Paul J. Bruillard MATH 220.970 Problem Set 6 Section

Paul J. Bruillard MATH 220.970 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, 20, 21b Section 3.2: 1f,i, 2e, 6, 12e,f,h, 13e, 17, 20, 21, 22, 26, 29, 30

Section 3.1

Problem 3.1.3 For the following functions, compute ﬁrst the image of f and then the image of the given subset X (No formal proofs are necessary.)

(b) f : → , f (x) = 2x + 1; X = [−1, 4]. R R ( 2n if n is even (c) f : Z → Z, f (n) = ; X = E. n if n is odd × × × 1 (e) f : R → R where R is the set of nonzero real numbers, f (x) = x ; X = (0, 1]. × x (i) f : R × R → R, f (x, y) = y ; X = {(x, 1) | x ∈ R}.

Remark 0.1. Since the problem explicitly stated that a formal proof was not required I will simply sketch the calculation.

Solution:

Part (b) We ﬁrst note that f (x) describes a line:

So graphically we see that its image is all of R and that the image of X is obtained as follows (image depicted in blue): P. J. Bruillard y

Algebraically, we have Im (f) = f (R) = {x ∈ R | ∃y ∈ R such that f (y) = x} x−1 However, x = f (y) = 2y + 1 is always solved by y = 2 and hence Im (f) = R. Similarly, the image of X under f is given by f (X) = {x ∈ R | ∃y ∈ [−1, 4] such that f (y) = x} = {2y + 1 ∈ R | ∃y ∈ [−1, 4]} = [−1, 9] Thus the image of f and the image of X under f are respectively given by:

Im (f) = R, f (X) = [−1, 9]

Part (c) Here we see that f is discrete, graphically it appears as a collection of points: y

From here one can read oﬀ the range of f. Indeed, each even number is doubled so f (2n) = 4n for n ∈ Z and each odd number is ﬁxed, i.e., f (2n + 1) = 2n + 1 for n ∈ Z. Therefore, the image of f may be expressed as Im (f) = {4n | n ∈ 2Z} ∪ {2m + 1 | m ∈ Z} 2 P. J. Bruillard

We additionally see from here that the image of E under f is given by

f (E) = 4Z = {4n | n ∈ Z} Collecting the results we see that the image of f and the image of E under f are respectively given by:

Im (f) = {4n | n ∈ Z} ∪ {2m + 1 | m ∈ Z} , f (E) = 4Z = {4n | n ∈ Z}

Part (e) Since f is a real-valued function with domain in R it is easily plotted y

× From the plot it is readily apparent that Im (f) = R = {x ∈ R | x 6= 0}. This can be seen 1 1 algebraically by noting that for y 6= 0 the equation x = y can be solved by x = y . This solution makes manifest the problem with y = 0, one would have to divide by zero to solve for x. The image of X = (0, 1] can similarly be determined. Indeed, as x → 0 f (x) → ∞ and at x = 1 we have f (1) = 1. Thus f (X) = [1, ∞). Therefore, the image of f and the image of X under f are respectively given by

× Im (f) = R , f (X) = [1, ∞)

Part (i) This function deﬁnes a nasty 2-dimensional surface:

2 −20 −2 0 −1 0 1 2 −2

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This plot is of course incomplete since as y → 0 the function diverges. However, we see immediately that the image of the function is all of R. To see this algebraically we would need x to solve y = z for any z ∈ R. This is of course possible by taking y = 1 and x = z. This analysis further shows that the image of X = {(x, 1) | x ∈ R} under f is R. Indeed, f (x, 1) = x for any x ∈ R. Consequently, the image of f and the image of X under f are respectively given by 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 you determine the image. No formal proof is necessary.) (√ x if x ≥ 0 (b) f (x) = √ − −x if x < 0 ( 4x if x > 0 (d) f (x) = x3 if x < 0 Remark 0.2. Just as in the previous problem, no proof is required, and so I will merely sketch the process utilizing graphs. Solution: Part (b) (√ x if x ≥ 0 Recall for this part f (x) = √ . From here we see that the image of f is always − −x if x < 0 a real number and f is deﬁned on all of the real numbers. Indeed, whenever x is negative we know that −x 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 2 values to a single point x ∈ R. Therefore, f is a function from R to R. Having convinced ourselves that f : R → R really is a function we need to determine its image. As the problem suggests, we start by graphing the function y

From here we see immediately that all “small numbers” are in the image of f but what about the large ones? Well, as x → ∞ we are taking the square-root of larger and larger numbers 4 P. J. Bruillard √ which still produces a larger√ and larger number. Indeed, calculus tells us that limx→∞ x = ∞ and similarly limx→−∞ − −x = −∞. 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 y

From here we see that Im (f) = R \{0} .

Problem 3.1.6 Let f : Z → Z be deﬁned by f (n) = 2n+1. What is wrong with the following proof that Im (f) = Z? y−1 y−1 Let y ∈ Z. Then f 2 = 2 2 + 1 = y. Hence y ∈ Im (f). Therefore, Z ⊂ Im (f). Since Im (f) ⊂ Z by deﬁnition, it follows that Im (f) = Z. Solution:

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) = 2n + 1 which is always odd. Thus we think that the image of f is all odd integers. Now to the examination of the “proof”. The proof says to take y ∈ Z and then compute y−1 y−1 1 f 2 . This is the problem with the proof. Indeed, if we take y = 2 then 2 = 2 which is 1 not an integer. Since f is only defined on the integers the expression f 2 is nonsense!

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Problem 3.1.9 Let f : R → Z be the greatest integer function. (a) Compute f ([0, 1]). Prove your answer. (b) Compute f ([−1, 1]). Prove your answer. Solution: Part (a) Claim 0.3. f ([0, 1]) = {0, 1}. Proof: First note that f (0) = 0 and f (1) = 1 whence {0, 1} ⊆ f ([0, 1]). On the other hand if x ∈ [0, 1] then x = 0 or x > 0. We’ve already seen that f (0) = 0 and so we need only focus on f (x) for 0 < x ≤ 1. In this cas,, by deﬁnition of f, we know that f (x) = 1. Consequently, f ([0, 1]) ⊂ {0, 1}.

Part (b) Claim 0.4. f ([−1, 1]) = {−1, 0, 1}. Proof: The proof here proceeds analogously 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 (x) ∈ {−1, 0, 1} for x ∈ [−1, 1]. Since we’ve dealt with x ∈ {−1, 0, 1} it suffices to consider x ∈ (−1, 0) and x ∈ (0, 1). In the first case, the definition of f tells us that f (x) = 0. While in the second case we have that f (x) = 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 3.1.15 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 example of sets A and B and a function f : A → B for which f (X ∩ Y ) 6= f (X) ∩ f (Y ) for some subsets X and Y of A. Solution: Part (a) Claim 0.5. f (X ∩ Y ) ⊆ f (X) ∩ f (Y ) Proof: This proof can be done by direction calculation, i.e., let x ∈ f (X ∩ Y ) and then show x ∈ f (X) ∩ f (Y ) by using the definition of the image of a set under a function. A quicker way to this result is to utilize Proposition 3.1.5. Indeed, by definition of intersection, we know that X ∩ Y ⊆ X. So Proposition 3.1.5 tells us that f (X ∩ Y ) ⊆ f (X). Similarly, the definition of intersection implies that X ∩ Y ⊆ Y . Once again turning to Proposition 3.1.5 allows us to deduce that f (X ∩ Y ) ⊆ f (Y ). 6 P. J. Bruillard

Thus for any x ∈ f (X ∩ Y ) we know that x ∈ f (X) and x ∈ f (Y ). In particular, the deﬁnition of the intersection tells us that x ∈ f (X) ∩ f (Y ). Since x was an arbitrary element in f (X ∩ Y ) we can conclude that f (X ∩ Y ) ⊆ f (X) ∩ f (Y ).

Part (b) An example is fairly easy to construct. For instance, one may take the function f : R → R given by f (x) = x2. h i From here one can show that f ([5, 6]) = [25, 36] = 52, 62 = (−5)2 , (−6)2 = 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 ) = [25, 36] and so f (X) ∩ f (Y ) = [25, 36]. Since 25 ∈ [25, 36] we see that [25, 36] 6= ∅.

Problem 3.1.16 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 example of sets A and B and a function f : A → B for which f (X) \ f (Y ) 6= f (X \ Y ) for some subsets X and Y of A. Solution: Part (a) Claim 0.6. f (X) \ f (Y ) ⊆ f (X \ Y ) Proof: Let z ∈ f (X) \ f (Y ). Then z ∈ f (X) and z∈ / f (Y ), by deﬁnition of the set diﬀerence. Therefore z = f (x) where x ∈ X but x∈ / Y . In particular, x ∈ X \ Y . Consequently z = f (x) ∈ f (X \ Y ). Since z was an arbitrary 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]. We then deﬁne f : R → R by f (x) = x2. 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 ).

Problem 3.1.18c, e For the following functions, compute the inverse of the given subsets of the codomain. (No proofs are necessary). (c) f : → , f (x) = cos (x); W = [−1, 1], W = {x ∈ | x ≥ 0}, W = . R R ( 1 2 R 3 Z n if n is even (e) f : Z → Z, f (n) = ; W1 = E, W2 = {1}, W3 = {6}, W4 = O, the set n − 1 if n is odd of odd integers. Remark 0.7. This problem explicitly states that proofs are not necessary and so I will not provide proofs, rather I will plot the functions and examine the graphs to determine the answer. 7 P. J. Bruillard

Solution:

Part (c) Recall that the function under consideration is f : R → R given by f (x) = cos (x). Plotting this function we see

We can now apply our graphical method√ for computing points in the preimage. For instance if we want to ﬁnd the preimage of 1/ 2 we draw a horizontal line at y = √1 and every time we 2 run into the graph we drop a perpendicular line to the x-axis. The places that these lines cross the x-axis is the preimage of that point.

−1 From here it becomes apparent that any f ([−1, 1]) = R. This is not surprising since Im (f) = [−1, 1] and f is deﬁned everywhere on R. −1 Similarly we can compute f (W2) and we ﬁnd

−1 f (W2) = {x ∈ R | f (x) ∈ W2} = {x ∈ R | cos (x) ≥ 0} [ (4n − 1) π (4n + 1) π = , 2 2 n∈Z

−1 Finally, we can determine f (Z). Since Im (f) = [−1, 1] we see that

−1 −1 −1 f (Z) = f ([−1, 1] ∩ Z) = f ({−1, 0, 1}) .

Returning to our graph we can get a feel for this set geometrically 8 P. J. Bruillard y

Algebraically, we would ﬁnd all x such that cos (x) = 0, 1, or −1. Examination of the above plot or the unit circle tells us that this set consists of all half integer multiplies of π. Indeed, we have nnπ o f −1 ( ) = | n ∈ Z 2 Z Collecting the results we have:

−1 f (W1) = R [ (4n − 1) π (4n + 1) π f −1 (W ) = , 2 2 2 n∈Z nnπ o f −1 ( ) = | n ∈ Z 2 Z Part (e) Here we need to compute the inverse of the subsets W1 = E, W2 = {1}, W3 = {6}, and W4 = O for the function f : Z → Z given by ( n if n is even f (n) = n − 1 if n is odd This proceeds just as in the previous two problem The only diﬀerence is that we need to ensure all of the points we ﬁnd are integers. Taking care when computing, one can show directly that

−1 f (W1) = Z −1 f (W2) = ∅ −1 f (W3) = {6, 7} f −1 (O) = ∅

The second and third results listed here can be understood by noting that f always produces an even number. Indeed, f (2n) = 2n and f (2n + 1) = 2n for any n ∈ Z. ( n if n is even Problem 3.1.19a Let f : Z → Z be deﬁned by f (n) = 2 . 2n + 4 if n is odd Compute f −1 ({11, 12, 13, 14, 15}). Claim 0.8. f −1 ({11, 12, 13, 14, 15}) = {5, 22, 24, 26, 28, 30}. Proof: 9 P. J. Bruillard

Here we need to compute f −1 for each element in the finite set {11, 12, 13, 14, 15}, the set containing the elements we compute will be the preimage. n So we note that 11 = f (n) means either 11 = 2 and n is even or 11 = 2m + 4 and m is odd. The second case cannot occur because 11 is not divisible by 2 and so n = 22. n Next, we try and solve 12 = 2 with n even or 12 = 2m + 4 with m odd. In the first case we can take n = 24, in the second case we have m = 6 − 2 = 4 which is not odd. Thus n = 24 is the only element leading to 12. Moving on to the next number we try to solve 13 = n/2 for n even or 13 = 2m + 4 for m odd. The second equation has no solution because 13 is not divisible by 2. The first equation is solved by n = 26. Next we take the equations 14 = n/2 for n even and 14 = 2m + 4 for n odd. The second equation says that m = 5 while the first equation is solved by n = 28. Finally, we should try and solve 15 = n/2 for n even and 15 = 2m + 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 by 2. Amalgamating these results gives the answer.

Problem 3.1.20 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). Proof: 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). By deﬁnition of intersection we can conclude that a ∈ f −1 (W ) ∩ f −1 (Z). Since a was arbitrary we have f −1 (W ∩ Z) ⊆ f −1 (W ) ∩ f −1 (Z).

Conversely, 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 arbitrary element of f −1 (W )∩f −1 (Z) we can conclude that f −1 (W )∩f −1 (Z) ⊆ f −1 (W ∩ Z).

Problem 3.1.21b Give an example of a function f : A → B for some A and b and a subset X of A such that X 6= f −1 (f (X)). Solution:

Take f : R → R deﬁned by f (x) = cos (x). −1 Then f ([0, 2π]) = [−1, 1] but we saw above that f ([−1, 1]) = R which is very deﬁnitely not equal to [0, 2π]. (So here A = B = R and X = [0, 2π].)

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1 Section 3.2 Problem 3.2.1f, i Determine which of the following functions are surjective. Give a formal proof of your answer (f) f : → , f (x) = x3 + x2. R R ( −x − 1 if x ≥ 0 (i) f : R → R, f (x) = x2 if x < 1 Solution:

We’ll ﬁrst plot the function to decide what to claim. A function f : R → R will be surjective if for any horizontal line we draw, it intersects the graph of the function at least once.

Part (f) y

From here it is pretty clear what we should claim. 3 2 Claim 1.1. f : R → R given by f (x) = x + x is surjective. Proof: The proof proceeds via the Intermediate Value Theorem (Theorem 3.1.4). Indeed we know that lim f (x) = ∞ x→∞ lim f (x) = −∞ x→−∞ Noting that f is continuous (it is a polynomial) we can apply the Intermediate Value Theorem to conclude that any 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 always contained in the codomain. Consequently, R = Im (f) and hence f is surjective.

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Part (i) Once again we plot the function to decide what to do. y

Here is appears as though [0, −1) is not in the range of f and so f is not surjective. Thus we claim ( −x − 1 if x ≥ 0 Claim 1.2. The function f : R → R deﬁned by f (x) = is not surjective x2 if x < 1 Proof: 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 ∃x ∈ R such that f (x) = 0. We know that x ≥ 0 or x < 0 and so we should consider these cases separately. Case 1: x ≥ 0. Then the deﬁnition of f tells us that 0 = f (x) = −x − 1. Since x ≥ 0 we know that −x ≤ 0 and hence −1 − x < 0. Since 0 is not less than zero we can conclude that x < 0 (case 2).

Case 2: x < 0. Then we know 0 = f (x) = x2. However, x > 0 implies that x2 > 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.

( n − 1 if n is even Problem 3.2.2e Determine if f : Z → Z deﬁned by f (n) = is surjective. 2n if n is odd Prove your answer Solution: In this setting f always produces an odd number (as can be seen by examining function). This motivates the claim ( n − 1 if n is even Claim 1.3. The function f : Z → Z deﬁned by f (n) = is not surjective. 2n if n is odd

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Proof:

To see this we’ll show that 4 ∈ Z \ Im (f). This is done by 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 by cases. If n is even then n = 2m for some m in Z. In this case 4 = f (n) = f (2m) = 2m − 1. 4 is even but the right hand side is odd (to see this reduce modulo 2), this is a contradiction. If n is odd then n = 2m + 1 for some m ∈ Z, and 4 = f (n) = f (2m + 1) = 2 (2m + 1) = 1 4m + 2. Dividing through by 4 shows that 2 = 1 − m ∈ Z, a contradiction. Since we always arrive at a contradiction we can conclude that our hypothesis is false and hence 4 ∈/ Im (f). Therefore f is not surjective.

Problem 3.2.6 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 example to show that equality does not in general hold in part (a). Solution: Part (a) Claim 1.4. B \ f (X) ⊆ f (A \ X) Proof: The proof proceeds by 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). Consequently, B \ f (X) ⊆ f (A \ X).

Part (b) 2 Consider the function f : R → [0, ∞) given by f (x) = x . Then in the context 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, ∞)

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Problem 3.2.12e,f,h Determine which of the following functions are injective. Give a formal proof of your answer

+ + x (e) f : R → R , where R is the set of positive real numbers and f (x) = e . (f) f : Z → Z × Z, f (n) = (n, n). × x (h) f : R × R → R, f (x, y) = y . Solution: Part (e) The idea here is to plot f. If any horizontal line that intersects the graph of f, intersects it exactly once, then f is injective. So plotting we see y

This looks as though it will pass this horizontal line test and so we claim

+ x Claim 1.5. f : R → R deﬁned by f (x) = e is injective. Proof:

To show this we suppose that x, y ∈ R such that f (x) = f (y). Then we have ex = ey. Taking logarithms we ﬁnd that x = ln (ex) = ln (ey) = y. So by deﬁnition, f is injective.

Part (f) This is an example of a type 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 by f in the plane we see the integer points on the line y = x. This motivates the claim

Claim 1.6. f : Z → Z × Z deﬁned by f (n) = (n, n) is injective. Proof:

This proceeds by 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 by returning to Deﬁnition 3.2.2.

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-axis will cross the function in more than one point. This motivates us to claim 14 P. J. Bruillard

+ x Claim 1.7. f : R × R → R defined by φ (x, y) = y is not injective. Proof: 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 many choices, we’ll take the following 4 2 (4, 2) 6= (2, 1) , f (4, 2) = = 2 = = φ (2, 1) 2 1

Problem 3.2.13e Determine if the following function is injective and provide a formal proof of ( n − 1 if n is even your answer. f : Z → Z, f (n) = 2n if n is odd

Solution: This function seems to ﬂip evens and odds (plug in a few numbers and try it out). Moreover, each of the functions n − 1 and 2n would be injective if they were considered alone (they are just lines). This motivates us to claim ( n − 1 if n is even Claim 1.8. f : Z → Z, f (n) = is injective. 2n if n is odd Proof:

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 2: m and n are both odd. Then 2n = f (n) = f (m) = 2m and hence m = n. Case 3: m is even and n is odd. Then we have m − 1 = f (m) = f (n) = 2n. 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 3.2.17 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 exercise 3.1.15)

Proof: In the previous section (exercise 3.1.15) we showed that f (X ∩ Y ) ⊆ f (X) ∩ f (Y ). So it 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 ∈ X and y ∈ Y such that f (x) = b = f (y). Since f is injective we can conclude that x = y and hence x ∈ X ∩ Y . Therefore, b = f (x) ∈ f (X ∩ Y ). Since b was a generic element of f (X) ∩ Y we see that f (X) ∩ f (Y ) ⊆ f (X ∩ Y ).

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Problem 3.2.20 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) 6= f (b), then a 6= b. (f) If a and b are in A and a 6= b, then f (a) 6= f (b). Solution: Many of these are just general features of functions e.g. (a) and (d). Thinking carefully about each statement gives the following (a) Not equivalent (b) Not equivalent (c) Equivalent (d) Not equivalent (e) Not equivalent (f) Equivalent

Problem 3.2.21 Let f : A → B be an injective function. Let X be a subset of A. Prove that f −1 (f (X)) = X (compare with exercise 3.1.21)

Proof: From exercise 3.1.21 (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) by deﬁnition of the preimage. Consequently f (x) = f (a) for some x ∈ X. However, f is injective and so x = a. Since x ∈ X by hypothesis, we can conclude that a ∈ X. Therefore f −1 (f (X)) ⊆ X and hence we may conclude that f −1 (f (X)) = X.

Problem 3.2.22 Let f : A → B be a surjective function. Let W be a subset of B. Prove that f f −1 (W ) = W (compare with exercise 3.1.22)

Proof: Applying exercise 3.1.22 (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 apply surjectivity 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 P. J. Bruillard

Problem 3.2.26 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 only if f is surjective.

Proof:

This is a biconditional statement and so there are two directions to prove. (⇒) Suppose that f is injective. Then by 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. There- fore Im (f) = B and so f is surjective.

(⇐) Suppose that f is surjective. Then Im (f) = B. In particular, |Im (f)| = |B| = |A|. However, every element of a gets mapped to exactly one element of f. That is, we have A = {a1, . . . , an} and Im (f) = {f (a1) , . . . , f (an)}. Since these sets have the same cardinality (we just showed this) it must be that the f (ai) are distinct i.e. f (ai) = f (aj) if and only if ai = aj. This is exactly the statement that f is injective.

Problem 3.2.29 Let A and B be sets and let f : A → B be a function. Prove that f is bijective if and only if for every b ∈ B, f −1 ({b}) is a single element subset of A.

Proof:

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}) 6= ∅ and so we may get c ∈ f −1 ({b}). By 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 every 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 every element b ∈ B has a preimage in A and hence f is surjective. To show that f is injective we suppose ∃x, y ∈ A such that f (x) = f (y). For later use we deﬁne b := f (x). Then x, y ∈ f −1 ({b}). However, by hypothesis f −1 ({b}) is a single element subset of A and so it must be that x = y, thereby proving that f is injective. Since f is both injective and surjective it must be bijective.

17 P. J. Bruillard

Problem 3.2.30 Let A and B be sets and let f : A → B be a surjective function. For each b ∈ B, −1 let Ab = f ({b}). Prove that the collection of sets P = {Ab | b ∈ B} is a partition of A. Proof: To show that this is partition we need to show three things. First we want to show that every set in the purported partition is nonempty. The sets in P are the Ab. So for any b ∈ B we have a set Ab ∈ P. Since f is surjective ∃a ∈ A such that f (a) = b, i.e., a ∈ Ab. In particular, Ab 6= ∅. [ Next we want to show that the union over sets in P gives all of A. That is we claim Ab = A. b∈B [ To show this note that Ab ⊆ A by deﬁnition and hence Ab ⊆ A. On the other hand if a ∈ A b∈B then f (a) ∈ B. So if we deﬁne b ∈ B by b = f (a) then Ab ∈ P. [ However, a ∈ Ab by construction and so a ∈ Ab ⊆ Ab. b∈B [ Since a was an arbitrary element of A we can conclude that A ⊆ Ab. b∈B

Finally, we need to show that the sets in P are pairwise disjoint. To do this we let Ab,Ac ∈ P with b 6= c. Then suppose that a ∈ Ab ∩ Ac. This is equivalent to saying a ∈ Ab and a ∈ Ac. Applying the deﬁnitions of Ab and Ac we see that f (a) = b and f (a) = c. Therefore, b = c. So we can conclude that Ab ∩ Ac 6= ∅ if and only if b 6= c.

Therefore P is a partition by the Deﬁnition 2.3.3.