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Unformatted text preview: 1 First we outline the proof that Ak+1 = Ak A . We define a map (a1 , a2 , . . . , ak+1 ) ((a1 , a2 , . . . , ak ), ak+1 ). This map is easily shown to be a bijection from Ak+1 to Ak A. Now we prove the main result by induction on k. For the base case, we simply note that the result holds trivially for k = 1. Now assume that Ak is countable so that there is a bijection g : N Ak . Also, since A itself is countable, there is a bijection f : N A. We define h : N2 A Ak by h(m, n) = (f (m), g(n)). h is easily shown to be a bijection, so Ak A is countable. From the result above, this implies that Ak+1 is countable. (More formally, composing h with a bijection from N to N2 yields a bijection from N to A2 . Generally this step is omitted in these types of arguments.) This establishes the induction step. By induction, the result holds for all k. 2 i=1 Given countable sets A1 , . . . , An , define Ai = A1 for all i > n. Then n Ai = i=1 Ai is countable since the countable union of countable sets is countable. 3 Let {Ai : i N} be a collection of sets that are each either countable or finite. If Ai is countable, let fi be a bijection from the natural numbers to Ai . If Ai  = n, then there is a bijection g : [n] Ai ; define a surjection fi : N Ai by fi (k) = g(k) if k n, g(1) if k > n. Then f : N2 Ai defined by f (m, n) = fm (n) is readily shown to be a surjeci=1 tion. By our lemma from class, Ai is either finite or countable. i=1 4 Let S be the set of all finite subsets of N. Clearly S is infinite and thus is at least countable. Let T be the set of all tuples of natural numbers (i.e. T = Ni ). i=1 We showed in class that T is countable. We can define a surjection from T to S by (n1 , . . . , nk ) {n1 , . . . , nk }. It follows that S is countable. 1 ...
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 Spring '10
 Math

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