 . Show that yn → x. Verify the statement in Example 5.2.4(a).

with n <n2 <., then sition 1.3.10, if a sequence in (X, d) converges to x, every subs to x. Unlike R it makes no sense to talk of monotonic sequences in ric space. So the results proved in $1.3 about monotonic sequences have here. A concept from our study of IR tha have great significance now, though its true prominence will not arise until 8 5.2.3. Definition. A sequence (xn) in (X, d) is called a Cauchy sequence if for e > 0 there is an integer N such that d(Xm, n)Ewhen m, n N. The discrete metric space (X, d) is said to be complete if every Cauchy sequence converges. 5.2.4. sequence in (X, d) (Exercise 2) n13 ubsequence converg an abstract mes t had little significance there, however, w Example. (a) As in R, every convergent sequence in (X, d) is a Cauchy (b) R is a complete metric space (1.3.13) as is RP (Exercise 3) (c) If we furnish Q with the metric it has as a subspace of R, then it is not a complete metric space. In fact just take a sequence (tn) in Q that converges to v2 in R. It is a Cauchy sequence in (Q, d) but does not converge to a point in Q. (d) The discrete metric space is complete. Indeed a sequence in the discrete space is a Cauchy sequence if and only if it is eventually constant. metrc 5.2.5. Proposition. If lxn) is a Cauchy sequence and some subsequence of converges to x, then xn → x. Proof. Suppose x,u → x and let € > 0. Choose an integer Ni such that d(Knk,x) < E/2 for n N, and choose an integer N2 such that d(tn, m /2 when m, n2 N2. Put N=max(N1, N2) and let n N. Fix any n. N. Since we have that nk-N1 and both n and nk are larger than N2 , we get that d(x,xn)-d(xAn)

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