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).