Adv Alegbra HW Solutions 42

Adv Alegbra HW Solutions 42 - 42 2.12 Let X = x 1 x m and Y...

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Unformatted text preview: 42 2.12 Let X = { x 1 , . . . , x m } and Y = { y 1 , . . . , y n } be fi nite sets, where the x i are distinct and the y j are distinct. Show that there is a bijection f : X → Y if and only if | X | = | Y | ; that is, m = n . Solution. The hint is essentially the solution. If f is a bijection, there are m distinct elements f ( x 1 ), . . . , f ( x m ) in Y , and so m ≤ n ; using the bijection f − 1 in place of f gives the reverse inequality n ≤ m . 2.13 ( Pigeonhole Principle ) (i) If X and Y are fi nite sets with the same number of elements, show that the following conditions are equivalent for a function f : X → Y : (i) f is injective; (ii) f is bijective; (iii) f is surjective. Solution. Assume that X and Y have n elements. If f is injec- tive, then there is a bijection from X to im f ⊆ Y . Exercise 2.12 gives | im f | = n . It follows that im f = Y , for there can be no elements in Y outside of im f , lest Y have more than n elements....
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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