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The function f : N0 −→ Q+ ; f (n) = rn , is a bijection. 4. Power sets and their cardinality For two sets X and Y , let Y X = {f : f : X −→ Y is a function}. 58 4. 11. Let X and Y be finite sets. Then Y X is finite and has cardinality |Y X | = |Y ||X| . Solution. Suppose that the distinct elements of X are x1 , . . , xm where m = |X| and those of Y are y1 , . . , yn where n = |Y |. A function f : X −→ Y is determined by specifying the values of the m elements f (x1 ), . . , f (xm ) of Y . Each f (xk ) can be chosen in n ways so the total number of choices is nm .

Then p = q1 since p1 = q0,1 , p = q2 since p2 = q1,2 , etc. So p cannot be in the list of qn ’s, contradicting the assumption that (0, 1] is countable. The method of proof used here is often referred to as Cantor’s diagonalization argument. In particular this shows that R is much bigger that the familiar subset Q ⊆ R, however it can be hard to identify particular elements of the complement R − Q. In fact the subset of all real algebraic numbers is countable, where such a real number is a root of a monic polynomial of positive degree, X n + an−1 X n−1 + · · · + a0 ∈ Q[X].

17. Let R ⊆ R2 be the rectangle centred at the origin O with vertices at A(2, 1), B(−2, 1), C(−2, −1), D(2, −1). B A O· C D A symmetry can send A to any of the vertices, and then the long edge AB must go to the longer of the adjacent edges. This gives a total of 4 such symmetries, thus | Sym(R)| = 4. Again we can describe symmetries in terms of their effect on the vertices. Here are the 4 elements of Sym(R) described in permutation notation. , a regular polygon with n sides all of the same length and n vertices V1 , V2 , .