By William A. Veech

Author William A. Veech, the Edgar Odell Lovett Professor of arithmetic at Rice collage, provides the Riemann mapping theorem as a different case of an life theorem for common overlaying surfaces. His specialize in the geometry of complicated mappings makes common use of Schwarz's lemma. He constructs the common protecting floor of an arbitrary planar area and employs the modular functionality to advance the theorems of Landau, Schottky, Montel, and Picard as results of the life of definite coverings. Concluding chapters discover Hadamard product theorem and leading quantity theorem.

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Is the composition of two uniformly continuous (Lipschitz) functions again uniformly continuous (Lipschitz)? 5. Suppose f: X -J>. n is uniformly continuous; show that if {x n } is a Cauchy sequence in X then {/(xn )} is a Cauchy sequence in n. Is this still true if we only assume thatfis continuous? ) 6. 14). Suppose that n is a complete metric space and thatf: (D, d) -J>. (n; p) is uniformly continuous, where D is dense in (X, d). Use Exercise 5 to show that there is a uniformly continuous function g: X -J>.

Z2- Z3 In any case S(Z2) = 1, S(Z3) tion having this property. 7 Definition. If Z1 E lC eo then (Z1' Z2' Z3' Z4)' (The cross ratio of Z1' Z2, Z3' and Z4) is the image of Z1 under the unique Mobius transformation which takes Z2 to 1, Z3 to 0, and Z4 to co. For example: (Z2' Z2' Z3' Z4) = 1 and (z, 1,0, co) = z. Also, if M is any Mobius map and W2, W3' W4 are the points such that MW2 = 1, MW3 = 0, MW4 = co then Mz = (z, W2, W3' w4). 8 Proposition. If Z2' Z3' Z4 are distinct points and T is any Mobius transformation then for any point Z l' Proof Let Sz = (z, Z2, Z3' Z4); then S is a Mobius map.

Hence there are real numbers ai' •.. •. , bn such that KeF = [ai' bdx •.. x[am bn]. 3(b)). Since IRn is complete and F is closed it follows that F is complete. Hence, again using part (d) of the preceding theorem we need only show that F is totally bounded. This is easy although somewhat "messy" to write down. Let E > 0; we now will write F as the union of n-dimensional rectangles each of diameter less than E. After doing this we m will have FeU B(Xk; E) where each Xk belongs to one of the aforementioned k-I 24 Metric Spaces and the Topology of C rectangles.