1.6. CONVERGENCE IN R 15 a. Then, since the sequence converges, given ε 0, there exists N ∈ N such that |an − a| ε 2 for all n ≥ N. Thus, if n, m ≥ N, we have |an − am| ≤ |an − a| + |am − a| ε 2 + ε 2 = ε and so (ak)k∈N is a Cauchy sequence. Suppose now that (ak)k∈N is a Cauchy sequence in R. Then, by Exercise 1.6.13, (ak)k∈N is a bounded sequence and hence by Lemma 1.6.10 has a convergent subsequence. Call the limit of this subsequence a. Then, since (ak)k∈N is Cauchy, it is clear that limk→∞ ak = a. Exercise 1.6.15. Show that if (an)n∈N and (bn)n∈N are Cauchy se- quences in R, then (an + bn)n∈N and (an · bn)n∈N are Cauchy sequences in R. Definition 1.6.16. Let S be a subset of R. Then x ∈ R is an accumu- lation point of S if, for all ε 0, we have ((x − ε, x + ε) \ {x}) ∩ S = ∅. Remark 1.6.17. Thus, x is an accumulation point of S if every interval around x contains points of S other than x. Of course, x does not have to be an element of S in order to be an accumulation point of S. Exercise 1.6.18. Find the accumulation points of the following sets: (i) S = (0, 1) (ii) S = {(−1)n + 1 n | n ∈ N} (iii) S = Q (iv) S = Z (v) S is the set of rational numbers whose denominators are prime. Lemma 1.6.19. Let S be a subset of R. Then every neighborhood of an accumulation point of S contains infinitely many points of S. Proof. Let x be an accumulation point of S. Given ε 0, there is a point x1 ∈ (x − ε, x + ε) ∩ S such that x1 = x. Let ε1 = |x − x1|. Then, there is a point x2 ∈ (x−ε1,x+ε1)∩S such that x2 = x. Iterating this procedure, we get an infinite set of elements in S which is contained in (x − ε, x + ε). Now here is a big-time theorem! Theorem 1.6.20 (Bolzano-Weierstrass). Let S be a bounded, infinite subset of R. Then S has an accumulation point. Proof. Pick an infinite sequence (ak)k∈N of distinct elements of S. Then, by Lemma 1.6.10, (ak)k∈N has a convergent subsequence, (bj)j∈N. If limj→∞ bj = b, then b is an accumulation point of S. Exercise 1.6.21. (i) Find an infinite subset of R which does not have an accumulation point. (ii) Find a bounded subset of R which does not have an accumulation point.

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