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Ii) There must be exactly one s ∈ n not in im f . Define g : k −→ n − 1 by g(j) = f (j) if 0 f (j) < s, f (j) − 1 if s < f (j) n. Then g is a surjection, so by the assumption that Q(k) is true, k n − 1, hence k + 1 In either case, we have established that Q(k) −→ Q(k + 1). By PMI, Q(n) is true for all n ∈ N0 . c) This follows from (a) and (b) since a bijection is both injective and surjective. n. 4. Suppose that X is a finite set and suppose that there are bijections m −→ X and n −→ X. Then m = n.

Let f : X −→ Y be a function. • f is an injection or one-one (1-1 ) if for x1 , x2 ∈ X, f (x1 ) = f (x2 ) =⇒ x1 = x2 . • f is a surjection or onto if for each y ∈ Y , there is an x ∈ X such that y = f (x). • f is a bijection or 1-1 correspondence if f is both injective and surjective. Equivalently, f is a bijection if and only if it has an inverse f −1 : Y −→ X. 2. A set X is finite if for some n ∈ N0 there is a bijection n −→ X. X is infinite if it is not finite. The next result is a formal version of what is usually called the Pigeonhole Principle.

6. The elements of S3 are the following, ι= 1 2 3 , 1 2 3 1 2 3 , 2 3 1 1 2 3 , 3 1 2 1 2 3 , 1 3 2 1 2 3 3 2 1 1 2 3 . 2 1 3 We can calculate the composition τ ◦σ of two permutations τ, σ ∈ Sn , where τ σ(k) = τ (σ(k)). Notice that we apply σ to k first then apply τ to the result σ(k). For example, 1 2 3 3 2 1 1 2 3 3 1 2 = 1 2 3 , 1 3 2 1 2 3 2 3 1 In particular, 1 2 3 3 1 2 = 1 2 3 1 2 3 = ι. −1 1 2 3 1 2 3 = . 2 3 1 3 1 2 Let X be a set with exactly n elements which we list in some order, x1 , x2 , .

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