Solutions, Chapter 3

E3.1 Prove that if f : A → B and 〈Ci : i ∈ I〉 is a system of subsets of A, then

f[⋃

i∈ICi

]

=⋃

i∈If [Ci].

x ∈ f

[

⋃

i∈I

Ci

]

iff x ∈ rng

(

f ↾⋃

i∈I

Ci

)

iff ∃y ∈⋃

i∈I

Ci[f(y) = x]

iff ∃i ∈ I∃y ∈ Ci[f(y) = x]

iff ∃i ∈ I[x ∈ rng(f ↾ Ci)]

iff ∃i ∈ I[x ∈ f [Ci]]

iff x ∈⋃

i∈I

f [Ci].

E3.2 Prove that if f : A → B and C, D ⊆ A, then f [C ∩ D] ⊆ f [C] ∩ f [D]. Give anexample showing that equality does not hold in general.

Take any x ∈ f [C ∩ D]. Choose y ∈ C ∩ D such that x = f(y). Since y ∈ C, we havex ∈ f [C]. Similarly, x ∈ f [D]. So x ∈ f [C] ∩ f [D]. Since x is arbitrary, this shows thatf [C ∩ D] ⊆ f [C] ∩ f [D].

For the required example, let dmn(f) = {a, b} with a 6= b and with f(a) = a = f(b).Let C = {a} and D = {b}. Then C ∩ D = ∅, so f [C ∩ D] = ∅, while f [C] = {a} = f [D]and hence f [C] ∩ f [D] = {a} 6= ∅. So f [C ∩ D] 6= f [C] ∩ f [D].

E3.3 Given f : A → B and C, D ⊆ A, compare f [C\D] and f [C]\f [D]: prove theinclusions (if any) which hold, and give counterexamples for the inclusions that fail tohold.

We claim that f [C]\f [D] ⊆ f [C\D]. For, suppose that x ∈ f [C]\f [D]. Choose c ∈ Csuch that x = f(c). Since x /∈ f [D], we have c /∈ D. So c ∈ C\D and hence x ∈ f [C\D],proving the claim.

The other inclusion does not hold. For, take the same f, C, D as for exercise E3.2.Then C\D = {a} and so f [C\D] 6= ∅. But f [C] = {a} = f [D], so f [C]\f [D] = ∅.

E3.4 Prove that if f : A → B and 〈Ci : i ∈ I〉 is a system of subsets of B, then

f−1[⋃

i∈ICi

]

=⋃

i∈If−1[Ci].

For any b ∈ B we have

b ∈ f−1

[

⋃

i∈I

Ci

]

iff f(b) ∈⋃

i∈I

Ci

iff ∃i ∈ I[f(b) ∈ Ci]

iff ∃i ∈ I[b ∈ f−1[Ci]]

iff b ∈⋃

i∈I

f−1[Ci].

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E3.5 Prove that if f : A → B and 〈Ci : i ∈ I〉 is a system of subsets of B, then

f−1[⋂

i∈ICi

]

=⋂

i∈If−1[Ci].

For any a,

a ∈ f−1

[

⋂

i∈I

Ci

]

iff f(a) ∈⋂

i∈I

Ci

iff ∀i ∈ I[f(a) ∈ Ci]

iff ∀i ∈ I[a ∈ f−1[Ci]]

iff a ∈⋂

i∈I

f−1[Ci].

E3.6 Prove that if f : A → B and C, D ⊆ B, then f−1[C\D] = f−1[C]\f−1[D].

For any a,

a ∈ f−1[C\D] iff f(a) ∈ C\D

iff f(a) ∈ C and f(a) /∈ D

iff a ∈ f−1[C] and a /∈ f−1[D]

iff a ∈ f−1[C]\f−1[D].

E3.7 Prove that if f : A → B and C ⊆ A, then

{b ∈ B : f−1[{b}] ⊆ C} = B\f [A\C].

First suppose that b is in the left side; but suppose also, aiming for a contradiction, thatb ∈ f [A\C]. Say b = f(a), with a ∈ A\C. Then a ∈ f−1[{b}], so a ∈ C, contradiction.

Second, suppose that b is in the right side. Take any a ∈ f−1[{b}]. Then f(a) = b,and it follows that a ∈ C, as desired.

E3.8 For any sets A, B define A△B = (A\B) ∪ (B\A); this is called the symmetricdifference of A and B. Prove that if A, B, C are given sets, then A△(B△C) = (A△B)△C.

Let D = A ∪ B ∪ C, A′ = D\A, B′ = D\B, and C′ = D\C. Then

A△B = (A ∩ B′) ∪ (B ∩ A′);

(A△B)′ = ((A ∩ B′) ∪ (B ∩ A′))′

= (A ∩ B′)′ ∩ (B ∩ A′)′

= (A′ ∪ B) ∩ (B′ ∪ A)

= (A′ ∩ B′) ∪ (A ∩ B).

These equations hold for any sets A, B. Now

A△(B△C) = (A ∩ (B△C)′) ∪ ((B△C) ∩ A′

= (A ∩ ((B′ ∩ C′) ∪ (B ∩ C))) ∪ (((B ∩ C′) ∪ (C ∩ B′)) ∩ A′)

= (A ∩ B′ ∩ C′) ∪ (A ∩ B ∩ C) ∪ (A′ ∩ B ∩ C′) ∪ (A′ ∩ B′ ∩ C).

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This holds for any sets A, B, C. Hence

(A△B)△C = C△(A△B)

= (C ∩ A′ ∩ B′) ∪ (C ∩ A ∩ B) ∪ (C′ ∩ A ∩ B′) ∪ (C′ ∩ A′ ∩ B)

= A△(B△C).

E3.9 For any set A letIdA = {〈x, x〉 : x ∈ A}.

Justify this definition on the basis of the axioms.

IdA = {y ∈ A × A : ∃x ∈ A[y = 〈x, x〉]}.

E3.10 Suppose that f : A → B. Prove that f is surjective iff there is a g : B → A suchthat f ◦ g = IdB. Note: the axiom of choice might be needed.

⇐: given b ∈ B, we have b = (f ◦ g)(b) = f(g(b)); so f is surjective.⇒: Assume that f is surjective. Let

A = {{(b, a) : a ∈ A, f(a) = b} : b ∈ B}.

Each member of A is nonempty; for let x ∈ A . Choose b ∈ B such that x = {(b, a) : a ∈A, f(a) = b}. Choose a ∈ A such that f(a) = b. So (b, a) ∈ x.

The members of A are pairwise disjoint: suppose x, y ∈ A with x 6= y. Choose b, cso that x = {(b, a) : a ∈ A, f(a) = b} and y = {(b, a) : a ∈ A, f(a) = c}. If u ∈ x ∩ y, thenthere exist a, a′ ∈ A such that u = (b, a), f(a) = b, and also (u = (c, a′), f(a′) = c. So byTheorem 3.3, b = c. But then x = y, contradiction.

Now by the axiom of choice, let C have exactly one element in common with eachmember of A . Then define

g = {(b, a) ∈ C : a ∈ A, f(a) = b}.

Now g is a function. For, suppose that (b, a), (b, a′) ∈ g. Let x = {(b, a′′) : a′′ ∈ A, f(a′′) =b}. Then (b, a), (b, a′) ∈ C ∩ x, so (b, a) = (b, a′). Hence a = a′.

Clearly g ⊆ B × A. Next, dmn(g) = B, for suppose that b ∈ B. Choose x ∈C ∩ {(b, a′′) : a′′ ∈ A, f(a′′) = b}; say x = (b, a) with a ∈ A, f(a) = b. Then x ∈ g and sob ∈ dmn(g).

Thus g : B → A. Take any b ∈ B, and let g(b) = a. So (b, a) ∈ g and hence f(a) = b.So f ◦ g = IdB .

E3.11 Let A be a nonempty set. Suppose that f : A → B. Prove that f is injective iffthere is a g : B → A such that g ◦ f = IdA.

First suppose that f is injective. Fix a ∈ A, and let

g = f−1 ∪ {(b, a) : b ∈ B\rng(f)}.

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Then g is a function. In fact, suppose that (b, c), (b, d) ∈ g. If both are in f−1, then(c, b(, (d, b) ∈ f , so f(c) = b = f(d) and hence c = d since f is injective. If (b, c) ∈ f−1

and b ∈ B\rng(f), the (c, b) ∈ f , so b ∈ rng(f), contradiction. If (b, c), (b, d) /∈ f−1, thenc = d = a.

Clearly then g : B → A. For any a ∈ A we have (a, f(a)) ∈ f , hence (f(a), a) ∈ f−1 ⊆g, and so g(f(a)) = a.

Second, suppose that g : B → A and g ◦ f = IdA. Suppose that f(a) = f(a′). Thena = (g ◦ f)(a) = g(f(a)) = g(f(a′)) = (g ◦ f)(a′) = a′.

E3.12 Suppose that f : A → B. Prove that f is a bijection iff there is a g : B → A suchthat f ◦ g = IdB and g ◦ f = IdA. Prove this without using the axiom of choice.

⇒: Assume that f is a bijection. By E3.11 there is a g : B → A such that g ◦ f = IdA.We claim that f ◦ g = IdB . Since f is a bijection, the relation f−1 is also a bijection. Nowfor any b ∈ B,

(f ◦ g)(b) = f(g(b)) = f(g(f(f−1(b)))) = f((g ◦ f)(f−1(b))) = f(f−1(b)) = b.

So f ◦ g = IdB , as desired.⇐: Assume that g is as indicated. Then f is injective, since f(a) = f(b) implies

that a = g(f(a)) = g(f(a′)) = a′. And f is surjective, since for a given b ∈ B we havef(g(b)) = b.

E3.13 For any sets R, S define

R|S = {(x, z) : ∃y((x, y) ∈ R ∧ (y, z) ∈ S)}.

Justify this definition on the basis of the axioms.

R|S = {(x, z) ∈ dmn(R) × rng(S) : ∃y((x, y) ∈ R ∧ (y, z) ∈ S)}.

E3.14 Suppose that f, g : A → A. Prove that

(A × A)\[((A× A)\f)|((A× A)\g)]

is a function.

Suppose that (x, y), (x, z) are in the indicated set, with y 6= z. By symmetry say f(x) 6= y.Then (x, y) ∈ [(A × A)\f ], so it follows that (y, z) ∈ g, as otherwise (x, z) ∈ [((A ×A)\f)|((A×A)\g)]. Hence (y, y) /∈ g, so (x, y) ∈ [((A×A)\f)|((A×A)\g)], contradiction.

E3.15 Suppose that f : A → B is a surjection, g : A → C, and ∀x, y ∈ A[f(x) = f(y) →g(x) = g(y)]. Prove that there is a function h : B → C such that h ◦ f = g. Define h as aset of ordered pairs.

Let h = {(f(a), g(a)) : a ∈ A}. Then h is a function, for suppose that (x, y), (x, z) ∈ h.Choose a, a′ ∈ A so that x = f(a), y = g(a), x = f(a′), and y = g(a′). Thus f(a) = f(a′),so g(a) = g(a′), as desired.

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Since f is a surjection it is clear that dmn(h) = B. Clearly rng(h) ⊆ C. So h : B → C.If a ∈ A, then (f(a), b(a)) ∈ h, hence h(f(a)) = g(a). This shows that h ◦ f = g.

E3.16 The statement

∀A ∈ A ∀B ∈ B(A ⊆ B) implies that⋃

A ⊆⋂

B

is slightly wrong. Fix it, and prove the result.

If A has a nonempty member and B is empty, the implication does not hold. Add thehypothesis B 6= ∅.

Suppose that a ∈⋃

A and B ∈ B; we want to show that a ∈ B. Choose A ∈ A suchthat a ∈ A. Since A ⊆ B, we have a ∈ B.

E3.17 Suppose that ∀A ∈ A ∃B ∈ B(A ⊆ B). Prove that⋃

A ⊆⋃

B.

Suppose that a ∈⋃

A ; we want to show that a ∈ B. Choose A ∈ A such that a ∈ A.Then choose B ∈ B such that A ⊆ B. Then a ∈ B. Hence a ∈

⋃

B.

E3.18 The statement

∀A ∈ A ∃B ∈ B(B ⊆ A) implies that⋂

B ⊆⋂

A .

is slightly wrong. Fix it, and prove the result.

If A is empty and⋂

B is nonempty, the statement is false. Fix it by adding the hypothesisthat A is nonempty.

Suppose that b ∈⋂

B and A ∈ A ; we want to show that b ∈ A. Choose B ∈ B suchthat B ⊆ A. Now b ∈ B since b ∈

⋂

B, so b ∈ A.

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