The University of Sydney

MATH 3901

Metric Spaces 2004

Tutorial 3

PROBLEM SET 3

1. Sketch (where possible) the following sets A, and decide whether A is an opensubset, or a closed subset, or neither, of the appropriate space Rn . Then foreach A, find Int A, A and FrA.

(i) A = ∪n∈N(n, n + 1), N = {0, 1, 2, . . . }(ii) A = set of all integers in R(iii) A =

{(x1, x2) ∈ R2 | x1x2 = 0

}(iv) A =

{(x1, x2) ∈ R2 | x1 rational

}(v) A =

{(x1, 0) ∈ R2 | 0 < x1 < 4

}Solution.

(i) The set A is as shown:

0 1 2 3 4 5 6

The set A is the union of open intervals and so it is open. The complementX \A is given by

X \A = (−∞, 0] ∪ {1, 2, 3, . . . },

and it is not open since any open ball containing, for example, 1 containspoints in A, i.e., points outside X \A. Hence A is not closed.Since A is open, Int A = A. The closure of A is A = [0, ∞). Since X \Ais closed, it follows that X \A = X \A and so

FrA = A ∩ (X \A) = {0, 1, 2, 3, · · · } .

(ii) The set A is as sketched:

−3 −2 −1 0 1 2 3

Since any open interval containing, for example, 0 contains points outsideA, it follows that A is not open. The complement of A is

X \A =⋃n∈Z

(n, n + 1) ,

2

which is the union of open intervals (n, n + 1). Thus X \ A is open sothat A is closed.

Clearly Int A = ∅. Since A is closed, A = A. Again it is clear thatX \A = R and so FrA = A ∩ (X \A) = A.

(iii) Note that x1x2 = 0 implies eitherx1 = 0 or x2 = 0 . Thus the set Ais as shown on the right. Clearly,any open ball containing (0, 0)in A contains points outside A.Hence A is not open. Next A isthe union of two closed subsets{(x1, x2) ∈ R2 | x1 = 0

}and{(x1, x2) ∈ R2 | x2 = 0

}.

Therefore A is closed.

Clearly Int A = ∅. Since A is closed, A = A. It is easily seen thatX \A = R2 and so

FrA = A ∩ (X \A) = A.

(iv) The set A cannot be sketched. Since any open ball containing (x1, x2)with x1 rational, contains points (y1, y2) with y1 irrational (i.e. pointsoutside A), it follows that A is not open. Similarly, the complement X \Aof A is not open and hence A is not closed.

Using the same reason, we see that Int A = ∅ and A = R2. Also X \A =R2 . Hence

FrA = A ∩ (X \A) = R2.

(v) The set A is as shown on theright. A is clearly not open. Thecomplement X \ A of A containsthe point (0, 0). Since any openball containing (0, 0) containspoints in A, it follows that X \Ais not open so that A is notclosed.

0 4

Clearly Int A = ∅, and

A ={(x1, 0) ∈ R2 | 0 ≤ x1 ≤ 4

}.

Now X \A = R2 and so

FrA = A ∩ (X \A) ={(x1, 0) ∈ R2 | 0 ≤ x1 ≤ 4

}.

3

2. Let A be an open subset of a metric space (X, d) and a ∈ A . Is A \ {a} openin X?

Solution.

We claim that A \ {a} = A ∩ (X \ {a}) . Clearly

A \ {a} ⊆ A ∩ (X \ {a}) .

Now let x ∈ A∩ (X \ {a}) . Then x ∈ A and x ∈ X \ {a} , so that x 6= a . Thusx ∈ A \ {a} and so

A ∩ (X \ {a}) ⊆ A \ {a} .

HenceA \ {a} = A ∩ (X \ {a}) ,

as claimed. Since both A and X \ {a} are open, we conclude that A \ {a} isopen.

3. Let (X, d) be a metric space. If A ⊆ B ⊆ X, prove that Int A ⊆ Int B.

Solution.

Let a ∈ Int A. Then there exists an open ball B(a; ε) such that B(a; ε) ⊆ A.Since A ⊆ B, it follows that B(a; ε) ⊆ B, and so a ∈ Int B. Hence Int A ⊆Int B.

4. Let x be a limit point of a subset A of a metric space (X, d). Prove that everyopen ball B(x; ε) contains an infinite number of points of A.

Solution.

Let x be a limit point of A and let B(x; ε) be any open ball. Then(B(x; ε) \ {x}

)∩A 6= ∅.

That is, there is a point x1 ∈ X such that x1 6= x and x1 ∈ B(x; ε). Letε1 = d(x, x1) < ε. Then the open ball B(x; ε1) also contains a point x2,say, of X with x2 6= x. Since x1 /∈ B(x; ε1), it follows that x1 6= x2. Letε2 = d(x, x2) < ε1. Next, by considering the open ball B(x; ε2), we obtain athird point x3 distinct from x, x1, x2; and so on. Hence every open ball B(x; ε)contains an infinite number of points of A.

5. Let (X, d) be a metric space and A ⊆ X. Prove that the derived set A′ of A isclosed.

Solution.

Let x be a limit point of A′. Then any open ball B(x; ε) contains a point y 6= xof A′. Let B(y; δ) be such that B(y; δ) ⊆ B(x; ε). Since y ∈ A′, it follows thatB(y; δ) contains an infinite number of points of A, by Question 4. Thus B(x; ε)contains an infinite number of points of A and so x ∈ A′. Hence (A′)′ ⊂ A′ sothat A′ is closed, by Theorem 2.7.

4

6. Let (X, d) be a metric space.

(i) If A ⊆ B ⊆ X, prove that A′ ⊆ B′.

(ii) If A and B are subsets of X, prove that (A ∪B)′ = A′ ∪B′.

Solution.

(i) Let x ∈ A′. Then for any open ball B(x; ε),(B(x; ε) \ {x}

)∩A 6= ∅.

Since A ⊆ B, it follows that(B(x; ε) \ {x}

)∩B ⊃

(B(x; ε) \ {x}

)∩A 6= ∅,

so that x ∈ B′. Hence A′ ⊆ B′.

(ii) Since A ⊂ A ∪ B and B ⊂ A ∪ B, it follows from (i) that A′ ⊂ (A ∪ B)′

and B′ ⊂ (A ∪B)′ so that

A′ ∪B′ ⊆ (A ∪B)′.

Suppose that x /∈ A′ ∪ B′. Then there exist open balls B(x; ε1) andB(x; ε2) such that (

B(x; ε1) \ {x})∩A = ∅

and (B(x; ε2) \ {x}

)∩B = ∅.

Let ε = min{ε1, ε2). Then we see that B(x; ε) is an open ball such that(B(x; ε) \ {x}

)∩ (A ∪B)

=[(

B(x; ε) \ {x})∩A

]∪

[(B(x; ε) \ {x}

)∩A

]⊆

[(B(x; ε1) \ {x}

)∩A

]∪

[(B(x; ε2) \ {x}

)∩A

]= ∅ ∪ ∅ = ∅.

so that x /∈ (A ∪B)′. Hence if x ∈ (A ∪B)′, then x ∈ A′ ∪B′; that is,

(A ∪B)′ ⊆ A′ ∪B′.

Consequently (A ∪B)′ = A′ ∪B′.

5

7. Let (X, d) be a metric space and A, B be two subsets of X. Prove that:

(i) If A ⊆ B, then A ⊆ B.

(ii) A =⋂{B ⊆ X | B is closed and A ⊆ B}. That is, A is the smallest

closed set containing A.(iii) A is closed if and only if A = A.

(iv) A ∪B = A ∪B.

(v) A ∩B ⊆ A ∩B.

Solution.

(i) Let x ∈ A . Then for any open set (or open ball) U containing x, U∩A 6= ∅.Since A ⊆ B, it follows that U ∩B 6= ∅ . Hence x ∈ B and so A ⊆ B .

(ii) LetF =

⋂{B ⊆ X | B is closed and A ⊆ B}.

Then F is closed. We want to show that A = F

We frist show that A ⊆ F . Suppose that x /∈ F . Then x ∈ X \ F . SinceF is closed, X \F is open so that there is an open ball B(x; ε) such that

B(x; ε) ⊆ (X \ F ),

that is, B(x; ε)∩F = ∅. Since A ⊆ F , it follows that B(x; ε)∩A = ∅ andso x /∈ A. Hence A ⊆ F.

Next, we show that F ⊆ A. Suppose that x /∈ A. Then there is an openball B(x; ε) with B(x; ε) ∩ A = ∅. Choose B = X \ B(x; ε). Then B isclosed, A ⊆ B and x /∈ B. Thus x /∈ F . Hence F ⊆ A.Consequently

A = F =⋂{B ⊆ X | B is closed and A ⊆ B}.

(iii) If A = A, then A is closed, since A is closed. Now suppose that A isclosed. Then, by (ii), A ⊆ A and so A = A.

(iv) Since A ⊆ A∪B and B ⊆ A∪B, it follows from (i) that A ⊆ A ∪B andB ⊆ A ∪B . Then

A ∪B ⊆ A ∪B .

Next, Since A ⊆ A and B ⊆ B, it follows that A∪B ⊆ A∪B and by (i),

A ∪B ⊆ A ∪B = A ∪B

since A ∪B is closed. Hence A ∪B = A ∪B.

(v) Since A ∩B ⊆ A and A ∩B ⊆ B, it follows from (i) that

A ∩B ⊆ A and A ∩B ⊆ B

and hence A ∩B ⊆ A ∩B.

6

8. Let (X, d) be a metric space and A ⊂ X. Prove that:

(i) A = Int A ∪ FrA.

(ii) A is closed if and only if FrA ⊆ A.

(iii) A is open if and only if FrA ⊆ X \A.

(iv) Fr(X \A) = FrA.

Solution.

(i) Since Int A ⊆ A ⊆ A, and FrA ⊆ A, it follows that

Int A ∪ FrA ⊆ A.

Let x ∈ A. If x ∈ Int A, then we are done. If x /∈ Int A, then every openball B(x; ε) must contain some points in the complement X \A. Then

x ∈ X \A and so x ∈ A ∩ (X \A) = FrA.

HenceA ⊆ Int A ∪ FrA

and thereforeA = Int A ∪ FrA.

(ii) Note that A is closed if and only if A = A. Hence the result followsimmediately from (i).

(iii) Note that A is open if and only if X \A is closed. Thus the result followsfrom (ii).

(iv) It follows from the definition and the fact that A = X \ (X \A).

9. Let A ={(x1, x2) ∈ R2 | x1, x2 both rationals

}. Show that A = R2 .

Deduce that R2 is separable.

Solution.

Since any open ball in R2 must contain points (x1, x2) with both x1 and x2

rationals, it follows that every point of R2 is a limit point of A . Hence

A = R2 .

SinceA = Q×Q =

⋃x∈Q

({x} ×Q)

and each {x} ×Q is countable, A, as a countable collection of a countable setis countable. Hence R2 is separable.

7

10. Prove that the metric space `p, with 1 ≤ p < ∞, is separable.

Solution.

Recall that the metric d in `p is defined by

d(x, y) ={ ∞∑

k=1

|xk − yk|p}1/p

,

for x = (xk) and y = (yk) in `p.

Let A be the set of all elements y in `p such that only a finite number of thecoordinates yk of y are non-zero, and the non-zero coordinates are all in Q.We first show that A is countable. For this, let An be the subset of A consistingof all those elements y = (yk) such that yk = 0 for all k > n + 1. Then, it iseasily seen that

A =∞⋃

n=1

An.

Since A1, A2, . . . are all countable, it follows that A is countable.Next we show that A is everydense in `p; that is, A = `p. Let x ∈ `p, y ∈ Aand set

y = {y1, y2, . . . , yn, 0, 0, . . . }.

Then

d(x, y) =

{n∑

k=1

|xk − yk|p +∞∑

k=n+1

|xk|p}1/p

.

Given any ε > 0, since x ∈ `p, there exists n such that

∞∑k=n+1

|xk|p <12

εp.

Fix n. Since Q is everywhere dense in R, for x1, x2, . . . , xn, we can choosey1, y2, . . . , yn in Q such that

|xk − yk| < ε/(2n)1/p or |xk − yk|p <εp

2n,

k = 1, 2, . . . , n. Hence for any x ∈ `p, there is y ∈ A such that

d(x, y) =

{n∑

k=1

|xk − yk|p +∞∑

k=n+1

|xk|p}1/p

<

{n∑

k=1

εp

2n+

εp

2

}1/p

={

nεp

2n+

εp

2

}1/p

= ε.

Hence A is everywhere dense in `p and so `p is separable.

8

11. Let (Y, dY ) a metric subspace of a metric space (X, d) and H ⊆ Y. Prove thatH is closed in Y if and only if there exists a closed subset C in X such thatH = C ∩ Y .

Solution.

Recall that H is closed in Y if the complement Y \H is open in Y .Suppose that H is closed in Y . Then G = Y \H is open in Y and so there isan open set U in X such that G = U ∩ Y . Let C = X \ U . Then C is closedin X and H = C ∩ Y .

Let H ⊆ Y . Suppose that there is a closed set C in X such that H = C ∩ Y .Let U = X \ C . Then U is open in X and Y \H = U ∩ Y and so G = Y \His open in Y . Hence H is closed in Y .

12. Let N = {1, 2, 3, . . . } be the set of all natural numbers in the Euclidean metricspace (R, d). Is the set {1} open in N? What are the open sets in N?

Solution.

Since {1} = (1/2, 3/2) ∩ N and (1/2, 3/2) is open in R, it follows that {1} isopen in N. Similarly, one can prove that every singleton set {a} in N is openin N and so every subset of N is open in N.

13. Let (X, d) be a metric space.

(i) Let Y ⊆ X be an open subset in X. If G ⊆ Y is open in Y , prove that Gis open in X.

(ii) Let Y ⊆ X be a closed subset in X. If H ⊆ Y is closed in Y , prove thatH is closed in X.

Solution.

(i) Since G is open in Y , there is an open set U in X such that G = U ∩ Y .Since Y is open in X, it follows that G, as an intersection of two opensets in X, is open in X.

(ii) Since H is closed in Y , there is a closed set C in X such that H = C ∩Y .Since Y is closed in X, it follows that H, as an intersection of two closedsets in X, is closed in X.