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Part B Set Theory What is a set? A set is a collection of objects. Can you give me some examples?

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Presentation on theme: "Part B Set Theory What is a set? A set is a collection of objects. Can you give me some examples?"— Presentation transcript:

1 Part B Set Theory What is a set? A set is a collection of objects. Can you give me some examples?

2 Section 6 Concept and Notation of Sets Tabular Form N={1, 2, 3, 4, … } Z={0, -1, 1, 2, -2, … } Q=? R=? C=? S={1, 2, 3, 4} T={fish, fly, a, 4}  ={ } ( is called the empty set) Set-Builder Form N={n: n is a natural number} Z={m: m is an integer} Q={p/q: p and q are integers and q  0} R={r: r is a real number} C={a+bi: a and b are real and i 2 =-1}

3 Elements of a Set 4  N means that: 4 is an element of N; 4 is a member of N; 4 belongs to N; 4 is contained in N; N contains 4.

4 Section 7 Subsets Definition 7.1 Let A and B be two sets. A is a subset of B iff every element of A is an element of B. Symbolically, A  B iff (  x)(x  A  x  B) Can you give me some examples? N  Z  Q  R  C

5 Important subsets of R Let a, b be two real numbers with a  b (a, b) = { x: x  R and a < x < b} Open interval [a, b] = { x: x  R and a  x  b} Closed interval (a, b] = { x: x  R and a < x  b} Half-open and half- closed interval [a, b) = { x: x  R and a  x < b} Half-closed and half- open interval (a, +  ) = {x : x  R and x > a} [a, +  ) = {x : x  R and x  a} (- , a) = {x : x  R and x < a} (- , a] = {x : x  R and x  a} (- , +  ) = R

6 Important Facts on Subsets A  A  A A  B and B  C  A  C Can you give proofs to them?

7 Equal Sets and Proper Subsets A = B iff A  B and B  A iff (  x)(x  A  x  B) Let A, B be two sets. A is a proper subsets of B, denoted by A B

8 Section 8 Intersection and Union of Sets Definition 8.1 Let A and B be sets.The intersection of A and B is the set A  B ={x: x  A and x  B}. A B A  B

9 Union of sets Definition 8.2 Let A and B be sets.The union of A and B is the set A  B ={x: x  A or x  B}. A B A  B

10 B\A Section 9 Complements Definition 9.1,2 Let A and B be sets. The complement of A in B is defined as the set B\A={x: x  B and x  A } B A

11 Example 9.2 Given that E={1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 2, 3, 4, 5}, B = { 4, 5, 6, 7} and C = { 8, 9, 10} A  B = A  B = C  B = A\B = B\A= A  B  C = {1, 2, 3, 4, 5, 6, 7} {4, 5}  (B and C are disjoint) {1, 2, 3} {6, 7} E Ex.2.3 1-9

12 Exercise (1, 5)  (3, 8) (1, 5)  (3, 8) (-10, 1]  [1, 4] (-10, 1]  [1, 4] (- , 3)  (-1, +  ) (- , 3)  (7, 100) R\Q R\(1, 5) (1,5 )\(3, 7) (3, 8)\[2, 9] (5, +  )\(1, 3] =(3, 5) =(1. 8) ={1} =(-10, 4] =R=R == =Set of all irrational numbers =(- , 1]  [5, +  ) =(1, 3] =  =(5, +  )

13 Section 10 Functions Definition f: A  B is a function from a set A to a set B iff f assigns every object in A a unique image in B. 12341234 abcdeabcde f A B Domain = A Range = B Codomain={a, b, c}

14 Group discussion Refer to Ex.2.4 Q.5, discuss on which are graphs of functions and state their domains, ranges and codomains. Determine which of the following are functions: 1.f: R  R is defined by f(x) = logx 2.g:R  R is defined by g(x)=  x 3.h:N  N is defined by h(x) = x/2 4.p:R  R is defined by p(x) = cosx 5.q: [-2, 3]  R is defined by q(x) =  (x 2 -2x – 3) Ex.2.4, Q.6

15 State the differences between the following functions f: Z  Z defined by f(x) = x 2 g:N  N defined by g(x)=x 2

16 Injective functions A function f: A  B is called an injection (injective function or one-to-one function) iff it doesn ’ t assign two distinct objects to the same image. Symbolically, (  x 1, x 2 )(x 1  x 2  f(x 1 )  f(x 2 ))  (  x 1, x 2 ) (f(x 1 ) = f(x 2 )  x 1 = x 2 )

17 Examples 1. Is the function f: N  N defined by f(x) = 2x injective? How to prove it? Proof: f(x 1 ) = f(x 2 )  2x 1 = 2x 2  x 1 = x 2  f is injective

18 2. Let a, b, c, d be real numbers and c  0. f: R\{-d/c}  R be a function defined by f(x)=(ax+b)/(cx+d). Show that if ad-bc  0, then f is injective. Proof: Let x 1, x 2  R\{-d/c}, and suppose that f(x 1 )=f(x 2 ), then (ax 1 +b)/(cx 1 +d)= (ax 2 +b)/(cx 2 +d)  (ad-bc)(x 1 -x 2 ) = 0  x 1 =x 2 (Since ad-bc  0)  f is injective.

19 3. Let f:C  C be a function satisfying f(az 1 +bz 2 )=af(z 1 )+bf(z 2 ) for any real numbers a and b and any z 1, z 2  C. (a)Show that f(0) = 0 (b)f is injective iff when f(z)=0 we have z=0. Proof: f(0)= f(0z 1 +0z 2 ) = 0f(z 1 )+0f(z 2 ) = 0 Proof: (  ) when f(z)=0, then f(0)=0=f(z)  z=0 since f is injective. (  ) If f(z 1 ) = f(z 2 ), then f(z 1 ) - f(z 2 )= 0  f(z 1 -z 2 ) = 0  z 1 -z 2 = 0  z 1 = z 2. Thus f is injective.

20 Which of the following functions are injective? Give proofs. 1. g(x) = x 2 + 1 2. f(x) = x/(1-x) 3. h(x) = (x + 1)/(x – 1) 4. k(x) = x 3 + 9x 2 +27x + 4 Ex. 2.4 Q.10

21 State the difference between the following functions h: Z  Z defined by h(x) = x + 1 and k: N  N defined by k(x) = x + 1

22 Surjective Functions A function f: A  B is called an surjection (surjective function or onto function) iff every element of B is an image of an element in A. Symbolically, (  b  B)(  a  A)(f(a) = b)

23 Examples Prove that f: R  R defined by f(x) = 3x + 2 is surjective. Proof: For any real number y, there exists a real number x = (y – 2)/3 such that f(x) = 3((y – 2)/3) + 2 = y Therefore f is surjective. ???? 5y5y f Group Discussion on Ex.2.4 Q.10 (  b  B)(  a  A)(f(a) = b)

24 2. Show that the function f: R  (0, 1] defined by f(x) = 1/(x 2 +1) is surjective. Proof: For any y  (0, 1], then there exists x=  ((1-y)/y)  R such that f(x)=1/((1-y)/y+1)=y. Therefore f is surjective. Ex.2.4 Q.10

25 Bijective Functions and their inverse functions Let f: A  B be a funcition. f is called a bijective function(or bijection) iff f is both injective and surjective. The inverse function f -1 : B  A of the function f is defined as f -1 = { (b, a) : (a, b)  f } Ex.2.4 Q.10

26 Example 1 12341234 a bcda bcd f AB 12341234 a bcda bcd f -1 B A

27 Example 2 Let f: R  R be a function defined by f(x) = 2x – 1. Then f is bijective. Since y = 2x – 1  x = (y + 1)/2 f -1 (x) = (x + 1)/2

28 Example 3 Let f: R +  R be a function defined by f(x) = log 10 x Then f is bijective. Since y = log 10 x  x = 10 y f -1 (x) = 10 x

29 Example 4 Let f: [0, +  )  [0, +  ) be a function defined by f(x) = x 2 Then f is bijective. Since y = x 2  x =  +  y, f -1 (x) = +  x

30 Graphs of a function & its inverse y=f(x) y=f -1 (x) x y y=x

31 Composite functions of f(x)and f -1 (x) f(f -1 (x))= f -1 (f (x))= X X Ex.2.4 Q.11 Ex.2.5 1-3 Ex.2.4 Q.11 Ex.2.5 1-3


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