By Robert B. Burckel

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**Extra resources for An Introduction to Classical Complex Analysis: 1**

**Example text**

Cover C with finitely many open disks whose closures lie in U and let V be their union. Thus C c V n X c V compact c U. If V is the family of all compact, relatively open subsets of V n X which contain C, then C = V by the last theorem. By the finite intersection property of compacta, since V = C c V , an intersection of finitely many members of W,which is again a member K of V, must already fall into the open set V. Thus C c K c V n X c U . K is compact and K is relatively open in n X, whence in V n X , whence in A', since V is open in @.

Then for any xl, x2 E R", y E S d ( x l ) I 11x1 - yllg(y) 5 [Ikl - x2II + 11x2 - yllIg(y) 5 Mlbl - x2II + 11x2 - rllg(y), that is, d(x,) - Mllx, - x211 is a lower bound for the set {Ilx, - yllg(y):y E S}. Consequently 4x1) - 11x1 - x2II 5 d(x2). Reversing the roles o f x1 and x2, we are therefore led to Id(x1) - 4x211 5 M l b l - x2II. (ii) For the function g = 1 we have dil({O})= 3. 33 Let X be a compact subset of C and C a component of X . Then C is the intersection of all the relatively clopen subsets of X which contain it.

E K. Then C, being closed in K\{x}, is closed in K, hence compact. Since x $ C, there is then an r > 0 sufficiently small that b ( x , r ) n C = 0 . Evidently C is then a component of the compact set K\D(x, r). 33 there is a compact set U which is relatively open in K\D(x, r ) , which contains C and which lies in the open set C\b(x, r ) . Then K\D(x, r)\U is relatively closed in K\D(x, r ) and so is compact, while K\U = [ K n b ( x , r ) ] u [K\D(x, r)\U]. This decomposition shows that K\U is compact.