220hw8solns

Mathematics 220 Homework 8 Due March 21

• There are 11 questions worth a total of 26. The first 6 questions are on material youshould know for the Test on March 14. Solutions for questions worth 0 points will beposted by March 9.

1. 6 marks (a) Give an example of a bijection from (0, 1) to (1,∞). Remember — youhave to prove it is a bijection–similarly for (b) and (c).

(b) Give an example of a surjection from R to Q.

(c) Give an example of a bijection from N to Z.

Solution:

Proof. (a) Let f : (0, 1)→ (1,∞) be defined by f(x) = 1/x. We must show it is bothinjective and surjective

• Let x, z ∈ (0, 1) and assume f(x) = f(z). Then 1/x = 1/z and so z/x = 1 andthus z = x. Hence f is injective.

• Let y ∈ (1,∞) and pick x = 1/y. Since 1 < y < ∞, it follows that 0 < x < 1.Further f(x) = f(1/y) = 1/(1/y) = y. Thus f is surjective.

Since f is injective and surjective, it is bijective.

2. 6 marks Let f : A → B be a function and if D ⊂ B define the inverse image of Dunder f by f−1(D) = {x ∈ A : f(x) ∈ D. Note that the set-valued function f−1(D) isdefined for any function f–it need not have an inverse.(a) If D1, D2 ⊆ B, prove that

f−1(D1 ∩D2) = f−1(D1) ∩ f−1(D2)

(b) If D1, D2 ⊆ B, prove that

f−1(D1 ∪D2) = f−1(D1) ∪ f−1(D2)

(c) If D ⊂ B, prove that

f−1(B −D) = A− f−1(D)

Solution:

Proof. (a) We first prove that LHS ⊆ RHS and then RHS ⊆ LHS.

• Let x ∈ f−1(D1 ∩ D2), so f(x) ∈ D1 ∩ D2. Hence f(x) ∈ D1 and f(x) ∈ D2.Since f(x) ∈ D1, x ∈ f−1(D1), and similarly since f(x) ∈ D2, x ∈ f−1(D2).Thus x ∈ f−1 ∩ f−1(D2).

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• Let x ∈ f−1(D1) ∩ f−1(D2), so x ∈ f−1(D1) and x ∈ f−1(D2). Since x ∈f−1(D1), f(x) ∈ D1, and similarly since x ∈ f−1(D2), f(x) ∈ D2. Thus f(x) ∈D1 ∩D2 and so x ∈ f−1(D1 ∩D2).

(c) We first prove that LHS ⊆ RHS and then RHS ⊆ LHS.

• Let x ∈ f−1(B −D), so f(x) ∈ B −D. Thus f(x) ∈ B and f(x) 6∈ D. Sincef(x) 6∈ D, we must have x 6∈ f−1(D). By definition of f−1 we must have x ∈ A.Hence x ∈ A and x 6∈ f−1(D) and thus x ∈ A− f−1(D).

• Let x ∈ A − f−1(D). Hence x ∈ A and x 6∈ f−1(D). Now f(x) ∈ B (sincef : A→ B) and f(x) 6∈ D. Thus f(x) ∈ B −D and so x ∈ f−1(B −D).

3. 0 marks 9.4

Solution: (a) is a function (since given x, y is determined uniquely by the formulay = 4x + 3).

(b) is not a function. For example, when x = 1, we get (y + 2)2 = 1, so y + 2 = ±1,that is , y = −1 or y = −3. Thus, both pairs (1,−1) and (1,−3) belong to therelation A2 × R, violating the definition of a function.

(c) This is also not a function. For example, take x = 0. Then both pairs (0, 2) and(0,−2) are in A3 × R, again violating the definition of a function.

4. 0 marks 9.6 (c) and (e).

Solution: (c). Let us first find the domain. A real number x is in the domain of f3if and only if 3x− 1 ≥ 0, that is, x ≥ 1/3. Thus, the domain of f3 is [1/3,∞). Nowlet us determine its range. That means, we have to determine for which real numbersy the equation

√3x− 1 = y has a solution. We have:

√3x− 1 = y

⇔ (3x− 1 = y2) ∧ (y ≥ 0)

⇔ 3x = y2 + 1, y ≥ 0.

We see that the solution to the equation√

3x− 1 = y exists if and only if y ≥ 0, sothe range of f3 is the set [0,∞).

(e). The domain is R − {3}, since f5 is not defined for x = 3, and this is the onlyvalue of x for whih it is not defined. Let us find the range. We have to determine for

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which real numbers y the equation xx−3

= y has a solution. We have:

x

x− 3= y

⇔ x = y(x− 3), x 6= 3

⇔ 3y = x(y − 1), x 6= 3

⇔ x =3y

y − 1, y 6= 1.

We see that for all real y except for y = 1, the pre-image x exists. Thus, the rangeof f5 is R− {1}.

5. 0 marks let A be a well-ordered set. Let f : P(A)→ A be the function that assigns toevery for B ⊆ A the smallest element of B.

(a) Is f is a function?

(b) Is f injective?

(c) Is f surjective?

Solution: This is, indeed, a function: let B be a subset of A. We need to showthat if b1 and b2 are both minimal elements of B, then b1 = b2. This was proved asTheorem 6.1 in the book.

If A has more than one element, then f is not injective. Indeed, let a, b ∈ A be twodistinct elements, with a < b. Then consider the sets B1 = {a}, and B2 = {a, b}.Then f(B1) = f(B2) = a.

The function f is surjective. For every element a ∈ A, we have f({a}) = a.

6. 0 marks (a) Prove that if f : A → B is an injective function, then f(C1 ∩ C2) =f(C1) ∩ f(C2) for all C1, C2 ⊆ A.

(b) Prove that if f : A → B is a surjective function, then f(f−1(D)) = D for everysubset D ⊆ B.

Solution: We know (from the workshop) that (a). f(C1 ∩ C2) ⊆ f(C1) ∩ f(C2)always. We need to show that if f is injective, then the reverse containment holds:f(C1 ∩C2) ⊇ f(C1) ∩ f(C2). Let y ∈ f(C1) ∩ f(C2). Then, by definition, there existx1 ∈ C1 and x2 ∈ C2 such that y = f(x1) = f(x2). Since f is injective, it followsthat x1 = x2, and then this is a common element of C1 and C2.

(b). We proved in lecture that f(f−1(D)) ⊆ D always. We need to show that if f issurjective, then every element y ∈ D is contained in f(f−1(D)). Since f is surjective,there exists x ∈ A such that f(x) = y. Since y ∈ D, this implies that x ∈ f−1(D),by definition of the pre-image. Then y = f(x) ∈ f(f−1(D)).

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7. 4 marks 9.28(b)

Solution:

• Let f : N → Z and g : Z → Z be defined by f(n) = n and g(n) = n2. Then(g ◦ f) : N→ Z is g(f(n)) = n2. Now g is not injective but g ◦ f is injective.

8. 6 marks 9.36

Solution:

• Proof of bijectiveness

Proof. We must show that f is both injective and surjective.

– Let x, z ∈ A and assume f(x) = f(z). Then

x

1− x=

z

1− z

x(1− z) = z(1− x) since x, z 6= 1x− xz = z − xz

x = z

And hence f is injective.

– Let y ∈ A = R − 1 and pick x = yy−1

. We must show that f(x) = y andthat x ∈ A.

f(x) = f

(y

y − 1

)=

yy−1

yy−1− 1

=y

y − (y − 1)= y multiply numer + denom by y − 1

Hence f(x) = y. Since y 6= 1 we know that x ∈ R. It suffices to show thatx 6= 1. Note that we always have y − 1 < y. If y > 1 then y − 1 > 0 andso x = y/(y − 1) > 1. On the other hand if y < 1 then y − 1 < 0 andx = y/(y − 1) < 1. Thus x 6= 1 for any legal value of y.

– The inverse of f(x) is

g : A→ A g(y) =y

y − 1

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We can check by considering g(f(x))

g(f(x)) =f(x)

f(x)− 1

=x

x−1x

x−1− 1

=x

x− (x− 1)= x

as required.

– Well - one could do this by carefully computing the composition. . . Or wecan think a bit. The previous part shows us that f ◦ f is the identity.Hence f ◦ f ◦ f = f ◦ (iA) = f . Thus

f(f(f(x))) = f(x).

9. 0 marks 9.48

Solution:

(a) Proof. Let a, b ∈ A and assume f(a) = f(b). Now g(f(a)) = iA(a) = a andg(f(b)) = iB(b) = b and since f(a) = f(b) we have a = b. Thus f is injective.

Now let a ∈ A. Pick b = f(a) ∈ B. Then g(b) = g(f(a)) = iA(a) = a. Hence gis surjective.

(b) Proof. Let A = {1, 2} and B = {x, y, z}. Define f(1) = x and f(2) = y. Defineg(x) = 1, g(y) = 2 and g(z) = 1. Then g(f(1)) = 1 and g(f(2)) = 2, so g◦f = iA,but f is not surjective. Similarly g ◦ f = iA but g is not injective.

(c) See previous example.

(d) Proof. Let b1, b2 ∈ B. Assume f is surjective, and g(b1) = g(b2). Since f issurjective, there are a1, a2 ∈ A such that f(a1) = b1 and f(a2) = b2. Theng(b1) = g(f(a1)) = a1 and g(b2) = g(f(a2)) = a2 and so by assumption a1 = a2.This implies that b1 = f(a1) = f(a2) = b2. So g is injective.

(e) For any b ∈ B we need to find a ∈ A such that f(a) = b. We will do this byshowing that we can pick a = g(b) and that f(a) = f(g(b)) = b.

Proof. Assume g is injective. Let b ∈ B and let a = g(b) and c = f(a). Thisimplies that g(c) = g(f(a)) = a. Since g is injective we have b = c. Hencef(a) = b and so given any b there is an a such that f(a) = b as required.

(f) Let f : A→ B and g : B → A with g ◦ f = iA. Then f is surjective if and onlyif g is injective.

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10. 4 marks Find an example of functions f : A→ B and g : B → C such that f and g ◦ fare both injective, but g is not injective.

Solution:

• Let f : [0, 1] → [−1, 1] be defined by f(x) = x and let g : [−1, 1] → [0, 1] bedefined by g(x) = x2. Then g ◦ f : [0, 1]→ [0, 1] is defined by g(f(x)) = x2.

• The function g is not injective because g(−1) = g(+1) = 1.

• The function f is injective. Let x, z ∈ [0, 1] and assume f(x) = f(z) then x = z.

• The function g ◦ f is injective. Let x, z ∈ [0, 1] and assume g(f(x)) = g(f(z))then x2 = z2. Hence x = ±z. But since x, z ≥ 0, it follows that x = z.

Thus we have given an example of f, g as required.

11. 0 marks Find an example of functions f : A→ B and g : B → C such that g and g ◦ fare both surjective, but f is not surjective.

Solution:

• Let f : [0, 1]→ [−1, 1] defined by f(x) = x and let g : [−1, 1]→ [0, 1] be definedby g(x) = x2. Then g ◦ f : [0, 1]→ [0, 1] is defined by g(f(x)) = x2.

• The function f is not surjective since there is no x so that f(x) = −1.

• The function g is surjective. Let y ∈ [0, 1] then pick x =√y. Now 0 ≤ x ≤ 1

and g(x) = g(√y) =

√y2 = y.

• Similarly, the function g ◦ f is surjective. Let y ∈ [0, 1] and pick x =√y. Then

0 ≤ x ≤ 1 and g(f(x)) = g(f(√y) = g(

√y) =

√y2 = y.

Thus we have given an example of f, g as required.

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