By Euler L.

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7. The set of all natural numbers N0 = {0, 1, 2, . } is infinite. Solution. Let us take the subset P = {1, 2, 3, . } and define a function f : N0 −→ P by f (n) = n + 1. 0O 1 1O 2 2O 3 ··· nO ··· n+1 ··· ··· If f (m) = f (n) then m + 1 = n + 1 so m = n, hence f is injective. If k ∈ P then k 1 and so (k − 1) 0, implying (k − 1) ∈ N0 whence f (k − 1) = k. Thus f is also surjective, hence bijective. 8. Show that there are bijections between the set of all natural numbers N0 and each of the sets S1 = {2n : n ∈ N0 }, S2 = {2n + 1 : n ∈ N0 }, S3 = {3n : n ∈ N0 }.

There are many important and interesting examples. 1. The following are all real arithmetic functions: (a) The ‘identity’ function id : Z+ −→ R; id(n) = n. 24. (c) For each positive natural number r, σr : Z+ −→ R; dr . σr (n) = d|n σ1 is often denoted σ; σ(n) is equal to the sum of the (positive) divisors of n. (d) The function given by δ : Z+ −→ R; δ(n) = 1 if n = 1, 0 otherwise. (e) The function given by η : Z+ −→ R; η(n) = 1. The set of all real (or complex) arithmetic functions will be denoted by AFR (or AFC ).

Let (G, ∗) act on X. Then for x ∈ X there is a bijection F : G/ StabG (x) −→ OrbG (x) between the set of cosets of StabG (x) in G and the orbit of x, defined by F (g StabG (x)) = gx. Moreover we have F ((t ∗ g) StabG (x)) = tF (g StabG (x)) (t ∈ G). Proof. We begin by checking that F is well defined. If g1 StabG (x) = g2 StabG (x), then ∈ StabG (x) and g1 x = g1 ((g1−1 g2 )x) = (g1 g1−1 g2 )x = g2 x. g1−1 g2 Hence F is well defined. , g −1 k ∈ StabG (x) which means that g StabG (x) = k StabG (x).