Topics in Advanced Calculus Preparatory Lessons for Real...
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Topics in Advanced Calculus Preparatory Lessons for Real Analysis 1 Leonor Aquino-Ruivivar, Ph.D. Department of Mathematics College of Science De La Salle University
Transcript of Topics in Advanced Calculus Preparatory Lessons for Real...
Topics in Advanced Calculus
Preparatory Lessons for Real Analysis 1
Leonor Aquino-Ruivivar, Ph.D.Department of Mathematics
College of ScienceDe La Salle University
Contents
1 The Real Number System 31.1 Properties of Real Numbers . . . . . . . . . . . . . . . . . . . . . . . 31.2 Bounded Sets and The Axiom of Completeness . . . . . . . . . . . . 81.3 Mathematical Induction . . . . . . . . . . . . . . . . . . . . . . . . . 13
2 The Euclidean Space Rk 192.1 Basic Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.2 Open and Closed Sets . . . . . . . . . . . . . . . . . . . . . . . . . . 25
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Introduction
The graduate course in Real Analysis assumes that the graduate student has therequisite knowledge and skills to understand the course content. These notes providea review of or an introduction to topics from both elementary and advanced calculuswhich the student needs to be familiar with. Included in these notes is a discussionof fundamental properties of the set of real numbers, the Axiom of Completeness andthe Archimedean property of real numbers, and the notion of denseness for the set ofrational numbers. The dot product in the Euclidean n-space is used to show that Rn
is a normed space. Fundamental concepts of the usual topology in the n-dimensionalspace Rn and the method of mathematical induction as a method of proof are alsoincluded.
At the end of each section or topic, exercises which require the application of thetopic/concepts and results that were presented are given. The readers are advised totry to work out the solutions first, before comparing their work with the solutionsprovided after the exercises.
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Chapter 1
The Real Number System
1.1 Properties of Real Numbers
The usual setting for the study of functions is the field R of real numbers. Theoperations on these set are governed by very specific rules and properties which thestudent applies to solve a specific problem.
The set of real numbers, denoted by R, is a nonempty set whose elements arecalled real numbers. We define two binary operations + and · in R, called additionand multiplication, respectively, satisfying the following properties:
The Field Axioms
F1. (Closure Properties)
A1. For all x, y ∈ R, x+ y ∈ R.
M1. For all x, y ∈ R, x · y ∈ R.
F2. (Commutative Properties)
A2. For all x, y ∈ R, x+ y = y + x.
M2. For all x, y ∈ R, x · y = y · x.
F3. (Associative Properties)
A3. For all x, y, z ∈ R, (x+ y) + z = x+ (y + z).
M3. For all x, y, z ∈ R, x · (y · z) = (x · y) · z.
F4. (Distributive Properties)
D1. a · (b+ c) = a · b+ a · cD2. (b+ c) · a = b · a+ c · a
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F5. (Existence of Identity Elements)
A4. There exists 0 ∈ R such that x+ 0 = 0 + x for all x ∈ R.
M4. There exists 1 ∈ R such that 1 · x = x · 1 = x for all x ∈ R.
F6. (Existence of Inverse Elements)
A5. For every x ∈ R, there exists a unique element −x ∈ R called the additiveinverse of x such that x+ (−x) = 0.
M5. For every x ∈ R, x 6= 0, there exists x−1 ∈ R called the multiplicativeinverse of x such that x · x−1 = 1.
Remark.
(a) The product xy of two numbers x and y when the operation · is performed isoften written as xy.
(b) The properties described above turn the set R into a structure called a field.
Theorem 1.1.1. The field axioms in R lead to the following consequences:
(a) If a+ c = b+ c, then a = b (Cancellation Law for Addition)
(b) a · 0 = 0 for all a ∈ R.
(c) (−a)(b) = (a)(−b) = −ab for all a, b ∈ R.
(d) (−a)(−b) = ab for all a, b ∈ R.
(e) If ac = bc and c 6= 0, then a = b. (Cancellation Law for Multiplication)
(f) If ab = 0, then either a = 0 or b = 0.
(g) 0 is the only additive identity, and 1 is the only multiplicative identity.
(h) For each a ∈ R, the additive inverse −a is unique. If a 6= 0, then the multiplicativeinverse a−1 is also unique.
The Order Axioms
The order axioms allow us to arrange the elements of R according to a particularorder. The set R has a nonempty subset denoted by R+, whose elements are calledthe positive real numbers. The following properties are satisfied:
O1. 0 6∈ R+.
O2. For all x, y ∈ R+, x+ y, xy ∈ R+. (Closure for Addition and multiplication)
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O3. (Trichotomy Property): For every x ∈ R+, exactly one of the following istrue:
(a) x = 0 (b) x ∈ R+ (c) −x ∈ R+
Note.
• For all x, y ∈ R, the notation y − x means y + (−x).
• The notation x < y, which is read as x is less than y, means y − x ∈ R+.
• The notation x > y, read as x is greater than y, is the same as y < x.
• The notation x ≤ y, read as x is less than or equal to y, means either x < y orx = y. The notation x ≥ y may be defined similarly.
• The order axioms allow for an ordering of the real numbers and turns R intoan ordered field.
Theorem 1.1.2. Let a, b, c ∈ R.
(a) If a < b and b < c, then a < c. (Transitive)
(b) If a < b, then a+ c < b+ c. (Addition Property)
(c) If a < b and 0 < c, then ac < bc. (Multiplication Property)
(d) If a < b and c < 0, then bc < ac. (Multiplication Property)
(e) If a 6= 0, then a2 = a · a > 0.
(f) 1 > 0
(g) If a > b and c > d, then a+ c > b+ d.
(h) If a > b > 0 and c > d > 0, then ac > bd.
(i) If a > b > 0, then b−1 > a−1.
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Proof.
(a) Since b− a and c− b are both in R+, it follows by closure that
(b− a) + (c− b) = (b− b) + (c− a) = 0 + (c− a) = c− a ∈ R+
which shows that a < c.
(b) We have (b + c) − (a + c) = (b − a) + (c − c) = b − a ∈ R+, which shows thata+ c < b+ c.
(c) Since (b− a) and c = c− 0 are both in R+, it follows by closure that
(b− a)(c) = bc− ac ∈ R+ ⇔ ac < bc
(d) This time, we have (b− a), − c ∈ R+, which shows that
(b− a)(−c) = ac− bc ∈ R+ ⇔ bc < ac
(e) By the trichotomy property, we have either a > 0 or a < 0. If a > 0, then using(c) with c = a, we have 0 < a ⇒ 0 · a = 0 < a · a = a2. If a < 0, we use (d) withc = a.
(f) Since 1− 0 = 1 ∈ R+, we have 0 < 1, or,equivalently, 1 > 0.
(g) We have a − b, c − d ∈ R+, so that (a − b) + (c − d) = (a + c) − (b + d) ∈ R+,and the desired conclusion follows.
(h) Since a > b and c > 0, we have ac > bc. Similarly, c > d and b > 0 gives usbc > bd. By transitivity, ac > bd.
(i) First, if a > 0, then a−1 > 0. Otherwise, we have 1 = aa−1 < 0 · a−1 = 0. Weshow that it is impossible for b−1 ≤ a−1. If b−1 = a−1, then by (c), we have a > bimplies 1 = aa−1 > bb−1 = 1, a contradiction. If a−1 > b−1, then by (h), we againhave 1 = aa−1 > bb−1 = 1. By trichotomy, we must have b−1 > a−1.
Definition 1.1.1. Let a ∈ R. The absolute value of a, denoted by |a|, is definedby
|a| =
a if a > 00 if a = 0−a if a < 0
Remark. It can be verified that |a| =√a2 for all a ∈ R.
The following theorem gives properties of the absolute value.
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Theorem 1.1.3. (Properties of the Absolute Value)Let a, b ∈ R. Then
(a) If |a| = |b| then either a = b or a = −b.
(b) |ab| = |a||b|
(c) If a 6= 0 then |a−1| = |a|−1.
(d) If b 6= o then |ab−1| = |a||b|−1.
(e) a ≤ |a|.
(f) ||a| − |b|| ≤ |a+ b| ≤ |a|+ |b|
We build the important subsets of R as follows: Starting with 1 ∈ R+, then bythe closure property for addition, we define 2 = 1 + 1 ∈ R+, 3 = 2 + 1 ∈ R+, 4 =3 + 1 ∈ R+, and so on. We then get 0 < 1 < 2 < 3 . . .. The set
N = { 1, 2, 2, 4, . . . }
is called the set of natural numbers. Clearly, this set has no identity element 0under addition, nor the additive inverses of its elements. If we add these numbers tothe set N, we form the set
Z = { 0, 1, − 1, 2, − 2, 3, − 3, 4, − 4, . . . }
of the integers. If a, b ∈ Z and b 6= 0, then the set of all numbers of the form ab−1,which we henceforth denote by a/b, is called the set Q of the rational numbers.All other elements of R which are not in Q are called irrational numbers.
Example 1.1.1. To show that x =√
2 is irrational, assume that it is. Then we can
find two integers a and b, b 6= 0, such that x =√
2 =a
b. Assume that (a, b) = 1. We
get
x2 = 2 =a2
b2⇔ a2 = 2b2
This shows that a2 is even and a is also even. Since the left-hand side is divisibleby 4,it follows that b2 and hence b is even, contradicting (a, b) = 1. Thus, ourassumption that x =
√2 ∈ Q is false.
Exercise 1.1.1.
1. If a ∈ R, prove the following:
(a) (−1)(a) = −a(b) (−a)2 = a2
(c) If a 6= 0, then (a−1)−1 = a.
(d) If a 6= 0, then (−a)−1 = −(a−1).
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2. If a, b ∈ R, prove that (ab)2 = a2b2.
3. If a, b ∈ R and a 6= 0, b 6= 0, show that (ab)−1 = a−1b−1.
4. If a, b ∈ R and a 6= 0, b 6= 0, show that ab 6= 0.
5. If a, b ∈ R and a ≥ b, then −b ≥ −a.
Solutions to the exercises:
1. (a) (−1)(a) + a = (−1)(a) + (1)(a) = (−1 + 1)(a) = 0× a = 0⇒ (−1)(a) = −a(b) We have (−a)2 = (−a)(−a) = a× a = a2
(c) Since a−1a = a × a−1 = 1, and since the inverse is unique, it follows that(a−1)−1 = a.
(d) We have (−a)[−(a−1)] = (a)(a−1) = 1⇒ (−a)−1 = [−(a−1)].
2. We have (ab)2 = (ab)(ab) = aabb = a2b2 by the associative and commutativeproperties of multiplication.
3. We have (ab)a−1b−1 = (aa−1)(bb−1) = 1 by the associative and commutative prop-erties of multiplication. Since the inverse is unique, we have (ab)−1 = a−1b−1.
4. Assume to the contrary that ab = 0. Since a 6= 0, a−1 exists. We have 0 = a−1×0 =a−1(ab) = (a−1a)b = 1 · b = b, and we have a contradiction. Hence, ab 6= 0.
5. Since a ≥ b and −1 < 0, we have a(−1) ≤ b(−1), or, equivalently, −b ≥ −a.
1.2 Bounded Sets and The Axiom of Completeness
The real numbers may be represented by the points on a line, and each point corre-sponds to a real number. Thus, the set of real numbers has no ”gaps”, i.e. betweenany two real numbers, we can always find another real number. This property isassured by the so-called completeness axiom.
Definition 1.2.1. (Maximum and Minimum Elements of a Set)Let S ⊂ R, S 6= ∅.
• If S has a largest element s0 in the sense that s ≤ s0 for all s ∈ S, then s0 iscalled the maximum of S and we write s0 = max S.
• If S has a smallest element s′ such that s′ ≤ s for all s ∈ S, then s′ is called theminimum of S and write s′ = min S.
Example 1.2.1.
(a) If S = { − 2, 3, 0,√
2, π,−e }, then max S = π and min S = −e.
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(b) If S = N, then min S = 1 but there is no maximum element.
(c) If S = Z, then S has neither maximum nor minimum element.
(d) If S = [1, 3) = { x ∈ R : 1 ≤ x < 3 } has minimum element 1 but no maximumelement.
Definition 1.2.2. (Upper and Lower Bounds)Let S be a nonempty subset of R.
(a) If M ∈ R such that s ≤ M for all s ∈ S, then M is an upper bound of S andS is bounded above. The smallest of the upper bounds of S is called the leastupper bound or supremum of S, denoted by sup S.
(b) If m ∈ R such that m ≤ s for all s ∈ S, then m is a lower bond for S andS is bounded below. The largest among the lower bounds of S is called thegreatest lower bound or infimum of S, denoted by inf S.
(c) If S has both an upper bound and a lower bound, then S is said to be bounded.
Remark.
• If U is an upper bound for a set S, then no element of S can be found to theright of U in the number line.
• Similarly, if L is a lower bound for a set S, then no element of S can be foundto the left of L in the number line.
• If S is a bounded subset of R, then all the elements of S can be found withinan interval [L,U ], where L and U are upper and lower bounds, respectively, forthe set S.
• If b is an upper bound for a set S, then any number bigger than b is also anupper bound for S. Similarly, if a is a lower bound for S, then any numbersmaller than a is a lower bound for S.
Example 1.2.2.
(a) Let S = [−1, 2) = { x ∈ R : − 1 ≤ x < 2 } is bounded. -1 and any real numberless than -1 are lower bounds, while 2 and all other real numbers greater than 2are upper bounds for S. We have inf S = −1 and sup S = 2.
(b) Let S = N. Then 1 is the infimum of S and any real number less than 1 is alower bound for S. There is no upper bound.
(c) Let S = { x ∈ Q : x <√
2 }. Then S has no lower bound, and√
2 is itssupremum.
(d) Let S = { 1/n : n ∈ N }
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(e) Let S =∞⋂n=1
(1− 1/n, 1 + 1/n)
Remark.
• Observe that inf S and sup S NEED NOT be in S unlike max S and min S.
• S may have an infimum even if it has no minimum, and a supremum even if ithas no maximum.
• If they exist for a set S, the minimum, maximum, infimum and supremum areunique.
• If max S exists, then max S = sup S. Similarly, min S = inf S if min S exists.
Exercise 1.2.1.
1. For each of the sets listed below, do the following:
• Find upper/lower bounds for the set. If these do not exist, write ”not boundedabove” or ”not bounded below”.
• Identify the supremum/infimum of the set, if these exist.
(a) { r ∈ Q : r2 < 4 }(b) { n+ (−1)n/n : n ∈ N }(c) { cos(nπ/3) : n ∈ N }
(d) { 1− 1/3n : n ∈ N }(e) { 1/n : n is prime }(f) { x2 : x ∈ R }
2. Let S be a nonempty bounded subset of R.
(a) Prove that inf S ≤ sup S.
(b) What can we conclude about S if inf S = sup S?
Solutions to the Exercises:
1. For the given sets, the supremum/infimum is indicated, if upper/lower boundsexist, respectively.
Set Infimum Supremum
{ r ∈ Q : r2 < 4 } −√
2√
2{ n+ (−1)n/n : n ∈ N } 0 no upper bound{ cos(nπ/3) : n ∈ N } −1 1{ 1− 1/(3n) : n ∈ N } 2/3 1{ 1/n : n is prime } 0 1/2{ x2 : x ∈ R } 0 no upper bound
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2. (a) Let a = inf S, b = sup S. For any x ∈ S, we have a ≤ x ≤ b, so that a ≤ b.
(b) Let a = inf S, b = sup S. If a = b, then for each x ∈ S, we have a ≤ x ≤ b⇒x = a = b. Hence, S consists of a single element.
Postulate 1.2.1. (The Axiom of Completeness)Let S be a nonempty subset of R. If S has an upper bound, then it must have a leastupper bound.
Remark. The Axiom of Completeness says that there is no gap between a set Swhich is bounded above and the set of its upper bounds. This axiom is used in manyinstances to establish results in Real Analysis. A statement of the completeness axiomfor sets with lower bounds is given in the following:
Corollary 1.2.0.1. Let S be a nonempty subset of R. If S has a lower bound, thenS has a greatest lower bound inf f .
Proof. Let T = { − s : s ∈ S }, and let x be a lower bound for S. This means thatfor each s ∈ S, we have x ≤ s. Thus, −x ≥ −s for all −s ∈ T . Since T has an upperbound,it must have a least upper bound, say −y. Then it is easily verified that y isa greatest lower bound for S.
The following example illustrates how the axiom of completeness is applied:
Example 1.2.3. Let S be a nonempty subset of R which is bounded above. If supS ∈ S, prove that sup S = max S.
Proof. Let x = sup S,so that s ≤ x for each s ∈ S. Since x ∈ S, it follows thatx = max S.
Another important result which a student of analysis needs to understand is thefollowing:
Theorem 1.2.1. (Archimedean Property of R)If a > 0 and b > 0, then there exists a positive integer n such that nb > a.
Proof. Suppose that no such n ∈ N exists. Then the set
S = { b, 2b, 3b, 4b, . . . }
is bounded above by a. By the Axiom of Completeness, S must have a least upperbound, say M . For any positive integer n, we have (n+1)b ≤M , so that nb ≤M−b,for all n ∈ N. This shows that M − b is an upper bound for S. But M − b < Mcontradicts the fact that M = sup S. Thus, there must be a positive integer n suchthat nb > a.
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Example 1.2.4. Let a, b ∈ R such that a ≤ b + 1/n for all n ∈ N. Prove thata ≤ b.Solution:
Suppose instead that a > b, so that a−b > 0. Moreover, 1 > 0, so by the Archimedeanproperty of the set of real numbers, there exists a positive integer n such that we have
n(a− b) > 1⇔ a− b > 1
n⇔ a > b+
1
n
contradicting the hypothesis on a and b. Hence, it must be true that a ≤ b.
The next result shows the ”abundance” of the rational numbers within the setof real numbers, in the sense that at least one (and hence infinitely many) rationalnumbers can be found between any two real numbers.
Theorem 1.2.2. (Denseness of Q)If a, b ∈ R and a < b, then there exists r ∈ Q such that a < r < b.
Proof. If a and b have opposite signs, then 0 ∈ Q lies between a and b. If 0 ≤ a < b,
thenb− a
2> 0, so by the Archimedean property, there exists n ∈ N such that
n
(b− a
2
)> 1, or, equivalently,
1
n<b− a
2
By the Archimedean property, there exists a positive integer k such thatk
n> a. Let
S = { k ∈ N : kn> a }, and let m be the least element of S. Then
m− 1
n≤ a, and
we havem
n=m− 1
n+
1
n≤ a+
1
n< a+
b− a2
=a+ b
2<b+ b
2= b
which shows thata <
m
n< b
Example 1.2.5. Let a, b ∈ R, a < b. Use the denseness of Q in R to show thatthere are infinitely many rationals between a and b.Solution:
By the denseness of Q, there exists a rational number q1 between a and b. Similarly,there is a rational number q2 between q1 and b, a rational number q3 between q2 andb, a rational number q4 between q3 and b, and, in general, for any positive integer n,there is a rational number qn+1 between qn and b. hence, there are infinitely manyrational numbers between any two real numbers.
Exercise 1.2.2.
1. Let S and T be nonempty bounded subsets of R such that S ⊆ T . Prove thatinf T ≤ inf S ≤ sup S ≤ sup T.
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2. If a ∈ R with a > 0, show that there exists n ∈ N such that 1/n < a < n.
3. Let A, B be bounded nonempty subsets of R, and let
S = { a+ b : a ∈ A, b ∈ B }
(a) Prove that sup S = sup A + sup B.
(b) Prove that inf S = inf A + inf B.
4. Show that sup{ r ∈ Q : r < a } = a for each a ∈ R.
Solutions to the Exercises:
1. Let x ∈ S. Since S ⊆ T , we have x ∈ T , so that inf T ≤ x ≤ sup T andinf S ≤ x ≤ sup S. If there exists y ∈ T\S such that y > sup S, then we havesup S < sup T. Otherwise, the two are equal. Similarly, if there exists z ∈ T\Ssuch that z < inf S, then inf T < inf S. Otherwise, they are equal. Finally, sinceit is clear that inf S ≤ sup S, we have the desired result.
2. Using the Archimedean principle with a > 0, 1 > 0, we know that there existsn1 ∈ N such that n1 = n1 · 1 > a. Similarly, there exists n2 ∈ N such that
n2 · a > 1, or, equivalently,1
n2
< a. Let n = max{ n1, n2 }. Then e have the
desired result1
n≤ 1
n2
< a < n1 ≤ n⇒ 1
n< a < b.
3. We only show the proof for (a), since the proof for (b) is analogous. For any a ∈ Aand any b ∈ B, we have
a+ b ≤ sup A + sup B⇒ sup S ≤ supA + sup B
To show equality, suppose sup S ¡
4. Let S = { r ∈ Q : r < a }. Since r < a for each rational in S, it follows thatsup S ≤ a. Suppose sup S < a. Then there exists a rational number q such thatsup S < q < a. However, q ∈ S, so this contradicts the definition of sup S. Hence,sup S = a.
1.3 Mathematical Induction
In this section, we will discuss a method of proof which is know as mathematicalinduction. This method is the proof of choice when proving mathematical statementswhich are supposed to be true for all except a finite number of natural numbers Someexamples of these statements are the following:
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(a)n∑
k=1
k2 =n(n+ 1)(2n+ 1)
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(b) For any positive integer n, the number 4n − 1 is divisible by 3.
Observe that both of these statements are supposed to be true for any positiveinteger n. Since the set N is infinite, it is not possible to prove either of the abovestatements by verifying the statement for each value of n. The method of mathe-matical induction as mathematical proof is well suited for proving statements such asthese two.
The basic proof by induction consists of two parts. Suppose P (n) is the statementabout positive integers n that we wish to prove. Then the two parts of the proof areas follows:
Part 1: Verify that P (n) is true for the smallest value of n. In most cases, thissmallest value is n = 1, but there are theorems where the starting value of n is someother integer.
part 2: Make the inductive assumption is true for some positive integer k, and showthat if this is the case, then P (k + 1) is true as well.
These two parts are sufficient to show that P (n) is true for all n. To see this, weknow that the statement is true for n = 1 (or some other starting value) by part 1of the proof. By part 2, it must all be true for n = 2, the next larger number. Sinceit is tru for n = 2, it must be true for n = 3. Hence, it must be true for n = 4 aswell, and so on. Hence, it must be true for all n ∈ N. A more formal discussion ofthe method of mathematical induction is given in the following:
Definition 1.3.1. A propositional function is a statement P (n) about integersthat may be true or false depending on the value chosen for the integer variable n.
Example 1.3.1. The following are examples of propositional functions:
(a) P (n) : 1 + 2 + 3 + · · ·+ n =n(n+ 1)
2
(b) P (n) : 11n − 4n is divisible by n.
(c) P (n) : 2n > n.
The Principle of Mathematical Induction is a means of proving that certain propo-sitional functions that can be defined for all positive integers are true for every positiveinteger. This principle is given in the following theorem:
Theorem 1.3.1. Let P (n) be a propositional function. If
(a) P (1) is true and
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(b) P (n+ 1) is true whenever P (n) is true
Then P (n) is true for every n ∈ N.
Proof. To show that P (n) is true for all n ∈ N, we let
B = { n ∈ N : P (n) is false }
and show that B is empty. Since the elements in B can be arranged in ascendingorder, B has a least element, say b. By (a) of the hypothesis, b > 1, so that b−1 ∈ N.Since b − 1 < b, it follows that b − 1 6∈ B,and hence the statement P (b − 1) is true.However, by (b) of the hypothesis, the statement P ((b − 1) + 1) = P (b) is true,contradicting the assumption that b ∈ B. Thus, B must be empty, and P (n) is truefor all n ∈ N.
We illustrate the principle of mathematical induction by proving the statementsin the preceding example.
Example 1.3.2. To prove that 2n > n for all n ∈ N, we have the following:
P (1) : 21 = 2 > 1
Assume that 2n > n for some positive integer n. Then 2n+1 = 2(2n) > 2n = n+n ≥n+ 1, which shows that 2n+1 > n+ 1, and P (n+ 1) is true.
Thus, the statement P (n) : 2n > n is true for all n ∈ N.
Example 1.3.3. To prove that 11n − 4n is divisible by 7 for all n ∈ N, we have:
P (1) : 111 − 41 = 11− 4 = 7 is divisible by 7.
Assume that P (n) is true for some positive integer n, i.e. assume 11n − 4n = 7q forsome integer q. We have
11n+1 − 4n+1 = 11n+1 − 4n+1 + 11 · 4n − 11 · 4n
= [11n+1 − 11 · 4n] + [11 · 4n − 4n+1] = 11(11n − 4n) + 4n(11− 4)
= 11(7q) + 4n(7) = 7(11q + 4n)
Since 11n+1 − 4n+1 = 7(11q + 4n), it is clear that P (n+ 1) is true.
This proves that the statement is true for any positive integer n.
Exercise 1.3.1. Use mathematical induction to prove that each of the followingstatements is true for all positive integers n.
1. 12 + 22 + 32 + · · ·+ n2 =n(n+ 1)(2n+ 1)
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2. 13 + 23 + 33 + · · ·+ n3 =n2(n+ 1)2
4
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3. 1 + 3 + 5 + 7 + · · ·+ (2n− 1) = n2
4. 1 +1
2+
1
4+
1
8+ · · ·+ 1
2n= 2
[1− 1
2n+1
]5.
1
1 · 2+
1
2 · 3+
1
3 · 4+ · · ·+ 1
n(n+ 1)=
n
n+ 1
6. n! ≤(n+ 1
2
)n
Solutions to the Exercises:
1. For n = 1, the left hand side is 12 = 1, while the right hand side isn(n+ 1)(2n+ 1)
6=
(1)(2)(3)
6= 1, so the equation holds for n = 1. Next assume inductively that the
formula is true for some integer k, i.e.
12 + 22 + · · ·+ k2 =k(k + 1)(2k + 1)
6
We then have
12 + 22 + · · ·+ k2 + (k + 1)2 =k(k + 1)(2k + 1)
6+ (k + 1)2 =
k(k + 1)(2k + 1) + 6(k + 1)2
6
=(k + 1)[k(2k + 1) + 6(k + 1)
6=
(k + 1)(2k2 + k + 6k + 6
6
=(k + 1)(2k2 + 7k + 6)
6=
(k + 1)(k + 2)(2k + 3)
6
=(k + 1)[(k + 1) + 1][2(k + 1) + 1]
6
The right-most expression above is precisely the value ofn(n+ 1)(2n+ 1)
6when
n = k + 1. This shows that P (n) is true when n = k + 1 as well.
2. For n = 1, the left hand side is 13 = 1, while the right-had side is12 · 22
4=
4
4= 1,
so P (1) holds. Assume the formula is true when n = k, so that
13 + 23 + · · · k3 =n2(n+ 1)2
4
16
If we add (k + 1)3 to both members of the above equation, then we get
k+1∑n=1
n3 =k2(k + 1)2
4+ (k + 1)3 =
k2(k + 1)2 + 4(k + 1)3
4
=(k + 1)2[k2 + 4(k + 1)]
4=
(k + 1)2(k2 + 4k + 4)
4=
(k + 1)2(k + 2)2
4
=(k + 1)2[(k + 1) + 1]2
4
The last expression above is precisely the right-hand side of the formula whenn = k + 1. This shows that whenever P (k) is true, then P (k + 1) is also true.
3. When n = 1, the left hand side is 1, while the right hand side is n2 = 12 = 1 aswell, so P (1) is true. Next assume that P (k) is true, i.e.
1 + 3 + · · ·+ (2k − 1) = k2
If we add 2(k + 1)− 1 = 2k + 1 to both members of this equation, we get
k∑n=1
(2n− 1) + (2k + 1) = k2 + (2k + 1) = k + 12
Observe that the right-most expression above is what we get when n = k+1. Thisshows that the formula also holds for n = k + 1 whenever it holds for n = k.
4. For n = 0, the left hand side is 1. For the right hand side, we have 2
(1− 1
2
)=
2(1/2) = 1. For n = 1, we have 1 +1
2=
3
2on the left-hand side, while on the
right-hand side, we have 2
(1− 1
4
)= 2× 3
4=
3
2, so the formula is true. Assume
the formula is true for n = k, so that
k∑n=0
1
n2= 2
(1− 1
2k+1
)
Adding1
2k+1to both members of the equation gives us
k+1∑n=0
1
n2= 2
(1− 1
2k+1
)+
1
2k+1= 2− 2
2k+1+
1
2k+1
= 2− 1
2k+1= 2
(1− 1
2k+2
)Observe that the right-most expression above is the value that we get when weuse n = k + 1. This shows that the formula holds when n = k + 1, and our proofis complete.
17
5. When n = 1, the left-hand side is1
1 · 2=
1
2, while the right-hand side is
n
n+ 1=
1
2,
and the formula holds. Assume that the formula holds when n = k, so that
1
1 · 2+
1
2 · 3+
1
3 · 4+ · · ·+ 1
k(k + 1)=
k
k + 1
We next show that P (k+1) is also true by adding1
(k + 1) · (k + 2)to both members
of the above equation, giving us
k+1∑n=1
1
n · (n+ 1)=
k
k + 1+
1
(k + 1) · (k + 2)=
k(k + 2) + 1
(k + 1)(k + 2)
=(k + 1)2
(k + 1)(k + 2)=
k + 1
(k + 1) + 1
This shows that P (k + 1) is also true.
6. For n = 1, the left-hand side is 1! = 1, while the right hand side is
(1 + 1
2
)2
= 1
and we have 1 ≤ 1. Assume that P (k) holds, i.e. k! ≤(k + 1
2
)k
. To show that
P (k+1) is true, we multiply both members of the above inequality by k+1, givingus
(k + 1)! ≤ (k + 1)
(k + 1
2
)k
=(k + 1)(k + 1)k
2=
(k + 1)k+1
2k
(k + 2)k+1 = (k + 1)k+1 + (k + 1)(k + 1)k +(k + 1)(k)
2(k + 1)k−1 + · · ·+ 1 ≥ 2(k + 1)k+1
(k + 2)k+1
2k+1≥ 2(k + 1)k+1
2k+1=
(k + 1)k+1
2k⇒ (k + 1)! ≤ (k + 1)k+1
2k≤(k + 2
2
)k+1
18
Chapter 2
The Euclidean Space Rk
In this chapter, we introduce metric and topological concepts associated with then-dimensional Euclidean space Rn. The reader will be introduced to the notion ofa distance function on these sets, and some of its important properties. Using thisdistance function, we define certain subsets of these Euclidean spaces, and introducethe notions of closed and open sets.
2.1 Basic Concepts
We first recall that a vector space V is a nonempty set with two operations called vec-tor addition and scalar multiplication defined on it, such that the following propertiesare satisfied: for vector addition, we require the properties of closure, commutativity,associativity, and the existence of identity and inverse elements. On the other hand,scalar multiplication must satisfy the properties f closure, associativity, left and rightdistributive properties, and the identity scalar property. We turn the n-dimensionalspace Rn into a vector space by defining the operations of vector addition and scalarmultiplication on these sets.
Definition 2.1.1. Let n be a positive integer. The n-dimensional Euclidean space, denoted by Rn, is the vector space which consists of all ordered n-tuples x =(x1, x2, . . . , xn) where xi ∈ R for i = 1, 2, . . . , n. The operations in Rn aredefined as follows:
Vector Addition
x+ y = (x1, x2, . . . , xn) + (y1, y2, . . . , yn) = (x1 + y1, x2 + y2, . . . , xn + yn)
Scalar Multiplication
c · x = c(x1, x2, . . . , xn) = (cx1, cx2, . . . , cxn)
Inner Product
x • y = (x1, x2, . . . , xn) • (y1, y2, . . . , yn) = x1y1 + x2y2 + · · ·+ xnyn
19
The fact that Rn is a vector space with respect to the vector addition and scalarmultiplication described above is given by the following theorem:
Theorem 2.1.1. Let x, y, z ∈ Rn and let a, b ∈ R. Then
(i) x+ y = y + x
(ii) (x+ y) + z = x+ (y + z)
(iii) x+ 0 = x, where 0 = (0, 0, . . . , 0)
(iv) x+ (−x) = 0, where −x = (−x1, − x2, . . . , − xn)
(v) a(x+ y) = ax+ ay
(vi) (a+ b)x = ax+ bx
(vii) a(bx) = (ab)x
(viii) 1 · x = x
The next theorem gives some properties of the inner or dot product defined on Rn.
Theorem 2.1.2. Let x, y ∈ Rn and let a ∈ R. Then
(a) x • x ≥ 0
(b) x • x = 0 if and only if x = 0
(c) x • y = y • x
(d) x • (y + z) = x • y + x • z
(e) (ax) • y = a(x • y)
Proof.
(a) x • x = x21 + x22 + · · ·+ x2n ≥ 0
(b) x • x = 0 if and only if x2i = 0 for i = 1, 2, . . . , n if and only if x = 0
(c) x • y = x1y1 + x2y2 + · · ·+ xkyn = y1x1 + y2x2 + · · ·+ ynxn = y • x
(d) x•(y+z) =k∑
i=1
xi(yi+zi) =n∑
i=1
(xiyi+xizi) =n∑
i=1
xiyi+n∑
i=1
xizi = x•y+x•z
(e) (ax) • y =n∑
i=1
(axi)(yi) =n∑
i=1
a(xiyi) = a
n∑i=1
xiyi = a(x • y)
20
We now define the norm of a vector x ∈ Rn.
Definition 2.1.2. Let x = (x1, x2, . . . , xn) ∈ Rn. The norm of x, denoted by ‖x‖,is defined to be
‖x‖ =√x • x =
√x21 + x22 + · · ·+ x2n =
√√√√ n∑i=1
x2i
Remark.
• If the vector x ∈ Rn is represented by a point in Rk, then the norm of xrepresents the distance of x from the origin 0.
• The norm can be interpreted as a function ‖ · ‖ : Rn → R which assigns to eachvector x the non-negative real number ‖x‖.
Example 2.1.1. Let k = 1, 2, 3. Draw the set of points x ∈ Rn satisfying each ofthe following conditions:
(a) ‖x‖ < 1 (b) ‖x‖ > 1 (c) ‖x‖ = 1
The next theorem shows that the norm satisfies the following properties:
Theorem 2.1.3. (Properties of the Norm)Let x, y ∈ Rn and let a ∈ R. Then
(a) ‖x‖ ≥ 0 (non-negativity)
(b) ‖x‖ = 0 if and only if x = 0
(c) ‖ax‖ = |a|‖x‖ )norm of a scalar product)
The above properties can be verified directly from the definition of the normin terms of the dot product. Hence, the formal proof is omitted but the reader isencouraged to try to work out the proofs.
The next two theorems contain the fundamental inequalities for the dot product andthe norm.
Theorem 2.1.4. (Cauchy-Schwarz Inequality)For all x, y ∈ Rn, we have
|a • b| ≤ ‖a‖‖b‖
Proof. If b = 0, then both sides of the above inequality are 0, and the inequality issatisfied. Assume b 6= 0. Define the function f by
f(t) = ‖(a− tb)‖2 = (a− tb) • (a− tb) = ‖a‖2 − 2ta • b+ t2‖b‖2
21
which is quadratic with respect to the real variable t. The minimum value of f(t)occurs when
t =a • b‖b‖2
and this minimum value is given by
f(a • b‖b‖2
) = ‖a‖2 − (a • b)2
‖b‖2
Since f(t) ≥ 0 for all t ∈ R, it follows that
‖a‖2 − (a • b)2
‖b‖2≥ 0⇔ ‖a‖2‖b‖2 ≥ (a • b)2 ⇔ |a • b| ≤ ‖a‖‖b‖
The Cauchy-Schwarz inequality is used to prove the following well-known and veryuseful result:
Theorem 2.1.5. (Triangle Inequality)For all x, y ∈ Rn, we have ‖a+ b‖ ≤ ‖a‖+ ‖b‖.
Proof. Using the Cauchy-Schwarz inequality, we get
‖a+ b‖2 = (a+ b) • (a+ b) = ‖a‖2 + 2a • b+ ‖b‖2
≤ ‖a‖2 + 2‖a‖‖b‖+ ‖b‖2 = (‖a‖+ ‖b‖)2
Taking the square root of both members gives the desired inequality.
Definition 2.1.3. Let x, ∈ Rn. We define the distance d(x, y) between x and y tobe
d(x, y) = ‖x− y‖
Example 2.1.2.
(a) If x = (1,−2, 4), y = (−1, 0, 3) ∈ R3, then
d(x, y) = ‖x− y‖ = ‖(2, − 2, 1)‖ =√
22 + (−2)2 + 12 =√
9 = 3
(b) Let a = (4, 2) ∈ R2. Then the set A = { x ∈ R2 ‖x‖ < ‖x− a‖ } consists of allpoints x which are closer to the origin than to a. These are the points that liebelow the perpendicular bisector of the line segment that joins a and the origin0 = (0, 0).
The next theorem gives the fundamental properties of the distance function d.
Theorem 2.1.6. Let x, y, z ∈ Rn. Then
22
(a) d(x, y) ≥ 0
(b) d(x, y) = 0 if and only if x = y.
(c) d(x, y) = d(y, x)
(d) d(x, y) ≤ d(x, z) + d(z, y)
Proof. Since the distance function is defined in terms of the norm, (a) and (b) followdirectly from (a) and (b) of Theorem 2.1.3.
(c) We have
d(x, y) = ‖x− y‖ =
√√√√ n∑i=1
(xi − yi)2 =
√√√√ n∑i=1
(yi − xi)2 = ‖y − x‖ = d(y, x)
(d) Using the triangle inequality for the norm, we have
d(x, y) = ‖x− y‖ = ‖(x− z) + (z − y)‖ ≤ ‖x− z‖+ ‖z − y‖ = d(x, z) + d(z, y)
Exercise 2.1.1.
1. If x = (a, b, c) ∈ R3, show that
(a) ‖x‖ ≤ |a|+ |b|+ |c| (b) sup { |a|, |b|, |c| } ≤ ‖x‖
2. Prove: If x, y ∈ Rn, then |‖x‖ − ‖y‖| ≤ ‖x± y‖ ≤ ‖x‖+ ‖y‖.
3. Prove: If x = (x1, x2, . . . , xk) ∈ Rk, then for i = 1, 2, . . . , k, we have
|xi| ≤ ‖x‖ ≤√k sup {|x1|, |x2|, . . . , |xk|}
4. Prove the Parallelogram Identity:
‖x+ y‖2 + ‖x− y‖2 = 2(‖x‖2 + ‖y‖2)
Geometrically, this equation states that the sum of the squares of the diagonals ofa parallelogram is equal to twice the sum of squares of its sides.
5. If x, y ∈ Rn, show that
‖x+ y‖2 = ‖x‖2 + ‖y‖2 if and only if x • y = 0
In this case, we say that x and y are perpendicular to each other or are orthogonalvectors.
23
6. Let n ≥ 3, x, y ∈ Rk, ‖x− y‖ = d > 0 and r > 0. Prove that
(a) If 2r > d, there are infinitely many z ∈ Rn such that ‖x− z‖ = ‖z − y‖ = r.
(b) If 2r = d, then there is exactly one z ∈ Rn satisfying the equation in (a).
(c) If 2r < d, then no such z exists.
7. If a = (4, 2), sketch the set of points x ∈ R2 such that
(a) ‖x‖+ ‖x− a‖ = 6 (b) ‖x‖+ ‖x− a‖ ≤ 4
Solutions to the Exercises:
1. (a) We have
(|a|+ |b|+ |c|)2 = |a|2 + |b|2 + |c|2 + 2|ab|+ 2|ac|+ 2|bc| ≥ a2 + b2 + c2 = ‖x‖2
|a|+ |b|+ |c| ≥ ‖x‖
(b) ‖x‖ =√a2 + b2 + c2 ≥ sup{
√a2,√b2,√c2} = sup{ |a|, |b|, |c|}
2. The inequality ‖x± y‖ ≤ ‖x‖+ ‖y‖ is clear from the Triangle inequality. For theother half of the inequality, we have
‖x‖ = ‖x− y + y‖ ≤ ‖x− y‖+ ‖y‖ ⇒ ‖x‖ − ‖y‖ ≤ ‖x− y‖‖y‖ = ‖y − x+ x‖ ≤ ‖x− y‖+ ‖x‖ ⇒ ‖y‖ − ‖x‖ ≤ ‖x− y‖
Combining the two inequalities gives us |‖x‖ − ‖y‖| ≤ ‖x − y‖ ≤ ‖x‖ + ‖y‖. Asimilar approach can be used to show that the inequality still holds if (x − y) isreplace by (x+ y).
3. It is clear that |xi| ≤ ‖x‖, i = 1, 2, . . . , n. On the other hand, ifM = sup{ |x1|, |x2|, . . . , |xn|},then we have
‖x‖2 =n∑
i=1
x2i =n∑
i=1
|xi|2 ≤ kM2 ⇒ ‖x‖ ≤√kM
4. We have
‖x+ y‖2 + ‖x− y‖2 = (x+ y) • (x+ y) + (x− y) • (x− y)
= x • x+ y • y + 2(x • y) + x • x+ y • y − 2(x • y)
= 2(x • x) + 2(y • y) = 2(‖x‖2 + ‖y‖2
)5. We have
‖x+ y‖2 = (x+ y) • (x+ y) = x • x+ y • y + 2x • y = ‖x‖2 + ‖y‖2
since the dot product x • y = 0.
24
6. The points z ∈ Rn which satisfy the equation ‖x−z‖ = ‖z−y‖ = r are equidistantfrom x and y. Hence,
(a) If 2r > d, then the points along the perpendicular bisector of the line segmentjoining x and y satisfy the equation. If 2r = d, then the midpoint of the linesegment joining x and y is the only point which satisfies the given equation.
(b) If 2r < d, then by the triangle inequality, we have ‖x − z‖ + ‖z − y‖ = 2r >‖x− y‖ = d and we have a contradiction. hence, there is no point z satisfyingthe given equation.
7. (a) The equation ‖x‖ + ‖x − a‖ = 6 describes an ellipse with foci at 0 and at aand with major axis of length equal to 6 units. The center of the ellipse is atC(2, 1).
(b) By (c) of the preceding item, there is no point x satisfying the inequality.
2.2 Open and Closed Sets
In this section, we introduce the reader to some topological concepts on the Euclideann-space. The concepts need getting used to, so the reader is advised to look at asmany examples as possible, and to work out as many of the exercises as he/she can.
Definition 2.2.1. Let x0 ∈ Rn, r > 0.
• The open ball with center at x0 and radius r is the set
B(x0, r) = { x ∈ Rn : d(x0, x) < r }
• The closed ball with center at x0 and radius r is the set
B(x0, r) = { x ∈ Rn : d(x0, x) ≤ r }
• The sphere with center at x0 and radius r is the set
S(x0, r) = { x ∈ Rn : d(x0, x) = r }
The appearances of balls and spheres differ according to the dimension of theEuclidean space being considered. Consider the following examples.
Example 2.2.1.
(a) If n = 1, then the open ball B(x0, r) is just the open interval (x0 − r, x0 + r).Correspondingly, the closed ball B(x0, r) is the closed interval [x0 − r, x0 + r].
(b) If n = 2, then B(x0, r) is the open disc interior to the circle with center x0 andradius r. The corresponding closed ball is the union of this open disc and thecircle serving as boundary of the disc.
25
(c) If n = 1, the sphere with center at x0 and radius r consists of the set { x0 −r, x0 + r }. If k = 2, the corresponding open sphere is the circle with center at x0and radius r.
If A is a subset of Rn, then the elements of Rn may be partitioned into threesubsets relative to the set A. These sets are defined as follows:
Definition 2.2.2. Let A ⊆ Rn and let x0 ∈ Rn. Then
• x0 is an interior point of A if there exists r > 0 such that B(x0, r) ⊆ A.
• x0 is a boundary point of A if for each r > 0, we have B(x0, r) ∩ A 6= ∅ andB(x0, r) ∩ Ac 6= ∅.
• x0 is an exterior point of A if there exists r > 0 such that B(x0, r) ⊆ Ac.
Remark. Since x0 ∈ B(x0, r), it follows that every interior point of a set A is anelement of A.
Example 2.2.2. For each of the following sets, determine the interior, the boundaryand the exterior points of the set.
(a) Let A = { 1n
: n ∈ N}.
(b) Let A = [0, 1]
(c) Let A = { x ∈ Q : 0 ≤ x ≤ 1 }
Solution:
• The only candidates for the interior points of A are its elements. For any1
n∈ A, the closest element of A is
1
n+ 1, and if we let r be any positive number
less than1
n− 1
n+ 1=
1
n(n+ 1), then the ball with center at
1
nand radius r
contains infinitely many points in R which are not in A. Thus,1
n6∈ Int A, and
A has no interior points. Every element of A is a boundary point, since for any
xn =1
n∈ A, and for any r > 0, the ball B(xn, r) intersects both A and A′.
Finally, every point not in A is an exterior point of A.
• The boundary points of A are 0 and 1, the interior points are the points in (0, 1),while the exterior points are those which belong to the set (−∞, 0) ∪ (1,∞).
• From the denseness of both the rationals and the irrationals within R, we knowthat every element of A is a boundary point. Moreover, every irrational numberin the interval [0, 1] is a boundary point. Hence, A has no interior point, andthe points in the set [0, 1] are the boundary points of A. Consequently, theexterior points of A are the elements of the set (−∞, 0) ∪ (1,∞).
26
Definition 2.2.3. Let A ⊆ Rn. Then
• The interior of A, denoted by A◦ or by Int A, is the set of all the interior pointsof A.
• The boundary of A, denoted by ∂A or by Bdy A, is the set of all the boundarypoints of A.
• The exterior of A, denoted by Ext A, is the set of all the exterior points of A.
Quick workout: Find the interior, boundary and exterior of each of the sets de-scribed in the preceding example.
Definition 2.2.4. Let A ⊆ Rn. We say that
• A is open if for each x0 ∈ A, there exists r > 0 such that B(x0, r) ⊆ A. Thus,each element of A is interior to A, and A◦ = A.
• A is closed if Ac = A′ is open.
Remark.
• Thus, a set A is open if every element of A is the center of an open ball whichlies completely inside A. As a result, an open set is expressible as a union ofopen balls.
• A set which is not open is not necessarily closed. For example, the subsetA = (1, 2] of R is not open since 2 ∈ A is not an interior point of A. However,A is also not closed, since Ac = A′ = (−∞, 1] ∪ (2,∞) is not open, as 1 ∈ Ac
is not an interior point. Thus, in this case, the given set A is neither open norclosed.
• In Mathematics, a condition about a set is said to be ”vacuously satisfied” bythe elements of the set if there is no element in the set which fails to satisfy thecondition. Thus, we say that the null set is open not because every element ofthe set is an interior point but because there is no element in the null set whichis not an interior point.
To get a better understanding of open and closed sets, study the following exam-ples:
Example 2.2.3.
(a) Let S = { (x, y) ∈ X : − 1 < x < 2 } ⊂ R2. Then every element of S is interiorto S and S is open.
(b) Let S = { 1nn ∈ N } ⊂ R. Then S◦ = ∅ and S is not open.
27
(c) A singleton is a set with exactly one element. To see that a singleton subsetA = { x }, x ∈ Rk is closed, consider the complement Ac = Rk\{ x }. For each
y ∈ Ac, let r =1
2d(x, y). Thus, the ball B(y, r) does not contain x, and must
there lie completely in Ac. Hence, y is interior to Ac, and since y is arbitrary,every element of Ac is an interior point, and Ac is open. Thus, A is closed.
(d) From the remarks above, the null set is open, so ∅c = Rk is closed. However, forevery x ∈ Rk and for any r > 0, we have B(x, r) ⊂ Rk, so Rk is likewise open.
(e) The closed interval [0, 1] is closed in R since its complement (−∞, 0)∪ (1, +∞)is open.
(f) The set S = { 1n
: n ∈ N } is not closed in R since Sc is not open. The element0 ∈ Sc is not an interior point of Sc. Thus, S is neither open nor closed.
(g) Any finite subset S of Rk is closed. To see this, it suffices to show that Sc isopen. To this end, let S = { x1, x2, . . . , xn }, and let x ∈ Sc. Let d =min { d(x, xi), xi ∈ S }. Thus, d is the minimum distance between x and theelements of the set S. If we let 0 < r < d, then the ball B(x, r) does not containany element of S, so that B(x, r) ⊂ Sc, that is, x is an interior point of Sc. Sincex is an arbitrary element of Sc, it follows that Sc is open, and hence S is closed.
The following theorem gives the fundamental properties of open sets. These propertiesallow us to construct new open sets from a given collection of open sets.
Theorem 2.2.1. (Properties of Open Sets)If X = Rn, then
(a) X, ∅ are both open in X.
(b) The union of any number of open sets in X is again an open set.
(c) The intersection of any finite number of open sets is again open.
(d) The open ball B(x0, r) is open.
Proof.
(a) Let x0 ∈ X. Then for any r > 0, the open ball B(x0, r) ⊂ X, so X is open. Thenull set ∅ is open since it satisfies the definition vacuously.
(b) Let {Ai : i ∈ I } be an arbitrary collection of open sets inX, and letA = ∪i∈I Ai.Let x0 ∈ A, so that x0 ∈ Aj for some j ∈ I. Since Aj is open there exists an openball B(x0, r) ⊂ Aj ⊂ A. This shows that A is open.
(c) Let A1, A2, . . . , An be open sets, and let A = ∩ni=1Ai. If x0 ∈ A, then x0 ∈ Ai fori = 1, 2, . . . , n. Since each Ai is open, there exists an open ball B(x0, ri) ⊂ Ai.Let r = min {r1, r2, . . . , rn }. Then B(x0, r) ⊂ B(x0, ri) ⊂ Ai for each i. Thisshows that B(x0, r) ⊂ A and A is open.
28
(d) Let y ∈ B(x0, r) and let d = ‖x0 − y‖. If r1 =r − d
2, then B(y, r1) ⊂ B(x0, r)
and B(x0, r) is open.
The properties in the preceding theorem have the following counterparts for closedsets:
Theorem 2.2.2. (Properties of Closed Sets)Let X = Rn.
(a) X, ∅ are closed sets in X.
(b) The intersection of an arbitrary number of closed sets is again closed.
(c) The union of a finite number of closed sets is again closed.
Proof. The proof of this theorem consists of looking at the complements of the indi-cated sets.
(a) By definition, a set is closed if its complement is open. Thus, X = Rk is a closedset since its complement, the empty set, is open. A similar proof can be madefor the null set.
(b) Let S = { Fi : i ∈ I } be an arbitrary collection of closed subsets of X. Thus,Gi = F c
i = X − Fi is open for each i ∈ I. By (b) of Theorem 2.2.1, we have⋃i∈I
Gi is open, and hence
(⋂i∈I
Fi
)c
=⋃i∈I
F ci =
⋃i∈I
Gi
is open, so that⋂
i∈I Fi is closed.
(c) Let F1, F2, . . . , Fn be a finite collection of closed subsets of X, so that eachGi = X − Fi = F c
i is open. We have(n⋃
i=1
Fi
)c
=n⋂
i=1
F ci =
n⋂i=1
Gi
is open by (c) of Theorem 2.2.1. Hence, the setn⋃
i=1
Fi is closed.
Definition 2.2.5. Let S be a subset of Rn and let x0 ∈ Rn.
29
(a) S is a neighborhood of x0 if there exists B(x0, r) ⊂ S.
(b) x0 is a a boundary point of S if every open set that contains x0 contains apoint in Sc = −S.
(c) x0 is an accumulation point (limit point, cluster point) of S if every neigh-borhood of x0 contains an element of A other than x0.
(d) The set of all accumulation points of S is called the derived set of S and isdenoted S ′.
(e) x0 ∈ S is called an isolated point of S if x0 is not an accumulation point of S.Thus, there exists a neighborhood U of x0 such that x0 ∈ U and U∩S−{x0} = ∅.
Example 2.2.4.
(a) Let S be the closed interval [0, 1]. If x ∈ S and r > 0, the ball B(x, r) containseither an element y ∈ (x− r, x) ∩ S or an element z ∈ (x, x+ r) ∩ S. This showsthat every neighborhood of x contains at least one element of S other than xitself, so that x ∈ S ′.
(b) Let S =
{1
n: n ∈ N
}. Since lim
n→∞
1
n= 0, we have S ′ = { 0 }. Moreover,
every element of S is a boundary point, so that
S◦ = ∅, Bdy S = S ∪ { 0 }, Ext S = R− (S ∪ { 0 })
(c) Let X = R and let S = Z be the set of all integers. If x ∈ S, consider the ballB(x, 0.5). It is clear that B(x, 0.5)∩ S −{ x } = ∅, so that x is an isolated pointof S. Since x is arbitrary, every element of S is an isolated point.
(d) Let S = (0, 1) = { x ∈ R : 0 < x < 1 }. If x ∈ S and 0 < r < 1, the ballB(x, r) = (x− r, x+ r) intersects S at infinitely many points other than x. Thisshows that x is not an isolated point of S. Since x is arbitrary, the open intervalS has no isolated points.
The following theorems give properties of the interior, the boundary and the derivedset of a subset S of Rk.
Theorem 2.2.3. Let A, B ⊂ Rn.
(a) The interior of A is the union of all the open subsets of A.
(b) A◦ is the largest open subset of A.
(c) (A◦)◦ = A◦
(d) (A ∩B)◦ = A◦ ∩B◦
30
Proof.
(a) Let { Ui : i ∈ I } be the collection of all the open subsets of A, and letU = ∪i∈I Ui. Let x ∈ U . Since U is open, there exists B(x, r) ⊂ U ⊂ A, so thatx ∈ A◦. This shows U ⊂ A◦. For the reverse inclusion, let y ∈ A◦. Then y is aninterior point of A and there exists B(y, r) ⊂ A. Since B(y, r) is an open subsetof A, it is one of the Ui’s, so that y ∈ U . This shows the inclusion Acirc ⊂ U .The two inclusions show U = A◦.
(b) From (a), since the union of open sets is open, it is clear that A◦ is an open subsetof A. Moreover, each open subset of A is a subset of A◦, so it is the largest opensubset of A.
(c) Since the interior of a set is the largest open subset of this set, it is clear thatthe interior of an open set is the set itself. Since A◦ is open, the desired resultfollows.
(d) We show both inclusions. Let x ∈ (A∩B)◦. Then there exists B(x, r) ⊂ A∩B.Since A ∩ B ⊂ A and A ∩ B ⊂ B, it follows that x ∈ B(x, r) ⊂ A andx ∈ B(x, r) ⊂ B and hence x ∈ A◦∩B◦. For the reverse inclusion, let y ∈ A◦∩B◦.There exist r1, r2 > 0 such that y ∈ B(y, r1) ⊂ A and y ∈ B(y, r2) ⊂ B. If welet r = min {r1, r2 }, then y ∈ B(y, r) ⊂ A ∩ B and y ∈ (A ∩ B)◦. This showsA◦ ∩B◦ ⊂ (A ∩B)◦ and equality follows.
The next theorem shows relationships among the boundary and the derived set ofa closed subset of Rn. When a set of statements are considered equivalent, it meansthat each statement implies all the other statements in the list. There are variousways of establishing an equivalence. One is by using a ”circular” proof, where (a)implies (b), (b) implies (c), and so on until the last statement implies the first. Inother cases, pairs of statements are shown to be equivalent. For this theorem, weapply the circular method of proof.
Theorem 2.2.4. Let S ⊂ Rn. The following are equivalent.
(a) S is closed.
(b) Bdy S ⊂ S.
(c) S ′ ⊂ S.
Proof.
(a) implies (b)Let S be a closed subset of Rn, so that S ′ is open. We want to show that Bdy S ⊂ S.To this end, let x ∈ Bdy S. Suppose x 6∈ S, so that x ∈ Sc. Since Sc is open, thereexists an open ball B(x, r) ⊂ Sc, so that B(x, r) ∩ S = ∅. This contradicts the
31
assumption that x is a boundary point of S, so that every open ball about x mustintersect S. Thus, the assumption that x 6∈ S is false. Since x is an arbitrary elementof bdy S, it follows that every element in the boundary of S is also an element of S.
(b) implies (c)Assume that Bdy S ⊂ S. Recall that Rn partitions into the interior, the boundaryand the exterior of the set S. Let x ∈ S ′. Since both the interior and the boundaryof S are subsets of S, it suffices to show that x 6∈ Ext S. Assume to the contrary thatx ∈ Ext S. Then there exists an open ball B(x, r) ⊂ Sc, so that B(x, r)∩S = ∅. Thiscontradicts the fact that x is an accumulation point of S. Hence, the assumptionthat x ∈ Ext S is false, so that x ∈ Bdy S ∪S◦ ⊂ S. Since x is an arbitrary elementof S ′, it follows that S ′ ⊂ S.
(c) implies (a)Assume that S ′ ⊂ S. To show that S is closed, we will show that Sc is open. Tothis end, let y ∈ Sc. Since S” ⊂ S, it follows that y 6∈ S ′, and hence there exists anopen ball B(y, r) ∩ S = ∅. This shows that B(y, r) ⊂ Sc. Since this is true for anyy ∈ Sc, it follows that Sc is open, and hence S is closed.
Definition 2.2.6. Let A ⊂ Rn. The closure of A, denoted by A, is the intersectionof all the closed subsets of Rk which contain A.
Example 2.2.5.
(a) Let A = (0, 1) ⊂ R. Then A = [0, 1].
(b) Let A = { 1, 2, 3, 4 }. Then A = A.
(c) If A = { 1nn ∈ N }, then A = ∂A = A ∪ { 0 }.
The next theorem gives the nature of the closure of a subset of Rk.
Theorem 2.2.5. Let A ⊂ Rn. Then
(a) A is the smallest closed set that contains A.
(b) A is closed if and only if A = A.
(c) A = A ∪ A′ = A ∪ ∂A = A◦ ∪ ∂A.
Proof.
(a) The intersection of any number of closed sets is closed, so A is a closed set. If Gis any closed set that contains A, then G is part of the collection of sets whoseintersection is A. Hence, A ⊂ G. Since G is an arbitrary closed superset of A, itfollows that A is the smallest closed set that contains A.
32
(b) Suppose A is closed. Since A ⊂ A, we know from (a) that A ⊂ A. However, byits definition, we know that A ⊂ A, so that A = A. For the converse, supposeA = A. Since A is closed, so is A.
(c) To prove that A = A ∪ A′, it suffices to show that A′ ⊂ A. Since A ⊂ A and Ais closed, it follows that A′ ⊂ A. Similarly, to show that A = A ∪ ∂A, it sufficesto show that ∂A ⊂ A. However, A ⊂ A and A is closed, so ∂A ⊂ A.
Exercise 2.2.1.
1. Show that the set R2 − {(0, 0)} is an open subset of R2.
2. Show that the set S = { (x, y) ∈ R2 : xy > 0 } is an open subset of R2.
3. Show that any open subset of R is either an open interval or is the union of disjointopen intervals.
4. In each of the following, determine if S is open, closed or neither open nor closedin the appropriate Euclidean space Rn.
(a) S = { x ∈ R : x(x− 1)2 > 0 }(b) S = { x ∈ R : x(x− 1)(x+ 1)2 < 0 }(c) S = { (x, y) ∈ R2 : x ≥ y2 }(d) S = { (x, y) ∈ R2 : x2 + y2 = 4 }(e) S = { (x, y) ∈ R2 : x < 0 and y < x3 }(f) S = { (x, y) ∈ R2 : x2 + y2 > 0 }(g) S = { (x, y) ∈ R2 : x = y }(h) S = { (x, y) ∈ R2 : x > 1 }(i) S = { (x, y) ∈ R2 : y < x2 and x ≤ y }(j) S = { (x, y) ∈ R2 : 0 < y < 2x− 3 }(k) S = { (x, y) ∈ R2 : y = 2x+ 1 }(l) S = { (x, y) ∈ R2 : x, y > 1 }
(m) S = { (x, y) ∈ R2 : x = 0, 0 < y < 1 }(n) S = { (x, y, z) ∈ R3 : x2 + y2 + z2 > 4 }(o) S = { (x, y, z) ∈ R3 : x2 + y2 ≤ 4 }
5. Determine all the limit points of each of the following sets.
33
(a) Z(b) (a, b] = { x : a < x ≤ b }
(c)
{1
n: n ∈ N
}(d) Q(e) { 2−m + 5−n : m, n ∈ N }
(f)
{(−1)n +
1
m: m, n ∈ N
}
(g)
{1
n+
1
m: m, n ∈ N
}(h) { x ∈ R2 : ‖x‖ > 1 }
(i) { x ∈ R2 : ‖x‖ ≥ 1 }
(j) { (x, y) ∈ R2 : x2 − y2 < 1 }
(k) { (x, y) ∈ R2 : x > 0 }
(l) { (x, y) ∈ R2 : x ≥ 0 }
6. Prove: If G ⊂ Rn, then the following are equivalent:
(a) G is open;
(b) G◦ = G
(c) G is a neighborhood of each of its elements
7. Give examples of
(a) an infinite subset of R with no accumulation points
(b) a non-empty subset of R which is contained in its derived set
(c) a subset of R which has infinitely many accumulation points but which doesnot contain any of them
8. Let A ⊂ Rn. Prove that
(a) A ⊂ A
(b) A = A
(c) A ∪B = A ∪B(d) ∅ = ∅
9. Let A ⊂ Rn. Prove the following:
(a) x ∈ A if and only if x is either an interior point or a boundary point of A.
(b) x ∈ A if and only if every neighborhood of x contains an element of A.
Solutions to the Exercises:
1. Let X = R2 and let S = X\{ 0 = (0, 0) }. To show that S id open, we will showthat for any x = (a, b) ∈ S, there exists r > 0 such that B(x, r) ⊂ S. To this end,let d =
√(a− 0)2 + (b− 0)2 =
√a2 + b2 > 0. If 0 < r < d, then 0 6∈ B(x, r), so
that B(x, r) ⊂ S. Since x is an arbitrary element of S, it follows that S is open.
2. Let P = (x, y) ∈ S. We consider two cases, namely: (a) x > 0, y > 0 and (b) x <0, y < 0. For (a), let 0 < r < min{ x, y }, while for (b), let 0 < r < min { |x|, |y| }.In both cases, it is clear that B(P, r) ⊂ S, and hence P is an interior point of S.Since P is an arbitrary element of S, it follows that S is open.
34
3. Let S be an open subset of X = R. For each x ∈ S, there exists rx > 0 such thatB(x, rx) = (x− rx, x+ rx) ⊂ S, and we can write
S = ∪x∈S B(x, rx).
Since S is a union of open intervals, S itself is an open interval.
4. (a) S = { x ∈ R : x(x− 1)2 > 0 }Since (x− 1)2 > 0 whenever x 6= 1, it follows that x > 0, so that the set S isthe infinite open interval S = (0, ∞). Clearly, this set is open, since for eachx ∈ S, we can choose 0 < r < x and the ball B(x, r) ⊂ S, making x an interiorpoint. Since this is true for any element of S, it follows that S is open.
(b) S = { x ∈ R : x(x− 1)(x+ 1)2 < 0 }Since (x + 1)2 > 0 whenever x 6= −1, it follows that the sign of the productx(x − 1)(x + 1)2 depends on the sign of the product x(x − 1). Clearly, theproduct is negative if the two factors have opposite signs. Thus, we have twocases, namely: (i) x > 0 and (x−1) < 0, giving us x ∈ (0, 1), or (ii) x < 0 and(x − 1) > 0, which has no solution. Thus, S is the open interval S = (0, 1),and S is an open set as shown in Exercise 3.
(c) S = { (x, y) ∈ R2 : x ≥ y2 }The set S is not open, since the points on the parabola x = y2 are not interiorpoints of S. The complement of S is the set Sc = { (x, y) ∈ R2 : x < y2 },which consists of the points not enclosed by the parabola x = y2, If A =(a, b) ∈ Sc, let d be the distance from A to the parabola, and let 0 < r < d.Then B(A, r) ⊂ Sc, and Sc is open. Thus, S is a closed subset of R2.
(d) S = { (x, y) ∈ R2 : x2 + y2 = 4 }S consists of points (x, y) on the circle with center at the origin and radius2. If a = (b, c) ∈ Sc, we consider two cases, namely: (i) b2 + c2 < 4 and (ii)b2 + c2 > 4. Let d =
√b2 + c2. In the first case, let 0 < r < 2− d, while in the
second case, let 0 < r < d− 2. In both cases, B(a, r) ⊂ Sc, so that Sc is open,and S is closed.
(e) S = { (x, y) ∈ R2 : x < 0 and y < x3 }The points not on the curve y = x3 are divided into the points whose coordi-nates satisfy y > x3 and the points satisfying the condition y < x3. It can beverified that the second set of points consists of the points to the right of thecurve y = x3. Thus, the set S consists of the points to the right of this curveAND to the left of the y-axis, since x < 0. The set S is shown below:
35
For any point a ∈ S, let d be the smallest among the distances of a from thecurve y = x3, the y-axis, and the x-axis, and let 0 < r < d. Then B(a, r) ⊂ S, and hence S is open.
(f) S = { (x, y) ∈ R2 : x2 + y2 > 0 }It can be verified that S is the same set described in Exercise 1. Thus, S is anopen set.
(g) S = { (x, y) ∈ R2 : x = y }The elements of S are the points on the line y = x. Let a ∈ Sc and let d bethe distance of a from the line y = x. If 0 < r < d, we have B(a, r) ⊂ Sc, andSc is open. This shows that S is closed.
(h) S = { (x, y) ∈ R2 : x > 1 }The set S consists of all points in R2 which lie to the right of the verticalline x = 1. Let P (x, y) ∈ S, so that x > 1, and let 0 < r < x − 1. ThenB(P, r) ⊂ S and S is open.
(i) S = { (x, y) ∈ R2 : y < x2 and x ≤ y }The set S consists of all points in R2 which lie on and above the line y = xAND below the parabola y = x2. The points in S which lie on the line y = xare boundary points of S, so S is not open . On the other hand, Sc is not openas well, since the points in the second quadrant which lie along the parabolay = x2 are in Sc and are boundary points of Sc. This shows that the set S isneither open nor closed.
(j) S = { (x, y) ∈ R2 : 0 < y < 2x− 3 }The elements of S are the points in the first quadrant which lie below the liney = 2x− 3. If P (x, y) ∈ S, let d be the smaller of the distances of P from thex-axis and the line y = 2x− 3, and let 0 < r < d. Then the ball B(P, r) ⊂ S,and P is an interior point of S. Since this is true for any element of S, itfollows that S is open.
(k) S = { (x, y) ∈ R2 : y = 2x+ 1 }The set S consists of all points on the line y = 2x+ 1. This set is closed, sinceSc is open. To see this, note that for any P ∈ Sc, if d is the distance of P fromthe line y = 2x + 1 and if 0 < r < d, then B(P, r) ⊂ Sc, and P is interior toSc. Since P is arbitrary, Sc is open.
(l) S = { (x, y) ∈ R2 : x, y > 1 }The set S consists of all points to the right of the line x = 1 and above theline y = 1. Clearly, this set is open, since for any P ∈ S, we can let r > 0 be apositive number less than the smaller of the distances of P from the two linesx = 1 and y = 1. The ball B(P, r) ⊂ S and hence P is an interior point of S.
(m) S = { (x, y) ∈ R2 : x = 0, 0 < y < 1 }The set S consists of all points in R2 which lie on the y-axis between the points(0, 0) and (0, 1). This set is not closed, since the points (0, 0) and (0, 1) areboundary points of S which are not in S. Now every element of S is clearly a
36
boundary point of S, so that S is not open as well. Hence, S is neither opennor closed.
(n) S = { (x, y, z) ∈ R3 : x2 + y2 + z2 > 4 }The set S consists of all points outside the sphere x2 + y2 + z2 = 4 in R3.For any point P ∈ S, let d be the distance of P from the origin, and let0 < r < d− 2. Then B(P, r) ⊂ S and S is open.
(o) S = { (x, y, z) ∈ R3 : x2 + y2 ≤ 4 }The set S consists of all point in R2 which lie on or inside the circle x2+y2 = 4.This set is closed, since Sc = { (x, y) ∈ R2 : x2 + y2 > 4 } is open. To seethis, let P ∈ Sc, and let d be the distance of P from the origin. If we let0 < r < d− 2, then B(P, r) ⊂ Sc, so that P is an interior point of Sc.
5. (a) S = ZEach element of S is an isolated point. On the other hand, if x ∈ Sc, thereexist two consecutive integers n and n + 1 such that n < x < n + 1. Letd = min{ x− n, n+ 1− x } and let 0 < r < d. Then B(x, r)∩ S = ∅, so thatx is not a limit point of S. Thus, S ′ = emptyset.
(b) S = (a, b] = { x : a < x ≤ b }Clearly, every element of S is a limit point. The real number b is also a limitpoint, since the ball B(b, r) = (b − r, b + r) will always intersect S for anyr > 0. Any real number x outside the closed interval [a, b] cannot be a limitpoint. If x < a, let d = a− x and if x > b, let d = x− b. Let 0 < r < d. ThenB(x, r) ∩ S = ∅, so x 6∈ S ′. This shows that S ′ = [a, b].
(c) S =
{1
n: n ∈ N
}Since lim
n→∞
1
n= 0, it follows that 0 ∈ S ′. If x =
1
n∈ S, let 0 < r < 1
n− 1
n+1.
Then B(x, r)∩S\{ x } = ∅, and x is not a limit point. Finally, if x 6∈ S, x 6= 0,
then there exists a positive integer n such that1
n+ 1< x <
1
n. Let d =
min
{x− 1
n+ 1,
1
n− x
}and let 0 < r < d. Then B(x, r) ∩ S = ∅, and x is
not a limit point of S. Thus, S ′ = { 0 }.(d) S = Q
Since the rationals and the irrationals are both dense in R, it follows that everyreal number is a limit point of S.
(e) S = { 2−m + 5−n : m, n ∈ N }As in (c), it is clear that 0 is a limit point. For a fixed m ∈ N, the number2−m is a limit point. Similarly, for a fixed n, 5−n is a limit point.
(f) S =
{(−1)n +
1
m: m, n ∈ N
}Since (−1)n = ±1 and lim
m→∞
1
m= 0, the limit points of S are ±1.
37
(g) S =
{1
n+
1
m: m, n ∈ N
}As in (e), the numbers
1
nand
1
mare limit points for fixed values of m and n.
(h) S = { x ∈ R2 : ‖x‖ > 1 }The limit points are all the elements of S. These include all the points in R2
except those in the interior of the unit circle |x| = 1.
(i) S = { (x, y) ∈ R2 : x2 − y2 < 1 }It is clear that every element of S is a limit point, and so are its boundarypoints. Thus, S ′ = { (x, y) ∈ R2 : x2 − y2 ≤ 1 }.
(j) S = { (x, y) ∈ R2 : x > 0 }As in the preceding item, the set S, consisting of all points to the right of they-axis, is open, so the limit points of S are all the elements of S together withits boundary points. We have S ′ = { (x, y) ∈ R2 : x ≥ 0 }.
(k) S = { (x, y) ∈ R2 : x ≥ 0 }This set is the closure of the set S in the preceding section. Thus, we alsohave S ′ = { (x, y) ∈ R2 : x ≥ 0 } = S.
6. (a)⇒ (b)Assume G is open. Then every element of G is interior to G, so that G ⊂ G◦.However, we always have the inclusion G◦ ⊂ G, so it follows that G = G◦.
(b)⇒ (c)Let x ∈ G = G◦. Thus, there exists r > 0 such that B(x, r) ⊂ G, and G is aneighborhood of x. Since x is an arbitrary element of G, this shows that G is aneighborhood of each of its elements.
(c)⇒ (a)Let G be a neighborhood of each of its elements. To show that G is open, we needto show that each element of G is an interior point. To this end, let x ∈ G. SinceG is a neighborhood of each of its elements, it follows that there exists r > 0such that B(x, r) ⊂ G, so that x is an interior point of G. Since x is arbitrary, itfollows that G is open.
This shows that the three conditions are equivalent.
7. The answers given below are just examples. There are many other possible answers.
(a) an infinite subset of R with no accumulation points: The set N of naturalnumbers has no accumulation or limit points, since the elements are equallyspaced along the real line, and any element of N cannot be a limit point since aball of radius less than 1 will not contain another element of the set. If x 6∈ N,then x lies between two consecutive natural numbers n and n + 1, and if wechoose 0 < r < min{ x− n, (n + 1)− x }, then B(x, r) ∩ N = ∅. This showsthat x is not a limit point as well.
38
(b) a non-empty subset of R which is contained in its derived set: Any openinterval S = (a, b) is a subset of its derived set, which is the closed intervalS ′ = [a, b].
(c) a subset of R which has infinitely many accumulation points but which doesnot contain any of them: Let x =
√y where y is a positive integer which is
not a perfect square. Thus, x is an irrational number. Let { xn } be thesequence where xn is the rational number formed by rounding off the decimalexpansion of x to n decimal places. Let S be the set of all the terms of thesesequences. Then the elements of S are rational numbers, and the numbersx are the accumulation points of S. This shows that S has infinitely manyaccumulation points, none of which is contained in S.
8. Prove the following:
(a) A ⊂ ABy definition, A is the intersection of all the closed supersets of A, so A ⊂ A.
(b) A = A
By its definition, A is the smallest closed set which contains A. But since Ais closed, we have the desired result.
(c) A ∪B = A ∪BSince A ∪B is the smallest closed set which contains A ∪ B and A ∪ B is aclosed set containing A ∪ B, it follows that A ∪B ⊂ A ∪ B. For the reverseinclusion, we have A ⊂ (A ∪ B), so A ⊂ A ∪B. Similarly, B ⊂ A ∪B, so itmust be true that A ∪B ⊂ A ∪B. The equality then follows immediately.
(d) ∅ = ∅Clearly, ∅ ⊂ emptyset. Since ∅ is closed, it must be the smallest closed setcontaining itself, so ∅ = ∅.
9. Prove the following:
(a) x ∈ A if and only if x is either an interior point or a boundary point of A.If x is an interior point of A, then x ∈ A ⊂ A. If x ∈ ∂A, then since A isa closed set containing A, it follows that x ∈ A. Conversely, suppose x ∈ A.Since A◦∪∂A is a closed set which contains A, it follows that A ⊂ A◦∪∂A, and(A◦ ∪ ∂A)c = Ext A ⊂ (A)c. Thus, A∩Ext A = ∅, and we have A = A◦ ∪ ∂A.
(b) x ∈ A if and only if every neighborhood of x contains an element of A.From (a), x ∈ A if and only if x is either an interior point or a boundary pointof A. If x is an interior point, then x ∈ A, so that every neighborhood of xcontains an element of A. If x is a boundary point, then by definition, everyneighborhood of x contains an element of A.
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SAMPLE QUIZZES
Below you will find practice quizzes for Chapters 1 and 2. After going through thediscussion in each chapter, and trying out the exercises provided in each section, tryanswering the corresponding practice quiz. Afterwards, compare your answers withthe provided solutions.
Sample Quiz Chapter 1
Directions: Answer the following as indicated. In each case, show your completesolution.
1. Give an example of each of the follow-ing and explain your answer:
(a) A bounded infinite set with nomaximum
(b) An infinite set with no infimum
(c) An unbounded countable set
(d) A bounded uncountable set
2. In each of the following, findsupS, inf S, maxS and minS, if theyexist. Explain your answers.
(a) S =
{n− 1
n: n ∈ N
}(b) S = { x ∈ Q+ : x2 ≤ 2 }(c) S = { 2−n : n ∈ N }
3. Use the definition of the absolute valueto show that for every x ∈ R, we have−|x| ≤ x ≤ |x|.(Hint: Consider thecases x ≥ 0 and x < 0 separately.)
4. Let m, n ∈ N with n ≤ m, and letK = { k ∈ N : kn ≤ m }.
(a) Show that K is a nonempty subsetof N.
(b) Use the Archimedean property ofthe real numbers to show that Khas an upper bound in R.
(c) Let q = maxK. Show that qn ≤m < (q + 1)n.
5. Show that the set 2N of even count-ing numbers is countable by defining aone-to-one correspondence between Nand 2N.
6. Let A, B be bounded non-empty sub-sets of R. Show that A ∪ B is alsobounded.
7. Prove: The intersection of two non-disjoint countable sets is countable.
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Solutions to Practice Quiz No. 1
1. The answers provided are just sample answers. There are other possible correctanswers.
(a) A bounded infinite set with no maximum: Let S =
{1− 1
n: n ∈ N
}is a
bounded infinite set with LUB and GLB 1 and 0, respectively. Since limn→∞
1−1
n= 1 and
1
n6= 0, the supremum 1 is never attained, so the set has no
maximum.
(b) An infinite set with no infimum: Let S = Z be the set of all integers. Clearly,this infinite set has no infimum
(c) An unbounded countable set: The set S = Z of all integers is an unboundedcountable set
(d) A bounded uncountable set: The interval S = (−1, 1) is a bounded infiniteset which is uncountable.
2. (a) S =
{n− 1
n: n ∈ N
}We have supS = 1, inf S = 0. Since 0 =
1− 1
1∈ S, it follows that 0 = minS.
On the other hand,n− 1
n< 1 for all n ∈ B, it follows that S has no maximum
element.
(b) S = { x ∈ Q+ : x2 ≤ 2 }We have sup S =
√2, inf S = 0. Since 0,
√2 6∈ S, it follows that S has no
maximum or minimum element.
(c) S = { 2−n : n ∈ N }The elements of S form a decreasing sequence, so supS = maxS = 2−1 =
1
2,
while inf S = 0. There is no minimum element.
3. We consider two cases. If x ≥ 0, then |x| = x > −|x|, since −|x| < 0. If x < 0,then |x| = −x > 0⇒ x = −|x| < |x|. In both cases, we have −|x| ≤ x ≤ |x|.
4. (a) Clearly, 1 ∈ K, so K is a non-empty subset of N.
(b) Since m,n ∈ N, the Archimedean property applies, and there exists a positiveinteger r such that rn > m. For each k ∈ K, we have kn ≤ m < rn, so thatk < r for each k ∈ K. This shows that r is an upper bound for S.
(c) Since q = maxK, it follows that q ∈ K and hence qn ≤ m. Since q + 1 6∈ K,it follows that (q + 1)n > m.
5. Consider the mapping φ : N→ 2N defined by φ(n) = 2n. It can be verified that φis a one-to-one correspondence between the two sets, and hence 2N is a countableset.
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6. For each x ∈ A, we have inf A ≤ x ≤ supA. Similarly, for each y ∈ B, wehave inf B ≤ y ≤ supB. Let L = min { inf A, inf B } and let U =max { sup A, sup B }. Clearly, for any x ∈ A ∪ B, we have L ≤ x ≤ U ,and A ∪B is a bounded set.
7. Let A and B be two non-disjoint countable sets. We consider two cases. If A ∩Bis finite, then clearly it is countable. If A ∩ B is infinite, the elements of theintersection can be treated as elements of A. Since A is countable, its elementscan be put into a one-to-one correspondence with N, so that the elements can bedenoted by an, where an is the element of A associated with the positive integer n.The elements of A ∩ B can then be rearranged according to increasing subscriptsas elements of A. If ank
is the k-th element of A ∩ B, then ankis the element of
the intersection associated with the positive integer k, and A ∩B is countable.
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Sample Quiz Chapter 2
Directions: Answer the following as indicated. In each case, show your completesolution.
1. Consider the following subsets of R2:
B = { (x, y) ∈ R2 | 4 ≤ x2 + y2 ≤ 9 }C = { (x, 0) ∈ R2 | − 2 ≤ x ≤ 2 }D = { (0, y) ∈ R2 | − 2 ≤ y ≤ 2 }
E = { (x, x) ∈ R2 | −√
2 ≤ x ≤√
2 }
F = { (x,−x) ∈ R2 | −√
2 ≤ x ≤√2 }
If A = B ∪ C ∪D ∪ E ∪ F , sketch each of the following sets: (6 points each)
(a) A (b) Int A (c) Bdy A
2. Let A ⊂ Rk.
(a) Aside from its complement being open, give two other equivalent conditionsfor A to be closed. (3 points each)
(b) Use either of the conditions in (a) to show that the set A = { (x, y) ∈R2 | x2 + y2 = 4 } is closed. (6 points)
3. Let A and B be subsets of Rk. In the following table, the row headings describethe two sets A and B. Fill up the slots with either True or False to describethe sets given in the column headings. (2 points each)
A, B A ∩B A ∪B Int Aclosedopen
not open
Bonus: Select one false statement from the above table and show why it is falseby giving a counterexample. (5 points)
4. Give examples of
(a) an infinite set A in R which has no accumulation point.
(b) a non-empty subset B of R such that B ⊆ B′.
(c) a non-empty subset C of R2 such that Int C = C.
(d) a non-empty subset D of R2 such that Bdy D = D.
(e) a non-empty subset E of R2 such that every element of E is an isolatedpoint of E.
In each case, explain your answer. (6 points each)
5. Prove: A subset A of Rn is open if and only if A ∩ ∂A = ∅. (Hint: A is open ifand only if A = A◦) (6 points)
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1. Set B is the set of all points in the plane that lie between the circles of radii 2 and3 centered at the origin. Set C is the line segment along the x-axis with endpoints(−2, 0) and (2, 0), while set D is the line segment along the y-axis from (0,−2) to(0, 2). Set E consists of all points on the line y = x from the point (−
√2,−√
2)to the point (
√2,√
2), while set F consists of all points on the line y = −x fromthe point (−
√2,√
2) to the point (√
2,−√
2).
(a) The set A = B ∪ C ∪D ∪ E ∪ F is shown below:
(b) Interior of A Boundary of A
2.
A, B A ∩B A ∪B Int Aclosed True True Falseopen True True True
not open False False False
False Statement: The union of two sets which are not open is not open. Considerthe sets A = (−2, 1], B = [0, 3). The two sets are both not open, but A ∪ B =(−2, 3) is open.
3. The following are just sample answers. There are many other possible correctanswers.
(a) an infinite set A in R which has no accumulation point: The set Z of integersis an infinite subset of R which has not accumulation point. Every element ofA is an isolated boundary point, and hence cannot be an accumulation point.If x is not an integer, then x lies between two consecutive integers k and k+1.If 0 < r < min{ x− k, k + 1− x }, then B(x, r) ∩ A = ∅. Hence, x is not anaccumulation point either.
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(b) a non-empty subset B of R such that B ⊆ B′: Let B = Q. Since B is dense inR, it follows that B′ = R and hence B ⊂ B′.
(c) a non-empty subset C of R2 such that Int C = C: Since the interior of a setC is the largest open set which is contained in C, it follows that if C = C◦ ifand only if C is open. Thus, for example, C = { (x, y) ∈ R2 | x > 0, y > 0 }is the set of all points in the first quadrant and is an pen set. Thus, C◦ = C.
(d) a non-empty subset D of R2 such that Bdy D = D: Let D = { (x, y) ∈R2 | x, y ∈ Z }. Every element of D is a boundary point, so ∂D = B.
(e) a non-empty subset E of R2 such that every element of E is an isolated pointof E: Let E be the set D described in (d). Then every element of E is anisolated point.
4. Let A be an open subset of Rn. Then A = A◦, so that A ∩ ∂A = A◦ ∩ ∂A = ∅.Conversely, suppose A◦∩∂A = ∅. Let x ∈ A. Then x 6∈ ∂A. Moreover, x 6∈ Ext A,so x ∈ A◦. This shows that A ⊂ A◦. Since A◦ ⊂ A, we have A = A◦, and A is anopen set.
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