BabyRudin-Chapter2.PDF

Principles of Mathematical Analysis

Notes and Problem Solutions

Yuqun Chen

New York, NY 10003

January 18, 2013

2

Chapter 2

BASIC TOPOLOGY

2.1 NOTES

COMPACT SETSA compact subset of a metric space need not be bounded.

Theorem 2A. The union of two non-disjoint, connected sets is connected.

Proof. Let E1 and E2 be two non-disjoint, connected sets in the metric spaceX , i.e., E1 ⊂ X,E2 ⊂ X,E1 ∩ E2 6= ∅. We want to show that E = E1 ∪ E2

is connected. We prove this by contradiction, i.e., suppose E is not connected,i.e., there exist two non-empty, separated sets A and B, i.e., A∩B = A∩B = ∅,such that E = A ∪B.

To show contradiction, letE1A = E1 ∩A,E1B = E1 ∩B,E2A = E2 ∩A,E2B = E2 ∩B.

Step 1 We first show that either E1A or E1B is empty. In other words, all thepoints of E1 must belong to one and only one of A and B. To show this, noticethat

E1A ⊂ A =⇒ E1A ⊂ A =⇒ E1A ⊂ A.The above, combined with A ∩B = ∅, results in E1A ∩B = ∅, which togetherwith E1B ⊂ B leads to

E1A ∩ E1B = ∅.Similarly, we can show

E1A ∩ E1B = ∅.In particular, E1A ∩E1B = ∅.

Since E1A∪E1B = E1∩ (A∪B) = E1∩E = E1 and E1A∩E1B = ∅, E1 canbe partitioned into two disjoint sets E1A and E1B . Because E1 is connected,one of these two sets must be empty.

3

4 CHAPTER 2. BASIC TOPOLOGY

Without loss of generality, let

E1A = E1, E1B = ∅ (2.1)

Step 2 The same conclusion can be drawn about E2A and E2B: one of them isempty and the other is equal to E2. We have here two cases to consider.

Case 1: E2A = E2, E2B = ∅

Since E1B = E1 ∩B = ∅ and E2B = E2 ∩B = ∅, B ∩E = B ∩ (E1 ∪E2) = ∅.This contradicts the assumption that E can be partitioned into two non-emptysets A and B.

Case 2: E2A = ∅, E2B = E2

Since E1A = E1, E2B = E2, we have

E1 ∩ E2 = E1A ∩ E2B

= (E1 ∩ A) ∩ (E2 ∩B)

= (E1 ∩ E2) ∩ (A ∩B)

Therefore, E1 ∩ E2 6= ∅ =⇒ A ∩ B 6= ∅ =⇒ A ∩ B 6= ∅. This contradicts ourassumption that E1 ∩ E2 6= ∅ and A ∩B = ∅.

In both cases, we derived contraditions. Therefore, E cannot be partitionedinto two non-empty, separated sets. Hence, E is connected.

2.2 EXERCISES

Problem 2.1 Prove that the empty set is a subset of every set.

Proof. Given any set S, we want to show that x ∈ ∅ =⇒ x ∈ S. This isequivalent to x /∈ S =⇒ x /∈ ∅. The latter statement is always true. Therefore,∅ ⊂ S.

Problem 2.2 A complex number z is said to be algebraic if there are integersa0, a1, ..., an, not all zero, such that

a0zn + a1z

n−1 + . . .+ an−1z + an = 0.Prove that the set of all algebraic numbers is countable. Hint: for every positiveinteger N there are only finitely many equations with

n+ |a0|+ |a1|+ . . .+ |an| = N .

Proof. Each algebraic equation can be characterized by the tuple (n, a0, ..., an).Since for every positive integer N there are only finitely many tuples

n+ |a0|+ |a1|+ . . .+ |an| = N,N > n.we can enumerate all of them beginning with N = 1. After listing tuples for

2.2. EXERCISES 5

N = 1, 2, ...,m, we next we list all tuples satisfying the above equation forN = m + 1. Thus for every tuple (n, a0, ..., an), there is a unique index in thelist. In other words, we just established a one-to-one correspondence betweeneach tuple (or algebraic equation) with a natural number. Hence, the set ofalgebraic equations is countable.

Problem 2.3 Prove that there exist real numbers which are not algebraic.

Proof. Since the set of algebraic real numbers is countable (Problem 2.2) andthe set of real numbers is not, there exist real numbers that are not algebraic.

Problem 2.4 Is the set of all irrational real numbers countable?

Solution No, because the R is uncountable, Q is countable and the set of allirrational real numbers is R−Q.

Problem 2.5 Construct a bounded set of real numbers with exactly three limitpoints.

Solution Let

S1 = {0 + 1

n| n = 1, 2, 3, ...}

S2 = {2 + 1

n| n = 1, 2, 3, ...}

S3 = {4 + 1

n| n = 1, 2, 3, ...}

Each set has exactly one limit point, 0, 2, and 4, respectively. Their union,S1 ∪ S2 ∪ S3, is bounded and has exactly 3 limit points.

Problem 2.6 Let E′ be the set of all limit points of a set E. Prove thatE′ is closed. Prove that E and E have the same limit points. (Recall thatE = E ∪ E′). Do E and E′ always have the same limit points?

Proof. We first prove that E′ is closed. For every limit point x of E′, i.e.,x ∈ (E′)′, and for every open ball B centered at x, there exists a point y 6= xsuch that y ∈ E′. One can always find an open ball B′ centered at y, such that


B′ ⊂ B, x /∈ B′. This can be done by choosing the radius r′ for the open ballB′ to be 1

2min(d(x, y), r − d(x, y)), where r is the radius for the open ball B.

Since y ∈ E′ and B′ is an open ball centered at y, there exists a pointp 6= y, p ∈ E such that p ∈ B′. But since x /∈ B′, x 6= p. And since B′ ⊂ B,p ∈ B. Therefore, there exists a point p ∈ E, p 6= x for every open ball centeredat x. Hence, x ∈ (E′)′ =⇒ x ∈ E′, which means E′ is closed.

Proof. We next prove that E′ = (E)′

Clearly, since E ⊂ E, E′ ⊂ (E)′. We want to show (E)′ ⊂ E′, i.e., to showfor every x ∈ (E)′, x ∈ E′.

Let x be a limit point of E. This means within every open ball B(x, r),centered at x and with radius r, there exists a point y ∈ E such that x 6= y, y ∈B(x, r). This point y is either in E or E′ (or both). If y ∈ E′, one can use thesame construction used in proving E′ is closed to prove the existence of a pointz ∈ E, z 6= x, z 6= y, and z ∈ B(x, r). Therefore, there always exists a pointp ∈ E, p 6= x in the open ball B(x, r). Hence, x is also a limit point of E, i.e.,x ∈ E′. Therefore, (E)′ ⊂ E′. QED.

Problem 2.7 Let A1, A2, A3, ... be subsets of a metric space.

2.7(a) If Bn =n⋃

i=1

Ai, prove that Bn =n⋃

i=1

Ai, for n = 1, 2, 3, ....

Proof. If we can show Bn′ =

n⋃

i=1

Ai′, we will have Bn =

n⋃

i=1

Ai, because E =

E ∪ E′. To do so, we want to show that

1. Bn′ ⊃

n⋃

i=1

Ai′, and

2. Bn′ ⊂

n⋃

i=1

Ai′.

Clearly, a limit point of Ai is also a limit point of Bn. Therefore, Bn′ ⊃

n⋃

i=1

Ai′ (1).

To prove (2), we pick an arbitrary limit point of Bn, call it x, and show x isa also limit point of at least one Ai, 1 ≤ i ≤ n. We prove this by contradiction.Suppose x is not a limit point of any Ai, ∀1 ≤ i ≤ n. Then there exists n openballs centered at x, N1, N2, ..., Nn, such that there does not exist another point(different from x) from Ai in Ni, i.e., Ni ∩ Ai ⊂ {x}, ∀1 ≤ i ≤ n.

2.2. EXERCISES 7

Now consider the intersection of all these open balls, N =n⋃

1

Ni. N is an

open ball centered at x and (N ∩n⋃

i=1

Ai) ⊂ {x}. Therefore, x cannot be a limit

point of Bn. Contradiction. Hence, every limit point of Bn is also a limit point

of at least one Ai, 1 ≤ i = 1 ≤ n, or Bn′ ⊂

n⋃

i=1

Ai′.

2.7(b) If B =∞⋃

i=1

Ai, prove that B ⊃∞⋃

i=1

Ai. And show that this inclusion can

be proper.

Proof. We want to show that every limit point of Ai is also a limit point of B.Let x be a limit point of an Ai, i ∈ Z+. For every open ball N , centered at x,there exists a point y, y 6= x, y ∈ Ai, y ∈ N . Since Ai ⊂ B, y ∈ B. Therefore, x

is also a limit point of B. Hence, B′ ⊃∞⋃

i=1

Ai′. Since E = E ∪ E′, B ⊃

∞⋃

i=1

Ai.

The inclusion can be proper. For example, let Ai = {1/i}, i = 1, 2, 3, ....

Then, Ai′ = ∅,

∞⋃

i=1

Ai′ = ∅. However, B′ = {0}.

Problem 2.8 Is every point of every open set E ⊂ R2 a limit point of E?Answer the same question for closed sets in R2.

Solution We prove this for a more general case E ⊂ Rk.

Every point of every open set E ⊂ Rk a limit point of E.

Proof. For every point x ∈ E ⊂ Rk, where E is an open set, there exists anopen ball B ⊂ Rk, centered at x, such that B ⊂ E. Since B is an open ball inRk and x ∈ B, x is a limit point of B. And since B ⊂ E, x is also a limit pointof E. Therefore, every point of E is a limit of E.

A point in a closed set E ⊂ Rk may not be a limit point of E.

Proof. Any finite set in Rk.

Problem 2.9 Let E◦ denote the set of all interior points of a set E. [E◦ iscalled the interior of E.]

2.9(a) Prove that E◦ is always open.


Proof. For every point p ∈ E◦, since p is an interior point of E, there exists anopen ball B(p, r), centered at p and with radius r, such that B(p, r) ⊂ E.

For every point q in B(p, r) such that q 6= p, we have d(p, q) < r. Letr′ = min(d(p, q), r − d(p, q)) and consider the open ball B′(q, r′), centered at qand with radius r′. For every point z ∈ B′(q, r′), we have d(q, z) < r′. Therefore,

d(p, z) ≤ d(p, q) + d(q, z) < d(p, q) + r′ ≤ d(p, q) + r − d(p, q) = r.Hence, every z ∈ B′ is also a point in B, or B′ ⊂ B ⊂ E. This means, q is aninterior point of E, or q ∈ E◦. Hence, B(p, r) ⊂ E◦. Therefore, every point inE◦ is an interior point of E◦, or E◦ is always open.

2.9(b) Prove that E is open iff E◦ = E.

Proof. Forward Direction If E is open, every point of E is an interior pointof E. Hence E ⊂ E◦. But since E◦ ⊂ E, E◦ = E.

Reverse Direction E◦ = E means every point of E is an interior point of E,therefore E is open.

2.9(c) If G ⊂ E and G is open, prove that G ⊂ E◦.

Proof. It is easy to show that since G ⊂ E, every interior point of G is alsoan interior point of E, i.e., G◦ ⊂ E◦. Since G is open, G = G◦. Therefore,G ⊂ E◦.

2.9(d) Prove that the complement of E◦ is the closure of the complement ofE, i.e., (E◦)c = Ec.

Proof. Step 1 We first show (E◦)c ⊂ Ec. For every point p ∈ (E◦)c, p /∈ E◦,i.e., p is not an interior point of E. In other words, there does not exist anopen ball B centered at p, such that B ⊂ E. Therefore, for every open ball Bcentered at p, there exists a point q ∈ Ec, such that q 6= p and q ∈ B. Thismeans that p is a limit point of Ec, or p ∈ Ec. Hence, (E◦)c ⊂ Ec.

Step 2 We next show (E◦)c ⊃ Ec. For every point p ∈ Ec, either p ∈ Ec orp is a limit point of Ec (or both). If p ∈ Ec, p /∈ E; hence p /∈ E◦; therefore,p ∈ (E◦)c. Otherwise, p must be a limit point of Ec. This means every openball centered at p contains infinitely many points of Ec. Therefore, p cannotbe an interior point of E, or p /∈ E◦. Again, we have p ∈ (E◦)c. Therefore,(E◦)c ⊃ Ec.

Combine the conclusions from both steps, we have (E◦)c = Ec.

2.9(e) Do E and E always have the same interiors?

Answer No. Let E = (0, 1) ∪ (1, 2). E = (0, 2). E◦ = (0, 1) ∪ (1, 2), while(E)◦ = (0, 2).

2.2. EXERCISES 9

2.9(f) Do E and E◦ always have the same closures?

Answer No. Consider the set Q of rational numbers in the metric space R1.Q◦ = ∅. Hence Q◦ = ∅. But Q = R.

Problem 2.10 Let X be an infinite set. For p ∈ X and q ∈ X , define

d(p, q) =

{

1 (if p 6= q)0 (if p = q).

(2.2)

Prove that d(p, q) is a metric. Which subsets of the resulting metric space areopen? Which are closed? Which are compact?

Solution We first prove d(p, q) is a metric.

Proof. There are three properties of a metric space d(p, q) must have (Defini-tion 2.15).

(a) d(p, q) > 0 if p 6= q; d(p, p) = 0;

This one is obvious from the definition.

(b) d(p, q) = d(q, p);

This one is also obvious.

(c) d(p, q) ≤ d(p, r) + d(r, q), for any r ∈ X .

The case for p = q is trivial. If p 6= q, d(p, q) = 1. But for any r ∈ X , p = rand q = r cannot simultaneosly be true. Therefore, d(p, r) + d(r, q) ≥ 1 =d(p, r).

Note that any open ball with a radius greater than or equal to 1 contains allpoints of X ; and any open ball with a radius less than 1 contains one and onlyone point (the center).

Every subset of X is open, because every point in the set is an interior point.(Choose an open ball centered at this point, with radius 0.5).

Every subset of X is closed, because there are no limit points in X .

Every finite subset of X is compact. But none of the infinite subsets of X arecompact: Given such a subset A ⊂ X , one can choose a open covering consistingof open balls with radius 0.5, centered at every point in A; this open coveringdoes not contain a finite subcovering of A because each open ball only coversone point.


Problem 2.11 For x ∈ R1 and y ∈ R1, define

d1(x, y) = (x− y)2,

d2(x, y) =√

|x− y|,d3(x, y) = |x2 − y2|,d4(x, y) = |x− 2y|,

d5(x, y) =|x− y|

1 + |x− y| .

Determine, for each of these, whether it is a metric or not.

Solution The properties of a metric d(x, y) for a space X are

(a) d(p, q) > 0 if p 6= q; d(p, p) = 0;

(b) d(p, q) = d(q, p);

(c) d(p, q) ≤ d(p, r) + d(r, q), for any r ∈ X .

With these in mind, we look at each metric.

1. d1(x, y) = (x− y)2 is not a metric.

d1 does not satisfy the triangular inequality (Property (c)). For example,d1(2, 0) = 4 > 2 = d1(2, 1) + d1(1, 0).

2. d2(x, y) =√

|x− y| is a metric.

It is trivial to show that d2 has Properties (a) and (b). To show that italso has Property (c), we use the basic inequality as follows.

|x− y| = |(x − z) + (z − y)|≤ |x− z|+ |z − y|≤ (

√

|x− z|+√

|z − y|)2

Therefre,√

|x− y| ≤√

|x− z|+√

|z − y, for all x, y, z ∈ R1.

3. d3(x, y) = |x2 − y2| is not a metric.

Property (a) does not hold because for any x 6= 0, d3(x,−x) = 0.

4. d4(x, y) = |x− 2y| is not a metric.

Property (a) does not hold because for any y 6= 0, d4(2y, y) = 0. Inaddition, property (b) does not hold, either.

5. d5(x, y) =|x−y|

1+|x−y| is a metric.

Obviously, properties (a) and (b) hold. We want to show d5 also satisfiesthe triangle inequality. Notice that d5(x, y) = 1 − 1

1+|x−y| , we need to

2.2. EXERCISES 11

prove that for any z ∈ R1,1 + 1

1+|x−y| ≥ 11+|x−z| +

11+|z−y

.

1

1 + |x− z| +1

1 + |z − y| =2 + |x− z|+ |z − y|

1 + |x− z|+ |z − y|+ |x− z||z − y|

≤ 2 + |x− z|+ |z − y|1 + |x− z|+ |z − y|

= 1 +1

1 + |x− z|+ |z − y|

≤ 1 +1

1 + |x− y|

The last inquality in the deductionn is due to |a+ b| ≤ |a|+ |b|.

Problem 2.12 Let K ⊂ R1 consist of 0 and the numbers 1/n, for n = 1, 2, 3....Prove that K is compact directly from the definition (without using the Heine-Borel theorem).

Proof. Let {Gα} be a collection of an open cover for K. Since 0 ∈ K, thereexists an index i such that 0 ∈ Gi. Since Gi is open, there exists an open ballin R1, centered at 0, with radius δ > 0, such that (−δ, δ) ∈ K. Given this δ,there exists an N ∈ Z+ such that for all n ≥ N , 1/n < δ. In other words, forall n ≥ N , 1/n ∈ Gi.

Let Fn be an open set in {Gα} such that Fn covers 1/n. Then Gi ∪N⋃

n=1Fn

is a finite open subcover for K. Therefore, K is compact.

Problem 2.13 Construct a compact set of real numbers whose limit pointsform a countable set.

Solution For each n ∈ Z1, let Sn = {1/n+ 12in(n+1) |i = Z+} and S =

∞⋃

n=1Sn.

Each Sn has exactly one limit point n and is closed because it contains thelimit point n. S′ = Z+ ⊂ S. Since S is closed and bounded in R1, S is compact.

Problem 2.14 Give an example of an open cover of the segment (0, 1) whichhas no finite subcover.

Solution Let Gn = ( 12n , 1). {Gn} is an open cover of (0, 1). But no finite

subcollection of {Gn} can cover (0, 1).


Problem 2.15 Show that Theorem 2.36 and its Corollary become false (in R1,for example) if the word ”compact” is replaced by ”closed” or by ”bounded.”

Solution

2.15(a) Let Gn = [2n,∞), n = 1, 2, 3, .... Gn is closed in R1 but not bounded.The intersection of any finite collection of {Gn} produces [2k,∞) for some k

and is therefore nonempty. However,∞⋂

n=1Gn = ∅.

2.15(b) Let Gn = (0, 2−n), n = 1, 2, 3, .... Gn is bounded but not closed inR1. The intersection of any finite collection of {Gn} produces (0, 2−k) for some

k and is therefore nonempty. However,∞⋂

n=1Gn = ∅.

Problem 2.16 Regard Q, the set of all rational numbers, as a metric space,with d(p, q) = |p− q|. Let E be the set of all p ∈ Q such that 2 < p2 < 3. Showthat E is closed and bounded in Q, but that E is not compact. Is E open in Q?

Solution

We first prove that E is closed and bounded in Q.

Proof. For every p ∈ E and q ∈ E, d(p, q) <√3−

√2. Hence, E is bounded.

Suppose x ∈ Q is a limit point of E. Since x ∈ Q, x2 6= 3 and x2 6= 2. Ifx2 > 3, there exists a rational number δ > 0, such that 3 < (x − δ)2. Hence,(x− δ, x+ δ) ∩ E = ∅. This contradicts the assumption that x is a limit pointof E. Therefore, x2 < 3. Similarly, we can prove x2 > 2. Hence, for every limitpoint x of E, x ∈ E. This proves that E is closed.

We next prove that E is not compact.

Proof. Let Gn consists of all rational numbers q ∈ Q such that 2 + 1n< q2 < 4,

for all n = 1, 2, 3, .... One can employ the same method used below to provethat Gn is open in Q. Therefore, }Gn} is an open cover of E.

However, there do not exist a finite number of indices i1, i2, ..., ik such thatthe collection {Gik} is a cover of E. Assume such a finite sequence can bechoosen and list them in an increasing order. Obviously, Gi1 ⊂ Gi2 ⊂ . . . ⊂ Gik .And Gik cannot cover the entire E.

We lastly prove that E is open in Q.

Proof. For every point x ∈ E, 2 < x2 < 3. Let x = n/m, where n ∈ Z+

and m ∈ Z+. We have 2 < (n/m)2 < 3. In particular, 3(m2 − n2) ≥ 1 and(n2 − 2m2) ≥ 1.

2.2. EXERCISES 13

There exists anN1 ∈ Z+ such that 3(m2−n2)k2 > m2+2mnk, or ( nm+ 1

k)2 <

3, for all k ≥ N1. Similarly, there exists an N2 ∈ Z+ such that (n2 − 2m2)k2 >2mnk −m2, or 2 < ( n

m+ 1

k)2, for all k ≥ N2.

Choose k = max(N1, N2), we have2 < ( n

m− 1

k)2 < ( n

m+ 1

k)2 < 3.

Hence ( nm

− 1k, nm

+ 1k) ∈ E. This proves that E is open in Q.

Problem 2.17 Let E be the set of all x ∈ [0, 1] whose decimal expansioncontains only the digits 4 and 7. Is E countable? Is E dense in [0, 1]? Is Ecompact? Is E perfect?

Solution No, no, yes, and no.

E is uncountable.

Proof. We use the same diagnonal argument used in proving the uncountabilityof R. Suppose E is countable and can be enumerated as e1, e2, ..., en, .... Weconstruct a number x such that the n-th digit of x is 4 if the n-th digit of enis 7; and 4 otherwise. Clearly, x should belong in E. Yet, it differs from everynumber in the sequence by at least one digit. Hence, x /∈ E. Contradiction.Therefore, E is uncountable.

E is not dense in [0, 1].

Proof. To show E is not dense in [0, 1], it suffices to show that 0.5, which is inthe interval, is at least 0.02 away from any point in E and cannot be a limitpoint of E.

E is compact.

Proof. To showE is compact, we only need to show that E is closed and boundedin [0, 1] and invoke Theorem 2.41. For every limit point x of E but x /∈ E,consider the first digit of x that is neither 4 nor 7. Let this digit be the k-thdigit, k ∈ Z+. For every number y ∈ E, we have d(x, y) > 0.2 ∗ 10−k = δ.Therefore, withint (x − δ, x+ δ) there do not exist any points from E. Hence,x cannot simultaneously be a limit point of E and exist outside E. Hence, E isclosed.

E is not perfect.

Proof. We need to show that not every point of E is a limit point. Consider0.4. There are no points of E in an open ball of 0.4, (0.4 − 0.01, 0.4 + 0.01).Hence, 0.4 is not a limit point of E. As a matter of fact, there does not exist alimit point of E in E.


Problem 2.18 Is there a nonempty perfect set in R1 which contains no rationalnumber?

Answer: Yes.

Proof. The Cantor set is just such a set: It is a perfect set that does not containany rational numbers. One can find online several papers on this topic. Theauthor found the following two papers particularly relevant to this problem.

• http://www.math.ucsb.edu/~yjshu/117/cantorset.pdf and

• http://www.missouriwestern.edu/orgs/momaa/

ChrisShaver-CantorSetPaper4.pdf.

Problem 2.19

(a) If A and B are disjoint closed sets in some metric space X , prove thatthey are separated.

(b) Prove the same for disjoint open sets.

(c) Fix p ∈ X, δ > 0, define A to be the set of all q ∈ X for which d(p, q) < δ,define B similarly, with > in place of <. Prove that A and B are separated.

(d) Prove that every connected metric space with at least two points is un-countable. Hint: Use (c).

Solution

2.19(a) If A and B are disjoint closed sets in some metric space X, prove that

they are separated.

Proof. Since A and B are closed in X , A = A and B = B. Since A and B aredisjoint, i.e., A ∩ B = ∅, A ∩ B = A ∩ B = ∅. In other words, A and B areseparated in X .

2.19(b) If A and B are disjoint open sets in some metric space X , prove thatthey are separated.

Proof. Let x be a limit point of A. If x ∈ B, there exists an open ball G(x, r),centered at x and with radius r, such that G ⊂ B. Then, there are infinitelymany points of A in G. Hence, A ∩ B 6= ∅. Therefore, A′ ∩ B = ∅. SinceA∩B = ∅, we have A∩B = ∅. Similarly, A∩B = ∅. Therefore, A and B areseparated in X .

2.2. EXERCISES 15

2.19(c) Fix p ∈ X, δ > 0, define A to be the set of all q ∈ X for whichd(p, q) < δ, define B similarly, with > in place of <. Prove that A and B areseparated.

Proof. From the definition of A and B, we have A ∩B = ∅. We want to showA′ ∩B = A ∩B′ = ∅.

Let x be a limit point of A. If d(p, x) > δ, choose a radius r = d(p, x) − δ.Consider the open ball G(x, r) centered at x and with radius r. For every pointy ∈ G(x, r), d(p, y) ≥ d(p, x) − d(y, x) > d(p, x) − r = δ, or y /∈ A. However,since x is a limit point of A, there are infinitely many points of A in G(x, r).Therefore, d(p, x) ≤ δ, or x /∈ B. This proves that A′ ∩B = ∅.

Similarly, we can prove A ∩B′ = ∅. Therefore, A and B are separated.

2.19(d) Prove that every connected metric space with at least two points isuncountable. Hint: Use (c).

Proof. Let p and q be two distinct points of a metric space X . For every realnumber r, 0 < r < d(p, q), construct the sets A and B as follows.

A = {x|d(p, x) < r}B = {x|d(p, x) > r}

From 2.19(c), we know A and B are separated. Therefore, in order for X to beconnected, there must exist a point x such that d(p, x) = r. Consider the set ofpoints

Y = {x|d(p, x) = r, ∀r : 0 < r < d(p, q)}.There are uncountably many points in the set as there are uncountably manyreal numbers bewteen 0 and d(p, q). Since Y ⊂ X , X is uncountable.

Problem 2.20 Are closures and interiors of connected sets always connected?(Look at subsets of R2.)

Solution Yes and No.

2.20(a) Connectedness is preserved under the closure operation

Proof. Let E be a connected set in a metric space X . If its closure, E is notconnected, there exists two separated sets A and B, such that E = A ∪B, andA ∩B = A ∩B = ∅. Obviously, since A ⊂ A, A ∩B = ∅.

Let EA = E ∩A and EB = E∩B. If we can show EA∩EB = EA∩EB = ∅,we will derive a contradition because E will be shown not to be connected.

Since EA ⊂ A ⊂ A, we have EA ⊂ A (Theorem 2.27). Using A ∩ B = ∅,we get EA∩B = ∅. And because EB ⊂ B, we have EA∩EB = ∅. Similary, wecan show EA ∩ EB = ∅. Therefore, E is a union of two separated sets. Hence,contradition.


2.20(b) Connectedness is not preserved under the interior operation

Proof. Consider the union of two discs in R2:D1 = {(x, y)|(x− 1)2 + y2 ≤ 1}D2 = {(x, y)|(x+ 1)2 + y2 ≤ 1} .

We can show both D1 and D2 are connected. Since D = D1 ∪D2 is the unionof two non-disjoint, connected sets, it is connected according to the theoremproved in the notes section.

Let I1 = D1◦ and I2 = D2

◦. We haveI1 = {(x, y)|(x − 1)2 + y2 < 1}I2 = {(x, y)|(x + 1)2 + y2 < 1} .

Therefore, I1 = D1 and I2 = D2. This leads to I1 ∩ I2 = I1 ∩ I2 = ∅. Hence, I1and I2 are separated.

Problem 2.21 Let A and B be separated subsets of some Rk, suppose a ∈ A,b ∈ B, and define

p(t) = (1 − t)a+ tbfor t ∈ R1. Put A0 = p−1(A), B0 = p−1(B). [Thus t ∈ A0 if and only ifp(t) ∈ A.](a) Prove that A0 and B0 are separated subsets of R1.(b) Prove that there exists t0 ∈ (0, 1) such that p(t0) /∈ A ∪B.(c) Prove that every convex subset of Rk is connected.

Solution

2.21(a) Prove that A0 and B0 are separated subsets of R1.

Proof. First, there does not exist t ∈ R1 such that p(t) ∈ A and p(t) ∈ B,because A and B are separated. Therefore, A0 ∩B0 = ∅.

We next prove that A0′ ∩B0 = ∅.

For every limit point t0 of A0, we want to show that p(t0) ∈ A′. To do so,consider every open ball B(p(t0), r), centered at p(t0) and with radius r. Chooseδ = r/|a− b|. Since t0 is a limit point of A0, there are infinitely many points ofA0 in (t0 − δ, t0 + δ). For every t ∈ (t0 − δ, t0 + δ),

d(p(t0), p(t)) = |(t− t0)(a− b)| < δ|a− b| < r.Therefore, there are infinitely many points of A in B(p(t0), r). Hence, p(t0) isalso a limit point of A, or p(t0) ∈ A′. Since A and B are separated, p(t0) /∈ B.Hence, t0 /∈ B0. Therefore, A0

′ ∩B0 = ∅.

Similarly, we can prove A0 ∩B0′ = ∅. These, combined with A0 ∩ B0 = ∅

prove that A0 and B0 are separated.

2.21(b) Prove that there exists t0 ∈ (0, 1) such that p(t0) /∈ A ∪B.

2.2. EXERCISES 17

Proof. Since p(0) = a and p(1) = b, we instead prove that there exists t0 ∈ [0, 1]such that p(t0) /∈ A ∪B.

Let A1 = A0 ∩ [0, 1] and B1 = B0 ∩ [0, 1]. Neither is empty, because 0 ∈ A1

and 1 ∈ B1. Furthermore, since we just proved that A0 and B0 are separatedin R1, so are A1 and B1.

If there does not exist a t0 ∈ [0, 1] such that p(t0) /∈ A ∪ B, the interval[0, 1] can be partitioned into two separated subsets A1 and B1. But, [0, 1] isconnected. Hence contradiction.

Therefore, there must exist t0 ∈ [0, 1] such that p(t0) /∈ A∪B. Since p(0) ∈ Aand p(1) ∈ B, t0 ∈ (0, 1).

2.21(c) Prove that every convex subset of Rk is connected.

Proof. A subset E is convex iff for any two points a ∈ E, b ∈ E, and for everyt ∈ [0, 1], p(t) = (1− t)a+ tb is a point of E.

If E can be partitioned into two disjoint, separated subsets A and B, then2.21(b) shows there exists a t0 ∈ (0, 1) such that p(t0) /∈ A∪B, or (t0) /∈ E. Thiscontradicts the definition of convexity. Therefore, E must be connected.

Problem 2.22 A metric space is called separable if it contains a countabledense subset. Show that Rk is separable. Hint: Consider the set of pointswhich have only rational coordinates.

Proof. Let E = {p = (p1, p2, ..., pk)|pi ∈ Q, 1 ≤ i ≤ k}. We prove that E is acountable dense subset of Rk. In order to do so, we want to show that for everypoint p /∈ E, p is a limit point of E.

For every open ball B(p, r) ∈ Rk, centered at p and with radius r, weconsider k open balls Bi(pi, r

′) ∈ R1, 1 ≤ i ≤ k, each centered at pi and withradius r′. Further, we choose r′ such that r′ = r√

k. For each open ball, there

exists a rational number qi 6= pi, qi ∈ Q, d(pi, qi) < r′. (This follows from theconstruction of the real numbers, if pi is irrational. If pi ∈ Q, we simply pick arational number 0 < δ < r′ and let qi = pi + δ.

Let q = (q1, q2, ..., qk). Clearly, q 6= p. And,

d(p, q) =√

Σ1≤i≤k(pi − qi)2

<√k · r′2

= r′√k

= r

Hence, q ∈ B(p, r).


We have just shown that for every open ball B(p, r), there exists a pointq ∈ Qk, q 6= p, q ∈ B. Therefore, every point p of Rk is a limit point of E.

Problem 2.23 A collection {Vα} of open subsets of X is said to be a base forX if the following is true: For every x ∈ X and every open set G ⊂ X such thatx ∈ G, we have x ∈ Vα ⊂ G for some α. In other words, every open set in X isthe union of a subcollection of {Vα}.

Prove that every separable metric space has a countable base. Hint: Takeall open balls with rational radius and center in some countable dense subset ofX .

Proof. Let E be a countable dense set in X and consider the set of open ballscentered on each point x ∈ E and with radius r ∈ Q+, Vx,r. F = {Vx,r, x ∈E, r ∈ Q+} is countable because it is a product of two countable sets. Alterna-tively, Fx = {Vx,r, r ∈ Q+} is a countable set. Therefore, F =

⋃

x∈E Fx is alsoa countable set (Theorem 2.12).

We prove that F is a countable base for X as follows. For any point p ∈ Xand an arbitrary open set G ⊂ X, x ∈ G, there exists an open ball B(p, r),centered at p, with radius r, such that B(p, r) ⊂ G.

Since E is a dense subset of X , p is either a point of E or one of its limitpoints. If p ∈ E, pick a rational radius r′ < r. Vp,r′ = B(p, r′) ⊂ B(p, r) ⊂ G.Hence, p ∈ Vp,r′ ⊂ G.

If p is a limit point of E, any open ball of p contains at least one point q ∈E, q 6= p. Consider B(p, r′), where r′ is an arbitrary rational number no greaterthan r

2 . It is easy to see that Vq,r′ = B(q, r′) ⊂ B(p, r), because given any pointx ∈ B(q, r′), d(x, p) ≤ d(x, q) + d(q, p) ≤ r. Hence, p ∈ Vq,r′ ⊂ B(p, r) ⊂ G.

We have just shown that for any point p ∈ X and an open set G ⊂ X, x ∈ G,there exists a Vq,r′ such that p ∈ Vq,r′ ⊂ G. Therefore F is a countable base forX .

Problem 2.24 LetX be a metric space in which every infinite subset has a limitpoint. Prove that X is separable. Hint: Fix δ > 0, and pick x1 ∈ X . Havingchosen x1, ..., xj ∈ X , choose xj+1 ∈ X , if possible, so that d(xi, xj+1) ≥ δ fori = 1, ..., j. Show that this process must stop after a finite number of steps,and that X can therefore be covered by finitely many neighborhoods of radiusδ. Take δ = 1/n(n = 1, 2, 3, ...), and consider the centers of the correspondingneighborhoods.

Proof. For any positive rational number r ∈ Q+, construct a sequence of openpoints xi ∈ X, i = 1, 2, ... in the following way. Pick an arbitrary point x1 ∈ X

2.2. EXERCISES 19

as the initial point. Given that points x1, ..., xn have been chosen, pick the nextpoint xn+1 ∈ X such that 0 < d(xn+1, xi) < r, ∀ i : 1 ≤ i ≤ n. This processcannot go on for ever because every infinite subset of X has a limit point (in X):That this limit must be arbitrarily close to an infinite number of points in thissequence is clearly impossible, given the construction of {xi}. This means, forany (positive) rational radius r, X can always be covered by a finite number ofopen balls Bi(r), each centered at point xi with a radius r. Denote the (finite)number of open balls so constructed by Bi(r). Let B(r) =

⋃

1≤i≤B(r) Bi(r). Br

is a finite open cover of X for a rational radius r. Let E(r) =⋃

1≤i≤B(r) xi.

And let E be the union of all E(r), i.e., E =⋃

r∈Q+ E(r). We show that E isa countable dense set in X .

Consider a point p ∈ X, p /∈ E. For every open ball B(p, r) centered atp with radius r, we choose a smaller open ball B(p, r′) with a rational radiusr′ ∈ Q+, r′ < r. Clearly, B(p, r′) ⊂ B(p, r). Since E(r′) consists of the centersof a finite number of open balls that together cover X , there exists an x ∈ E(r′)such that d(x, p) < r′ or, x ∈ B(p, r′). Since B(p, r′) ⊂ B(p, r), x ∈ B(p, r).Since p /∈ E, x 6= p. Therefore, p is a limit point of E.

We have just shown that a point of X is either a point of E or its limit point(or both). Therefore, E is dense in X . Since E is a countable union of finitesets, E is countable. Hence, X is separable.

Problem 2.25 Prove that every compact metric space K has a countable base,and that K is therefore separable. Hint: For every positive integer n, there arefinitely many neighborhoods of radius 1/n whose union covers K.

Proof. We divide the problem into two cases: K is finite and K is infinite.

Case 1: K is finiteThe finite case is straightforward: Make each point in K the center of a sequenceof open balls, each with radius 1/n, n = 1, 2, ...,∞. Note however, there doesnot exist a countable dense set in K. Hence, K is NOT separable.

Case 2: K is infiniteFor every positive integer n, consider open balls centered at every point p ∈ Kwith radius 1/n, i.e., B(p, 1

n). Clearly,

⋃

p∈K B(p, 1n) is an open cover of K.

(The union of any open sets is open.) Since K is compact, there exists a finitesubset cover of K. There may be more than one such covers. We pick one ofthem, and without loss of generality, label the open balls in the subset coverBi(n), i = 1, 2, ...,M(n), and the respective centers Ci(n), where M(n) is anumber depends on n. We want to show that

V = {Bi(n)|n ∈ Z+, 1 ≤ i ≤ M(n)}is a a countable base for K.

Given every x ∈ K and every open set G ⊂ K such that x ∈ G, there existsan open ball N , centered at x such that N ⊂ G, according to the definition of


an open set. Let r be the radius of N and choose a positive integer m such that0 < 2/m < r. Since there is a cover of K with a finite number of open balls ofradius 1/m, there exists an open ball Bi(m) ∈ V with radius 1/m and centeredat ci(m), such that x ∈ Bi(m). In addition, since 2/m < r and Bi(m) containsthe center of N (x), Bi(m) ⊂ N . In other words, there exists Vα ∈ V such thatx ∈ Vα ⊂ G for every x ∈ K and every open set G ⊂ K,x ∈ G. This proves thatV is a countable base for K. The separability of K follows from Theorem 2.2.

Theorem separable. A metric space that is infinite in size and second-countable,

i.e., having a countable base, is also separable.

Proof. Let K be a second-countable metric space and V = {Vi|i ∈ Z+} be acountable base for K. There are infinitely many points in K. We construct asequence of points {xi|i ∈ Z+} by picking an arbitrary point xi from each baseVi, i = 1, 2, ...,∞. We want to show that {xi} is a countable dense set in K.

Step 1 We first show that the range of {xi} has infinitely many and thereforecountably many points. Suppose the range of {xi} has a finite number of points.There exists a finite number of indices, i1, i2, ..., in, such that for every xi thereexists a j, 1 ≤ j ≤ n so that xi = xij . Since there are infinitely many points inK, there exists a point y ∈ K such that y 6= xij , 1 ≤ j ≤ n. Choose a d ∈ R+

such thatd < min(d(xij , y)), 1 ≤ j ≤ n.

Hence, d < d(xi, y), ∀i ∈ Z+.

Consider an open ball B(y, d), centered at y and with radius d. Clearly,y ∈ B(y, d) ⊂ K and xi /∈ B(y, d), ∀i ∈ Z+. Therefore, no base set Vi from Vcan satisfy the condition Vi ⊂ B(y, d), because xi ∈ Vi and xi /∈ B(y, d). Thiscontradicts our assumption that V is a countable base for K. Therefore, therange of {xi} has infinitely many points. Since {xi} has at most countably manypoints, due to its construction, the range of {xi} has countably many points.

Step 2 We now show that {xi} is dense in K: Every point in K is either apoint in {xi} or one of its limit points. Consider a point p in K but not in {xi}.We want to show p is a limit point of {xi}.

Since V is a countable base for K, for neighborhood G of p, p ∈ G ⊂ K,there exists an open set with index i ∈ Z+, Vi ∈ V , such that p ∈ Vi ⊂ G.Since xi ∈ Vi and p /∈ {xi}, there exists a point from {xi} but different from pin every neighborhood of p. Therefore, p is a limit point of {xi}. This provesthat {xi} is dense in K.

Problem 2.26 Let X be a metric space in which every infinite subset hasa limit point. Prove that X is compact. Hint: By Exercises 23 and 24, Xhas a countable base. It follows that every open cover of X has a countable

2.2. EXERCISES 21

subcover {Gn}, n = 1, 2, 3, .... If no finite subcollection of {Gn} covers X , thencomplement Fn of G1 ∪ . . . ∪Gn is nonempty for each n, but

⋂

Fn is empty. IfE is a set which contains a point from each Fn, consider a limit point of E, andobtain a contradiction.

Proof. By Exercises 23 and 24, X has a countable base. It follows that everyopen cover of X has a countable subcover {Gn}, n = 1, 2, 3, ....

Let Hn =n⋃

i=1

Gi and Fn = Hnc. Since Gn is open for all n, Hn is also open

for all n (Theorem 2.24). Therefore, Fn is closed for all n. Furthermore, sinceHn ⊂ Hn+1 for all n, we have

F1 ⊃ F2 ⊃ . . . Fn ⊃ Fn+1 ⊃ . . ..

If no finite subcollection of {Gn} covers X , then Fn is nonempty for eachn. We produce a sequence of points {x1, x2, ..., xn, ...} as follows. First, pickx1 from F1 (since Fn is nonempty for all n). Assume x1, x2, ..., xn have beenchosen. We pick xn+1 from Fn+1 such that xn+1 6= xi, 1 ≤ i ≤ n. Either thisprocess has to stop at n = N , which means FN is finite and has at most Npoints or it can go on forever.

Case 1: FN is finite and has at most N points.

In this case, we can pick from V an open cover Ci for each xi, 1 ≤ i ≤ n such

that Ci ⊂ X . (Note that every open cover from V is a subset of X .) HN ∪N⋃

i=1

Ci

is a finite subcover for X .

Case 2: The process goes on forever.

We produce an infinite sequence x1, x2, x3, ..., such that xi 6= xj , for all i 6= j.Since {xn} is an infinite subset of X , it has a limit point x ∈ X . We want to

show that x ∈∞⋂

n=1Fn.

Since F1 ⊃ F2 ⊃ . . . Fn ⊃ Fn+1 ⊃ . . ., the subset {xn, xn+1, . . .} ∈ Fn andhas the same limit point x. Since Fn is closed, x ∈ Fn, for all n. Therefore,

x ∈∞⋂

n=1Fn. However,

∞⋂

n=1Fn = (

∞⋃

n=1Gn)

c = Xc = ∅. We have a contradiction

here.

Hence, there exists a finite subcollection of {Gn} that covers X .

Problem 2.27 Define a point p in a metric space X to be a condensation point

of a set E ⊂ X if every open ball of p contains uncountably many points of E.

Suppose E ⊂ Rk, E is uncountable, and let P be the set of all condensationpoints of E. Prove that P is perfect and that at most countably many ponts ofE are not in P . In other words, show that P c ∩ E is at most countable. Hint:


Let {Vn} be a countable base of Rk, let W be the union of those Vn for whichE ∩ Vn is at most countable, and show that P = W c.

Proof. Firstly, let us prove that P is closed. Pick any limit point of P , labelit x. For every open ball of x, B(x, r), with radius r, there exists a point y,y ∈ B(x, r), y ∈ P and y 6= x. Since y 6= x, 0 < d(x, y) < r. Consider an openball B(y, r′), r′ < min(d(x, y), r− d(x, y)). For any point q ∈ B(y, r′), d(x, q) ≤d(x, y) + d(y, q) < d(x, y) + r′ < r, therefore q ∈ B(x, r) and B(y, r′) ⊂ B(x, r).But since r′ < min(d(x, y), r − d(x, y)) ≥ d(x, y), x /∈ B(y, r′).

Because y ∈ P , there are uncountably many points of E in B(y, r′). There-fore, there are uncountably many points of E in B(x, r) or, in every open ballof x. Hence, x ∈ P .

Next, we prove that every point of P is its limit point. Suppose this isuntrue. Then there exists a point x ∈ P such that x is not a limit point of P .This means there exists an open ball of x, Bx, such that Bx ∩ P = {x}. Nowfor every point y ∈ Bx, y 6= x, we have y /∈ P . Therefore, there exists an openball of y, B(y), such that B(y) contains at most countably many points of E.

Since Rk has a countable base {Vα}, there exists an open set Vα(y) such thaty ∈ Vα(y) and Vα(y) ⊂ B(y). The second condition means that Vα(y) containsat most countably many points of E.

Let Q =⋃

y∈BxVα(y). We have Bx ⊂ {x} ∪Q. But since there are at most

countably many Vα(y) and each of them contains at most countably many pointsof E, Q and therefore Bx contains at most countably many points of E.

However, because x ∈ P , Bx should contain uncountably many points of E.This contradicts the conclusion we just made. Therefore, every point in P is itslimit point. Hence P is perfect.

We now prove that at most countably many points of E are not in P . Sup-pose this is untrue and let S = P c ∩ E, there exist uncountably many pointsof E in S. For each point x ∈ S, since x /∈ P , there exists an open ball of x,Bx, which contains at most countably many points in E. Since S ⊂ Rk, Bx isan open set, and Rk has a countable base {Vα}, there exists a Vα(x), such thatx ∈ Vα(x) ⊂ Bx. Obviously, Vα(x) contains at most countably many points ofE.

Since the range of {Vα(x)} contains at most countably many distinct Vα, itfollows that

⋃

x∈S Vα(x) contains at most countably many points of E. But thisand the fact that S ⊂ ⋃

x∈S Vα(x) contradict our assumption that S containsuncountably many points of E. Therefore, our assumption is wrong and S (orP c ∩ E) contains at most countably many points of E.

An alternative proof We use the hint in the book.

Proof. Let {Vn} be a countable base of Rk, let W be the union of those Vn forwhich E ∩ Vn is at most countable. We can show that P = W c.

2.2. EXERCISES 23

We first show that P ⊂ W c. For every p ∈ P , p cannot belong to any Vn

from W . Suppose otherwise, i.e., p belongs to Vn ∈ W . Since Vn is open, thereexists an open ball B centered at p, such that p ∈ B ⊂ Vn. Since there areat most countably many points of E in Vn, there are at most countably manypoints of E in B. Hence, p /∈ P . Contradiction. Therefore, p ∈ W c. AndP ⊂ W c.

We next prove that P ⊃ W c. This amounts to proving that for every pointp /∈ P , p /∈ W c (or p ∈ W ). If p is not a condensation point of E, there existsan open ball B centered at p such that B contains at most countably manypoints of E. Since {Vn} is a countable base for Rk, there exists a Vn such thatp ∈ Vn ⊂ B. Furthermore, there exists another open ball B′, centered at p,such that p ∈ B′ ⊂ Vn ⊂ B. There are at most countably many points of E inVn and consequently B′. Therefore, p ∈ W (or p /∈ W c).

Since P ⊂ W c and P ⊃ W c, P = W c. It is easy to show that W contains atmost countably many points of E, because it is the union of at most countablenumber of sets, each of which contains at most countably many points of E.Therefore, there are at most countably many points of E that are not in W c

(or P ).

To show P is perfect, we first note that P is the complement of an open setW and is therefore closed. To show that every point of P is a limit point of P ,we suppose the contrary: Let x ∈ P be an isolated point of P . There existsan open ball B centered at x such that B ∩ P = {x}. Since P = W c, we have(B − {x}) ⊂ W . Therefore, there are at most countably many points of E inB. This contradicts the assumption that x ∈ P (or x is a condensation point ofE). Hence, x must be a limit point of P . This proves that P is perfect.

Problem 2.28 Prove that every closed set in a separable metric space is theunion of a (possibly empty) perfect set and a set which is at most countable.(Corollary: Every countable closed set in Rk has isolated points.) Hint: UseExercise 27.

Proof. Let E be a closed set in a separable metric space X , and P be the setof all condensation points of E. Let A = P ∩ E and B = P c ∩ E. Clearly,E = A ∪ B and A ∩ B = ∅. We want to show that A is perfect and B has atmost countably many points.

We first use the proof for Problem 2.27 to show that P is perfect and |B|is at most countable. Since the only property specific to Rk that we used inthe proof is that Rk is separable, the conclusions of Problem 2.27 are entirelyapplicable to any separable metric space X . Therefore, P is a perfect set and|B| is at most countable.

Since the intersection of a finite number of closed sets are closed, A is closed.We next show that every point in A is its limit point. Consider a point p ∈ A.


Since A = P ∩ E, p is a condensation point of E: For every open ball N of p,there exist uncountably many points of E that are in N . Of these points, thereexists at least one point q, such that q 6= p and q /∈ B, because |B| is at mostcountable. Since E = A ∪ B, this point belongs to A. Therefore, p is a limitpoint of A. Since A closed, it is a perfect set.

Problem 2.29 Prove that every open set in R1 is the union of an at mostcountable collection of disjoint segments. Hint: Use Exercise 22.

Proof. Let E be an arbitrary open set in R1. Every point x ∈ E is an interiorpoint. There exists an open ball B(x) centered at x, such that B(x) ⊂ E. InR1, an open ball is simply a segment, i.e., B(x) = (x− δ, x+ δ), δ > 0,

In Problem 2.23 we proved that R1 has a countable base in the form ofV = {(q − 1

n, q + 1

n) | n ∈ Z+}, q ∈ Q.

Therefore, there exists a segment in V , Vα(x) = (q(x)− 1n(x) , q(x) +

1n(x)), such

that x ∈ Vα(x) ⊂ B(x) ⊂ E.

LetF =

⋃

x∈E

Vα(x).

We want to show F = E.

Clearly, E ⊂ F . And for every y ∈ F , there exists a Vα(x) ∈ V such thaty ∈ Vα(x) ⊂ B(x) ⊂ E. Therefore, F ⊂ E. Hence, E = F .

Since F is the union of at most countably many segments (Vα(x)), F (andtherefore E) is the union of at most countably many segments. These segmentsmay not be disjoint. But since the union of any two overlapping segments inR1 is also a segment in R1, F can be reduced to the union of at most countablymany disjoint segments.

Problem 2.30 Imitate the proof of Theorem 2.43 to obtain the following result:

1. If Rk =⋃∞

1 Fn, where each Fn is a closed subset of Rk, then at least oneFn has a nonempty interior.

2. Equivalent Statement: IfGn is a dense open subset ofRk, for n = 1, 2, 3, ...,then

⋂∞1 Gn is not empty (in fact, it is dense in Rk).

(This is a special case of Baire’s theorem; see Exercise 22, Chapter 3, for thegeneral case.)

Solution

2.2. EXERCISES 25

1. If Rk =⋃∞

1 Fn, where each Fn is a closed subset of Rk, then at least oneFn has a nonempty interior.

Proof. Suppose that each Fn has an empty interior. We show that⋃∞

1 Fn

does not cover Rk, thereby deriving a contradiction.

Since Fn is closed and has an empty interior, Fn 6= Rk, for otherwise Rk

is the interior or Fn. Hence, there exists a point x ∈ Rk such that x /∈ F1.Since F1 is closed, Rk−F1 is open. There exists an open ball B(x, r), withradius r and centered at x, such that x ∈ B(x, r) ⊂ Rk − F1. Let V1 =B(x, r/2). Let y be an arbitrary point in F1. Since B(x, r) ⊂ (X − F1),d(x, y) ≥ r. Therefore, B(y, r/2) ∩ B(x, r/2) = Φ. Hence, y cannot be alimit point of V , or V1 ∩ F1 = Φ.

Starting with V1, we inductively define Vn as follows. Given an open setVn such that Vn ∩ (

⋃n

1 Fi) = Φ, we let V ′ = Vn ∩ (X − Fn+1). SinceFn+1 is assumed to have empty interior, Vn cannot be entirely containedin Fn+1. Therefore, V ′ is nonempty. Pick a point x ∈ V ′ and follow thesame construction as V1, we have an open set Vn+1, Vn+1∩(

⋃n+11 Fi) = Φ.

Since each Vn is an open ball, Vn is bounded in Rk. Hence Vn is a compactsubset ofRk. Since {Vn} satisfies the finite intersection condition,

⋂∞n=1 Vn

is nonempty or, there exists a point p ∈ Vn for all n > 0. Our constructionof Vn satisfies that for each n > 0, Vn ∩ (

⋃n

1 Fi) = Φ. Therefore therecannot exist an m > 0 such that p ∈ Fm, for otherwise this will contradictthe fact that Vn ∩ (

⋃m

i=1 Fi) = Φ.

We just derived a contradiction by assuming each Fn has an empty interior.Therefore at least one Fn has a nonempty interior.

2. Equivalent Statement: If Gn is a dense open subset ofRk, for n = 1, 2, 3, ...,then

⋂∞1 Gn is not empty (in fact, it is dense in Rk).

Proof. We need to show that one can derive a statement (either 1 or 2)from the other.

We derive the second statement assuming the first statement is true.

Define Fn = Rk − Gn. Since Gn is open, Fn is closed. And since⋂∞

1 Gn = Φ, We have

∞⋃

n=1

Fn =

∞⋃

n=1

(Rk −Gn)

= Rk −∞⋂

n=1

Gn

= Rk

The first statement says at least one Fn has a nonempty interior. Thiscontradicts the fact that each Fn, being the complement of a dense open


set Gn, has an empty interior. Thus we conclude the second statementmust be true.

We derive the first statement assuming the second statement is true.

Suppose each Fn has an empty interior. We want to show that eachGn = Rk − Fn is a dense open set in Rk and derive a contradition usingthe first statement.

First of all, Gn is the complement of a closed set and hence is open. Sinceeach Fn has an empty interior, Gn must be dense in Rk. Hence

⋂∞n=1 Gn

is nonempty. Since,

∞⋃

n=1

Fn = Rk − (Rk −∞⋃

n=1

Fn)

= Rk −∞⋂

n=1

(Rk − Fn)

= Rk −∞⋂

n=1

Gn

We have⋃∞

n=1 Fn 6= Rk which contradicts the assumption in first state-ment.

3. Extra A finite intersection of dense open sets in X is also dense and openin X .

Proof. Let Gi, 1 ≤ i ≤ n be a finite collection of dense open sets in X . Wewant to show that G =

⋂n

i=1 Gi is also dense and open in X .

Being a finite intersection of open sets, G is open. Let F = Rk −G. F isclosed. We want to show F has an empty interior (or F ◦ = Φ).

Since Gi is dense in X , we have for any point x ∈ X and any neighborhoodof x, N(x), N(x) ∩ Gi 6= Φ; otherwise, we will have x /∈ Gi for some1 ≤ i ≤ n, contradicting the assumption that each Gi is dense in X . Givenan arbitrary neighborhood N(x) of an arbitrary point x ∈ X , consider thefolllowing finite intersection.

B(N(x)) = N(x) ∩ (n⋂

i=1

Gi)

=

n⋂

i=1

(N(x) ∩Gi)

Since eachN(x)∩Gi contains x, B(N(x)) contains x and hence is nonempty.Furthermore, being the finite intersection of open sets, B(N(x)) is open.Since B(N(x)) ⊂ N(x) and B(N(x) 6= Φ, every neighborhood of x ∈ Xmust have nonempty intersection with G. Hence either x ∈ G or x is alimit point of G. In other words, G is dense in X .

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