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Math 323: Homework 8 Solutions

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Math 323: Homework 8 Solutions Math 323: Homework 8 Solutions David Glickenstein March 14, 2013 7.12a) Let f : R R and g : R R. We de…ned f + g by (f + g)(x) = f (x) + g (x) : It is not true that if f and g are bijective,! then the sum!f + g is bijective. Consider f (x) = x and g (x) = x: Both are clearly bijective (they are their own inverses). But the sum f + g is equal to the constant function zero, which is clearly not bijective. 7.12b) Let f : R R and g : R R. We de…ned fg by (fg)(x) = f (x) g (x) : It is not true that if f and g are bijective, then! their product!fg is bijective. Consider f (x) = g (x) = x: Then fg (x) = x2; which is neither injective nor surjective. 7.20) Suppose f : A B and suppose C A and D B: ! 1 a) The statement f (C) D i¤ C f (D) is true. Suppose f (C) D and consider x C: Then 1 1 2 f (x) D; so x f (D) : Conversely, suppose C f (D) ; and let y f (C) : Thus there exists x C 2 2 1 2 2 such that y = f (x) : Since x f (D) ; we know that f (x) D; i.e., y D: 2 2 2 1 b) If f is injective and D rng(f) (in particular, if f is bijective), then f (C) = D i¤ C = f (D). 1 Proof. By part (a), f (C) D i¤ C f (D) ; thus we need that if f is injective, then D f (C) i¤ 1 1 f (D) C: First suppose that D f (C) : Let x f (D) : Then f (x) D; and hence f (x) f (C) ; 2 2 2 i.e., there exists x0 C such that f (x0) = f (x) : Since f is injective, x = x0; so x C: 2 1 2 Conversely, suppose f (D) C: Let y D: Since D rng(f); there exists x A such that f (x) = y: 1 2 2 Since x f (D) ; x C: Thus y = f (x) f (C) : 2 2 2 Extra problem: Consider the following relation on [ ; ]: x y i¤ x = y or x; y ; : 2 f g a) We …rst show that this is an equivalence relation. Notice that it is re‡exive, since x = x: It is symmetric, since if x y then x = y or x; y ; , which is the same as if we switch x and y: To prove transitivity, we …rst assume x y and y z:2We f considerg two cases. First suppose that x = ; : Then y = x; and so y = ; : Thus z = y= x and x z: Now suppose x ; ; then2 f we mustg have y ; and hence2 f z g ; ; thus x z: 2 f g 2b) fWe describeg the equivalence2 f g classes. There is one equivalence class for every number between and and a single equivalence class containing and : One can think of this as taking the interval[ ; ] and gluing one end to the other. c) Show that f ([x]) = (cos x; sin x) is a well-de…ned function A R2; where A is the set of equivalence classes of . ! We need to show that if x y; then (cos x; sin x) = (cos y; sin y) : Certainly if x = y this is true. Now suppose x = and y = : We see that (cos ; sin ) = ( 1; 0) = (cos ( ) ; sin ( )) : Since is symmetric, this is su¢ cent. d) Show that if we set the codomain to be 2 B = (cos t; sin t) R : t R ; 2 2 1 then f : A B is a bijection. Hint: you can use inverse trig functions, but be careful of where they exist and what their! domains and ranges are! First we show that f is injective. Suppose f ([x]) = f ([y]), so cos x = cos y and sin x = sin y: For x [ ; ] ; cos x = cos y only if x = y: If x = y; we are done. Suppose x = y: Then we see that sin2x = sin y = sin ( x) = sin x: So we must have sin x = 0; so x = : Thus [x] = [y] : We now show that f is surjective. Since cos t and sin t are periodic with period 2; for any t R, if we let 2 t 1 x = t 2 + ; 2 2 we have sin x = sin t; cos x = cos t; and t 1 t t 1 = t 2 + t 2 < t 2 = ; 2 2 2 2 2 so f ([x]) = (cos t; sin t) and f is surjective. 8.3c) Proposition 1 The sets S = [0; 1) and T = (0; 1) are equinumerous. Proof. We can always insert one element into an in…nite set by taking out a countable set and shifting it. We can de…ne the following function f : T S by ! 1 x if x = n+1 for all n N 1 61 2 f (x) = n if x = n+1 for some n N 1 8 1 2 n f g < 0 if x = 2 One can show it is bijective directly, or: see that it has an inverse function g : S T de…ned by ! 1 y if y = n+1 for all n N and y = 0 1 6 1 2 6 g (y) = n+2 if y = n+1 for some n N 8 1 2 < 2 if y = 0 : 8.3e) Proposition 2 The sets S = (0; 1) and T = R are equinumerous. Proof. There are several ways to get a bijection. The key idea is that we need a function that takes a 1 …nite interval to an in…nite interval. These are functions like tan x; x ; and rational numbers. Also, one can compose several bijections together to get a bijection with the appropriate domain. So, for instance, f (x) = tan x gives a bijection between ; and R, and so one can then scale and shift by precomposing 2 2 with the function g (x) = x to get the function tan x : Here are some bijections S T : 2 2 ! f1 (x) = tan x 2 x 1 f (x) = 2 2 x (1 x) 1 1 x 1 if x 0; 4 4 2 f (x) = 1 if x 3 ; 1 3 8 x 3 4 4 2 <> 16x 8 if x 1 ; 3 2 4 4 One can prove these are bijections by considering:> regions where they are increasing and decreasing, or sometimes by …nding inverse functions (for instance, the inverse to f3 is 1 + 1 if y < 4 y 4 g (y) = 1 + 3 if y > 4 3 8 y 4 1 (y + 8) if x [ 4; 4] < 6 2 : 2 . More Solutions: 7.30) Suppose g : A C and h : B C: If h is bijective, then there exists a function f : A B such that g = h f: ! ! ! 1 1 Proof. Since h is bijective, there is a function h : C B: If we de…ne f to be h g; then ! 1 h f = h h g = C g = g: 3.
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