22 1. THE CONSTRUCTION OF REAL AND COMPLEX NUMBERS Proof. The first half of the proof is identical to the proof of Theorem 1.6.14. Suppose now that (zk)k∈N is a Cauchy sequence in C. Let zk = ak + bki, where ak,bk R. Then |zm zn|2 = (am an)2 + (bm bn)2. It follows immediately that (ak)k∈N and (bk)k∈N are Cauchy sequences in R. If limk→∞ ak = a and limk→∞ bk = b, then limk→∞ zk = z where z = a + bi. Exercise 1.9.12. Show that every bounded sequence in C has a con- vergent subsequence. Definition 1.9.13. Let S be a subset of C. Then z is an accumulation point of S if, for all ε 0, we have (Bε(z) \ {z}) S = ∅. Remark 1.9.14. Thus, z is an accumulation point of S if every open ball around z contains points of S other than z. Of course, z does not have to be an element of S in order to be an accumulation point of S. Exercise 1.9.15. Find the accumulation points of the following sets: (i) S = {z C | |z| = 1} (this is the unit circle in C) (ii) S = {z C | Re z Im z} (iii) S = {a + bi | a, b Q} (iv) S = {a + bi | a, b Z}. Exercise 1.9.16. (i) Let S be a subset of C. Show that every neighborhood of an accumu- lation point of S contains infinitely many points of S. (ii) (Bolzano-Weierstrass theorem for C) Prove that any bounded infinite set in C has an accumulation point in C. Theorem 1.9.17 (Heine-Borel). Let S be a closed and bounded subset of C. Given a collection {Ui}i∈I of open sets such that S i∈I Ui, there exists a finite subcollection U1,...,Un of {Ui}i∈I such that S U1 ∪· · ·∪Un. Proof. For the purposes of this proof, we treat C as R2. We prove it for S = [a, b] × [c, d] where a, b, c, d R and a b and c d and leave the general case as an exercise. Take a point x0 [a, b] and consider the set {x0} × [c, d]. We take an open set N C containing {x0} × [c, d]. We claim that there exists an open interval I around x0 such that I × [c, d] N. For each point in (x0,y) {x0}× [c, d], choose ry 0 such that the open square (x0 ry,x0 + ry) × (y ry,y + ry) N. By intersecting these squares with {x0} × R and projecting on the second coordinate, we get a collection of open intervals of the form {x0} × (y ry,y + ry) that cover {x0} × [c, d]. By the Heine-Borel theorem in R, there exists a finite subcollection of these open intervals that covers the interval {x0}×[c, d]. Hence the corresponding collection of open squares also covers {x0} × [c, d]. Let r be the minimum of the ry from this finite collection. Then I = (x0 r, x0 + r) is the interval we sought. Now let {Uj}j∈J be an open covering of S. For each x [a, b], the collection {Uj}j∈J covers {x} × [c, d]. As we did above, we choose a finite
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