By Herbert Amann

ISBN-10: 3764371536

ISBN-13: 9783764371531

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Then there is a function ϕn : N → N such that ϕn (0) = ν(n) and ϕn ν(m) = ν ϕn (m) for all m ∈ N. Deﬁne ψ: N→N , m → ν ϕn (m) . Then ψ(0) = ν ϕn (0) = ν ν(n) and also m∈N. ψ ν(m) = ν ϕn (ν(m)) = ν ν(ϕn (m)) = ν ψ(m) , Thus we have shown that n ∈ N implies ν(n) ∈ N . Once again, (N1 ) implies N = N. 5) is unique. For n ∈ N, suppose that ψn : N → N is a function such that ψn (0) = ν(n) and m∈N, ψn ν(m) = ν ψn (m) , and deﬁne Mn := m ∈ N ; ϕn (m) = ψn (m) . From ϕn (0) = ν(n) = ψn (0) we deduce that 0 ∈ Mn .

14(a). 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!

Show the following: (a) f | A = f ◦ i. (b) (f | A)−1 (B) = A ∩ f −1 (B), B ⊆ Y . 22 4 I Foundations Relations and Operations In order to describe relationships between elements of a set X it is useful to have a simple set theoretical meaning for the word ‘relation’: A (binary) relation on X is simply a subset R ⊆ X × X. Instead of (x, y) ∈ R, we usually write xRy or x ∼ y. R A relation R on X is reﬂexive if xRx for all x ∈ X, that is, if R contains the diagonal ∆X := (x, x) ; x ∈ X . It is transitive if (xRy) ∧ (yRz) = ⇒ xRz .

### Analysis/ 1 by Herbert Amann

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