Section 3.1. ; X = (0, 1]. (i) f : R R R, f (x, y) = x y

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1 Paul J. Bruillard MATH Problem Set 6 An Introduction to Abstract Mathematics R. Bond and W. Keane Section 3.1: 3b,c,e,i, 4bd, 6, 9, 15, 16, 18c,e, 19a, 0, 1b Section 3.: 1f,i, e, 6, 1e,f,h, 13e, 17, 0, 1,, 6, 9, 30 Section 3.1 Problem For the following functions, compute first the image of f and then the image of the given subset X (No formal proofs are necessar.) (b) f : R R, f () = { + 1; X = [ 1, 4]. n if n is even (c) f : Z Z, f (n) = n if n is odd ; X = E. (e) f : R R where R is the set of nonzero real numbers, f () = 1 ; X = (0, 1]. (i) f : R R R, f (, ) = ; X = {(, 1) R}. Remark 0.1. Since the problem eplicitl stated that a formal proof was not required I will simpl sketch the calculation. Part (b) We first note that f () describes a line: So graphicall we see that its image is all of R and that the image of X is obtained as follows (image depicted in blue):

2 Algebraicall, we have Im (f) = f (R) = { R R such that f () = } However, = f () = + 1 is alwas solved b = 1 and hence Im (f) = R. Similarl, the image of X under f is given b f (X) = { R [ 1, 4] such that f () = } = { + 1 R [ 1, 4]} = [ 1, 9] Thus the image of f and the image of X under f are respectivel given b: Im (f) = R, f (X) = [ 1, 9] Part (c) Here we see that f is discrete, graphicall it appears as a collection of points: From here one can read off the range of f. Indeed, each even number is doubled so f (n) = 4n for n Z and each odd number is fied, i.e., f (n + 1) = n + 1 for n Z. Therefore, the image of f ma be epressed as Im (f) = {4n n Z} {m + 1 m Z}

3 We additionall see from here that the image of E under f is given b f (E) = 4Z = {4n n Z} Collecting the results we see that the image of f and the image of E under f are respectivel given b: Im (f) = {4n n Z} {m + 1 m Z}, f (E) = 4Z = {4n n Z} Part (e) Since f is a real-valued function with domain in R it is easil plotted From the plot it is readil apparent that Im (f) = R = { R 0}. This can be seen algebraicall b noting that for 0 the equation 1 = can be solved b = 1. This solution makes manifest the problem with = 0, one would have to divide b zero to solve for. The image of X = (0, 1] can similarl be determined. Indeed, as 0 f () and at = 1 we have f (1) = 1. Thus f (X) = [1, ). Therefore, the image of f and the image of X under f are respectivel given b Im (f) = R, f (X) = [1, ) Part (i) This function defines a nast -dimensional surface:

4 This plot is of course incomplete since as 0 the function diverges. However, we see immediatel that the image of the function is all of R. To see this algebraicall we would need to solve = z for an z R. This is of course possible b taking = 1 and = z. This analsis further shows that the image of X = {(, 1) R} under f is R. Indeed, f (, 1) = for an R. Consequentl, the image of f and the image of X under f are respectivel given b Im (f) = R, f (X) = R Problem 3.1.4b,d Which of the following are functions from R to R? For those that are functions, find their image. (Drawing a graph will help ou determine the image. No formal proof is necessar.) { if 0 (b) f () = if < 0 { 4 if > 0 (d) f () = 3 if < 0 Remark 0.. Just as in the previous problem, no proof is required, and so I will merel sketch the process utilizing graphs. Part (b) { if 0 Recall for this part f () =. From here we see that the image of f is alwas if < 0 a real number and f is defined on all of the real numbers. Indeed, whenever is negative we know that is positive and so at no point does computing f require us to take the square-root of a negative number. Moreover, f passes the vertical line test, i.e., it does not assign values to a single point R. Therefore, f is a function from R to R. Having convinced ourselves that f : R R reall is a function we need to determine its image. As the problem suggests, we start b graphing the function From here we see immediatel that all small numbers are in the image of f but what about the large ones? Well, as we are taking the square-root of larger and larger numbers 4

5 which still produces a larger and larger number. Indeed, calculus tells us that lim = and similarl lim =. Thus Im (f) = R Part (d) f is not defined at 0, a point in the purported domain and hence f is not a function. Since f is not a function there is no need to compute its image. f would be a function if it was considered as f : R \ {0} R. In this case the graph of the function is From here we see that Im (f) = R \ {0}. Problem Let f : Z Z be defined b f (n) = n + 1. What is wrong with the following proof that Im (f) = Z? ( ) ( ) Let Z. Then f 1 = =. Hence Im (f). Therefore, Z Im (f). Since Im (f) Z b definition, it follows that Im (f) = Z. The first thing we should ask ourselves is: Do we believe that Im (f) = Z?. The answer is, of course, no. Indeed, from the definition of f we see that f (n) = n + 1 which is alwas odd. Thus we think that the image of f is all odd integers. Now ( to) the eamination of the proof. The proof sas to take Z and then compute f 1. This is the problem with the proof. Indeed, if we take = then 1 = 1 which is not an integer. Since f is onl defined on the integers the epression f ( 1 ) is nonsense! 5

6 Problem Let f : R Z be the greatest integer function. (a) Compute f ([0, 1]). Prove our answer. (b) Compute f ([ 1, 1]). Prove our answer. Part (a) Claim 0.3. f ([0, 1]) = {0, 1}. First note that f (0) = 0 and f (1) = 1 whence {0, 1} f ([0, 1]). On the other hand if [0, 1] then = 0 or > 0. We ve alread seen that f (0) = 0 and so we need onl focus on f () for 0 < 1. In this cas,, b definition of f, we know that f () = 1. Consequentl, f ([0, 1]) {0, 1}. Part (b) Claim 0.4. f ([ 1, 1]) = { 1, 0, 1}. The proof here proceeds analogousl to part (a). Indeed, f ( 1) = 1, f (0) = 0, and f (1) = 1. So we know that { 1, 0, 1} f ([ 1, 1]). So it suffices to show that f () { 1, 0, 1} for [ 1, 1]. Since we ve dealt with { 1, 0, 1} it suffices to consider ( 1, 0) and (0, 1). In the first case, the definition of f tells us that f () = 0. While in the second case we have that f () = 1. Therefore φ ([ 1, 1]) { 1, 0, 1}. Since we ve shown both containments we can conclude that f ([ 1, 1]) = { 1, 0, 1}. Problem Let A and B be sets and X and Y be subsets of A. Let f : A B be a function. (a) Prove that f (X Y ) f (X) f (Y ). (b) Give an eample of sets A and B and a function f : A B for which f (X Y ) f (X) f (Y ) for some subsets X and Y of A. Part (a) Claim 0.5. f (X Y ) f (X) f (Y ) This proof can be done b direction calculation, i.e., let f (X Y ) and then show f (X) f (Y ) b using the definition of the image of a set under a function. A quicker wa to this result is to utilize Proposition Indeed, b definition of intersection, we know that X Y X. So Proposition tells us that f (X Y ) f (X). Similarl, the definition of intersection implies that X Y Y. Once again turning to Proposition allows us to deduce that f (X Y ) f (Y ). 6

7 Thus for an f (X Y ) we know that f (X) and f (Y ). In particular, the definition of the intersection tells us that f (X) f (Y ). Since was an arbitrar element in f (X Y ) we can conclude that f (X Y ) f (X) f (Y ). Part (b) An eample is fairl eas to construct. For instance, one ma take the function f : R R given b f () =. From here one can show that f ([5, 6]) = [5, 36] = [ 5, 6 ] [ = ( 5), ( 6) ] = f ([ 6, 5]). So if we take X = [5, 6] and Y = [ 6, 5] then we have X Y = and so f (X Y ) =. However, f (X) = f (Y ) = [5, 36] and so f (X) f (Y ) = [5, 36]. Since 5 [5, 36] we see that [5, 36]. Problem Let A and B be sets and X and Y subsets of A. Let f : A B be a function. (a) Prove that f (X) \ f (Y ) f (X \ Y ) (b) Give an eample of sets A and B and a function f : A B for which f (X) \ f (Y ) f (X \ Y ) for some subsets X and Y of A. Part (a) Claim 0.6. f (X) \ f (Y ) f (X \ Y ) Let z f (X) \ f (Y ). Then z f (X) and z / f (Y ), b definition of the set difference. Therefore z = f () where X but / Y. In particular, X \ Y. Consequentl z = f () f (X \ Y ). Since z was an arbitrar element in f (X) \ f (Y ) we can conclude that f (X) \ f (Y ) f (X \ Y ). Part (b) Here we ll take A = B = R, X = [ 1, 0], and Y = [0, 1]. f () =. Then X \ Y = [ 1, 0) and so f (X \ Y ) = (0, 1]. On the other hand f (X) \ f (Y ) = [0, 1] \ [0, 1] =. So f (X) \ f (Y ) f (X \ Y ). We then define f : R R b Problem c, e For the following functions, compute the inverse of the given subsets of the codomain. (No proofs are necessar). (c) f : R R, f () = { cos (); W 1 = [ 1, 1], W = { R 0}, W 3 = Z. n if n is even (e) f : Z Z, f (n) = n 1 if n is odd ; W 1 = E, W = {1}, W 3 = {6}, W 4 = O, the set of odd integers. Remark 0.7. This problem eplicitl states that proofs are not necessar and so I will not provide proofs, rather I will plot the functions and eamine the graphs to determine the answer. 7

8 Part (c) Recall that the function under consideration is f : R R given b f () = cos (). Plotting this function we see We can now appl our graphical method for computing points in the preimage. For instance if we want to find the preimage of 1/ we draw a horizontal line at = 1 and ever time we run into the graph we drop a perpendicular line to the -ais. The places that these lines cross the -ais is the preimage of that point. From here it becomes apparent that an f 1 ([ 1, 1]) = R. This is not surprising since Im (f) = [ 1, 1] and f is defined everwhere on R. Similarl we can compute f 1 (W ) and we find f 1 (W ) = { R f () W } = { R cos () 0} = [ ] (4n 1) π (4n + 1) π, n Z Finall, we can determine f 1 (Z). Since Im (f) = [ 1, 1] we see that f 1 (Z) = f 1 ([ 1, 1] Z) = f 1 ({ 1, 0, 1}). Returning to our graph we can get a feel for this set geometricall 8

9 Algebraicall, we would find all such that cos () = 0, 1, or 1. Eamination of the above plot or the unit circle tells us that this set consists of all half integer multiplies of π. Indeed, we have { nπ } f 1 (Z) = n Z Collecting the results we have: f 1 (W 1 ) = R f 1 (W ) = [ (4n 1) π, n Z { nπ } f 1 (Z) = n Z ] (4n + 1) π Part (e) Here we need to compute the inverse of the subsets W 1 = E, W = {1}, W 3 = {6}, and W 4 = O for the function f : Z Z given b { n if n is even f (n) = n 1 if n is odd This proceeds just as in the previous two problem The onl difference is that we need to ensure all of the points we find are integers. Taking care when computing, one can show directl that f 1 (W 1 ) = Z f 1 (W ) = f 1 (W 3 ) = {6, 7} f 1 (O) = The second and third results listed here can be understood b noting that f alwas produces an even number. Indeed, f (n) = n and f (n + 1) = n for an n Z. Problem a Let f : Z Z be defined b f (n) = Compute f 1 ({11, 1, 13, 14, 15}). Claim 0.8. f 1 ({11, 1, 13, 14, 15}) = {5,, 4, 6, 8, 30}. 9 { n if n is even n + 4 if n is odd.

10 Here we need to compute f 1 for each element in the finite set {11, 1, 13, 14, 15}, the set containing the elements we compute will be the preimage. So we note that 11 = f (n) means either 11 = n and n is even or 11 = m + 4 and m is odd. The second case cannot occur because 11 is not divisible b and so n =. Net, we tr and solve 1 = n with n even or 1 = m + 4 with m odd. In the first case we can take n = 4, in the second case we have m = 6 = 4 which is not odd. Thus n = 4 is the onl element leading to 1. Moving on to the net number we tr to solve 13 = n/ for n even or 13 = m + 4 for m odd. The second equation has no solution because 13 is not divisible b. The first equation is solved b n = 6. Net we take the equations 14 = n/ for n even and 14 = m + 4 for n odd. The second equation sas that m = 5 while the first equation is solved b n = 8. Finall, we should tr and solve 15 = n/ for n even and 15 = m + 4 for m odd. The first equation has a unique solution, n = 30. While the second equation has no solution in Z because 15 is not divisible b. Amalgamating these results gives the answer. Problem Let f : A B be a function. Let W and Z be subsets of B. Prove that f 1 (W Z) = f 1 (W ) f 1 (Z). To establish this result we ll show both inclusions beginning with f 1 (W Z) f 1 (W ) f 1 (Z). To show this we let a f 1 (W Z). Then f (a) W Z and so f (a) W and f (a) Z. Therefore, a f 1 (W ) and a f 1 (Z). B definition of intersection we can conclude that a f 1 (W ) f 1 (Z). Since a was arbitrar we have f 1 (W Z) f 1 (W ) f 1 (Z). Conversel, we let a f 1 (W ) f 1 (Z). Then a f 1 (W ) and a f 1 (Z). Therefore, f (a) W and f (a) Z. In particular, f (a) W Z Therefore, a f 1 (W Z). Since a was an arbitrar element of f 1 (W ) f 1 (Z) we can conclude that f 1 (W ) f 1 (Z) f 1 (W Z). Problem 3.1.1b Give an eample of a function f : A B for some A and b and a subset X of A such that X f 1 (f (X)). Take f : R R defined b f () = cos (). Then f ([0, π]) = [ 1, 1] but we saw above that f 1 ([ 1, 1]) = R which is ver definitel not equal to [0, π]. (So here A = B = R and X = [0, π].) 10

11 1 Section 3. Problem 3..1f, i Determine which of the following functions are surjective. Give a formal proof of our answer (f) f : R R, f () = { if 0 (i) f : R R, f () = if < 1 We ll first plot the function to decide what to claim. A function f : R R will be surjective if for an horizontal line we draw, it intersects the graph of the function at least once. Part (f) From here it is prett clear what we should claim. Claim 1.1. f : R R given b f () = 3 + is surjective. The proof proceeds via the Intermediate Value Theorem (Theorem 3.1.4). Indeed we know that lim f () = f () = lim Noting that f is continuous (it is a polnomial) we can appl the Intermediate Value Theorem to conclude that an real number is in the image of f. Therefore R Im (f). On the other hand Im (f) R since R is the codomain of f and the image is alwas contained in the codomain. Consequentl, R = Im (f) and hence f is surjective. 11

12 Part (i) Once again we plot the function to decide what to do. Here is appears as though [0, 1) is not in the range of f and so f is not surjective. Thus we claim { 1 if 0 Claim 1.. The function f : R R defined b f () = is not surjective if < 1 To show that f is not surjective we ll show that 0 is not in the image of f. This will be done via contradiction and so we suppose that 0 Im (f). Then R such that f () = 0. We know that 0 or < 0 and so we should consider these cases separatel. Case 1: 0. Then the definition of f tells us that 0 = f () = 1. Since 0 we know that 0 and hence 1 < 0. Since 0 is not less than zero we can conclude that < 0 (case ). Case : < 0. Then we know 0 = f () =. However, > 0 implies that > 0 and so we have 0 > 0, a contradiction. In either case we arrive at a contradiction and so we are forced to conclude that 0 / Im (f). In particular, f is not surjective. Problem 3..e Determine if f : Z Z defined b f (n) = Prove our answer { n 1 if n is even n if n is odd is surjective. In this setting f alwas produces an odd number (as can be seen b eamining function). This motivates the claim { n 1 if n is even Claim 1.3. The function f : Z Z defined b f (n) = is not surjective. n if n is odd 1

13 P. J. Bruillard To see this we ll show that 4 Z \ Im (f). This is done b contradiction and so we suppose that 4 Im (f). Then n Z such that f (n) = 4. Since the even and odd integers partition the set of integers we know that n is either even or odd. Thus we proceed b cases. If n is even then n = m for some m in Z. In this case 4 = f (n) = f (m) = m 1. 4 is even but the right hand side is odd (to see this reduce modulo ), this is a contradiction. If n is odd then n = m + 1 for some m Z, and 4 = f (n) = f (m + 1) = (m + 1) = 4m +. Dividing through b 4 shows that 1 = 1 m Z, a contradiction. Since we alwas arrive at a contradiction we can conclude that our hpothesis is false and hence 4 / Im (f). Therefore f is not surjective. Problem 3..6 Let A and B be sets and let X be a subset of A. Let f : A B be a surjective function. (a) Prove that B \ f (X) f (A \ X) (b) Give an eample to show that equalit does not in general hold in part (a). Part (a) Claim 1.4. B \ f (X) f (A \ X) The proof proceeds b direct calculation. Suppose that b B \ f (X), then b B and b / f (X). Since f is surjective, a A such that f (a) = b. However, b / f (X) and so a / X. Therefore, a A \ X and b = f (a) f (A \ X). Consequentl, B \ f (X) f (A \ X). Part (b) Consider the function f : R [0, ) given b f () =. Then in the contet of the problem we have A = R and B = [0, ). Now let X = [0, 1] A and note that f (X) = [0, 1]. Then B \ f (X) = (1, ). However, X \ X = (, 0) (1, ) and so f (A \ X) = (0, ) 13

14 Problem 3..1e,f,h Determine which of the following functions are injective. proof of our answer (e) f : R R +, where R + is the set of positive real numbers and f () = e. (f) f : Z Z Z, f (n) = (n, n). (h) f : R R R, f (, ) =. Give a formal Part (e) The idea here is to plot f. If an horizontal line that intersects the graph of f, intersects it eactl once, then f is injective. So plotting we see This looks as though it will pass this horizontal line test and so we claim Claim 1.5. f : R R + defined b f () = e is injective. To show this we suppose that, R such that f () = f (). Then we have e = e. Taking logarithms we find that = ln (e ) = ln (e ) =. So b definition, f is injective. Part (f) This is an eample of a tpe of map known as a diagonal embedding and takes the integers and lifts them to a diagonal line. Indeed, if we view the parametric curve defined b f in the plane we see the integer points on the line =. This motivates the claim Claim 1.6. f : Z Z Z defined b f (n) = (n, n) is injective. This proceeds b direct calculation, indeed let m, n Z such that f (n) = f (m). Then we have (n, n) = (m, m) and hence n = m. The result follows b returning to Definition 3... Part (h) We saw a plot of this surface in the previous section. Staring at this plot for a long time we see that it would fail to be injective, i.e., a plane perpedicular to the z-ais will cross the function in more than one point. This motivates us to claim 14

15 Claim 1.7. f : R R + R defined b φ (, ) = is not injective. Since we are claiming that the function is not injective it suffices to find distinct points in the domain of f that map to the same point in the image. There are man choices, we ll take the following (4, ) (, 1), f (4, ) = 4 = = 1 = φ (, 1) Problem 3..13e Determine if { the following function is injective and provide a formal proof of n 1 if n is even our answer. f : Z Z, f (n) = n if n is odd This function seems to flip evens and odds (plug in a few numbers and tr it out). Moreover, each of the functions n 1 and n would be injective if the were considered alone (the are just lines). This motivates us to claim { n 1 if n is even Claim 1.8. f : Z Z, f (n) = is injective. n if n is odd To see this we suppose that m, n Z such that f (n) = f (m). Up to relabeling there are three cases to consider. Case 1: m and n are both even. Then n 1 = f (n) = f (m) = m 1 and hence m = n. Case : m and n are both odd. Then n = f (n) = f (m) = m and hence m = n. Case 3: m is even and n is odd. Then we have m 1 = f (m) = f (n) = n. However, this cannot occur since the left hand side if even but the right hand side is odd. In all cases we see that m = n and hence f is injective. Problem Let A and B be sets and X and Y subsets of A. Let f : A B be an injective function. Prove that f (X Y ) = f (X) f (Y ). (compare with eercise ) In the previous section (eercise ) we showed that f (X Y ) f (X) f (Y ). suffices to show that f (X) f (Y ) f (X Y ). To do this we let b f (X) f (Y ). Then b f (X) and b f (Y ). So X and Y such that f () = b = f (). Since f is injective we can conclude that = and hence X Y. Therefore, b = f () f (X Y ). Since b was a generic element of f (X) Y we see that f (X) f (Y ) f (X Y ). So it 15

16 Problem 3..0 Let f : A B be a function. Which of the following statements are equivalent to the statement f is injective. (a) f (a) = f (b) when a = b. (b) f (a) = f (b) and a = b for all a, b A. (c) If a and b are in A and f (a) = f (b), then a = b. (d) If a and b are in A and a = b, then f (a) = f (b). (e) If a and b are in A and f (a) f (b), then a b. (f) If a and b are in A and a b, then f (a) f (b). Man of these are just general features of functions e.g. (a) and (d). Thinking carefull about each statement gives the following (a) Not equivalent (b) Not equivalent (c) Equivalent (d) Not equivalent (e) Not equivalent (f) Equivalent Problem 3..1 Let f : A B be an injective function. Let X be a subset of A. Prove that f 1 (f (X)) = X (compare with eercise 3.1.1) From eercise (see above), we know that X f 1 (f (X)). Thus it suffices to prove that f 1 (f (X)) X. To this end we let a f 1 (f (X)). Then f (a) f (X) b definition of the preimage. Consequentl f () = f (a) for some X. However, f is injective and so = a. Since X b hpothesis, we can conclude that a X. Therefore f 1 (f (X)) X and hence we ma conclude that f 1 (f (X)) = X. Problem 3.. Let f : A B be a surjective function. Let W be a subset of B. Prove that f ( f 1 (W ) ) = W (compare with eercise 3.1.) Appling eercise 3.1. (see above), we know that f ( f 1 (W ) ) W. Thus it remains to show W f ( f 1 (W ) ). To do this we let b W and appl surjectivit of f to conclude that there is an a A such that f (a) = b. Because f (a) W we can conclude that a f 1 (W ). Therefore, b = f (a) f ( f 1 (W ) ) and hence W f ( f 1 (W ) ). 16

17 Problem 3..6 Let f : A B be a function where A and B are finite sets such that A = B. Prove that f is injective if and onl if f is surjective. This is a biconditional statement and so there are two directions to prove. ( ) Suppose that f is injective. Then b definition of injective each element of A goes to a unique element of B. That is, Im (f) = A. However, A = B and so Im (f) = B. On the other hand, Im (f) B and these finite sets have the same number of elements. Therefore Im (f) = B and so f is surjective. ( ) Suppose that f is surjective. Then Im (f) = B. In particular, Im (f) = B = A. However, ever element of a gets mapped to eactl one element of f. That is, we have A = {a 1,..., a n } and Im (f) = {f (a 1 ),..., f (a n )}. Since these sets have the same cardinalit (we just showed this) it must be that the f (a i ) are distinct i.e. f (a i ) = f (a j ) if and onl if a i = a j. This is eactl the statement that f is injective. Problem 3..9 Let A and B be sets and let f : A B be a function. Prove that f is bijective if and onl if for ever b B, f 1 ({b}) is a single element subset of A. Since this is a biconditional statement there are two directions to prove. ( ) Suppose that f is bijective and let b B. Then since f is surjective we know a A such that f (a) = b. In particular a f 1 ({b}). This tells us that f 1 ({b}) and so we ma get c f 1 ({b}). B definition of this inverse set we know f (c) = b. On the other hand, f (a) = b and so we have f (a) = f (c). Since f is injective we can conclude that c = a. Therefore, f 1 ({b}) = {a}. ( ) Suppose that for ever element b B, the set f 1 ({b}) is a single element subset of A. Then given b B, a A such that f 1 ({b}) = {a}. Therefore, f (a) = b. In particular, this implies that ever element b B has a preimage in A and hence f is surjective. To show that f is injective we suppose, A such that f () = f (). For later use we define b := f (). Then, f 1 ({b}). However, b hpothesis f 1 ({b}) is a single element subset of A and so it must be that =, thereb proving that f is injective. Since f is both injective and surjective it must be bijective. 17

18 Problem Let A and B be sets and let f : A B be a surjective function. For each b B, let A b = f 1 ({b}). Prove that the collection of sets P = {A b b B} is a partition of A. To show that this is partition we need to show three things. First we want to show that ever set in the purported partition is nonempt. The sets in P are the A b. So for an b B we have a set A b P. Since f is surjective a A such that f (a) = b, i.e., a A b. In particular, A b. Net we want to show that the union over sets in P gives all of A. That is we claim A b = A. b B To show this note that A b A b definition and hence A b A. On the other hand if a A b B then f (a) B. So if we define b B b b = f (a) then A b P. However, a A b b construction and so a A b A b. b B Since a was an arbitrar element of A we can conclude that A Finall, we need to show that the sets in P are pairwise disjoint. To do this we let A b, A c P with b c. Then suppose that a A b A c. This is equivalent to saing a A b and a A c. Appling the definitions of A b and A c we see that f (a) = b and f (a) = c. Therefore, b = c. So we can conclude that A b A c if and onl if b c. Therefore P is a partition b the Definition.3.3. b B A b. 18