Analysis/ 1 by Herbert Amann

By Herbert Amann

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4) is injective. For ϕ ∈ {0, 1}N , let A(ϕ) := ϕ−1 (1) ∈ P(N). Then χA(ϕ) = ϕ. 4) is surjective. 6. 12 Corollary The sets {0, 1}N and P(N) are equinumerous. Exercises 1 Let n ∈ N× . Prove that any injective function from {1, . . , n} to itself is bijective. (Hint: Use induction on n. Let f : {1, . . , n + 1} → {1, . . , n + 1} be an injective function and k := f (n + 1). Consider the functions ⎧ j=k , ⎪ ⎨ n+1 , k, j =n+1 , g(j) := ⎪ ⎩ j otherwise , together with h := g ◦ f and h | {1, . . 6 Countability 2 51 Prove the following: (a) Let m, n ∈ N× .

N + 1)! + (n + 1) is prime. Hence there are arbitrarily large gaps in the set of prime numbers. (b) Show that there is no greatest prime number. (Hint: Suppose that there is a greatest prime number and let {p0 , . . , pm } be the set of all prime numbers. I. Stake has finally found a mathematical proof of Thomas Jefferson’s assertion that ‘all men are created equal’: Proposition If M is a finite set of men and a, b ∈ M , then a and b are equal. Proof We prove the claim by induction on the number of men in M : (a) If M contains exactly one man, then the claim is obviously true.

Thus we have shown that 0 ∈ N and that n ∈ N implies n + 1 ∈ N . By induction, that is, by (N1 ), we conclude that N = N. (b) To prove uniqueness we suppose that there are m ∈ N× and k, k , , such that km + = k m + and

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