In fact m 1 and m 2 can be interchanged in the proposition with no impact on

In fact m 1 and m 2 can be interchanged in the

This preview shows page 230 - 233 out of 239 pages.

only; it does not interfere with the meaning of the proposition. In fact m 1 and m 2 can be interchanged in the proposition with no impact on its meaning. This means that the additional assumption that m 1 < m 2 is only an assumption about our choice of notation and can therefore be made without loss of generality . This is so common that one often simply writes ‘WLOG, assume...’, in place of ‘without loss of generality, assume...’.
3.3. FUNCTIONS 231 Images and Pre-Images We now consider not only how a function f : A B relates elements of the domain A to elements of the codomain B , but also how f relates the subsets of the domain A to subsets of the codomain B . Definition 3.3.6. Let f : A B be a function, and let S A . The image of the set S under the function f is the set f ( S ) = { y B | ∃ x S , f ( x ) = y } . The set f ( A ) is called the image (or range) of f . Caution The notation f ( S ) is not meant to suggest that the function f is applied to the set S in the same sense that it is applied to an element x . It should be noted that the notation f ( S ) when S A stands for a very di ff erent concept than the notation f ( x ) when x A . Indeed, when S A , f ( S ) is a set; it is { y B | ∃ x S , f ( x ) = y } . On the other hand, when x A , you will recall from the beginning of this section that f ( x ) stands for the unique element y B for which ( x , y ) f . Thus, we must be cautious to consider the context when interpreting this notation. Proposition 3.3.14. Let f : A B be a function. f is surjective if and only if f ( A ) = B . Proof. Let f : A B be a function. Assume f is surjective. Since B is the universe of discourse in which f ( A ) is defined, we have f ( A ) B . Let y B . Since f is surjective, x A , f ( x ) = y . That is, y f ( A ). Therefore, B f ( A ); thus f ( A ) = B . Conversely, assume f ( A ) = B . Let y B . Then y f ( A ), since f ( A ) = B . That is, x A , f ( x ) = y . Therefore, y B , x A , f ( x ) = y . Therefore, f is surjective. Therefore, f is surjective if and only if f ( A ) = B .
232 CHAPTER 3. RELATIONS Example 3.3.15. Let f : R R be the function given by x R , f ( x ) = x 2 . Then f (( - 1 , 2)) = [0 , 4). Proof. Let y f (( - 1 , 2)). Then x ( - 1 , 2), y = f ( x ). For such an x , we have y = f ( x ) = x 2 and - 1 < x < 2. Now, since 0 x 2 , we have 0 y . Since - 2 < - 1 < x , we have - 2 < x . Then - 2 < x < 2, and by exercise 127 in section 1.1, we then have x 2 < 4. Now, 0 y and y < 4, so y [0 , 4). Therefore, f (( - 1 , 2)) [0 , 4). Conversely, let y [0 , 4). Putting x = y , gives us y = x 2 = f ( x ). Then 0 x 2 < 4. Now, since x = y , we have 0 x ; hence - 1 < x . Suppose 2 x . Then 4 2 x and 2 x x 2 ; hence 4 x 2 . Since this is a contradiction, we must have x < 2. Therefore, x ( - 1 , 2). Therefore, x ( - 1 , 2), y = f ( x ), which means y f (( - 1 , 2)). Therefore, [0 , 4) f (( - 1 , 2)). Thus, f (( - 1 , 2)) = [0 , 4). Example 3.3.16. Consider the function f : Z Z given by x Z , f ( x ) = ( x if x is even 2 x if x is odd . Then f ( h 3 i ) = h 6 i .

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture