Some Basic Notations Of Set Theory References There are some good books about set theory we write them down We wish the reader can get more 1 2 3 4 Set Theory and Related Topics by Seymour Lipschutz Set Theory by Charles C Pinter Theory of sets by Kamke Naive set by Halmos 2 1 Prove Theorem 2 2 Hint a b c d means a a b c c d Now appeal to the definition of set equality Proof It is trivial Suppose that a b c d it means that a a b c c d It implies that a c c d and a b c c d So if a 6 c then a c d It implies that c a which is impossible Hence a c Similarly we have b d 2 2 Let S be a relation and let D S be its domain The relation S is said to be i reflexive if a D S implies a a S ii symmetric if a b S implies b a S iii transitive if a b S and b c S implies a c S A relation which is symmetric reflexive and transitive is called an equivalence relation Determine which of these properties is possessed by S if S is the set of all pairs of real numbers x y such that a x y Proof Write S x y x y then we check that i reflexive ii symmetric and iii transitive as follows It is clear that D S R 1 i Since x x x x S That is S is reflexive ii If x y S i e x y then y x So y x S That is S is symmetric iii If x y S and y z S i e x y and y z then x z So x z S That is S is transitive b x y Proof Write S x y x y then we check that i reflexive ii symmetric and iii transitive as follows It is clear that D S R i It is clear that for any real x we cannot have x x So S is not reflexive ii It is clear that for any real x and y we cannot have x y and y x at the same time So S is not symmetric iii If x y S and y z S then x y and y z So x z wich implies x z S That is S is transitive c x y Proof Write S x y x y then we check that i reflexive ii symmetric and iii transitive as follows It is clear that D S R i Since it is impossible for 0 0 S is not reflexive ii Since 1 2 S but 2 1 S S is not symmetric iii Since 0 1 S and 1 0 S but 0 0 S S is not transitive d x2 y 2 1 Proof Write S x y x2 y 2 1 then we check that i reflexive ii symmetric and iii transitive as follows It is clear that D S 1 1 an closed interval with endpoints 1 and 1 i Since 1 D S and it is impossible for 1 1 S by 12 12 6 1 S is not reflexive ii If x y S then x2 y 2 1 So y x S That is S is symmetric iii Since 1 0 S and 0 1 S but 1 1 S S is not transitive e x2 y 2 0 Proof Write S x y x2 y 2 1 then S automatically satisfies i reflexive ii symmetric and iii transitive f x2 x y 2 y Proof Write S x y x2 x y 2 y x y x y x y 1 0 then we check that i reflexive ii symmetric and iii transitive as follows It is clear that D S R 2 i If x R it is clear that x x S So S is reflexive ii If x y S it is clear that y x S So S is symmetric iii If x y S and y z S it is clear that x z S So S is transitive 2 3 The following functions F and G are defined for all real x by the equations given In each case where the composite function G F can be formed give the domain of G F and a formula or formulas for G F x a F x 1 x G x x2 2x Proof Write G F x G F x G 1 x 1 x 2 2 1 x x2 4x 3 It is clear that the domain of G F x is R b F x x 5 G x x x if x 6 0 G 0 0 Proof Write G F x G F x if x 6 5 G x 5 x 5 x 5 0 if x 5 It is clear that the domain of G F x is R 2 2x if 0 x 1 x if 0 x 1 c F x G x 1 otherwise 0 otherwise Proof Write 2 4x if x 0 1 2 0 if x 1 2 1 G F x G F x 1 if x R 0 1 It is clear that the domain of G F x is R Find F x if G x and G F x are given as follows d G x x3 G F x x3 3x2 3x 1 Proof With help of x 1 3 x3 3x2 3x 1 it is easy to know that F x 1 x In addition there is not other function H x such that G H x x3 3x2 3x 1 since G x x3 is 1 1 e G x 3 x x2 G F x x2 3x 5 3 Proof Write G x x G F x 1 2 2 11 4 1 F x 2 then 2 11 x2 3x 5 4 which implies that 2F x 1 2 2x 3 2 which implies that F x x 2 or x 1 2 4 Given three functions F G H what restrictions must be placed on their domains so that the following four composite functions can be defined G F H G H G F H G F Proof It is clear for answers R F D G and R G D H Assuming that H G F and H G F can be defined prove that associative law H G F H G F Proof Given any x D F then H G F x H G F x H G F x H G F x H G F x So H G F H G F 2 5 Prove the following set theoretic identities for union and intersection 4 a A B C A B C A B C A B C Proof For the part A B C A B C Given x A B C we have x A or x B C That …