# let C be the set of complex number and d-CxC to R be defined d(z1-z2).docx

4 Feb 2023
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### let C be the set of complex number and d-CxC to R be defined d(z1-z2).docx

• 1. let C be the set of complex number and d:CxC to R be defined d(z1,z2)=0when z1=z2 but Iz1I+Iz2I when z1not equal z2 show that (C,d) is a complete metric? Solution The verification that (C,d) is a metric space is fairly straightforward. Metric spaces have to satisfy (1) non-negativity, (2) identity of indiscernables, (3) symmetry, and (4) the triangle inequality. 1) for z1 = z2, d(z1,z2) = 0 = 0 (obviously) for z1 ? z2, d(z1,z2) = |z1| + |z2| = 0 + 0 = 0 (since abs. values are non-negative) 2) d(z1,z2) = 0 if and only if z1 = z2 (by definition of 'd') 3) d(z1,z2) = |z1| + |z2| = |z2| + |z1| = d(z2,z1) 4) d(z1,z3) = |z1| + |z3| = |z1| + |z2| + |z2| + |z3| = d(z1,z2) + d(z2,z3) Completeness: A metric space is complete if every Cauchy sequence converges within the space. A Cauchy sequence is a sequence of points where the distance between two points gets arbitrarily small as the sequence progresses. Formally, for every epsilon > 0, there exists N such that for all m,n > N, d(z_m,z_n) < epsilon. If we have a sequence consisting of some point in the metric space called 'z' repeated forever, then this trivially converges to z, which is in the metric space. Now let's deal with sequences that aren't single values repeated. We will now show that every other type of sequence converges to 0. Since d(z_m,z_n) < epsilon for all epsilon > 0, then by definition, |z_m| + |z_n| < epsilon for all epsilon > 0. This means that |z_m| < epsilon and |z_n| < epsilon. Since epsilon can be made arbitrarily small, both z_m and z_n must converge to 0 (they are bounded above and below by values converging to 0). More intuitively, since the distances are getting smaller and smaller, it must by that z1 and z2 are both getting closer to 0 (by the way that 'd' is defined).