By Herbert Amann

ISBN-10: 3764371536

ISBN-13: 9783764371531

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N Proof Let Xj , j = 0, 1, . . , n be countable sets, and X := j=0 Xj . By definition n−1 X= j=0 Xj × Xn , so it suffices to consider the case n = 1. Thus we suppose X := X0 × X1 with X0 and X1 countably infinite. Write X0 = { yk ; k ∈ N } and X1 = { zk ; k ∈ N }, and set xj,k := (yj , zk ) for j, k ∈ N. 3) again to define a bijection from X to N. 9 is no longer correct if we allow ‘infinite products’ of countable sets. To make this claim more precise, we need to explain first what an ‘infinite product’ is.

An ∈ X, n − 1 operation symbols and an arbitrary number of (correctly nested) parentheses, for example, K7 := (a1 a2 ) (a3 a4 ) (a5 (a6 a7 )) . We will prove by induction on n that Kn = · · · (a1 a2 ) · · ·) a3 ) n∈N. an , an−1 For n = 3, the claim is true by definition of associativity. Our induction hypothesis is Kk = · · · (a1 a2 ) · · ·) a3 ) ak ak−1 for all expressions Kk of length k ∈ N with 3 ≤ k ≤ n. Now let Kn+1 have length n + 1. Then there are , m ∈ N× such that + m = n + 1 and Km .

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 . If xRy = ⇒ yRx holds, then R is symmetric. Let Y be a nonempty subset of X and R a relation on X. Then the set RY := (Y × Y ) ∩ R is a relation on Y called the restriction of R to Y . Obviously xRY y if and only if x, y ∈ Y and xRy. Usually we write R instead of RY when the context makes clear the set involved.

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Analysis/ 1 by Herbert Amann


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