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Unformatted text preview: Prove that R is an infnite set. Proof. Suppose For a contradiction that R is fnite. By defnition, that means it is either the empty set or has a fnite cardinality. Since 1 ∈ R , R 6 = ∅ . ThereFore, there is n ∈ Z + such that  R  = n . Equivalently, there is a Function f : N n → R with is a bijection. Since bijections have inverses, there is a Function f1 : R → N n which is a bijection, hence an injection. Notice that N n +17 ⊂ R . ThereFore, we can restrict f to N n +17 to get a Function f ± N n +17 : N n +17 → N n which is an injection (since the restriction oF an injection is an injection). Lemma 10.1.4 says that iF there is an injection From N m to N n then m ≤ n . In our case, m = n + 17 and we conclude that n + 17 ≤ n , a contradiction. ² Note: in the above, the number 17 was chosen arbitrarily; any positive number would work....
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 Spring '06
 Knutson
 Math, Counting, Natural number, Finite set, Bijection

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