solving the part 2 EXERCISE 2 [20 points] (i) (5 points)LeLF={C:A(z) = ai2+ bi,
ID: 3281732 • Letter: S
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solving the part 2
EXERCISE 2 [20 points] (i) (5 points)LeLF={C:A(z) = ai2+ bi, i - 1,2,.., N} where the ais and bis are complex numbers, be a contractive IFS with attractor A. Write down in complex notation, explicity, an IFS whose attractor is T(A) where T-cz+d. (ii) Analyze the topology of each of the following sets S. That is, state for each whether it is open, closed, bounded, and/or compact, (as a member of the space in which it sits) and in each case give your reasons. (a)(5 points) S = {rational numbers r : 0Explanation / Answer
(a) The set of rational numbers is not open because in every neighbourhood of a rational number there are infinitely many irrational numbers. The set is also not closed since it's complement is not open. So, it is not compact as well because compact set is a set which is closed and bounded. The set contains rationsl numbers which are between 0 and 1 so it is bounded above and below.
(b) R as subset of R2 means that it contains all the lines in xy plane. since there are infinitely many lines which can be drawn in a plane so R is open set and not closed and it is unbounded so it is not compact oo.
(c) S = [0,1) is neither open nor closed because in the neighbourhood of 0 there exists a point outside S, so S is not open. S' = (-infinity, 0) union [1, infinity) which is itself not open, so S is not closed either. Thus, is is not compact. Moreover, S is bounded below by 0 and above by 1. So S is bounded set.
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