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By Herbert Amann

ISBN-10: 3764371536

ISBN-13: 9783764371531

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F (n) , g(n + 1) = Vn+1 g(0), . . , g(n) , n∈N. 12) We want to show that f = g, that is, f (n) = g(n) for all n ∈ N. The condition f (0) = g(0) ( = a) starts the induction. For the induction hypothesis we assume that f (k) = g(k) for 0 ≤ k ≤ n. 12) it follows that f (n + 1) = g(n + 1). 9 we have that f (n) = g(n) for all n ∈ N, that is, f = g. (b) We turn to the existence of the function f . We first claim that, for each n ∈ N, there is a function fn : {0, 1, . . , n} → X such that fn (0) = a , fn (k) = fk (k) , fn (k + 1) = Vk+1 fn (0), .

B) For m, n ∈ N, the binomial coefficient n m n m n! m! (n−m)! := ∈ N is defined by m≤n, , 0, m>n. Prove the following: (i) (ii) n m n n−m = n m−1 n m + (iii) n n k=0 k (iv) m n+k k=0 n . , 1 ≤ m ≤ n. n+1 m = n =2 . = n+m+1 n+1 . Remark The formula (ii) makes calculating small binomial coefficients easy when they are written down in the form of a Pascal triangle. In this triangle, the symmetry (i) and the equation (iv) are easy to see. k n=0 n=1 n=2 n=4 n=5 1 q 6 q = 2 k 10 q 3 1 k 4 10 = = 4 k 1 5 = 5 1 q q q Simplify the sum n S(m, n) := k=0 m + n + k n+1−k m + n + k + 1 n−k 2 2 − k k for m, n ∈ N.

C) f is surjective ⇐ ⇒ ∃ h : Y → X such that f ◦ h = idY . 4 Let f : X → Y be a function. Show that the following are equivalent: (a) f is injective. (b) f −1 f (A)) = A, A ⊆ X. (c) f (A ∩ B) = f (A) ∩ f (B), A, B ⊆ X. 5 Determine the fibers of the projections prk . 6 Prove that, for each nonempty set X, the function P(X) → {0, 1}X , A → χA is bijective. 7 Let f : X → Y be a function and i : A → X the inclusion of a subset A ⊆ X in X. Show the following: (a) f | A = f ◦ i. (b) (f | A)−1 (B) = A ∩ f −1 (B), B ⊆ Y .

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Analysis/ 1 by Herbert Amann


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