# Cartesian Products and Relations Denition Cartesian product If and are sets the Cartesian product of and is the set ab and The following points are worth special attention The Cartesian product of PDF document - DocSlides

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In each ordered pair the 64257rst component is an element of and the second component is an element of Example Cartesian product If and ab cd then ab cd ab cd Determining If and are 64257nite sets then 57527 because there are choices for the ID: 22731

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## Presentations text content in Cartesian Products and Relations Denition Cartesian product If and are sets the Cartesian product of and is the set ab and The following points are worth special attention The Cartesian product of

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Cartesian Products and Relations Deﬁnition (Cartesian product) If and are sets, the Cartesian product of and is the set a,b ):( ) and ( The following points are worth special attention: The Cartesian product of two sets is a set, and the elements of that set are ordered pairs. In each ordered pair, the ﬁrst component is an element of , and the second component is an element of Example (Cartesian product) If {{ }} and a,b c,d , then a,b )) c,d )) a,b )) c,d )) Determining If and are ﬁnite sets, then || because there are choices for the ﬁrst component of each ordered pair and, for each of these, choices for the second component of the ordered pair. Cartesian Product is not commutative For the sets and one paragraph above, (( a,b (( a,b (( c,d (( c,d This example shows that, in general, . The underlying reason is that if and are non-empty and one set, say , contains an element which is not in , then contains an ordered pair with ﬁrst component equal to , but contains no such ordered pair. The condition that and are non-empty is required because of the following Proposition. Proposition CPR1 .If is a set, then and Proof We argue by contradiction using the deﬁnition of Cartesian product: Suppose and consider ( x,y . Then, by deﬁnition of Cartesian product, ,a contradiction. Therefore, the set must be empty. The proof that is similar, and is left as an exercise Proposition CPR2 .If and are sets, if and only if ,or or Proof )If then substituting for gives .If or , then by Proposition CP1, ) Suppose that and are non-empty sets and . +et . Since , there exists an element , so that ( x,y . Since ,we have that ( x,y (too). ,y the deﬁnition of Cartesian product, . Therefore, . Similarly . Thus, It is sometimes true that the Cartesian product distributes over other set operations similarly to the distributive law of multiplication over addition.

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Proposition CPR3 . +et A,B and be sets. Then, )=( )- )=( )- )( =( )- )( =( ). Proof We prove part ( ) and leave the proofs of the remaining parts as an exercise .We have ( x,y and ) and ( or [( ) and ( )] or [( ) and ( )] (by a distributive law of logic) [( x,y ]or[( x,y x,y ). Exercise . Investigate, and prove or disprove as appropriate, similar statements involving the set operations relative complement ( ), and symmetric di0erence. Deﬁnition(relation). relation from a set to a set is a subset of .1 (binary) relation on is a subset of It is important to remember that a relation is a set or ordered pairs. There need be no relationship between the components of the ordered pairs- any set of ordered pairs is a relation. 2sually, however, we choose which ordered pairs belong to the relation so that components are related in some way, so we think of the relation as somehow representing the connection. For example, if Gary,Jing,Keika and 4554 4556 4577 then Gary, 4556) Jing, 4554) Keika, 4577) is a relation from to that pairs each 28ic 9ath instructor in set and her:his 28ic telephone extension in set Counting relations. Since any subset of is a relation from to , it follows that if and are ﬁnite sets then the number of relations from to is 2 =2 || ;ne way to see this is as the number of subsets of . 1 direct way to count is the same way one counts subsets: observe that for each of the || ordered pairs in there are two possibilities, either the ordered pair belongs to the relation or it doesn’t, so by the rule of product the number of relations from to is 2 || twos). Similarly, if is a ﬁnite set then the number of relations on is 2 || +et ,... 1= . ,y the above, there are 2 100 relations on . The number of these that contain the pairs (1 1) (2 2) ,..., (1= 1=) is 1 10 90 =2 90 : each of the 1= speciﬁed pairs must be in the relation (1 way to do this), and there are two possibilities – in or not – for each of the remaining ?= pairs. Similar reasoning shows that the number of relations on that contain none of (1 2) ( 5) (7 6) is 2 97 . The number of relations on that contain (2 7) or (4 ?) is 2 100 98 (total minus the number that contain neither ordered pair). 1 di0erent way of counting these gives the equivalent expression 2 99 A2 99 98 (the number that contain (2 7) plus number that contain (4 ?) minus the number that contain both). Finally, the number of relations on that contain either (2 7) or (4 ?) but not both is 2 98 A2 98 (the number that contain (2 7) and not (4 ?) plus the number that contain (4 ?) and not (2 7)).

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Example(lessthanorequaltorelation) The relation on the set given by (1 1) (1 2) (1 ) (2 2) (2 ) ( ) is the set of all ordered pairs ( a,b ) of elements of such that and we can think of the set as representing the Bless than or equal toC relation. Inﬁx notation for relations. If is a relation on and ( a,b , we sometimes use the inﬁx notation aRb and say B is related to (under )C. If is not related to under , we sometimes use the inﬁx notation with a slash through the Example (subset relation, inﬁx notation). +et a,b,c and let be the relation on ) (the power set of , i.e. the set of all subsets of ) deﬁned by That is, a subset of is related to a subset of under exactly when is a subset of . The symbol can be regarded as a synonym for the symbol or, alternatively, the symbol could be regarded as the name of the set of all ordered pairs ( X,Y ) where X,Y ∈P ) and is a subset of Example (recursively deﬁned relations). Delations are sets (of ordered pairs), and thus can sometimes be deﬁned recursively. For example, let be the relation on (the positive integers) deﬁned by: ,1SIS: 1 7- DEC2DSI;F: For all x,y ,if xRy then ( A1) A 7). 1fter generating a few terms, it is not diGculy to guess and prove that a,b =7 . The statement to be proved is ): An ordered pair a,b belongs to if and only if =7 We ﬁrst prove by induction on that if and =7 , then a,b ,1SIS ( = 1): (1 7) by deﬁnition of . Thus, the statement is true for =1. IFH2CTI;F IYP;TIESIS: For some 1, suppose that if =7 then ( k,n IFH2CTI;F STEP: Suppose =7( A 1). Then 7=7 , so by the induction hypothesis ( k,m 7) . ,y the deﬁnition of R, k,m 7) A1 ,m 7A7) Thus, if =7( A 1), then ( A1 ,m Therefore, by induction, for all ,if and =7 , then ( a,b To complete the proof, we show by induction on that if a,b then =7 ,1SIS ( = 1): ,y deﬁnition of , the only ordered pair in with ﬁrst component equal to1is(1 7). Since 7 = 7 1, the statement is true for =1. IFH2CTI;F IYP;TIESIS: For some 1, suppose that if ( k,n , then =7 IFH2CTI;F STEP: Suppose ( A1 ,m . ,y deﬁnition of , this can happen only if k,m 7) . ,y the induction hypothesis, 7=7 . Ience =7 A7=7( A 1). Thus, if ( A1 ,m , then =7( A 1). Therefore, by induction, for all ,if( a,b then =7 For a more diGcult example, consider the relation on deﬁned by: ,1SIS: (1 2) (1 ) DEC2DSI;F: For all x,y ,if( x,y then ( A1 ,y A2) A1 ,y A ) Exercise : Prove by induction that k,n :2

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Functions Deﬁnition (function) .1 function from a set to a set is a relation from to with the property that for every element there exists one and only one element such that ( a,b Deﬁnition (image, value, preimage) .If is a function from to , then we use the notation . From the deﬁnition of a function if , then can be viewed as an assignment, to each element , of a unique element in .If( a,b then we denote the assignment of to by writing ) and calling the image of under , or the value of at (a) - the element is called a preimage of . (Fote that it is preimage rather than the preimage- more than one element of could map to .) It is common usage to say B maps to C. This expression arises from the usual arrow diagram where each element of is Loined by an arrow to the element of assigned to it. 2nfortunately, this tends to lead to the confusion that the elements of are somehow assigned to the elements of which is backwards! It is the elements of that are assigned to the elements of It is important to keep the following facts straight. Every element of has some element of assigned to it. Fo element of is assigned more than one element of , each is assigned exactly one . There is no guarantee that di0erent elements of are assigned di0erent elements of . When we say that each element of is assigned a unique element of , we mean that each element of is assigned one and only one elemnt of . This does not mean that if , then )- it is quite possible that )= )(We have a special name for functions with the property that implies ): 1-1.) There is no guarantee that any particular element of is assigned to any element of . (We have a special name for functions with the property that every element of is the image of at least one element of : onto.) Deﬁnition (domain, inputs, codomain, range) . ,efore we can talk about functions, we need names for the obLects we want to talk about: The set is called the domain , and the elements of are called inputs to (so the domain is where the inputs live). The set is called the codomain The subset of consisting of the elements which are values of (i.e., are assigned to some element in ) is called the range of . (Think: the values of range over the elements in this set.) The range of is the set )= and ) for some Example (function, domain, codomain, range, image, preimage) . +et be the set of all faculty and students at 28ic, and let be the set of all amounts of money in dollars and cents. +et be the relation from to where ( a,b person owes amount to the library. Since for every person there is a unique amount of money that s:he owes to the library (possibly M=), is a function. The domain of is its codomain is , and its range is the set of all amounts of money that are owed (each by at least one person). If (Nary, M1 7?) , then (Nary) = M1 7?, the image of Nary

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is M1 7?, a pre-image of M1 7? is Nary, and the amount M1 7? belongs to the range of (Fote: any person who owes M1 7? to the library is also a pre-image of M1 7?.) The above example demonstrates a function which can not be deﬁned by Bgiving a formulaC for ). In the deﬁnition of function and are Lust sets – there don’t have to be any numbers anywhere – so it may be very diGcult to give a formula. Example (function, domain, codomain, range, image, preimage) . +et deﬁned by )=2 . (Decall that, for a real number the ceiling of , denoted is the smallest integer which is greater than or equal to . Ience is a function.) The domain of is the set of real numbers. The codomain is also the set of real numbers. The range of is the set of even integers: since is an integer, )=2 is an even integer. Thus, the range is a subset of the even integers. To see that every even integer t, ) is a value of , observe that for )=2 =2 Exercises . We leave as an exercise for the reader to determine the range of the function deﬁned by )= . (Decall that for a real number the ﬂoor of , denoted is the largest integer which is less than or equal to .) For more exercises, ﬁnd the range of deﬁned by )= , and show that the range of deﬁned by )= A5 A5is ,... Counting functions . +et and be ﬁnite sets, say ,a ,...,a and ,b ,...,b . We count the number of functions from to . ,y the deﬁnition of function, for each there is exactly one such that ( a,b . Thus, there are choices for the element to be paired with choices for the element to be paired with , and so on. In general, for each choice for there are choices for the element such that ( a,b . ,y the rule of product, the number of functions from to is therefore terms, all equal to ), which equals (or ). Deﬁnition (image of a set, preimage of a set) . The notions of image and preimage can be generaliOed to sets. Suppose is a function. If , then the image of under is the set )= ) for some That is, ) is the set whose elements are the images under of the elements in .If , then the preimage of under is the set )= That is, ) is the set of elements in whose image under is in Example (image of a set, preimage of a set) . +et ,B a,b,c,d and be given by (1 ,d (2 ,a ( ,c (5 ,a (7 ,c Then )= d,a,c , and d,a,c )= . Fotice that this shows that if )= then it need not be the case that )= - it is however, true that if )= , then . (To see this, let . then ) for some . Ience ).) We leave it as an exercise to prove that equality occurs, that is )) = , if and only if there is no element such that ). 1s a further exercise , prove that for we have ))

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with equality if and only if is a subset of the range of (that is, every element of is the image of some element of ). ;bserve that if is a function then, by deﬁnirtion of function )= and ) is the range of (which is a subset of ). +et . ,y deﬁnition of preimage of a subset of the codomain we have )= if and only if there is no element with .Thus )= and, in the example above )= Example (ﬁnding the range) . +et be deﬁned by )= .To determine the range of , we begin by testing a few values of (=) = =- If (= 2), then 2 (= 1) so = 1 and = =, hence )=1- (1 2)=2- If (1 1) then 2 (1 2) = 2 and = 1, hence )= - (1) = 5- If (1 2), then 2 (2 )so = and = 2, hence )=7. ( 2)=6. ,ased on these computations, it seems reasonable to guess that the range of is .We prove that this is the case. First, observe that ) is an integer for every ,so . To show the opposite inclusion, let . We must ﬁnd such that )= .If is even, say =2 t,t , then t/ 2) = t/ t/ =2 .If is odd, say =2 A1 ,t , then for any t/ ,t/ 2A1 2) we have 2 t,t A 1), hence )=( A1)A =2 A1= . Ience the range of is .1san exercise , show that the image of the set of natural numbers is )= Example(ﬁndingpreimages) . +et be deﬁned by )= . We determine the preimage of { and of . To determine { ) we need to ﬁgure out the preimages of each element of { . We have )= =) =). Similarly, )=1 [1 ). Thus, { )= =) [1 ). The set ) can be determined in the same way. For we have )=2 =2 [2 n, A1) [2 n/ (2 A1) ). Therefore, )= =0 [2 n/ (2 A1) )=[= ) [2 1) [5 ) [ ) )heorem F1 . +et be a function, ,A and ,B . Then (a) )= )- (b) ). (c) )= ). (d) )= ). Proof We prove (b) and leave the others as exercises. +et ). then ) for some . Since if and only if and , we have for some and ) for some . Therefore, ) and ), i.e. ). To see that strict containment can occur in part (b) of Theorem F1, consider the function }→{ where (1 1) (2 1) ( 2). Take and . Then )= )= whereas )= }∩{

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Deﬁnition (equality of functions) . Two functions and are e ual if )= ) for every .If and are equal, we write There is a subtle point hidden in the deﬁnition of equality of functions. For two functions to be equal, they must have the same domain, the same codomain, and give the same value for the same input. Thus, technically, the function deﬁned by )= is not equal to the function deﬁned by )= because these two functions do not have the same codomain. Deﬁnition (restriction of a function, extension of a function) . +et be a function. For the restriction of to is the function deﬁned by )= ) for all .If an extension of to is any function for which )= ) for every The restriction of a function is the new function obtained by retricting the allowed inputs for to a subset of its domain . 1n extension of is any function that is identical to on the inputs in , and is also deﬁned for the inputs in . ;bserve that can, and probably does, have more than one extension. This is why we sa an extension, rather than the extension. ;bserve that if is an extension of , then Example (restriction) . +et be deﬁned by )=2 . The restriction of to the integers is the function where )= for all (because if is an integer then ). Countingrestrictionsandextensions . +et ,B s,t,u,v,w,x,y,z and be (1 ,t (2 ,x ( ,x (5 ,s (7 ,t . The number of extensions of to a function from ,..., to is 6 , since there are 6 choices for the image of each of 6 6 and ? (in an extension the images of 1 5 and 7 must be the same as for ). The range of is s,t,x , and the number of subets such that also has range is 1 , since : must contain 5 (since )= ), must contain at least one of 1 and 7 (since ) must contain at least one of 2 and (since )= ). Deﬁnition (converse of a relation) . +et be a relation from to . The converse of is the relation from to deﬁned by b,a ):( a,b . (This is the relation obtained by reversing the components of each ordered pair in .) Motivating question . Suppose is a function. Then, by deﬁnition, is relation from to . It is natural to wonder when the converse relation, , is a function (from to ). Example ( not a function) . +et ,B a,b,c,d and (1 ,a (2 ,b ( ,c . Then, a, 1) b, 2) c, ) is not a function because it contains no ordered pair with ﬁrst component equal to . (Equivalently, the preimage of some 1-element subset of is the empty set.) The same situation will arise for any function from a set to a set where the range is a proper subset of the codomain. Ience, for to have a chance at being a function, it must be true that )= . (These are called onto functions.)

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Example ( not a function) . +et ,B a,b and (1 ,a (2 ,b ( ,b . Then, a, 1) b, 2) b, ) is not a function because it contains two ordered pairs with ﬁrst component equal to . The same situation will arise for any function from a set to a set where some two elements of have the same image. (That is, there exist ,a where but )= )). Equivalently, the preimage of some 1-element subset of contains two or more elements.) Ience, for to have a chance at being a function, it must be true that if ,a and then ). (These are called 1-1 functions.) We will develop some results about the functions with the properties suggested by the two examples above, and after that we will return to the question of when the converse of a function from to is a function from to Deﬁnition (1-1 function) . 1 function is called one!to!one (1-1, or inLective, or an inLection) if )= ) implies , for all x,y . ( That is, is 1-1 if and only if di0erent elements of have di0erent images in . Equivalently, is 1-1 if and only if every element of is the image under of at most one element of .) Proposition F2 .If and are ﬁnite sets and is a 1-1 function, then |≤| Proof Suppose . If no two elements of have the same image under , then the range of contains exactly elements. Since the range of is a subset of ,wehave ≤| Proving 1-1 . To prove a function is 1-1, start with B Assume )= ).C and then argue, using what you are given about , that (the last clause is ... . Therefore is 1!1. ) Equivalently, you could prove the contrapositive: start with B Assume and argue until you can conclude with B Then C. Example (proving a function is 1-1) . +et be deﬁned by )=7 4. We prove is 1-1. 1ssume )= ). Then 7 4=7 4. In turn, this implies 7 =7 and . Therefore is 1-1. Disproving 1-1 . To prove that a function is not 1-1, ﬁnd distinct elements and in the domain so that )= ). ,y doing this, you have demonstrated that the implication ( )= )) ) is False, and so )= ) implies , for all x,y , is False. Example (proving a function is not 1-1) . +et be deﬁned by (1)=1, and for ,f ) is the largest prime factor of . Computing a few values of yields (2) = 2 ,f ( ) = and (5) = 2. Since (5) = (2) and 5 = 2, the function is not 1-1. ,dvice about investigating functions re- 1-1 . If it is not possible to easily identify elements and in the domain so that )= ), then start trying to prove that is 1-1. If isn’t 1-1, then the proof will break down at some point, and this will lead to the required elements. For example, let be )= 15 A74=( 4) A6. Fow, )= 4) A6=( 4) A6 4) =( 4) . To prove is

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1-1, it is necessary to obtain , but the last equality gives 4= 4), which leads to or 15. The last of these equations gives the elements and required to show is not 1-1: Choose = = (say) and 15 = 15. Then and )=74= ) (check the arithmeticP), so is not 1-1. Counting 1-1 functions . Suppose and . We count the number of 1-1 functions from to . ,y Proposition F2, if m>n there are none. 1ssume Suppose ,,a ,...,a . There are choices for the image of and, for each of these there are 1 choices for the image of 2 choices for the image of , and so on until, ﬁnally, there are 1) = A1 choices for the image of . Thus, by the rule of product, the number of 1-1 functions from to is 1) A1) = )P. ;ne to one functions allow us to describe a situation when equality holds in theorem F1 (b). Proposition F3 . +et be a 1-1 function and ,A . Then )= ). Proof ,y Theorem F1 (b), ). Thus, it remains to prove the opposite inclusion. Suppose is 1-1, and let ). Then ) for some and ) for some . Since is 1-1, )= . Ience, ) for some , that is, ). Thus, ), as required. Compare the part of the argument where the property that is 1-1 was used to the paragraph following Theorem F1. To see that the converse of Proposition F is false, consider where ) = = for all Deﬁnition (onto function) . 1 function is called onto (or surLective, or a surLection) if for every element there is an element such that )= . (That is, is onto if and only if every element of is the image under of some element of Equivalently, is onto if and only if every element of is the image under of at least one element of .) Proposition F. .If and are ﬁnite sets and is an onto function, then |≥| Proof Suppose is onto. Then every element of has at least one preimage. ,y the deﬁnition of function, each element of is a preimage for a unique element of . Thus, | 1= Proving onto . To prove that a function is onto you must argue that for every (equivalently, an arbitrarily chosen ) there is an such that )= . Thus, the ﬁrst line of the proof is B Let . We must ﬁnd such that )= C. What you do next depends on the description of and of the set . 2sually, you BsolveC )= for in terms of (as written it gives in terms of ). Then, you verify that what you have is an element of . Finally, you substitute this back into and show that it gives The last line of the proof is B (ence if ... , then )= , and so is onto. C.

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Example (proving onto) . We show that the function deﬁned by )= ?A( A ) is onto. +et . Then )= ?A( A ) A ) A? A = A ?. Since ,x A = and we can disregard the negative square root to obtain A A ?. We must verify that this belongs to the domain. Since we have A? ? and A? . Therefore A A? Ience, if A A ?, then )= ?A(( A A?)A ) ?A( A?) and so is onto. Disproving onto . To prove that a function is not onto you must ﬁnd an element which is not ) for any . Iow you do this depends on . In general, it is useful to try to prove that is onto as above. If it isn’t onto, then you will reach a point where either you can’t solve for in terms of , or you will succeed in doing this but the only possibilities you ﬁnd are not elements of . In either case you are done. What you are doing is assuming that is onto and showing that leads to (logically implies) a contradiction. Since only a False statement logically implies a contradiction, it must be that the assumption that is onto is False. Example (disproving onto) . Consider the function deﬁned by )= 2) 1=. +et . Then )= 2) 1= = 2) A 1=. If is chosen so that A1= = (say 11) then there is no real number so that 2) A 1= because ( 2) = for all real numbers . Thus, there is no so that )= 11, so is not onto. Example(disprovingonto) . Consider the function deﬁned by )=2 A . +et . Then )= A = =( ) . ,ut ( ) for all : In particular if = = then ( ) 2= 2. We now have that if ) = =, then 2. Thus, there is no such that )==,so is not onto. Onto depends on the codomain . ;bserve that the function in the example above is onto as a function from to Exercise : What about from to For ﬁnite sets and , it has been straightforward to count the number of functions from to , and also the number of 1-1 functions from to . We do not presently have the techniques necessary to count the number of onto functions from to . This will have to wait until we have covered the Principle of Inclusion and Exclusion, later on. Iowever, we state and use the result now. Fact F1 . +et and be sets with and . Then, the number of onto functons from to is =0 1) Example (counting using onto functions) . We count the number of ways that a collection of 7 labelled (distiguishable) containers can hold a collection of 6 labelled (dis- tinguishable) balls, if no container is left empty. This is the same as the number of functions from the set of balls onto the the set of containers ( ) is the container that holds ball ), which is =0 1) (7 = 126===. The number of ways in which the ﬁrst containers hold the ﬁrst 7 balls and the last 2 containers hold the last balls (and, still, no container is left empty) equals the number of functions from a set of 7 labelled balls to the a set of labelled containers, times the 1=

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number of functions from a set of labelled balls to the a set of 2 labelled containers. This is =0 1) ( =0 1) (2 = 665. Corollary F2 . +et and be ﬁnite sets. If is both 1-1 and onto, then Proof Since is 1-1, we have |≤| by Proposition F2. Since is onto, we have |≥| by Proposition F5. The result now follows. Deﬁnition (1-1 correspondence) . 1 function is called a 1!1 correspondence (or biLective, or a biLection) if it is both 1-1 and onto. ,y the deﬁnitions of 1-1 and onto, a function is a 1-1 correspondence if and only if every element of is the image of exactly one element of under . To determine if a function is a 1-1 correspondence, use the methods discussed above to check whether it is 1-1 and onto. (;f course, you can stop once it fails to have one of these properties.) Corollary F6 is an important counting principle. It says that if two ﬁnite collections of obLects can be put into 1-1 correspondence, then there are the same number of obLects in each collection. For example, there is a 1-1 correspondence between the subsets of ,a ,...,a and the set of binary sequences ...b of length : It is given by the function deﬁned by ) is the binary sequence ...b where for =1 ,...,n 1if =if X. To see that is 1-1, note that if )= ) then the deﬁnition of implies that and contain exactly the same elements. To see that is onto, let ...b and construct ∈P ) by the rule = 1. If follows from the deﬁnition of that )= ...b . Ience, the number of subsets of equals the number of binary sequences of length , which equals 2 (two choices for each position). Exercises . Suppose that . Prove: (a) If is a 1-1 function, then it is a 1-1 correspondence (i.e. is also onto). (b) If is an onto function, then it is a 1-1 correspondence (i.e. is also 1-1). Proposition F3 . +et be a function. (a) If is 1-1, then any restriction of to a subset is 1-1. (b) If is onto, then for any , any extension of to a function is onto. Proof (a): +et x,y and suppose )= ). Since , we have x,y . Since is 1-1, )= ). Therefore is 1-1. (b): +et . Since is onto, there exists such that )= Deﬁnition (composition of functions) . +et A,B and be sets, and and be functions. The composition of and (or the composite function) is the function ( ): deﬁned by )( )= )) for every element in 11

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Function composition is not commutative . In general, order matters. For and as above, the composition of and is usually not the same as the composition of and . For example, let be deﬁned by )= and be deﬁned by )= A . Then, )( )= )) = )= A and )( )= )) = A )=( A ) A6 A?. Since ( )(=) = and ( )(=) = ?, we have . It is also possible that, depending on the sets A,B and , the function is deﬁned but the function is not. We leave it as an exercise to ﬁnd an example that demonstrates this. Deﬁnition (identity function on a set) . The identity function on the set is the function 1 deﬁned by 1 )= for every element of The integer = is an identity element with respect to addition of real numbers: A== =A for all . Similarly, 1 is an identity with respect to multiplication of nonOero real numbers: 1 1= for all −{ . The identity function on a set acts in a similar way with respect to function composition. Proposition F4 . +et and be sets and be a function. Then (a) , and (b) 1 Proof We prove (a) and leave (b) as an exercise . +et . Then, )= (1 )) = ). Ience, Even though function composition is not commutative, it is associative. We now prove this. Proposition F5 . +et A,B,C , and be sets and B, g , and be functions. Then )=( Proof ,y deﬁnition of function composition, both ) and ( are functions from to , so they have the same domain and codomain. +et . Then, ))( )= (( )( )) = ))=( )( )) = (( )( ), as required. Proposition F16 . +et A,B and be sets and and be functions. Then, (a) If and are 1-1, then is 1-1. (b) If and are onto, then is onto. (c) If and are 1-1 correspondences, then is a 1-1 correspondence. Proof We prove (a) and leave the proof of (b) for an exercise. Statement (c) is an immediate consequence of (a) and (b). Suppose ( )( )=( )( ). Then )) = )). Since is 1-1, )= ). Since is 1-1, . therefore, is 1-1. 12

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Exercises : ﬁnd examples to demonstrate that the converse of each statement in propo- sition F1= is false, but the following statement is true: If is a 1-1 correspondence, then is 1-1 and is onto (but need not be onto and need not be 1-1). 1lso, ﬁnd an example to demonstrate that the converse of the above implication is False (never mind the part in brackets). We now return to the question of when the converse of a function is itself a function. Proposition F11 . +et be a function. Then is a function if and only if is 1-1 and onto. Proof ) Suppose is a function. ,y the deﬁnition of function, for every Bf contains exactly one ordered pair with ﬁrst component equal to . Thus, for every contains exactly one ordered pair with second component equal to . That is, is 1-1 and onto. ) Suppose is 1-1 and onto. Then, for every Bf contains exactly one ordered pair with second component equal to . This means that for every Bf contains exactly one ordered pair with ﬁrst component equal to . That is, is a function. Suppose is a 1-1 and onto function. Then, by Proposition F11, is a function. Since ( and is a function, Proposition F11 implies that the function is (also) 1-1 and onto. In the arithmetic of real numbers, the inverse of is because A( ) equals the additive identity, Oero. For = =, the multiplicative inverse of is 1 /x because (1 /x equals the multiplicative identity, one. The inverse of a function is analogous, and is deﬁned below. Deﬁnition (invertible function) . 1 function is called invertible if there exists a function such that =1 and =1 Suppose is a 1-1 correspondence. Then, by Proposition F11, is a function. Consider the composite function .For we have ( )= )) = by deﬁnition of converse. Thus, =1 . Similarly, for we have ( )( )= )) = . Thus, =1 . Therefore, a 1-1 and onto function is invertible. Proposition F12 . Suppose is invertible. Then the function such that =1 and =1 is unique. Proof . Suppose and are functions such that =1 and =1 and =1 and =1 . We need to show . +et . Then )= (1 ))=( )( )=( ))( )=(( )( )=(1 )( )=1 )) = ). Thus, . This comepletes the proof. Deﬁnition (inverse) . +et be an invertible function. The inverse of is the unique function such that =1 and =1 . It is customary to denote the inverse of by In the case where is 1-1 and onto, we have above that is invertible and 1

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Ho not confuse the notation for the inverse of a function, (which exists only sometimes) and the notation for the preimage of a subset of the codomain (which always exists). They use the same symbols, but the former of these is a function and the latter is a set. It should be clear from the context which obLect is under discussion. Proposition F13 . +et be a function. Then, is invertible if and only if it is 1-1 and onto. Proof ) Suppose is 1-1 and onto. We have noted above that in this case is invertible and ) Suppose that is invertible. Then, there is a function such that =1 and =1 Suppose )= ). Then )) = )). The +IS of this equation is )( )=1 )= and the DIS is ( )( )=1 )= . Thus, , and is 1-1. +et . Since is a function, there exists such that )= .,y deﬁnition of inverse function, we have )= ))=( )( )=1 )= Therefore is onto. This completes the proof. Combining Propositions F12 and F1 gives that if is an invertible function, then . ,y deﬁnition of converse, this means that has the property that, for and )= )= . (Compare this to the last paragraph in the proof of Proposition F1 .) Some (inverse) functions are deﬁned in exactly this way – for example for and ,wehave log 10 )= 1= Proving is invertible and ﬁnding the inverse . There are two ways to prove that a function is invertible. ;ne way is to write down the right function and then check that =1 and . 1nother way is to use Proposition F1 and check if is 1-1 and onto. In the proof that is onto, one starts with C+et and suppose )= .C and ultimately derives BThen ).C. ,y our discussion in the previous paragraph, is the inverse function. Thus, a description of the inverse function is a byproduct of the proof that is onto. Fotice that the deﬁnition of an invertible function is symmetric in and (a.k.a. ). Thus, if is invertible, so is and ( Example (proving is invertible and ﬁnding . +et be deﬁned by )=2 ln 4. We show that is invertible and ﬁnd . Suppose )= ). Then, we have 2 ln 4=2 ln ln )=2 ln ln )= ln ln ln( , since the exponential and logarithm functions are inverses. Thus, is 1-1. Fow, suppose . Then, )= ln 4= ln )=( A4) +7) For the quantity +7) , the domain of . Ience, if +7) then )= +7) )=2 ln +7) 4=2( A4) 4= ,so is onto. 9oreover is described by )= +7) 15