By Serge Lang (auth.)

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There exists a point W E S such that d(S, v) = Iw - vi. 7. Let K be a compact set of complex numbers, and let S be a closed set. There exist elements d(K, S) [Hint: Consider the function ZH Zo E = Izo - K and Wo E S such that wol. 8. Let S be compact. Let r be a real number > O. There exists a finite number of open discs of radius r whose union contains S. Proof. Suppose this is false. Let Z 1 E S and let D 1 be the open disc of radius r centered at Zl' Then Dl does not contain S, and there is some Z2 E S, Z2 =1= Zl' Proceeding inductively, suppose we have found open discs D1, ...

The first quadrant, consisting of all numbers + iy with x > 0 and y > 0 is open, and drawn on Fig. 9. z= x Figure 8 18 COMPLEX NUMBERS AND FUNCTIONS (a) [I, §4] (b) Figure 9 On the other hand, the set consisting of the first quadrant and the vertical and horizontal axes as on Fig. 9(b) is not open. The upper half plane by definition is the set of complex numbers z= x + iy with y > O. It is an open set. Let S be a subset of the plane. A boundary point of S is a point oc such that every disc D(oc, r) centered at oc and of radius r > 0 contains both points of S and points not in S.

We say that the family covers S if S is contained in this union, that is, every Z E S is contained in some Vi' We then say that the family {V;}iel is an open covering of S. If J is a subset of I, we call the family {Vj}jeJ a subfamily, and if it covers S also, we call it a subcovering of S. 9. Let S be a compact set, and let {V;}iel be an open covering of S. Then there exists a finite subcovering, that is, a finite number of open sets ViI"" ,Vi" whose union covers S. Proof. 8, for each n there exists a finite number of open discs of radius l/n which cover S.