Sets and Mappings
This article presumes that readers have a basic understanding of sets and mappings; however, we will briefly review these concepts to establish clear notation While some content in this section is more appropriately placed after our discussion on natural numbers (§1.3), we include it here for the sake of convenience.
2 1 The Set of Real Numbers
Asetis a collection Aof objects, calledelements If xis an element ofA we writex∈A Ifxis not an element ofA, we writex /∈A LetA,Bbe sets.
If every element of A is an element ofB, we say Ais a subsetof B, and we write A⊂B Equivalently, we sayB is asupersetofA and we writeB ⊃A.
When we write A ⊂ B or A ⊃B, we allow for the possibility A = B, i.e.,
Theunionof setsAandBis the setCwhose elements lie inAor lie inB; we writeC=A∪B, and we sayC equalsAunionB Theintersectionof sets
AandBis the setCwhose elements lie inAand lie inB; we writeC=A∩B and we sayCequalsAinterB Similarly, one defines the unionA 1 ∪ .∪A n and the intersectionA 1 ∩ .∩A n of finitely many setsA 1 , , A n
In set theory, for any infinite collection of sets A₁, A₂, , their union, denoted as ∞ₙ₌₁ Aₙ, includes all elements that exist in at least one of the sets Conversely, their intersection, ∞ₙ₌₁ Aₙ, contains only the elements that are present in all the sets When two sets A and B have no elements in common, they are referred to as disjoint, indicated by A ∩ B being empty or A ∩ B = ∅, where ∅ represents the empty set By definition, the empty set is considered a subset of every set The set of elements in A that are not in B is expressed as A \ B = {x ∈ A: x ∉ B}, known as the complement of B in A Notably, if A is a subset of B, then A \ B is empty In many cases, A is implied from the context, leading to the notation Bᶜ for the complement of B.
We will have occasion to useDe Morgan’s law,
We leave this as an exercise Of course these also hold for finitely many sets
In set theory, the product of two sets A and B is represented as A×B, comprising all ordered pairs (a, b) where a belongs to A and b belongs to B A relation between sets A and B is defined as a subset f of A×B When this relation is a mapping, it means that for every element a in A, there is exactly one corresponding element b in B such that the pair (a, b) is included in f In this scenario, it is standard to denote this relationship as b = f(a) and express the mapping as f: A → B.
In a mapping denoted as f : A → B, the set A represents the domain while set B is known as the codomain The range of the function, expressed as f(A) = {f(a) : a ∈ A}, is a subset of B Specifically, a function is defined as a mapping where the codomain consists of real numbers, meaning the outputs of f are real-valued.
A mapping f :A →B is injectiveif f(a) =f(b) implies a =b, whereas f : A→B issurjective if every elementb of B equals f(a) for some a∈A, i.e., if the range equals the codomain A mapping that is both injective and surjective isbijective.
The Set R
If f : A → B and g : B → C are mappings, their composition is the mappingg◦f :A→Cgiven by (g◦f)(a) =g(f(a)) for alla∈A In general, g◦f =f◦g.
In mathematics, two mappings f: A → B and g: B → A are considered inverses if g(f(a)) = a for every element a in A and f(g(b)) = b for every element b in B A mapping f is deemed invertible if it possesses an inverse g Notably, a mapping f is invertible if and only if it is bijective, meaning it is both injective and surjective.
1.1.3.Show that a mappingf :A→B is invertible iff it is bijective.
1.1.4.Letf :A→Bbe bijective Show that the inverseg:B→Ais unique.
We are ultimately concerned with one and only one set, the set R of real numbers The properties ofRthat we use are
Throughout, we use ‘real’ to mean ‘real number’, i.e., an element ofR.
Thearithmetic propertiesstart with the fact that realsa,bcan beaddedto produce a reala+b, thesumofaandb The rules for addition area+b=b+a anda+ (b+c) = (a+b) +c, valid for all realsa,b, andc There is also a real
0, calledzero, satisfyinga+ 0 = 0 +a=afor all realsa, and each realahas anegative−asatisfyinga+ (−a) = 0 As usual, we writesubtractiona+ (−b) as a−b.
In the realm of real numbers, multiplication can be performed to yield a product, denoted as ab, where a and b are real numbers The fundamental properties of multiplication include commutativity, expressed as ab = ba, and associativity, illustrated by the equation a(bc) = (ab)c, which holds true for all real numbers a, b, and c Additionally, there exists a unique real number known as one, represented by 1, which satisfies the equations a1 = 1a = a for any real number a Furthermore, for every real number a that is not equal to zero, there is a reciprocal, denoted as 1/a, which fulfills the condition a(1/a) = 1 Division is commonly expressed as a(1/b) or simply a/b.
Addition and multiplication are interconnected through the property a(b+c) = ab + ac for all real numbers a, b, and c, under the assumption that 0 = 1 This leads to the conclusion that there is a unique real number, denoted as 0, which satisfies the equation 0 + a = a + 0 = a for all real numbers a If another real number also satisfied this equation, it would result in 0 = 0 + 0, reinforcing that 0 must equal itself Additionally, there exists a unique real number that represents one, with the property that 0a = 0 for all a These principles define the fundamental arithmetic properties of real numbers.
4 1 The Set of Real Numbers
The ordering properties begin with the subset \( R^+ \) of \( R \), representing the set of positive numbers that remains closed under addition and multiplication; specifically, if \( a, b \in R^+ \), then both \( a + b \) and \( ab \) are also in \( R^+ \) When a number \( a \) is positive, we denote it as \( a > 0 \) or \( 0 < a \), indicating that \( a \) is greater than zero or that zero is less than \( a \) Conversely, \( R^- \) denotes the set of negative numbers, defined as \( R^- = -R^+ \), which consists of the negatives of all elements in \( R^+ \) These foundational rules for ordering apply to the sets involved.
The sets R−, {0}, and R+ are mutually exclusive and collectively encompass all real numbers We denote a > b and b < a to indicate that a − b > 0 This implies that a is negative if and only if 0 > a, and if a > b, then adding c to both sides results in a + c > b + c Specifically, for any two real numbers a and b, one of the following holds: a < b, a = b, or a > b, which illustrates the fundamental ordering properties of real numbers.
The ordering properties establish that 0 is less than 1, indicating that 1 is positive Additionally, if a is less than b and c is greater than 0, then the product ac is less than bc Furthermore, if 0 is less than a and a is less than b, it follows that the square of a is less than the square of b Lastly, if a is less than b and b is less than c, then it can be concluded that a is less than c In this context, we use the symbols ≤ to denote less than or equal to, and ≥ for greater than or equal to, while also defining a as nonnegative if a is greater than or equal to 0, and nonpositive if a is less than or equal to 0.
In mathematics, a set of real numbers, denoted as S, has an upper bound M if every element x in S is less than or equal to M Conversely, a number m is considered a lower bound for S if m is less than or equal to every element x in S For instance, the numbers 1 and 2 serve as upper bounds for the sets J = {x: 0 < x < 1} and I = {x: 0 ≤ x ≤ 1}, while 0 and -1 act as lower bounds for these sets A set S is classified as bounded above if it possesses an upper bound and bounded below if it has a lower bound A set is deemed bounded if it is both bounded above and below.
Not every set of real numbers possesses upper or lower bounds; for instance, the set of all real numbers, R, is unbounded in both directions A notable example of an unbounded set is the natural numbers, denoted as N, which lack an upper limit.
Fig 1.1.Upper and lower bounds forA.
A set of real numbers, denoted as S, can possess multiple upper bounds When there exists an upper bound M for S that is less than or equal to any other upper bound b of S, we refer to M as the least upper bound, or supremum, of S, denoted as M = sup S.
The supremum, or least upper bound, of a set is uniquely determined when it exists, as there cannot be more than one least upper bound For instance, if M serves as an upper bound for set I, it follows that M is greater than or equal to every element x within I.
The least upper bound for the set I is 1, denoted as 1 = supI, since M is greater than or equal to 1 In contrast, for the set J, if M is less than 1, the value c = (1 + M)/2 falls between M and 1, indicating that c is an element of J Therefore, M cannot serve as an upper bound for J.
1 is the least upper bound forJ, or 1 = supJ.
A realmthat is a lower bound forSand satisfiesm≥bfor all other lower boundsb is called a greatest lower boundor an infimumorinfforS, and we
1Ifaandbare least upper bounds, then,a≤banda≥b.
1.2 The SetR 5 writem= infS Again the inf, whenever it exists, is uniquely determined As before, it follows easily that 0 = infI anf 0 = infJ.
Thecompleteness property of Rasserts that every nonempty set S ⊂ R that is bounded above has a sup, and every nonempty set S ⊂ R that is bounded below has an inf.
We introduce the symbols ∞ and −∞, representing infinity and minus infinity, respectively, with the ordering rule −∞ < x < ∞ for all real numbers x When a set S is not bounded above, we denote its supremum as sup S = ∞, and if it is not bounded below, we denote its infimum as inf S = −∞ For instance, we have sup R = ∞ and inf R = −∞, and in §1.4, we demonstrate that sup N = ∞ Additionally, it is important to note that the empty set ∅ is a subset of R.
Another convenient abbreviation is to write sup∅=−∞, inf∅=∞ Clearly, whenS is nonempty, infS≤supS.
With this terminology, the completeness property asserts thatevery subset of R , bounded or unbounded, empty or nonempty, has a sup and has an inf; these may be reals or ±∞.
It is important to note that ±∞ are not real numbers but rather useful notations The ordering properties of ±∞ indicate that −∞ is less than any real number x, which in turn is less than +∞ This understanding allows us to establish certain arithmetic properties for ±∞.
Note that we have not defined∞ − ∞, 0ã ∞,∞/∞, or c/0.
Letabe an upper bound for a setS If a∈S, we say ais amaximumof
S, and we write a= maxS For example, with I as above, maxI = 1 The max of a set Sneed not exist; for example, according to the Theorem below, maxJ does not exist.
Similarly, let a be a lower bound for a set S If a ∈ S, we say a is a minimum of S, and we writea = minS For example, minI = 0 but minJ does not exist.
Theorem 1.2.1 Let S ⊂ R be a set The max of S and the min of S are uniquely determined whenever they exist The max of S exists iff the sup of
S lies inS, in which case the max equals the sup The min of S exists iff the inf of S lies inS, in which case the min equals the inf.
6 1 The Set of Real Numbers
The first statement is derived from the second, as the supremum (sup) and infimum (inf) are uniquely defined If supS belongs to the set S, then it serves as the maximum of S Conversely, if a maximum exists in S, it must be greater than or equal to the supremum, since the supremum is the least upper bound of the set.
In mathematical analysis, the supremum (supS) serves as an upper bound for a set S, while the maximum (maxS) is an element of S This relationship establishes that maxS is less than or equal to supS, leading to the conclusion that if supS is both less than or equal to and greater than or equal to maxS, then maxS must equal supS A similar reasoning applies to the infimum, resulting in an analogous derivation.
The Subset N and the Principle of Induction
The inequality supA ≤ sup(A+B) - y indicates that supA is finite and that y must be less than or equal to sup(A+B) - supA for all y in B This establishes that sup(A+B) - supA serves as an upper bound for B Consequently, since supB is defined as the least upper bound, we derive that supB ≤ sup(A+B) - supA, leading to the conclusion that sup(A+B) is greater than or equal to supA + supB Given that we also know sup(A+B) is less than or equal to supA + supB, we can ultimately state that sup(A+B) equals supA + supB.
To verify inf(A+B) = infA+ infB, use reflection and what we just
finished to write inf(A+B) =−sup[−(A+B)] =−sup[(−A) + (−B)]
This completes the derivation of the addition property.
Every assertion that follows in this book depends only on the arithmetic, ordering, and completeness properties of R , just described.
1.2.1.Show thata0 = 0 for all reala.
1.2.2.Show that there is a unique real playing the role of 1 Also show that each real ahas a unique negative −a and each nonzero reala has a unique reciprocal.
1.2.4.Show that negative times positive is negative, negative times negative is positive, and 1 is positive.
1.2.5.Show thata < bandc∈ Rimplya+c < b+c,a < bandc >0 imply ac < bc,a < b andb < cimplya < c, and 0< a < bimpliesaa < bb.
1.2.6.Leta, b≥0 Show that a≤biffaa≤bb.
1.2.7.Verify the properties of sup and inf listed above.
1.3 The Subset N and the Principle of Induction
B S is closed under addition by 1:x∈S impliesx+ 1∈S.
For example, R + is inductive The subset N ⊂ R of natural numbers or naturalsis the intersection of all inductive subsets of R,
8 1 The Set of Real Numbers
Then, Nitself is inductive Indeed, since 1∈S for every inductive setS, we conclude that 1 ∈
{S : S ⊂ Rinductive} = N Similarly, n ∈ N implies n∈S for every inductive set S Hence,n+ 1∈S for every inductive setS. hence,n+ 1∈
{S:S⊂ Rinductive}=N This shows thatNis inductive. From the definition, we conclude thatN ⊂Sfor any inductiveS ⊂ R For example, sinceR + is inductive, we conclude thatN ⊂ R + , i.e., every natural is positive.
The set of natural numbers \( N \) is the only inductive subset of itself For instance, the set \( S = \{1\} \cup (N + 1) \) is a subset of \( N \) because \( N \) is inductive It is evident that \( 1 \in S \) Furthermore, if \( x \in S \), then \( x \in N \), which leads to \( x + 1 \in N + 1 \) and consequently \( x + 1 \in S \), confirming that \( S \) is inductive Therefore, we conclude that \( S = N \) or \( \{1\} \cup (N + 1) = N \), indicating that \( n - 1 \) is a natural number for every natural number \( n \) other than 1.
The conclusions above are often paraphrased by sayingN is the smallest inductive subset of R ,and they are so important they deserve a name.
Theorem 1.3.1 (Principle of Induction) If S ⊂ R is inductive, then,
In this article, we demonstrate that there are no natural numbers between 1 and 2 by defining the set S as {1} ∪ {n ∈ N : n ≥ 2} Since 1 is included in S, we consider two cases for any n in S: if n = 1, then n + 1 = 2 is also in S; if n ≥ 2, then n + 1 > n ≥ 2, confirming that n + 1 is a natural number and belongs to S Therefore, S is an inductive set and a subset of natural numbers, leading us to conclude that S equals N Consequently, this proves that for all natural numbers n, there are no integers between n and n + 1.
Nis closed under addition and multiplication by any natural To see this,
fix a natural n, and let S = {x: x+n∈ N }, so, S is the set of all reals x whose sum withnis natural Then, 1∈Ssincen+ 1∈ N, andx∈S implies x+n∈ N implies (x+n) + 1 = (x+ 1) +n∈ N impliesx+ 1∈S Thus,
The set N is inductive and is the smallest set satisfying this property, which implies that N is a subset of S and that for all m in N, the sum m + n is also in N Therefore, we conclude that N is closed under addition, denoted as N + N ⊂ N A similar argument can be made for closure under multiplication, which is left as an exercise.
In the sequel, when we apply the principle of induction, we simply say ‘by induction’.
To show that a given setSis inductive, one needs to verifyAandB Step
B is often referred to as the inductive step, even though, strictly speaking, induction is bothAandB, because, usually, most of the work is in establishing
B Also, the hypothesis in B, x ∈ S, is often referred to as the inductive hypothesis.
Induction can be illustrated through the classification of natural numbers as even or odd A natural number is defined as even if it belongs to the set 2N = {2n : n ∈ N}, while a natural number is odd if it can be expressed as n + 1, where n is even We assert that every natural number is either even or odd To demonstrate this, let S represent the union of the set of even natural numbers and the set of odd natural numbers For instance, 2 can be expressed as 2 × 1, confirming that it is indeed even.
1 is odd Hence, 1∈S Ifn∈S and n= 2k is even, then,n+ 1 is odd since(n+ 1) + 1 =n+ 2 = 2k+ 2 = 2(k+ 1) Hence,n+ 1∈S Ifn∈S andnis
The subset S is shown to be inductive by demonstrating that if n belongs to S, then n + 1 also belongs to S, confirming that S is closed under addition by 1 Consequently, we conclude that S equals the set of natural numbers N This implies that every natural number is either even or odd, and the established parity rules, such as the sum of two even numbers being even, are upheld.
Let A be a nonempty set with n elements if there exists a bijection between A and the set {k ∈ N: 1 ≤ k ≤ n}, commonly represented as {1, 2, , n} Conversely, if A is empty, we define the number of elements in A as zero.
A set A is considered finite if it contains a specific number of elements, denoted as n; otherwise, it is classified as infinite Key implications of this definition include that for disjoint sets A and B, with n and m elements respectively, their union A∪B will have n+m elements Additionally, if A is a finite subset of the real numbers R, both the maximum (max A) and minimum (min A) values will exist Notably, we denote the larger and smaller values between two numbers a and b as max(a, b) and min(a, b) Interestingly, max A and min A can also exist for infinite subsets of R According to Theorem 1.3.2, if S is a nonempty subset of the natural numbers N, then the minimum value of S (min S) will exist.
The infimum \( c \) of a set \( S \) is finite because \( S \) is bounded below Since \( c + 1 \) is not a lower bound, there exists an element \( n \in S \) such that \( c \leq n < c + 1 \) If \( c = n \), then \( c \) is the minimum of \( S \) Alternatively, if \( c < n \), it follows that \( n - 1 < c < n \), indicating that \( n \) cannot be a lower bound for \( S \) This implies \( n > 1 \), and there must be an element \( m \in S \) that lies between \( n - 1 \) and \( n \), but there are no natural numbers in that interval.
The two other subsets, mentioned frequently, are theintegers Z=N ∪{0}∪
Zis is closed under subtraction, while Qis is closed under all four arithmetic operations except for division by zero In the case of natural numbers, the integers in the set 2Z={2n:n∈ Z} are classified as even, and an integer n is considered odd if n + 1 is even.
To establish the function f: N → R, we utilize induction to demonstrate that f(1) = a and f(n + 1) = a f(n) for all n, leading us to define f(n) = a^n Consequently, we find that a^1 = a and a^(n + 1) = a^n a for all n Moreover, since the set {n ∈ N : (ab)^n = a^n b^n} is inductive, it follows that (ab)^n = a^n b^n for all n in N.
Now, (−1) n is 1 or−1 according to whethern∈ N is even or odd, a >0 implies a n >0 for n ∈ N, and a > 1 implies a n >1 for n ∈ N These are easily checked by induction.
If a = 0, we extend the definition of a n to n ∈ Z by setting a 0 = 1 and a −n = 1/a n for n ∈ N Then (Exercise 1.3.10), a n+m = a n a m and (a n ) m =a nm for all integersn,m.
Let a > 1 Then, a n = a m with n, m ∈ Z only when n = m Indeed, n−m ∈ Z, and a n − m =a n a − m =a n /a m = 1 But a k >1 for k∈ N, and a k = 1/a − k jand odd integersp,q, then,q= 2 k − j p= 2ã2 k − j − 1 p is even, a contradiction On the other hand, ifj > k, then,pis even Hence, we must have k=j This establishes the uniqueness ofk.
To demonstrate the existence of \( k \), we can assume \( n \in \mathbb{N} \) and, if necessary, multiply by a negative value For odd \( n \), we set \( k = 0 \) and \( p = n \) In the case of even \( n \), we define \( n_1 = n/2 \), which is a natural number less than \( 2n - 1 \) If \( n_1 \) is odd, we then assign \( k = 1 \) and \( p = n_1 \).
If \( n_1 \) is even, then \( n_2 = n_1 / 2 \) is a natural number less than \( 2^{n-2} \) If \( n_2 \) is odd, we set \( k = 2 \) and \( p = n_2 \) If \( n_2 \) is even, we continue dividing \( n_2 \) by 2 This iterative process generates a sequence of natural numbers \( n_1, n_2, \ldots \) such that \( n_j < 2^{n-j} \) Since this procedure completes in fewer than \( n \) steps, there exists some natural number \( k \) or 0 for which \( p = n / 2^k \) is odd.
The Completeness Property
The set of natural numbers, denoted as N, does not have an upper bound If N were to have an upper bound, denoted as c, it would imply that c is the least upper bound or supremum However, since c−1 is not an upper bound, there exists a natural number n (where n ≥ 1) that exceeds c−1, resulting in n + 1 being greater than c and also belonging to N This situation contradicts the definition of c as an upper bound Therefore, we conclude that N is not bounded above, and in mathematical terms, the supremum of N is infinity.
Let S = {1/n : n ∈ N } be the reciprocals of all naturals Then, S is bounded below by 0, hence,Shas an inf We show that infS= 0 First, since
The definition of the infimum indicates that 0 serves as a lower bound, meaning that infS is always greater than or equal to 0 Additionally, for any positive constant c, we can find a natural number k such that k exceeds 1/c, given that the supremum of natural numbers is infinite By multiplying this inequality by the positive value c/k, we derive that c must be greater than 1/k Since 1/k belongs to the set S, this demonstrates that c cannot be a lower bound for S Therefore, it follows that any proposed lower bound for S must be reconsidered.
S must be less or equal to 0 Hence, infS= 0.
The two results just derived are so important we state them again.
As a consequence, sinceZ ⊃ N, it follows that supZ=∞ SinceZ ⊃(− N) and inf(A) =−sup(−A), it follows that infZ ≤inf(− N) =−supN=−∞, hence, infZ=−∞.
Aninterval is a subset ofRof the following form:
Intervals of the form (a, b), (a,∞), (−∞, b), (−∞,∞) areopen, whereas those of the form [a, b], [a,∞), (−∞, b] are closed When −∞ < a < b a}, (−∞, b] ={x:x≤b}, and (−∞,∞) =R.
Forx∈ R, we define|x|, theabsolute valueofx, by
Then, x≤ |x|for all x, and, fora >0,{x:−a < x < a} ={x:|x|< a} {x:x < a} ∩ {x:x >−a},{x:x a}={x:|x|> a}. The absolute value satisfies the following properties:
14 1 The Set of Real Numbers
We leave the first two as exercises The third, the triangle inequality, is derived using |x| 2 =x 2 as follows:
Since a ≤ b iff a 2 ≤ b 2 for a, b nonnegative (Exercise 1.2.6), the triangle inequality is established.
Frequently, the triangle inequality is used in alternate forms, one of which is
This follows by writing|x|=|(x−y) +y| ≤ |x−y|+|y| and transposing|y| to the other side Another form is
We show how the completeness property can be used to derive the existence of√
Let S be the set defined as {x : x ≥ 1 and x² < 2} Since 1 is an element of S, it is nonempty For any x in S, we find that x must satisfy x = x₁ ≤ x ≤ x₂ < 2, indicating that S is bounded above by 2 Consequently, S has a supremum, which we denote as a We assert that a² = 2 To support this assertion, we eliminate the possibilities of a² being less than 2 or greater than 2, leading us to the conclusion that a² must indeed equal 2, acknowledging that every real number is either positive, negative, or zero.
So, suppose thata 2 2 since, forb andxpositive, b 2 > x 2 iffb > x.
Now suppose thata 2 >2 Then,b= (a 2 −2)/2ais positive, hence, there is a naturalnsatisfying 1/n < bwhich impliesa 2 −2a/n >2 Hence,
=a 2 −2a n + 1 n 2 >2, so,a−1/nis an upper bound forS This shows thatais not theleastupper bound, contradicting the definition ofa Thus, we are forced to conclude that a 2 = 2
A reala satisfyinga 2 = 2 is called asquare root of 2 Since (−x) 2 =x 2 , there are two square roots of 2, one positive and one negative From now on, the positive square root is denoted √
2 Similarly, every positivea has a positive square root, which we denote√ a In the next chapter, after we have developed more material, a simpler proof of this fact will be derived.
More generally, for everyb >0 andn≥1, there is a uniquea >0 satisfying a n =b, thenth roota=b 1/n ofb Now, forn≥1,k≥1, andm∈ Z,
= (b m ) k =b mk , hence, by uniqueness of roots, (b m ) 1/n = (b mk ) 1/nk Thus, forr=m/nratio- nal, we may setb r = (b m ) 1/n , defining rational powers of positive reals. Since√
2∈ Q,R \ Qis not empty The reals inR \ Qare theirrationals.
In fact, both the rationals and the irrationals have an interlacing ordensity property.
Theorem 1.4.3 If a < b are any two reals, there is a rational s between them, a < s < b, and there is an irrationalt between them,a < t < b.
To see this, first, choose a natural n satisfying 1/n < b−a Second let
S ={m ∈ N :na < m}, and let k = infS = minS Since k ∈ S, na < k.
For the second assertion, choose a naturalnsatisfying 1/n√
2na < m}, and letk= minT Since k∈T,k >√
Approximation of reals by rationals is discussed further in the exercises.
1.4.1.Show thatx≤ |x|for all x and, fora >0, {x: −a < x < a} ={x:
16 1 The Set of Real Numbers
1.4.2.For all x ∈ R, |x| ≥ 0, |x| > 0 if x = 0, and |x| |y| = |xy| for all x, y∈ R.
1.4.3.By induction, show that|a 1+a 2+ã ã ã+a n | ≤ |a 1 |+|a 2 |+ã ã ã+|a n | forn≥1.
1.4.4.Show that everya >0 has a unique positive square root.
1.4.5.Show thatax 2 +bx+c= 0, a= 0, has two, one, or no solutions inR according to whether b 2 −4ac is positive, zero, or negative When there are solutions, they are given byx= (−b±√ b 2 −4ac)/2a.
1.4.6.By induction, show that (1+a) n ≤1+(2 n −1)aforn≥1 and 0≤a≤1. Also show that (1 +a) n ≥1 +naforn≥1 anda≥ −1.
1.4.7.Fora, b≥0, show that a n ≥b n iffa≥b Also show that everyb >0 has a unique positiventh root for all n≥1 (use Exercise 1.4.6and modify the derivation for√
1.4.8.Show that the realtconstructed in the derivation of Theorem 1.4.3 is irrational.
1.4.9.Let a be any real Show that, for each > 0, no matter how small, there are integersn= 0,msatisfying a−m n
(Let {a} denote the fractional part of a, consider the sequence {a}, {2a},
{3a}, , and divide [0,1] into finitely many subintervals of length less than
Since there are infinitely many terms in the sequence, at least 2 of them must lie in the same subinterval.)
(Consider the two cases |a−m/n| ≥ 1 and |a−m/n| ≤1, separately, and look at the minimum ofn 2 |f(m/n)|with f(x) =x 2 −2.)
2 Then,ais irrational, and there is a positive realc satisfying a−m n
(Factorf(a) =a 4 −2a 2 −1 = 0, and proceed as in the previous exercise.)
1.4.12.Forn∈ Z \ {0}, define|n| 2= 1/2 k where kis the number of factors of 2 inn Also define|0| 2= 0 Forn/m∈ Qdefine|n/m| 2=|n| 2 /|m| 2 Show that | ã | 2:Q → Ris well defined and satisfies the absolute value properties
Sequences and Limits
A sequence of real numbers is defined as a function f: N → R, typically represented as (a_n), where a_n = f(n) denotes the nth term of the sequence For instance, the sequences derived from the formulas a_n = n, b_n = 2n, c_n = 2^n, and d_n = 2^(-n) + 5n illustrate this concept with their respective sequences (a_n), (b_n), (c_n), and (d_n) While we will later explore sequences of sets (Q_n) and sequences of functions (f_n), this discussion focuses solely on sequences of real numbers.
It is essential to differentiate between a sequence, denoted as (a_n) or the function f, and the set {a_n}, which represents the range f(N) of f A sequence is an ordered collection, such as (a_1, a_2, a_3, ), rather than just a set like {a_1, a_2, a_3, } For convenience, sequences can also start from the index n=0, treating them as functions on N ∪ {0}; for instance, the sequence (1, 2, 4, 8, ) can be expressed as a_n = 2^n for n ≥ 0 Specific examples of sequences are typically constructed through induction, as shown in Exercise 1.3.9, though we will not reiterate that construction for each sequence discussed.
This section focuses on the behavior of sequences as the index \( n \) approaches infinity, commonly known as the "limiting behavior" of sequences For instance, we can analyze specific sequences to illustrate this concept.
(d n ) = (2,3/2,17/12,577/408, ), where, in the last 4 sequence,d 1 = 2, d 2 = (d 1 + 2/d 1 )/2,d 3 = (d 2 + 2/d 2 )/2, d 4 = (d 3 + 2/d 3 )/2, and so on What are the limiting behaviors of these sequences?
As n increases, the sequence (a n) is organized in ascending order, with a n remaining less than or equal to 1 for all n ≥ 1 Notably, as n becomes sufficiently large, the terms a n (n−1)/n = 1−1/n approach 1, indicating that the supremum of the sequence is 1 Therefore, it is reasonable to conclude that the limit of the sequence (a n) is equal to 1.
The sequence \( (b_n) \) oscillates between 1 and -1, indicating that it does not converge to a single real number Instead, it appears to have two distinct limits: 1 and -1.
The third sequence exhibits a subtle behavior, as it is established that √x < x for x > 1, leading to a decreasing arrangement of terms This suggests that the sequence (c_n) converges towards its lower limit, denoted as L = inf{c_n : n ≥ 1} While this convergence is ultimately confirmed, the precise nature of this approach is not immediately evident.
3This notion makes sense for finite sets also: Afinite sequence(a 1 , , a n ) of reals is a functionf:{1, , n} → R.
4Decimal notation, e.g., 17 = (9 + 1) + 7, is reviewed in the next section.
18 1 The Set of Real Numbers
The limiting behavior of the fourth sequence remains unclear, although computing the first nine terms suggests it converges rapidly However, due to the approximate nature of these computations, we cannot definitively establish a single real number as the limit of the sequence (d_n) For further exploration of the sequence (d_n), refer to Exercise 1.5.12 and Exercise 1.6.5.
It is important to realize that
To compute the limit, we often encounter varying complexities in mathematical situations In simpler cases, such as sequences (a n) or (b n), the concept of "limit" may seem intuitive and require no detailed explanation However, relying on an ad hoc approach can lead to challenges, prompting mathematicians to adopt a more systematic method This shift towards a universal definition of "limit" has proven to be highly effective and is now the standard practice in mathematics.
In this article, we explore the concept of limits in two phases: initially focusing on monotone sequences and subsequently addressing general sequences To handle cases where sequences may possess multiple limits, we introduce the auxiliary concept of a "limit point" as outlined in Exercise 1.5.9 We will now proceed with the formal development of these concepts.
A sequence \((a_n)\) is defined as decreasing if \(a_n \geq a_{n+1}\) for all natural numbers \(n\) When the infimum of the sequence, denoted as \(L = \inf\{a_n : n \geq 1\}\), is approached as \(n\) approaches infinity, we express this as \(a_n \to L\) Similarly, a sequence is increasing if \(a_n \leq a_{n+1}\) for all \(n \geq 1\), and if the supremum \(L = \sup\{a_n : n \geq 1\}\) is approached, we also write \(a_n \to L\) In both scenarios, we can refer to the limit of the sequence \((a_n)\) as \(L\), denoting it as \(\lim_{n \to \infty} a_n = L\).
Note that since sups and infs are uniquely determined, we say ‘the limit’ instead of ‘a limit’ Thus, n∞ lim
1.5 Sequences and Limits 19 since sup{n 2 :n≥1}=∞.
A sequence is considered monotone if it is either increasing or decreasing, allowing for the definition of limits for all monotone sequences Additionally, a sequence is termed constant if it exhibits both increasing and decreasing behavior, taking the form (a, a, ), where 'a' represents a fixed real number.
If a monotone sequence \( (a_n) \) approaches a nonzero limit \( a \), there exists a natural number \( N \) such that \( a_n = 0 \) for all \( n \geq N \) Assuming \( (a_n) \) is an increasing sequence and \( a > 0 \), it follows that \( a = \sup\{a_n : n \geq 1\} \), indicating that \( a/2 \) cannot be an upper bound for \( (a_n) \) Consequently, there exists a natural number \( N \) such that \( a_N > a/2 > 0 \) Since the sequence is increasing, it can be concluded that \( a_n \geq a_N > 0 \) for all \( n \geq N \).
(a n ) is increasing anda