model3 - n = n 1 is injective but not surjective Now deFne...

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MODEL ANSWERS TO THE THIRD HOMEWORK 1. 7. There are many ways to prove this. Perhaps the simplest pro- ceeds as follows. Consider trying to construct a permutation f : S -→ S . Order the elements of S from 1 to n . We Frst have to decide where to send the Frst element of S . We have n choices where to send it. Now we have to decide where to send the second element of S . In this case we only have n - 1 choices where to send the second element, since we cannot send it to the image of the Frst element. In the end this gives us n ( n - 1)( n - 2) . . . 1= n ! possibilities. 8. We prove these parts in a slightly unusual order. (b). Suppose that A is Fnite. Then every injection is a surjection, by deFnition. (a) Now suppose that σ is a surjection. Let τ be a right inverse of σ . Then σ is a left inverse of τ , so that τ is injective. But then, by part (a), τ is bijective. It follows that σ is bijective as well. (c) We have already seen that if A = N , then σ
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Unformatted text preview: ( n ) = n +1 is injective but not surjective. Now deFne σ ( n ) = ± n-1 if n ± = 0 if n = 0. Then τ is clearly surjective, but not injective as τ (0) = τ (1) = 0. 9. Let A = { a } , B = { b, c } and C = { d } . DeFne σ : A-→ B by the rule σ ( a ) = b and τ : B-→ C by the rule τ ( b ) = τ ( c ) = d . Then the composition τ ◦ σ : A-→ C sends a to d . Thus the composition is bijective and yet σ is not surjective and τ is not injective. 2. Suppose that A n are countable sets, n ∈ N . We want to prove that A = ² n ∈ N A n is countable. We might as well assume that this is a disjoint union. Pick injections f n A n-→ N . DeFne a map f : A-→ N × N , by the rule f ( a ) = ( n, f n ( a )) where a ∈ A n . It is clear that f is an injection, and the result is an easy application of Schr¨oder-Bernstein. 1...
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This note was uploaded on 07/15/2008 for the course MATH 111 taught by Professor Long during the Spring '08 term at UCSB.

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