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Example 2/24 8 Let R be a binary relation on Z defined by Mappings - PowerPoint PPT Presentation

Jan 19, 2024 •451 likes •525 views

Example 2/24 8 Let R be a binary relation on Z defined by Mappings x R y if y = x 2 1 . Lectures 9 and 10 (Chapter 18) 3 On every vertical line exactly 1 point. 3 2 2 3 / department of mathematics and computer science /

  1. Example 2/24 8 Let R be a binary relation on Z defined by Mappings x R y if y = x 2 � 1 . Lectures 9 and 10 (Chapter 18) 3 On every vertical line exactly 1 point. � 3 � 2 2 3 / department of mathematics and computer science / department of mathematics and computer science Mapping (1) Mapping (1) 3/24 4/24 The relation R between A and B is a mapping (function) if The relation R between A and B is a mapping (function) if 8 x [ x 2 A : 9 1 8 x [ x 2 A : 9 1 y [ y 2 B : x R y ] } ] y [ y 2 B : x R y ] } ] | {z | {z 9 y [ y 2 B : x R y ] 9 y [ y 2 B : x R y ] at least one: at least one: ^ ^ at most one: 8 y 1 ,y 2 [ y 1 , y 2 2 B : ( x R y 1 ^ x R y 2 ) ) y 1 = y 2 ] at most one: 8 y 1 ,y 2 [ y 1 , y 2 2 B : ( x R y 1 ^ x R y 2 ) ) y 1 = y 2 ] Exercise Determine if the following relations are mappings: y 1 x = 1. R 1 ✓ R ⇥ R defined by x R 1 y iff y = x 2 Yes y 2 2. R 2 ✓ R ⇥ R defined by x R 2 y iff y 2 = x No 3. R 3 ✓ R + ⇥ R + defined by x R 3 y iff y 2 = x Yes 4. R 4 ✓ R ⇥ R defined by x R 4 y iff y = 2 x _ y = 3 x No / department of mathematics and computer science / department of mathematics and computer science

  2. Mapping (2) Image (1) 5/24 6/24 F : A ! B means “ F ✓ A ⇥ B and F is a mapping (function)” A B A : domain of F F : A ! B F ( A 0 ) The image F ( A 0 ) of A 0 under F is B : range of F A 0 the set of all end-points in B of We usually write y = F ( x ) instead of x F y arrows starting in A 0 . Property of mapping F : A ! B : 8 x [ x 2 A : 9 1 y [ y 2 B : F ( x ) = y ]] F ( A 0 ) def = { b 2 B | 9 x [ x 2 A 0 : F ( x ) = b ] } zero Example: outgoing 8 arrows F : Z ! Z with F ( x ) = x 2 � 1 x y so F ( { 0 , 2 } ) = { � 1 , 3 } Then: 3 exactly F ( ; ) = ; one two F ( Z ) = { � 1 , 0 , 3 , 8 , 15 , 24 , . . . } outgoing arrow � 3 � 2 2 3 or more outgoing arrows / department of mathematics and computer science / department of mathematics and computer science Image (2) Example 7/24 8/24 A B F : A ! B F ( A 0 ) A 0 Let F : A ! B be a mapping, and let S, T ✓ A . Prove that F ( S ) \ F ( T ) ✓ F ( S \ T ) . [Proof on blackboard. (Also available as detailed example of the construction of the proof of a property involving sets and mappings Properties of ‘image’: from Course Material section of the website)] x 2 A 0 | val = = F ( x ) 2 F ( A 0 ) val = 9 x [ x 2 A 0 : F ( x ) = y ] y 2 F ( A 0 ) = val NB: x 2 A 0 = 6 = = F ( x ) 2 F ( A 0 ) ; it may happen that F ( x ) 2 F ( A 0 ) and at the same time x 62 A 0 . ( Counterexample on p. 264 of the book.) / department of mathematics and computer science / department of mathematics and computer science

  3. Example: F ( S ) \ F ( T ) ✓ F ( S \ T ) Example: F ( S ) \ F ( T ) ✓ F ( S \ T ) 9/24 10/24 A B Proof: S T F ( S ) F ( T ) (1) var y ; y 2 F ( S ) \ F ( T ) According to the property of ✓ , we need to prove that all elements of x y F ( S ) \ F ( T ) are also elements of F ( S \ T ). (2) y 2 F ( S ) ^ ¬ ( y 2 F ( T )) Let y 2 F ( S ) \ F ( T ) ; we need to establish that y 2 F ( S \ T ) . To this end, it (3) 9 x [ x 2 S : F ( x ) = y ] (Prop. image) suffices, by the property of image, to prove the existence of x 2 S \ T such (4) Pick an x with x 2 S and F ( x ) = y that F ( x ) = y . (5) x 2 T Note that, from y 2 F ( S ) \ F ( T ) it follows, by the property of \ , that y 2 F ( S ) Properties of ‘image’: (6) F ( x ) 2 F ( T ) (Prop. image) and y 62 F ( T ) . So, by the property of image, there exists x 2 S such that x 2 A 0 | val = = F ( x ) 2 F ( A 0 ) (7) y 2 F ( T ) val = 9 x [ x 2 A 0 : F ( x ) = y ] F ( x ) = y . y 2 F ( A 0 ) = (8) False It therefore remains to prove that x 2 S \ T , which, according to the property (9) ¬ ( x 2 T ) of \ and since x 2 S , amounts to proving that x 62 T . x 2 S ^ ¬ ( x 2 T ) (10) To prove x 62 T , we suppose that x 2 T , and derive a contradiction. From x 2 S \ T (Prop. \ ) (11) x 2 T it follows, by the property of image, that F ( x ) 2 F ( T ) . Hence, since (12) 9 x [ x 2 S \ T : F ( x ) = y ] (Prop. image) F ( x ) = y , we have that y 2 F ( T ) . NB: the other direction does not hold. Since also y 62 F ( T ) (see above), we have thus arrived at a (13) y 2 F ( S \ T ) Exercise: give a counterexample. contradiction. (14) F ( S ) \ F ( T ) ✓ F ( S \ T ) / department of mathematics and computer science / department of mathematics and computer science Source (1) Source (2) 11/24 12/24 Property of ‘source’: A B val F : A ! B x 2 F ( B 0 ) = = F ( x ) 2 B 0 F ( B 0 ) The source F ( B 0 ) of B 0 is the B 0 set of all starting points in A of Example: arrows with their end-point in B 0 . Let F : A ! B and U, V ✓ B . Prove that F ( U ) [ F ( V ) = F ( U [ V ) . For all x 2 U : F ( B 0 ) def = { a 2 A | F ( a ) 2 B 0 } x 2 F ( U ) [ F ( V ) val = = { Property of [ } F : A ! B A B F ( U ) 8 U x 2 F ( U ) _ x 2 F ( V ) Example: val = = { Property of source } F : Z ! Z with F ( x ) = x 2 � 1 F ( x ) 2 U _ F ( x ) 2 V 3 val = = { Property of [ } Then: F ( { 1 , . . . , 10 } ) = { � 3 , � 2 , 2 , 3 } F ( x ) 2 U [ V F ( V ) V � 3 � 2 2 3 val = = { Property of source } x 2 F ( U [ V ) / department of mathematics and computer science / department of mathematics and computer science

  4. Source (3) Surjection 13/24 14/24 A B Lemma F : A ! B If F : A ! B en A 0 ✓ A , then F ( F ( A 0 )) ◆ A 0 . F is a surjection if every b 2 B has at least one incoming arrow. Proof: see book pp. 264–265. NB: A 0 can be really smaller than F ( F ( A 0 )) : Property of surjection: F : Z ! Z with F ( x ) = x 2 � 1 , A 0 = { 2 } . 8 8 y [ y 2 B : 9 x [ x 2 A : F ( x ) = y ]] Then F ( { 2 } ) = { 3 } Examples: 3 and I f : R ! R with f ( x ) = x 2 is not a surjection. F ( F ( { 2 } )) = { � 2 , 2 } 6 = { 2 } . I f : R ! R + [ { 0 } with f ( x ) = x 2 is a surjection. � 3 � 2 2 3 / department of mathematics and computer science / department of mathematics and computer science Composed mapping Example 15/24 16/24 Let A , B and C be sets. Lemma Let F : A ! B and G : B ! C be mappings. If F : A ! B and G : B ! C are both surjections, then G � F is a The composed mapping G � F (pronounce: ‘ G after F ’) is defined by surjection too. ( G � F )( x ) = G ( F ( x )) for all x 2 A . Proof: var z ; z 2 C 9 y [ y 2 B : G ( y ) = z ] ( 8 -elim with z 2 C on ‘ G is a surjection’) Pick a y 2 B with G ( y ) = z x F ( x ) G ( F ( x )) 9 x [ x 2 A : F ( x ) = y ] ( 8 -elim with y 2 B op ‘ F is a surjection’) F G Pick an x 2 A with F ( x ) = y ( G � F )( x ) = G ( F ( x )) = G ( y ) = z 9 x [ x 2 A : ( G � F )( x ) = z ] G � F 8 z [ z 2 C : 9 x [ x 2 A : ( G � F )( x ) = z ]] / department of mathematics and computer science / department of mathematics and computer science

  5. Example Injection 17/24 18/24 Lemma A B F : A ! B If F : A ! B and G : B ! C are both surjections, then G � F is a surjection too. F is an injection if every b 2 B has at most one incoming arrow. Proof: To prove that G � F is a surjection, according to the definition of surjection we need to establish that for all z 2 C there exists x 2 A such that ( G � F )( x ) = z . Property of injection: To this end, let z 2 C . Since G is a surjection, there exists y 2 B such 8 x 1 ,x 2 [ x 1 , x 2 2 A : F ( x 1 ) = F ( x 2 ) ) x 1 = x 2 ] that G ( y ) = z , and hence, since F is a surjection, there exists x 2 A such that F ( x ) = y . Example: It follows, using the definition of � , F ( x ) = y and G ( y ) = z , that I f : R ! R with f ( x ) = x 2 is not an injection. ( G � F )( x ) = G ( F ( x )) = G ( y ) = z . I f : R + [ { 0 } ! R with f ( x ) = x 2 is an injection. / department of mathematics and computer science / department of mathematics and computer science Example Example 19/24 20/24 Lemma Lemma If F : A ! B and G : B ! C are both injections, then so is G � F . If F : A ! B and G : B ! C are both injections, then so is G � F . Proof: Proof: To prove that ( G � F ) is an injection, according to the definition of var x 1 , x 2 ; x 1 , x 2 2 A injection we need to establish for all x 1 , x 2 2 A that ( G � F )( x 1 ) = ( G � F )( x 2 ) ( G � F )( x 1 ) = ( G � F )( x 2 ) implies x 1 = x 2 . F ( x 1 ) , F ( x 2 ) 2 B ( 8 -elim+ 9 ⇤ -elim on Prop. mapping+Leibniz) So, let x 1 , x 2 2 A and suppose that ( G � F )( x 1 ) = ( G � F )( x 2 ) ; we G ( F ( x 1 )) = G ( F ( x 2 )) (Def. of composed mapping G � F ) show that x 1 = x 2 . G ( F ( x 1 )) = G ( F ( x 2 )) ) F ( x 1 ) = F ( x 2 ) ( 8 -elim on ‘ G is an injection’) From ( G � F )( x 1 ) = ( G � F )( x 2 ) it follows, by the definition of � , that F ( x 1 ) = F ( x 2 ) G ( F ( x 1 )) = G ( F ( x 2 )) . F ( x 1 ) = F ( x 2 ) ) x 1 = x 2 ( 8 -elim on ‘ F is an injection’) Hence, since G is an injection, we have that F ( x 1 ) = F ( x 2 ) . x 1 = x 2 8 x 1 ,x 2 [ x 1 , x 2 2 A : ( G � F )( x 1 ) = ( G � F )( x 2 ) ) x 1 = x 2 ] Therefore, since F is an injection, we have that x 1 = x 2 . / department of mathematics and computer science / department of mathematics and computer science

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