131aW02-3b

131aW02-3b - Math 131a Handout#3 We will assume two...

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Unformatted text preview: Math 131a Handout #3 We will assume two fundamental properties of N : ( N1 ) Every non-empty subset S of N has a least element. ( N2 ) The Fundamental Theorem of Arithmetic: every number n ∈ N has a unique factorization n = 2 a 1 3 a 2 ... where 0 ≤ a k ∈ N S { } . Theorem 0.1. Suppose that S is an infinite subset of a countable set T. Then S is countably infinite (i.e., S ≈ N ) . Proof. First assume that T = N . We define a function f : N → T by induction. From ( N1) we may let f (1) = min S. Let us suppose that we have defined f ( n- 1) (where n > 1) . Since S is assumed infinite, S \{ f (1) ,... ,f ( n- 1) } is non-empty, and we may use ( N1) to define f ( n ) = min S \{ f (1) ,... ,f ( n- 1) } . It is evident that f (1) < f (2) < ... and in particular f is 1-1 . To see that f is onto we have to show that if p ∈ S, then there is an n such that f ( n ) = p. First observe that for all n ∈ N , n ≤ f ( n ) . To see this note that 1 ≤ f (1) since 1 is the least element in all of N . Suppose that we know that n ≤ f ( n ) . Then n ≤ f ( n ) < f ( n +1) implies that n +1 ≤...
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131aW02-3b - Math 131a Handout#3 We will assume two...

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