By Herbert Amann

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2 Verify the following equalities using induction: (a) n k=0 k = n(n + 1)/2, n ∈ N. (b) n k=0 k2 = n(n + 1)(2n + 1)/6, n ∈ N. 3 Verify the following inequalities using induction: (a) For all n ≥ 2, we have n + 1 < 2n . (b) If a ∈ N with a ≥ 3, then an > n2 for all n ∈ N. 4 Let A be a set with n elements. Show that P(A) has 2n elements. 44 I Foundations 5 (a) Show that m! (n − m)! divides n! for all m, n ∈ N with m ≤ n. (Hint: (n + 1)! = n! (n + 1 − m) + n! ) (b) For m, n ∈ N, the binomial coeﬃcient n m n m n!

Then we deﬁne the operation on Funct(X, Y ) induced from by (f g)(x) := f (x) x∈X . g(x) , It is clear that is associative or commutative whenever the same is true of If Y has an identity element e with respect to , then the constant function X→Y , . x→e is the identity element of Funct(X, Y ) with respect to . Henceforth we will use the same symbol for the operation on Y and for the induced operation on Funct(X, Y ). From the context it will be clear which function the symbol represents. We will soon see that this simple and natural construction is extremely useful.

7 Proposition Any subset of a countable set is countable. Proof (a) Let X be a countable set and A ⊆ X. We are done if A is ﬁnite (see Exercise 9), so we can assume that A is inﬁnite, in which case X must be countably inﬁnite. That is, there are a bijection ϕ from X to N and a bijection ψ := ϕ |A from A to ϕ(A). Therefore we can assume, without loss of generality, that X = N and A is an inﬁnite subset of N. (b) We deﬁne recursively a function α : N → A by α(0) := min(A) , α(n + 1) := min m ∈ A ; m > α(n) .