By Saxl Dr.J.
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3 Permutations A permutation of A is a bijection f : A → A. One notation is f= 1 2 1 3 3 4 4 2 5 8 6 7 7 6 8 . 5 The set of permutations of A is a group under composition, the symmetric group sym A. If |A| = n then sym A is also denoted Sn and |sym A| = n!. Sn is not abelian — you can come up with a counterexample yourself. We can also think of permutations as directed graphs, in which case the following becomes clear. Proposition. Any permutation is the product of disjoint cycles. 1 For our function f above, we write f = (1)(2 3 4)(5 8)(6 7) = (2 3 4)(5 8)(6 7).
Proof. Let A be the set of algebraic numbers and T the set of transcendentals. Then R = A ∪ T , so if T was countable then so would R be. Thus T is uncountable. 7 Bigger sets The material from now on is starred. Two sets S and T have the same cardinality (|S| = |T |) if there is a bijection between S and T . One can show (the Schr¨oder-Bernstein theorem) that if there is an injection from S to T and an injection from T to S then there is a bijection between S and T . For any set S, there is an injection from S to P (S), simply x → {x}.
Proof. This is clear for finite S. Hence assume S is infinite. If f : S → N is an injection then f (S) is an infinite subset of N. Hence ∃ a bijection g : f (S) → N. Thus gf : S → N is a bijection. An obvious result is that if S is countable and ∃ an injection f : S → S then S is countable. Proposition. Z is countable. Proof. Consider f : Z → N, f: x→ 2x + 1 −2x if x ≥ 0 if x < 0. This is clearly a bijection. Proposition. Nk is countable for k ∈ N. Proof. The map (i1 , . . , ik ) → 2i1 3i2 . .