Solutions 11 MT1100 From Euclid to Mandelbrot 1. (i) Any bijection f between A and B will do as an answer here. For example the function f : A -> B given by f(x) = (a - 1)x + 1 is a bijection. (ii) A simple example of a bijective function mapping between A and B is given by f(x) ={ x 1 2 2 0 x < 1 3 2 . 2. There are lots of possible answers to this question. Here is one solution: (i) A= {2, 4, 6, 8, ... }, B = {1, 2, 3, 4, . . . }. (ii) A= {2, 4, 6, 8, . . . }, B = {1, 3, 5, 7. . . }. (iii) A= {2, 4, 6, 8, ... }, B = {1, 3, 5, 7, . . . }. (iv) A = Q, B = R | Q. 3. If A C B, then the function f : A -> B defined by f(x) = x is injective. Hence | A| ? | B|. 4. Since (0, 1) [ [0, 1], we find that |(0, 1)| ? [0, 1]| by the previous question. The map f : [0,1] -> (0, 1) defined by f(x) = (1+x) (for example) is an injection, so |[0, 1]| ? |(0, 1)|. The Cantor-Bernstein-Schroeder Theorem now implies that [0, 1]| = |(0, 1)|, as required. 5. Let f : N -> T be a mapping. We need to show that f cannot be a bijection. Let f(n) = an,1, an,2, an,3, . . . Define a sequence b1, b2, . .. € T by 1 if an,n = 0, bn = { 2 if an,n = 1, 0 if an,n = 2. 1
This sequence is not in the image of f, since it differs from f(n) in its nth term. So f is not surjective, and hence f is not bijective as required. 2