The University of Sydney

MATH 3901

Metric Spaces 2004

Tutorial 1

PROBLEM SET 1

1. If A, B and C are subsets of X, prove that

A ∪ (B ∩ C) = (A ∪B) ∩ (A ∪ C).

Solution.

We first show that

A ∪ (B ∩ C) ⊆ (A ∪B) ∩ (A ∪ C).

Let x ∈ A ∪ (B ∩ C). Then x ∈ A or x ∈ B ∩ C. If x ∈ A, then in this case,x ∈ A∪B and x ∈ A∪C so that x ∈ (A∪B)∩ (A∪C). On the other hand, ifx ∈ B ∩ C, then x ∈ B and x ∈ C so that, in this case, again x ∈ A ∪ B andx ∈ A ∪ C and thus x ∈ (A ∪B) ∩ (A ∪ C). Hence we have shown that

A ∪ (B ∩ C) ⊆ (A ∪B) ∩ (A ∪ C).

Next we show that

A ∪ (B ∩ C) ⊇ (A ∪B) ∩ (A ∪ C).

Let x ∈ (A ∪B) ∩ (A ∪ C). Then x ∈ A ∪B and x ∈ A ∪ C and so

(i) x ∈ A or x ∈ B, and

(ii) x ∈ A or x ∈ C.

If x ∈ A, then x ∈ A∪ (B ∩C). If x /∈ A, then it follows from (i) and (ii) thatx ∈ B and x ∈ C so that x ∈ B ∩ C and thus x ∈ A ∪ (B ∩ C). Hence

A ∪ (B ∩ C) ⊇ (A ∪B) ∩ (A ∪ C).

Consequently,A ∪ (B ∩ C) = (A ∪B) ∩ (A ∪ C).

2

2. Let X be a nonempty set and A and B be two subsets of X. Prove that

X \ (A ∪B) = (X \A) ∩ (X \B).

Solution.

We first show that

X \ (A ∪B) ⊆ (X \A) ∩ (X \B).

Let x ∈ X \ (A ∪B). Then x /∈ A ∪B so that x /∈ A and x /∈ B. Therefore

x ∈ X \A and x ∈ X \B.

Thusx ∈ (X \A) ∩ (X \B)

and henceX \ (A ∪B) ⊆ (X \A) ∩ (X \B).

Next we show that

X \ (A ∪B) ⊇ (X \A) ∩ (X \B).

Let x ∈ (X \ A) ∩ (X \ B). Then x ∈ X \ A and x ∈ X \ B and so x /∈ A andx /∈ B. Thus x /∈ A ∪B and therefore

x ∈ X \ (A ∪B).

HenceX \ (A ∪B) ⊇ (X \A) ∩ (X \B).

ConsequentlyX \ (A ∪B) = (X \A) ∩ (X \B).

3. In R, let A1 = (−1, 1), A2 = (−1/2, 1/2), . . . , Ak = (−1/k, 1/k), . . . . Show

that∞⋂

k=1

Ak = {0}.

Solution.

Clearly

{0} ⊆∞⋂

k=1

Ak.

On the other hand, suppose that

a ∈∞⋂

k=1

Ak.

We need to show that a = 0. If a 6= 0, then let K ∈ N be such that 1/K < |a|.Then a /∈ (−1/K, 1/K); for otherwise |a| < 1/K, contradicting the choice ofK. Thus

a /∈∞⋂

k=1

Ak,

a contradiction. Hence a = 0 and so∞⋂

k=1

Ak = {0}.

3

4. Let X and Y be any two nonempty sets and let f : X → Y be any mapping.Then, for any subsets A and B of X, prove that

(i) f(A ∪B) = f(A) ∪ f(B)(ii) f(A ∩B) ⊆ f(A) ∩ f(B)(iii) f(B \A) ⊇ f(B) \ f(A)

Solution.

(i) Since A ⊆ A ∪ B and B ⊆ A ∪ B, it follows that f(A) ⊆ f(A ∪ B) andf(B) ⊆ f(A ∪B) and so

f(A) ∪ f(B) ⊆ f(A ∪B) .

Next

y ∈ f(A ∪B) =⇒ there is x ∈ A ∪B such that y = f(x)

=⇒ there is x ∈ A or x ∈ B such that y = f(x)

=⇒ y = f(x) ∈ f(A) or y = f(x) ∈ f(B)

=⇒ y ∈ f(A) ∪ f(B) .

Thusf(A ∪B) ⊆ f(A) ∪ f(B) .

and hencef(A ∪B) = f(A) ∪ f(B) .

(ii) If x ∈ f(A ∩ B), then there is x ∈ A ∩ B such that y = f(x) so thatthere is x ∈ A and x ∈ B with y = f(x). Thus y = f(x) ∈ f(A) andy = f(x) ∈ f(B) and so y = f(x) ∈ f(A) ∩ f(B). Hence

f(A ∩B) ⊆ f(A) ∩ f(B).

OR: Since A ∩ B ⊆ A and A ∩ B ⊆ B, it follows that f(A ∩ B) ⊆ f(A)and f(A ∩B) ⊆ f(B). Hence

f(A ∩B) ⊆ f(A) ∩ f(B).

[Note that y ∈ f(A) ∩ f(B) means that there is a ∈ A and b ∈ B suchthat y = f(a) = f(b), but a may not be in B and b may not be in A.Thus equality does not hold in general.]

(iii) Suppose that y ∈ f(B) \ f(A). Then y ∈ f(B) and y /∈ f(A). Thus thereis x ∈ B \A such that y = f(x). That is, y = f(x) ∈ f(B \A) and hence

f(B \A) ⊇ f(B) \ f(A).

[Note that there may be element b ∈ B \ A and a ∈ A with f(b) = f(a)so equality does not hold in general]

4

5. Let X and Y be any two nonempty sets and let f : X → Y be any mapping.Then, for any subsets A and B of Y , prove that

(i) f−1(A ∪B) = f−1(A) ∪ f−1(B)(ii) f−1(A ∩B) = f−1(A) ∩ f−1(B)(iii) f−1(B \A) = f−1(B) \ f−1(A)

Solution.

(i) We have

x ∈ f−1(A ∪B) ⇐⇒ f(x) ∈ A ∪B

⇐⇒ f(x) ∈ A or f(x) ∈ B

⇐⇒ x ∈ f−1(A) or x ∈ f−1(B)

⇐⇒ x ∈ f−1(A) ∪ f−1(B) .

Hencef−1(A ∪B) = f−1(A) ∪ f−1(B) .

(ii) We have

x ∈ f−1(A ∩B) ⇐⇒ f(x) ∈ A ∩B

⇐⇒ f(x) ∈ A and f(x) ∈ B

⇐⇒ x ∈ f−1(A) and x ∈ f−1(B)

⇐⇒ x ∈ f−1(A) ∩ f−1(B) .

Hencef−1(A ∩B) = f−1(A) ∩ f−1(B) .

(iii) We have

x ∈ f−1(B \A) ⇐⇒ f(x) ∈ B \A

⇐⇒ f(x) ∈ B and f(x) /∈ A

⇐⇒ x ∈ f−1(B) and x /∈ f−1(A)

⇐⇒ x ∈ f−1(B) \ f−1(A) .

Hencef−1(B \A) = f−1(B) \ f−1(A) .

5

6. Let X and Y be any two nonempty sets and let f : X → Y be any mapping.Then for any subset A of X and any subset B of Y , prove that

(i) A ⊆ f−1(f(A)

), and equality holds if f is injective.

(ii) f(f−1(B)

)⊆ B, and equality holds if f is surjective.

Solution.

(i) For x ∈ A, clearly f(x) ∈ f(A) so that x ∈ f−1(f(A)

)and thus, whether

or not f is injective,A ⊆ f−1

(f(A)

).

Now suppose that f is injective. Let x ∈ f−1(f(A)

). Then f(x) ∈ f(A)

and so there is a ∈ A such that f(x) = f(a). Since f is injective, it followsthat x = a so that x ∈ A. Hence

f−1(f(A)

)⊆ A,

and thereforeA = f−1

(f(A)

).

(ii) For y ∈ f(f−1(B)

), there is x ∈ f−1(B) such that y = f(x). As x ∈

f−1(B), we see that f(x) ∈ B and so y ∈ B. Thus, whether or not f issurjective,

f(f−1(B)

)⊆ B.

Now suppose that f is surjective. Let y ∈ B. Since f is surjective, thereis x ∈ X such that y = f(x). As y = f(x) ∈ B, we see that x ∈ f−1(B)and so

y = f(x) ∈ f(f−1(B)

).

HenceB ⊆ f

(f−1(B)

),

and thereforef(f−1(B)

)= B.

6

7. If 0 < a ≤ b, show thata

1 + a≤ b

1 + b.

Solution.

Method 1: Since a ≤ b, it follows that a+ab ≤ b+ab and so a(1+b) ≤ b(1+a).Since 0 < 1 + a and 0 < 1 + b, we have

a

1 + a≤ b

1 + b.

Solution.

Method 2: If 0 < a ≤ b, then 1/a ≥ 1/b so that 1 +1a≥ 1 +

1b

> 0. Then

11 + 1/a

≤ 11 + 1/b

,

which givesa

1 + a≤ b

1 + b.

Solution.

Method 3: If a ≤ b, then 1 + a ≤ 1 + b so that1

1 + a≥ 1

1 + b. Then

− 11 + a

≤ − 11 + b

and so 1− 11 + a

≤ 1− 11 + b

,

which givesa

1 + a≤ b

1 + b.

Solution.

Method 4: Consider the function

f(x) =x

1 + x

for x > 0. Thenf ′(x) =

1(1 + x)2

so that f ′(x) > 0 for x > 0. Hence if 0 < a ≤ b, then

a

1 + a≤ b

1 + b.