By Stanislaw Saks
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Extra resources for Analytic Functions
In fact, since ZE An, there exists a point p E An such that 31 Connected sets in the plane. 11) we have p EOk for a certain value k�n; hence the points p and a can be joined in the set O"CAnCAn by a sequence of points with characteristic number less than l/k� 1/n. 13), the point z can also be joined with the point a < 1/n, and consequently every pair of points of An can be joined in the set An by such a sequence. 12) is a connected subset of the set F; and since a EAn and bEAn in An by a. sequence of points with characteristic number for every n, it follows that also aEA and bEA, and we are led to a contradiction of the assumption that the points a and b belong to different components of the set F.
We can deduce further from this the connectedness of every P(z0;r1,r1), where we can assume again (cf. § 8, p. 20) that z0 =t= oo. Let z1 and z 1 be arbitrary points of this annulus and, for brevity, let R 1= lz 1 Z0 ! 2) is therefore also a continuum; as w� perceive at once,· it con z 1 and z 1, and is itself contained in the annulus P(z0;r1,r1). 8, every annul1,1,s P(z0;rur2) is a connected set and hence a region. In particular, all annular neighbourhoods are regions. Olosed annuli, since they are the closures of open annuli (cf.
Zn), defined on the Cartesian product (cf. Introduction, § 13) Z=Z1 x Z2 x .. X Zn. , Znt:Zn, we can also write F(z), where zt:Z1 x Z2 x ... x Zn. F(z1) I< e for every pair of points definition presupposes We denote by z2 that e(z17z2) F z11z2 e>O there e(z17z2)< 11 exists a number 11> 0 such that the inequality of the set is finite-valued). implies Z (this · the distance between the points z1 and in agreement with the definitions in the Introduction, §§ 8, 13. ,Zn are F on the set Z closed, then every finite and = Z1 x Z2 x ...