... Z1(v). LEMMA 2.8. Let v be a minimal geometric tripotent in an atomic complex neutral SFS space Zsatisfying FE, STP and ERP. Then Z(v) is atomic and satisfies FE. PROOF. We show first that FE holds in Z1(v). Let K be a proper norm ...
... Z1, Z2 e 2, the category of TF objects. Then Z1 - QZ, and Z2 – QZ2 are dense maps and QZ1 X QZ2 is also TFD. If we know that Z1 X Z2 – QZ, X QZ2 is dense, then Q(Z1 X Z2) = QZ, X QZ2. Since the composite of dense maps is dense, it ...
... z1|/|z2| < p for some p > 0. Outside the set S, there is no problem; the above inequality is satisfied even for r = 3. Inside the set S, we consider the following three subsets: H1 = {(21, z2)||z1 – i22"| < was "}, H2 = {(21, 22) |z1 ...
... (Z1) must be contained in a 4-component of G(d5(K')); that is, hi(Z1) C F(K')' or hi(Z1) C Z, for j = 1 or 2. Assume hi(Z1) C F(K')' first. Since Z1 generates A (H1), h1(A (H1)) C F(K')'. The image of F(K')' under the canonical ...
... Z1 that would raise the standards set by the Mach III. By 1969, mock-ups were ready. But then Honda stole the show with its original CB750, a motorcycle that did everything Kawasaki had planned for the Z1. So the Z1 project came to a ...
... (z1, z2), L'b(z1, z2)) for all asz1, z2) and b(z1, z2). What we need is the following special case of [16, Theorem 1]. LEMMA 1. Let f = (f(zi,z)),ez. be a diagonal sequence satisfying the system of equations (2.8) Uisk(z1, z2) = c(k)Vsk(z1 ...
... Z1.a4-1/2 To prove (A.1) it is sufficient to prove that the random variables 1 1 Z1.041/2(1 – gi) and ...( - 1) are independent. (A.2) 81 1,0--1/2 Let us note that gi = g = Z1/2.1/2 is an arcsine distributed random variable. In fact, we ...
... Z1-R. AMA Awards Night is open to the public. AMA members who show their current membership card become eligible to win the Z1-R, donated by the Kawasaki Motors Corporation. Non members may obtain a membership at the door and become ...
... z1 and z2 of the form (1.1) f(zi,z) = zi/b/z1, z2) for i = 1,2, where p(z1, z2) is an fps with to O,0) # 0 for i = 1,2. Then there exists a unique pair of fps F(zi,z) = (F1(zi,z), F2(zi,z)), which is also of the form (1.1), with (1.2) ...
... Z1–p Zi-p+, - Zi-p+, § { Zp X Z, and X Z1–p - Z1–p {(1 – š) = - - Z! + Z1–p to X Z1–p + Zp + Z1–p to But, again by (A.4), the random variable Zi-p+a - Zi-p+o + Zp — 1 Zp Zp is independent from Z, + Zi-p1a, hence, from Z1–p Z1– ...