Real Numbers
Chapter 1
Real Numbers 1.1 1.1.1
The Completeness Property of R Suprema and Infima
Firstly, we shall require a definition for the upper and lower bound for a set of real numbers. Definition 1. Let S be a nonempty subset of S. 1. The set S is bounded above if there exists a number u ∈ R, called the upper bound of S, such that for every s ∈ R we have that s ≤ u. 2. The set S is bounded below if there exists a number w ∈ R, called the lower bound of S, such that for every s ∈ R we have that s ≥ w. 3. A set is bounded if it is bounded both above and below. A set is unbounded if it is not bounded. It may be worth stressing that a set bounded either above or below, but not above and below, is still unbounded. Also, if an upper bound u exists on a set, then the set will have infinitely many upper bounds, as u plus some arbitrary real number is still an upper bound. The same applies to lower bounds. The points of interest, however, are the least elements of the set of upper bounds, and the greatest elements of the set of lower bounds. Definition 2. Let S be a nonempty subset of R. 1. The upper bound u of S is a supremum (least upper bound) if it satisfies (a) u is an upper bound of S. (b) u ≤ v where v is any upper bound of S. 2. The lower bound w is an infimum (greatest lower bound) if it satisfies (a) w is a lower bound of S. (b) t ≤ w where w is any lower bound of S. Since the supremum and infimum are unique we can speak unambiguously about the supremum or infimum from here on. The supremum and infimum of a set S will be denoted by sup S,
inf S
We may make a number of equivalent statements about the upper bound u of a set S. Thus we may also make a number of alternate formulations for the supremum of a set. Only two such formulations will be made, and are presented in the two lemmas that follow. 1
Lemma 1. A number u is a supremum of a nonempty subset S of R if and only it satisfies 1. s ≤ u for all s ∈ S 2. If v < u, then there exists an s0 ∈ S such that v < s0 . Lemma 2. An upper bound u of a nonempty subset S of R is the supremum of S if and only if for every ε > 0 there exists an sε ∈ S such that u − ε < sε . Proof. Let u = sup S, where S is nonempty, and let ε > 0. Then u − ε < u, and since u is the supremum of S it follows that u−ε cannot be an upper bound of S. Hence, there exists some sε ∈ S such that u−ε < sε ≤ u. Conversely, suppose that for every ε > 0 there exists some sε ∈ S such that u − ε < sε . Let v < u so that we may put ε := u − v. Then v = u − ε, and consequently v < sε . Hence, v is not an upper bound of S, and u = sup S. Axiom 1 (Completeness Property of R). Every nonempty subset of R that has an upper bound also has a supremum in R. This is the Supremum Property of R. Analogously, every nonempty subset of R that has a lower bound also has an infimum in R.
1.2 1.2.1
Applications of the Supremum Property Functions
We consider the range of a function when talking about the upper or lower bound applied to a function. A function f : D → R, where D is the domain of f , is bounded above if the set f (D) = {f (x) : x ∈ D} is bounded above in R. Essentially, if f is bounded above then there exists some B ∈ R such that f (x) ≤ B for all x ∈ D. Similarly, the function f is bounded below if f (D) is bounded below. The function f is bounded if f (D) is bounded above and below; that is, there exists some B ∈ R such that kf (x)| ≤ B for all x ∈ D.
1.2.2
The Archimedian Property
Theorem 1 (Archimedean Property). If x ∈ R, then there exists nx ∈ N such that x ≤ nx . Proof. Suppose, to the contrary, that n ≤ x for all n ∈ N. Then x is an upper bound for N, and by the Completeness Property, N has a supremum u ∈ R. But u − 1 < u, so that u − 1 is not an upper bound of N, and so there must exist some m ∈ N such that u − 1 < m. But this implies that u < m + 1 ∈ N, which contradicts that u is an upper bound. Hence, our assertion that n ≤ x for all n ∈ N is false, and we have that x ≤ n for all n ∈ N. The Archimedean Property implies that the set of natural numbers N is not bounded.1 Corollary 1. If S := {1/n : n ∈ N}, then inf S = 0. Proof. Since S 6= ∅ is bounded below by 0, we know from the Completeness Property that it has an infimum w := inf S ≥ 0. By the Archimedean Property, for any ε > 0 there exists n ∈ N such that 1/ε < n, which implies that 1/n < ε. Hence 0 ≤ w ≤ 1/n < ε Since ε > 0 is arbitrary, w := inf S = 0. Corollary 2. If t > 0, there exists nt ∈ N such that 0 < 1/nt < t. 1 The
set of naturals is unbounded because it is not bounded above, but we can find a lower bound of the natural numbers
2
Proof. Since inf{1/n : n ∈ N} = 0 and t > 0, then t is not the lower bound of the set {1/n : n ∈ N}. Thus, there exists nt ∈ N such that 0 < 1/nt < t. Corollary 3. If y > 0, there exists ny ∈ N such that ny − 1 ≤ y ≤ ny Proof. Let Ey := {m ∈ N : y < m} ⊆ N be the subset of reals greater than some y. The Archimedean Property ensures that this set is nonempty as there must be at least one m such that y < m. Also, the Well-Ordering Property of N ensures that Ey has a least element, say ny . But then ny − 1 ∈ / Ey , so that ny − 1 ≤ y < n y
1.2.3
Density of Rational Numbers
It can be shown that there exists a positive real number x such that x2 = 2. This is the positive square root of 2, which is an irrational number. Since the existence of at √ least one irrational number is proven by this result, we ask whether other irrational numbers apart from 2 exist. In fact, there are more irrational numbers than rational numbers since the set of rationals is countable and the set of irrationals is uncountable (not shown yet). The set of rationals is said to be dense in R in the sense that given any two real numbers there exists a rational number between them. Theorem 2 (The Density Theorem). If x, y ∈ R and x < y, then there exists an r ∈ Q such that x < r < y. Proof. Assume, without loss of generality, that x > 02 . Since x < y we have that y − x > 0, and it follows that there must exist some n ∈ N such that 0 < 1/n < y − x. Thus nx + 1 < ny. But for nx > 0, there must exist some m ∈ N such that m − 1 ≤ nx < m, from which we obtain m ≤ nx + 1 < m + 1. Hence we have nx < m < ny, and with r := m/n, we have our final result x < r < y. The density of irrational numbers in R can also be shown: Corollary 4. If x, y ∈ R and x < y, then there exists an irrational number z such that x < z < y. √ √ Proof. Applying the Density Theorem to the real numbers x/ 2 and y/ 2, we obtain a rational r 6= 0 such that x y √ a},
(−∞, b) := {x ∈ R : x < b}
Here, the set (a, ∞) has no upper bound, while the set (−∞, b) has no lower bound. The infinite closed intervals are the sets [a, ∞) := {x ∈ R : x ≥ a},
1.3.1
(−∞, b] := {x ∈ R : x ≤ b}
Characterization of Intervals
If two points x, y, where x < y, belong to some interval I, then any point t, where x < t < y, belongs to I. Extending this idea, if x, y belong to I then the interval [x, y] belongs to I. A subset of R possessing this property must then be an interval. Theorem 3 (Characterization Theorem). If S ⊆ R such that it contains at least two points with the property x, y ∈ S,
x < y ⇒ [x, y] ⊆ S
then S is an interval.
1.3.2
Nested Intervals
A sequence of intervals In , where n ∈ N, is nested if . . . ⊆ In+1 ⊆ In ⊆ . . . ⊆ I2 ⊆ I1 An important property of R is that every nested sequence of closed bounded intervals in R share a common point. Theorem 4 (Nested Intervals Property). If In = [an , bn ], where n ∈ N, is a nested sequence of closed bounded intervals, then there exists some ξ ∈ R such that ξ ∈ In for all n. Proof. Since the intervals are nested we have In ⊆ I1 for all n ∈ N, and so an ≤ b1 for all n ∈ N. The nonempty set {an : n ∈ N} is then bounded above, and we let ξ := sup{an : n ∈ N}. Consequently, an ≤ ξ for all n ∈ N. We claim that ξ ≤ bn for all n. This is established by showing that for a particular n, the number bn is an upper bound of the set {ak : k ∈ N}. There are two cases to consider: 1. If n ≤ k then since Ik ⊆ In , we have ak ≤ bk ≤ bn . 2. If k < n, then since In ⊆ Ik , we have ak ≤ an ≤ bn . Thus we conclude that ak ≤ bn for all k, so that bn is an upper bound of the set {ak : k ∈ N}. Hence, an ≤ ξ ≤ bn for each n ∈ N, we have that ξ ∈ In for all n ∈ N. Theorem 5. If In := [an , bn ], where n ∈ N, is a nested sequence of closed bounded intervals such that the lengths bn − an of In staisfy inf{bn − an : n ∈ N} = 0 then the number ξ contained in In for all n is unique. 4
1.3.3
The Uncountability of R
Theorem 6. The set R is uncountable. Proof. It is sufficient to show that the unit interval I := [0, 1] is an uncountable set as this would imply that R is an uncountable set3 . Suppose, to the contrary, that I is countable. Then we can enumerate the set as I = {x1 , x2 , . . . , xn , . . .}. Select first a closed subinterval I1 ⊆ I such that x1 ∈ / I1 . Then select a closed subinterval I2 ⊆ I1 such that x2 ∈ / I2 , and so on until we obtain the closed intervals I1 ⊆ I2 ⊆ · · · ⊆ In ⊆ · · · such that In ⊆ I and xn ∈ / In for all n ∈ N. But the Nested Intervals Property implies that there exists ξ ∈ I such that ξ ∈ In for all n, and since xn ∈ / In for all n, ξ 6= xn for all n ∈ N. The enumeration is not a complete listing of the elements of I as claimed. Hence, I is an uncountable set.
3 If
R were countable, then the subset I would be countable
5
6
Chapter 2
Sequences and Series 2.1 2.1.1
Sequences and Their Limits Sequences
Definition 3. A sequence of real numbers (a sequence in R) is a function whose domain is defined on the set N = {1, 2, . . .}, and whose range is contained in the set R. In other words, a sequence X in R is a function X : N → R that assigns to every natural number n = 1, 2, . . . a uniquely determined real number xn . The values of xn are the terms (or elements) of the sequence. A sequence may be denoted by any one of the following notations: X,
(xn ),
(xn : n ∈ N)
It should be noted that the notation (xn : n ∈ N) denotes an ordered set, while the notation {xn : n ∈ N} denotes an unordered set in the range of the sequence. Sequences may be listed, but are more frequently defined by specifying a formula for the general nth term xn .
2.1.2
The Limit of a Sequence
The limit of a sequence is the most basic of limit concepts in real analysis. Definition 4. A sequence X = (xn ) in R is said to converge to a limit x ∈ R if for every ε > 0 there exists some K(ε) ∈ N such that for all n ≥ K(ε) |xn − x| < ε If a sequence X = (xn ) has a limit x, then we denote this by lim X = x,
or
lim(xn ) = x,
or xn → x
If a sequence has a limit, then the sequence is said to be convergent. Otherwise, if the limit of the sequence does not exist, then the sequence is said to be divergent. Now that we have established the existence of a limit for a convergent sequence, the next question to ask is whether or not the limit of a particular convergent sequence is unique; that is, whether the sequence has only one limit or possibly many limits. Theorem 7 (Uniqueness of Limits). A sequence in R has at most one limit. 7
Proof. Suppose that x0 and x00 are both limits the the sequence (xn ) in R. It is sufficient to show that x0 = x00 . It follows from the definition that for every ε > 0 there exists a K 0 such that |xn − x0 | < ε/2 for all n ≥ K 0 , and there exists a K 00 such that |xn − x00 | < ε/2 for all n ≥ K 00 . Let K > K 0 , K 00 so that for all n ≥ K we have |xn − x0 | < ε/2 and |xn − x00 | < ε/2. Applying the Triangle Inequality |x0 − x00 | = |x0 − xn + xn − x00 | ≤ |x0 − xn | + |xn − x00 |
0 is arbitrary, we conclude that x0 − x00 = 0, and hence x0 = x00 . There are several ways of saying that a sequence (xn ) converges to x: Theorem 8. Let X = (xn ) be a sequence in R, and let x ∈ R. Then the following statements are equivalent: 1. X converges to x. 2. For every ε > 0, there exists a K ∈ N such that for all n ≥ K, the terms xn satisfy |xn − x| < ε. 3. For every ε > 0 there exists a K ∈ N such that for all n ≥ K, the terms xn satisfy x − ε < xn < x + ε. 4. For every ε-neighbourhood Vε (x) := {u ∈ R : |u − x| < ε} of x, there exists a K ∈ K such that for all n ≥ K, the terms xn belong to Vε (x). The final statement of the previous theorem is equivalent to saying that for every ε-neighbourhood Vε (x) of x, all but a finite number of terms x1 , x2 , . . . , xK−1 of X belong to Vε (x). Those terms belonging to Vε (x) are xK , xK+1 , xK+2 , . . .. Clearly, there are infinitely many of them. To show that that a sequence X = (xn ) does not converge to a limit x ∈ R, it is sufficient to show that there is at least one ε0 > 0 such that irrespective of how large K is chosen, there exists an nK satisfying nK ≥ K such that |xNK − x| ≥ ε0 .
2.1.3
Tails of a Sequence
If the first m ∈ N terms of a sequence X are removed, then the resulting sequence Xm will converge if and only if the original sequence X converges. This idea will be expressed in a theorem, but we shall first define the m-tail Xm of a sequence X: Definition 5. If X = (x1 , x2 , . . . , xn , . . .) is a sequence in R, and if m ∈ N is given, then the m-tail of X is the sequence Xm := {xm+n : n ∈ N} = (xm+1 , xm+2 , . . .) Now we show that the m-tail Xm of a sequence X converges if and only if the original sequence X converges: Theorem 9. Let X = (xn : n ∈ N) be a sequence in R, and let m ∈ N. Then the m-tail Xm = (xm+n : n ∈ N) of X converges if and only if X converges, and converges to the limit lim Xm = lim X.
2.1.4
Further Examples
Theorem 10. Let (xn ) be a sequence in R, and let x ∈ R. If (an ) is a sequence of positive reals with lim(an ) = 0, and if there exists some constant C > 0 such that for some m ∈ N |xn − x| ≤ Can , then lim(xn ) = x. 8
for all n ≥ m
Proof. Let ε > 0. Since lim(an ) = 0 there exists K ∈ N such that for all n ≥ K an = |an − 0|
0 is arbitrary, it follows that lim(xn ) = x.
2.2
Limit Theorems
Definition 6. A sequence X = (xn ) in R is bounded if there exists a real number M > 0 such that |xn | ≤ M for all n ∈ N. In other words, a sequence (xn ) is bounded if and only if the set of terms {xn : n ∈ N} is a bounded subset of R. It should be evident that a convergent sequence is bounded. Theorem 11. A convergent seqience of real numbers is bounded. Proof. Suppose that lim(xn ) = x. Let ε := 1. Then there exists K = K(1) such that |xn − x| < 1 for all n ≥ K. Then, if n ≥ K, and using the Triangular Ineqality |xn | = |xn − x + x| ≤ |xn − x| + |x| < 1 + |x| Thus, we can set M := sup{|x1 |, |x2 |, . . . , |xK−1 |, 1 + |x|} and it follows that |xn | ≤ M for all n ∈ N. We may apply the operations of addition, subtraction, multiplication, and division to sequences of real numbers. Definition 7. Let X = (xn ) and Y = (yn ) be sequences of real numbers. Then 1. X ± Y := (xn ± yn ) 2. XY := (xn yn ) 3. X/Y := (xn /yn ) 4. cX := (cxn ) for c ∈ R Applying these arithmetic operations to convergent sequences result in new sequences whose limits can be determined. Theorem 12. 1. Let X = (xn ) and Y = (yn ) be sequences in R that converge to limits lim X = x and lim Y = y respectively, and let c ∈ R. Then the sequences X ± Y , XY , and cX converge to the limits x ± y, xy, and cx respectively. 2. If X = (xn ) converges to x and Z = (zn ) is a sequence of nonzero real numbers that converges to z 6= 0, then the quotient sequence X/Z converges to x/z. We can extend these results by Mathematical Induction to a finite number n of convergent sequences A = (zn ), B = (bn ), . . . , Z = (zn ). Thus, the sum A + B + · · · + Z = (an + bn + · · · + zn ) is also a convergent sequence with limit lim(an + bn + · · · + zn ) = lim(an ) + lim(bn ) + · · · + lim(zn ) Similar results may be obtained for the product of a finite number n of convergent sequences. A convergent sequence, each of whose terms is positive, will converge to a limit which is also positive. 9
Theorem 13. If X = (xn ) is a convergent sequence of real numbers, and if xn ≥ 0 for all n ∈ N, then lim(xn ) ≥ 0. Proof. Suppose, to the contrary, that the sequence (xn ) with xn converges with limit x = lim(xn ) < 0. Then ε := −x is positive, and since X converges to x, there exists a K ∈ K such that x − ε < xn < x + ε for all
n≥K
In particular, for n = K we have xK < x + ε = x + (−x) = 0. But this contradicts the hypothesis that xn ≥ 0 for all n ∈ N. This contradiction implies that the assumption x < 0 is incorrect, and we conclude that x ≥ 0. Theorem 14. If X = (xn ) and Y = (yn ) are convergent sequences in R with xn ≤ yn for all n ∈ N, then lim(xn ) ≤ lim(yn ). Theorem 15. If X = (xn ) is a convergent sequence and a ≤ xn ≤ b for all n ∈ N, then a ≤ lim(xn ) ≤ b. The next result, known as the Squeeze Theorem is an important result. Essentially, it says that that if a sequence is squeezed between two sequences that converge to the same limit, then it must also converge to the same limit. Theorem 16 (Squeeze Theorem). Suppose that X = (xn ), Y = (yn ), and Z = (zn ) are sequences in R such that xn ≤ yn ≤ zn for all n ∈ N and that X, Y converge with limits lim(xn ) = lim(zn ). Then, the sequence Y is convergent and lim(xn ) = lim(yn ) = lim(zn ) Proof. TODO... Theorem 17. Let the sequence X = (xn ) in R converge to lim(xn ) = x ∈ R. Then the sequence (|xn |) of absolute values converges to lim(|xn |) = |x|. Proof. As a consequence of the Triangular Inequality, for all n ∈ N ||xn | − |x|| ≤ |xn − x| Then, since (xn ) converges to x, (|xn |) must converge to |x|. Theorem 18. Let the sequence X = (xn ) in R converge to lim(xn ) √ = x, and suppose that xn ≥ 0. Then √ √ the sequence ( xn ) of positive square roots converges to lim( xn ) = x. Proof. TODO... Theorem 19 (The Ratio Test). Let (xn ) be a sequence of positive reals such that the limit L := lim(xn+1 /xn ) exists. If L < 1, then (xn ) converges to the limit lim(xn ) = 0. Proof. TODO...
2.3
Monotone Sequences
Thus far we have employed several techniques to show that a sequence X := (xn ) in R is convergent. However, most of these techniques require that we already know the value of the limit. There may, however, be instances where we do not have a candidate for the limit of a sequence. In the following sections we establish several results that may be employed to show that a sequence is convergent even though the value of the limit is not known. 10
Definition 8. Let X := (xn ) be a sequence in R. Then, 1. X is increasing if its terms satisfy x1 ≤ x2 ≤ · · · ≤ xn ≤ xn+1 ≤ · · · 2. X is decreasing if its terms satisfy x1 ≥ x2 ≥ · · · ≥ xn ≥ xn+1 ≥ · · · Definition 9 (Monotone). A sequence in R is monotone if it is neither increasing nor decreasing. The following theorem establishes the existence of the limit of a bounded monotone sequence, since if a sequence is convergent, then a limit must exist. It also provides a simple means for determining the value of the limit of the sequence provided we can evaluate the supremum or infimum. Theorem 20 (Monotone Convergence Theorem). A monotone sequence X = (xn ) in R is convergent if an only if it is bounded. Furthermore, 1. If X is increasing, then lim(xn ) = sup{xn : n ∈ N} 2. If X is decreasing, then lim(xn ) = inf{xn : n ∈ N} Proof. For the forward implication, suppose that the sequence X is convergent. But, by Theorem 3.2.2 in Bartle, a convergent sequence is bounded. Conversely, for the reverse implication, suppose that the sequence X is a bounded monotone sequence. Since the sequence is monotone, the sequence must be either increasing or decreasing. We consider both cases: 1. If X is bounded increasing, then there exists some M ∈ R such that xn ≤ M for all n ∈ N. Since the sequence is bounded above, the Completeness Property of R ensures that a supremum exists, say x∗ = sup{xn : n ∈ N}. We must show that x∗ = lim X. Since x∗ is the supremum, there must exist some xK ∈ R such that for any ε > 0 given x∗ − ε < xK ≤ xn < x∗ < x∗ + ε for n ≥ K Thus |xn − x∗ |varepsilon for all n ≥ K, and since ε > 0 is arbitrary, we conclude that (xn ) converges to x∗ . 2. If Y = (yn is bounded decreasing, then X = −Y = (−yn ) is a bounded increasing sequence. Then lim X = − lim Y = sup{−yn : n ∈ N}. But, from an earlier result sup{−yn : n ∈ N} = − inf{yn : n ∈ N}, and so lim Y = − lim X = inf{yn : n ∈ N}.
Again, the Monotone Convergence Theorem states that a bounded monotone sequence is convergent, so a limit must exist. It further states that every bounded increasing sequence converges to the supremum of the underlying set, and every bounded decreasing sequence converges to the infimum of the underlying set. 11
2.4
Subsequences and the Balzano-Weierstrass Theorem
A subsequence is a selection of terms from a sequence such that the selected terms form a new sequence. Definition 10. Let X = (xn ) be a sequence in R and let n1 < n2 < · · · < nk < · · · be a strictly increasing sequence of natural numbers. Then the sequence X 0 = (xnk ) given by (xn1 , xn2 , . . . , xnk , . . .) is a subsequence of X. It is natural to ask what properties of a sequence a subsequence of that sequence may inherit. Primarily, we are interested in the convergence and limit of a sequence, in which case, is there a relationship between the convergence and limit of a sequence and any of its subsequences? Simply, subsequences of the same convergent sequence are themselves convergent, and converge to the same limit as the sequence. This is stated in the following theorem: Theorem 21. If a sequence X = (xn ) in R converges to x ∈ R, then any subsequence X 0 = (xnk ) of X also converges to x. Proof. Let ε > 0 be given, and let K(ε) ∈ N be such that n ≥ K(ε). Then |xn − x| < ε. Since n1 < n2 < · · · < nk < · · · is an increasing sequence in N, by Induction, nk ≥ k. Thus, if k ≥ K(ε), then nk > k > K(ε), and consequently |xnk − x| < ε. Hence, the sequence (xnk ) converges to x. Theorem 22. Let X = (xn ) be a sequence in R. The the following statements are equivalent: 1. The sequence X does not converge to x ∈ R. 2. There exists some ε0 > 0 such that for any k ∈ N there exists some nk ∈ N such that nk ≥ k and |xnk − x| ≥ ε0 . 3. There exists some ε0 > 0 and a subsequence X 0 = (xn ) of X such that |xnk − x| ≥ ε0 for all k ∈ N. We know that subsequences of a convergent sequence must converge to the same limit. We also know that convergent sequences are bounded. The following theorem shows how we can use these facts as criteria for divergence. Theorem 23 (Divergence Criteria). If a sequence X = (xn ) in R. Then X is divergent if either 1. X has two convergent subsequences X 0 and X 00 whose limits are not equal, or 2. X is unbounded.
2.4.1
The Existence of Monotone Subsequences
While not every sequence is monotone, it can be shown that every sequence has at least one subsequence that is monotone. Theorem 24 (Monotone Subsequence Theorem). Every sequence in R has a subsequence of X that is monotone. Proof. TODO... 12
2.4.2
The Bolzano-Weierstrass Theorem
Theorem 25 (The Bolzano-Weierstrass Theorem). A bounded sequence of real numbers has a convergent subsequence. Proof. Let X = (xn ) be a bounded sequence in R. By the Monotone Subsequence Theorem there exists a subsequence X 0 = (xnk ) of X that is monotone. But the subsequence X 0 is also bounded, so that by the Monotone Convergence Theorem, the subsequence X 0 is convergent. It should be noted that a bounded sequence may have various subsequences which converge to different limits or even diverge. A simple example is the sequence ((−1)n ), whose subsequence (−1) converges to −1, while (1) converges to 1. Subsequences of subsequences are also subsequences of the original sequence. Theorem 26. Let X = (xn ) be a bounded sequence in R. If every convergent subsequence of X converges to x ∈ R, then X converges to x.
2.5
The Cauchy Criterion
The Monotone Convergence Theorem allowed us to determine the convergence of a sequence without first knowing the value of the limit of the sequence. However, we are still incumbered by the fact that the Monotone Convergence Theorem requires us to know whether the sequence is monotone or not. In this section we introduce the so-called Cauchy Criterion, which will provide us with a condition which implies convergence without requiring that we know the value of the limit or whether the sequence is monotone in advance. The main result of this section will be a theorem which will assert that a sequence in R is convergent if an only if it is a Cauchy sequence. The definition of a Cauchy sequence follows: Definition 11. A sequence X = (xn ) of real numbers is said to be a Cauchy sequence if for every ε > 0 there exists a natural number H(ε) such that for all n, m ∈ N with the condition n, m ≥ H(ε), the terms xn , xm satisfy |xn − xm | < ε. Before presenting the Cauchy Convergence Criterion, the following two lemmas will be useful in proving, firstly, that a convergent sequence is a Cauchy sequence, and secondly, that a Cauchy sequence is bounded: Lemma 3. If X = (xn ) is a convergent sequence of real numbers, then X is a Cauchy sequence. Proof. By definition, if x := lim X, then given any ε > 0 there exists a K(ε/2) ∈ N such that if n ≥ K(ε/2), then |xn − x| < ε. If we let H(ε) := K(ε/2) and n, m ≥ H(ε), then by the Triangular Inequality |xn − xm | = |(xn − x) + (x − xm )| ≤ |xn − x| + |xm − x| < ε/2 + ε/2 = ε Since ε > 0 is arbitrary, it follwos that the sequence (xn ) is a Cauchy sequence. Lemma 4. A Cauchy sequence of real numbers is bounded. Proof. Let X := (xn ) be a Cauchy sequence, and let ε := 1. If n ≥ H := H(1), then |xn − xH | < 1, and by the Triangular Inequality |xn − xH | ≤ |xn | − |xH | < 1 ⇒ |xn | ≤ |xH | + 1 for all n ≥ H. If we set M := sup{|x1 , x2 , . . . , |xH−1 |, |xH | + 1} Hence, |xn ≤ M | for all n ∈ N; that is, the Cauchy sequence (xn ) is bounded. We now make use of these two results in proving the Cauchy Convergence Criterion: Theorem 27 (Cauchy Convergence Criterion). A sequence of real numbers is convergent if and only if it is a Cauchy sequence. Proof. TODO...
13
14
Chapter 3
Limits 3.1
Limits of Functions
Informally, a function f has a limit L at a point c if the values of f (x) are close to L when x is close to c. By close we mean infinitesimally close but not equal to c. A formal ε-δ definition of the limit of a function follows, but it must be kept in mind that we require the function f to be defined at points near to c even if f is not defined at c. Before formally defining the limit of a function, it may be necessary to explore the meaning of close to a point c, as we have used it in our informal discussion. Definition 12. A point c ∈ R is called a cluster point of the a set A ⊆ R if for every δ > 0 there exists at least one point x ∈ A, x = 6 c such that |x − c| < δ. We could similarly say that c ∈ R is a cluster point of A ⊆ R if every δ-neighbourhood Vδ (c) = (c−δ, c+δ) of the point c contains at least one point of A distinct from c. Note that c need not belong to A. Theorem 28. A number c ∈ R is a cluster point of A ⊆ R if and only if there exists a sequence (an ) in A such that lim(an ) = c, and an 6= c for all n ∈ N. Proof. For the forward implication, let c be a cluster point of A. Then for any n ∈ N we may choose δ = 1/n such that every 1/n-neighbourhood V1/n (c) of c contains at least one point an of A distinct from c. Since an ∈ A, an 6= c, and |an − c| < 1/n, this implies that lim(an ) = c. Conversely, if there exists a sequence (an ) in the set A {c} with lim(an ) = c, then for every δ > 0 there must exist some K ∈ N such that if n ≥ K then |an − c| < δ; that is, an ∈ Vδ (c), an 6= c. Since the δ-neighbourhood Vδ (c) of the point c contains the points an for n ≥ K belonging to A but distinct from c, c is a cluster point of A.
3.1.1
The Definition of the Limit
In the following definition of the limit of a function f at a point c it is important, again, to remember that the function f need not be defined at the point c. Definition 13. Let A ⊆ R, and let c be a cluster point of A. Then, a function f : A → R has a limit L ∈ R at the point c, written L = lim f (x) x→c
if for any ε > 0 there exists a δ(ε) > 0 such that if x ∈ A and 0 < |x − c| < δ, then |f (x) − L| < ε. If the limit of f at c does not exist, then f diverges at c. In reading the above definition, it is useful to visualise 0 < |x − c| < δ as the set of all x ∈ A within the neighbourhood (c − δ, c + δ) but not including c itself (as implied by the inequality 0 < |x − c|). The following theorem expresses our definition in the language of neighbourhoods: 15
Theorem 29. Let f : A → R, and let c be a cluster point of A. Then the following statements are equivalent: 1. limx→c f (x) = L 2. Given any ε-neighbourhood Vε (L) of L, there exists a δ-neighbourhood Vδ (c) of c such that if x 6= c is any point in Vδ (c) ∩ A, then f (x) belongs to Vε (L). Now that we have formally defined the limit L of a function f at a point c, it is natural to ask whether the limit L is unique. Theorem 30. Let f : A → R and c be a cluster point of A. Then f can have at most one limit at c.
3.1.2
Sequential Criterion for Limits
Theorem 31 (Sequential Criterion). Let f : A → R, and let c be a cluster point of A. Then the following statements are equivalent: 1. limx→c f = L 2. For every sequence (xn ) in A that converges to c such that xn 6= c for all n ∈ N, the sequence (f (xn )) converges to L. Proof. TODO...
3.1.3
Divergence Criteria
Theorem 32 (Divergence Criteria). Let A ⊆ R, and let f : A → R and c ∈ R be a cluster point of A. 1. If L ∈ R, then f does not have a limit L at c if and only if there exists a sequence (xn ) in A with xn 6= c for all n ∈ N such that the sequence (xn ) converges to c but the sequence (f (xn )) does not converge to L. 2. The function f does not have a limit at c if and only if there exists a sequence (xn ) in A with xn 6= c for all n ∈ N such that the sequence (xn ) converges to c but the sequence (f (xn )) does not converge in R.
3.2
Limit Theorems
The theorems in this section are useful in calculating the limits of functions. Definition 14. Let A ⊆ R and c ∈ R be a cluster point of A. If f : A → R, then f is bounded on a neighbourhood of c if there exists a δ-neighbourhood Vδ (c) of c and a constant M > 0 such that |f (x)| ≤ M for all x ∈ A ∩ Vδ (c) Theorem 33. If A ⊆ R, and f : A → R has a limit at c ∈ R, then f is bounded on some neighbourhood of c. Theorem 34. Let A ⊆ R, and let f, g be functions defined on A to R. Then for all x ∈ A (f ± g)(x) := f (x) ± g(x),
(f g)(x) := f (x)g(x)
Furthermore, if b ∈ R, then for all x ∈ A (bf )(x) := bf (x) Finally, if h(x) 6= 0, then for all x ∈ A f f (x) (x) := g g(x) 16
Theorem 35. Let A ⊆ R, and let c ∈ R be a cluster point of A. If f, g are functions on A to R, and b ∈ R, then 1. If limx→c f = L and limx→c g = M lim (f ± g) = L ± M,
x→c
lim (f g) = LM,
x→c
lim (bf ) = bL
x→c
2. If h : A → R with h(x) 6= 0 for all x ∈ A, and if limx→c h = H 6= 0, then f L lim = x→c h H Theorem 36. Let A ⊆ R, and let c ∈ R be a cluster point of A. If f : A → R and a ≤ f (x) ≤ b for all x ∈ A, x 6= c, and if limx→c f exists, then a ≤ limx→c f ≤ b. Theorem 37 (Squeeze Theorem). Let A ⊆ R, and let c be a cluster point on A. If f, g, h : A → R and f (x) ≤ g(x) ≤ h(x) for all x ∈ A, x 6= c, and if limx→c f = L = limx→c h, then limx→c g = L. Theorem 38. TODO...
17
18
Chapter 4
Continuous Functions 4.1
Continuous Functions
Definition 15. Let A ⊆ R, and let c ∈ R. If f : A → R, then f is continuous at c if, given any number ε > 0, there exists a δ > 0 such that if x is any point of A satisfying |x − c| < δ, then |f (x) − f (c)| < ε. If f is not continuous at the point c, then it is discontinuous at c. We can similarly formulate a definition of continuity at a point in terms of neighbourhoods: Theorem 39. A function f : A → R is continuous at a point c ∈ A if and only if given any ε-neighbourhood Vε (f (c)) there exists a δ-neighbourhood Vδ (c) such that if x is any point of A ∩ Vδ (c), then f (x) belongs to Vε (f (c)); that is, f (A ∩ Vδ (c)) ⊆ Vε (f (c)) TODO: Read remarks to theorem carefully... Theorem 40 (Sequential Criterion for Continuity). A function f : A → R is continuous at the point c ∈ A if and only if for every sequence (xn ) in A that converges to c, the sequence (f (xn )) converges to f (c). Theorem 41 (Discontinuity Criterion). Let A ⊆ R, and let c ∈ A. If f : A → R, then f is discontinuous at c if and only if there exists a sequence (xn ) in A such that (xn ) converges to c, but the sequence (f (xn )) does not converge to f (c). Thus far we have discussed continuity at a point. What about continuity on a set? A function is continuous on a set if it is continuous at every point on the set. Definition 16. Let A ⊆ R and let f : A → R. If B is a subset of A, then f is continuous on the set B if f is continuous at every point of B.
4.2 4.2.1
Combination of Continuous Functions Continuity
The following theorem applies to the continuity of combinations of functions at a point in R: Theorem 42. Let A ⊆ R and b ∈ R. If f, g : A → R are continuous at c ∈ A, then 1. f ± g, f g, and bf are continuous at c 2. If h : A → R is continuous at c and h(x) 6= 0 for all x ∈ A, then f /h is continuous at c. The following theorem applies to the continuity of combinations of functions on a subset of R: 19
Theorem 43. Let A ⊆ R and b ∈ R. If f, g : A → R are continuous on A, then 1. f ± g, f g, and bf are continuous on A 2. If h : A → R are continuous at A and h(x) 6= 0 for all x ∈ A, then f /h is continuous on A.
4.2.2
Composition of Functions
Theorem 44. Let A, B ⊆ R and let f : A → R and g : B → R be functions such that f (A) ⊆ B. If f is continuous at c ∈ A, and g is continuous at b = f (c) ⊆ B, then the composition g ◦ f : A → R is continuous at c. Theorem 45. Let A, B ⊆ R, and let f : A → R be continuous on A and g : B → R be continuous on B. If f (A) ⊆ B, then the composite function g ◦ f : A → R is continuous.
20
Real Numbers 1.1 1.1.1
The Completeness Property of R Suprema and Infima
Firstly, we shall require a definition for the upper and lower bound for a set of real numbers. Definition 1. Let S be a nonempty subset of S. 1. The set S is bounded above if there exists a number u ∈ R, called the upper bound of S, such that for every s ∈ R we have that s ≤ u. 2. The set S is bounded below if there exists a number w ∈ R, called the lower bound of S, such that for every s ∈ R we have that s ≥ w. 3. A set is bounded if it is bounded both above and below. A set is unbounded if it is not bounded. It may be worth stressing that a set bounded either above or below, but not above and below, is still unbounded. Also, if an upper bound u exists on a set, then the set will have infinitely many upper bounds, as u plus some arbitrary real number is still an upper bound. The same applies to lower bounds. The points of interest, however, are the least elements of the set of upper bounds, and the greatest elements of the set of lower bounds. Definition 2. Let S be a nonempty subset of R. 1. The upper bound u of S is a supremum (least upper bound) if it satisfies (a) u is an upper bound of S. (b) u ≤ v where v is any upper bound of S. 2. The lower bound w is an infimum (greatest lower bound) if it satisfies (a) w is a lower bound of S. (b) t ≤ w where w is any lower bound of S. Since the supremum and infimum are unique we can speak unambiguously about the supremum or infimum from here on. The supremum and infimum of a set S will be denoted by sup S,
inf S
We may make a number of equivalent statements about the upper bound u of a set S. Thus we may also make a number of alternate formulations for the supremum of a set. Only two such formulations will be made, and are presented in the two lemmas that follow. 1
Lemma 1. A number u is a supremum of a nonempty subset S of R if and only it satisfies 1. s ≤ u for all s ∈ S 2. If v < u, then there exists an s0 ∈ S such that v < s0 . Lemma 2. An upper bound u of a nonempty subset S of R is the supremum of S if and only if for every ε > 0 there exists an sε ∈ S such that u − ε < sε . Proof. Let u = sup S, where S is nonempty, and let ε > 0. Then u − ε < u, and since u is the supremum of S it follows that u−ε cannot be an upper bound of S. Hence, there exists some sε ∈ S such that u−ε < sε ≤ u. Conversely, suppose that for every ε > 0 there exists some sε ∈ S such that u − ε < sε . Let v < u so that we may put ε := u − v. Then v = u − ε, and consequently v < sε . Hence, v is not an upper bound of S, and u = sup S. Axiom 1 (Completeness Property of R). Every nonempty subset of R that has an upper bound also has a supremum in R. This is the Supremum Property of R. Analogously, every nonempty subset of R that has a lower bound also has an infimum in R.
1.2 1.2.1
Applications of the Supremum Property Functions
We consider the range of a function when talking about the upper or lower bound applied to a function. A function f : D → R, where D is the domain of f , is bounded above if the set f (D) = {f (x) : x ∈ D} is bounded above in R. Essentially, if f is bounded above then there exists some B ∈ R such that f (x) ≤ B for all x ∈ D. Similarly, the function f is bounded below if f (D) is bounded below. The function f is bounded if f (D) is bounded above and below; that is, there exists some B ∈ R such that kf (x)| ≤ B for all x ∈ D.
1.2.2
The Archimedian Property
Theorem 1 (Archimedean Property). If x ∈ R, then there exists nx ∈ N such that x ≤ nx . Proof. Suppose, to the contrary, that n ≤ x for all n ∈ N. Then x is an upper bound for N, and by the Completeness Property, N has a supremum u ∈ R. But u − 1 < u, so that u − 1 is not an upper bound of N, and so there must exist some m ∈ N such that u − 1 < m. But this implies that u < m + 1 ∈ N, which contradicts that u is an upper bound. Hence, our assertion that n ≤ x for all n ∈ N is false, and we have that x ≤ n for all n ∈ N. The Archimedean Property implies that the set of natural numbers N is not bounded.1 Corollary 1. If S := {1/n : n ∈ N}, then inf S = 0. Proof. Since S 6= ∅ is bounded below by 0, we know from the Completeness Property that it has an infimum w := inf S ≥ 0. By the Archimedean Property, for any ε > 0 there exists n ∈ N such that 1/ε < n, which implies that 1/n < ε. Hence 0 ≤ w ≤ 1/n < ε Since ε > 0 is arbitrary, w := inf S = 0. Corollary 2. If t > 0, there exists nt ∈ N such that 0 < 1/nt < t. 1 The
set of naturals is unbounded because it is not bounded above, but we can find a lower bound of the natural numbers
2
Proof. Since inf{1/n : n ∈ N} = 0 and t > 0, then t is not the lower bound of the set {1/n : n ∈ N}. Thus, there exists nt ∈ N such that 0 < 1/nt < t. Corollary 3. If y > 0, there exists ny ∈ N such that ny − 1 ≤ y ≤ ny Proof. Let Ey := {m ∈ N : y < m} ⊆ N be the subset of reals greater than some y. The Archimedean Property ensures that this set is nonempty as there must be at least one m such that y < m. Also, the Well-Ordering Property of N ensures that Ey has a least element, say ny . But then ny − 1 ∈ / Ey , so that ny − 1 ≤ y < n y
1.2.3
Density of Rational Numbers
It can be shown that there exists a positive real number x such that x2 = 2. This is the positive square root of 2, which is an irrational number. Since the existence of at √ least one irrational number is proven by this result, we ask whether other irrational numbers apart from 2 exist. In fact, there are more irrational numbers than rational numbers since the set of rationals is countable and the set of irrationals is uncountable (not shown yet). The set of rationals is said to be dense in R in the sense that given any two real numbers there exists a rational number between them. Theorem 2 (The Density Theorem). If x, y ∈ R and x < y, then there exists an r ∈ Q such that x < r < y. Proof. Assume, without loss of generality, that x > 02 . Since x < y we have that y − x > 0, and it follows that there must exist some n ∈ N such that 0 < 1/n < y − x. Thus nx + 1 < ny. But for nx > 0, there must exist some m ∈ N such that m − 1 ≤ nx < m, from which we obtain m ≤ nx + 1 < m + 1. Hence we have nx < m < ny, and with r := m/n, we have our final result x < r < y. The density of irrational numbers in R can also be shown: Corollary 4. If x, y ∈ R and x < y, then there exists an irrational number z such that x < z < y. √ √ Proof. Applying the Density Theorem to the real numbers x/ 2 and y/ 2, we obtain a rational r 6= 0 such that x y √ a},
(−∞, b) := {x ∈ R : x < b}
Here, the set (a, ∞) has no upper bound, while the set (−∞, b) has no lower bound. The infinite closed intervals are the sets [a, ∞) := {x ∈ R : x ≥ a},
1.3.1
(−∞, b] := {x ∈ R : x ≤ b}
Characterization of Intervals
If two points x, y, where x < y, belong to some interval I, then any point t, where x < t < y, belongs to I. Extending this idea, if x, y belong to I then the interval [x, y] belongs to I. A subset of R possessing this property must then be an interval. Theorem 3 (Characterization Theorem). If S ⊆ R such that it contains at least two points with the property x, y ∈ S,
x < y ⇒ [x, y] ⊆ S
then S is an interval.
1.3.2
Nested Intervals
A sequence of intervals In , where n ∈ N, is nested if . . . ⊆ In+1 ⊆ In ⊆ . . . ⊆ I2 ⊆ I1 An important property of R is that every nested sequence of closed bounded intervals in R share a common point. Theorem 4 (Nested Intervals Property). If In = [an , bn ], where n ∈ N, is a nested sequence of closed bounded intervals, then there exists some ξ ∈ R such that ξ ∈ In for all n. Proof. Since the intervals are nested we have In ⊆ I1 for all n ∈ N, and so an ≤ b1 for all n ∈ N. The nonempty set {an : n ∈ N} is then bounded above, and we let ξ := sup{an : n ∈ N}. Consequently, an ≤ ξ for all n ∈ N. We claim that ξ ≤ bn for all n. This is established by showing that for a particular n, the number bn is an upper bound of the set {ak : k ∈ N}. There are two cases to consider: 1. If n ≤ k then since Ik ⊆ In , we have ak ≤ bk ≤ bn . 2. If k < n, then since In ⊆ Ik , we have ak ≤ an ≤ bn . Thus we conclude that ak ≤ bn for all k, so that bn is an upper bound of the set {ak : k ∈ N}. Hence, an ≤ ξ ≤ bn for each n ∈ N, we have that ξ ∈ In for all n ∈ N. Theorem 5. If In := [an , bn ], where n ∈ N, is a nested sequence of closed bounded intervals such that the lengths bn − an of In staisfy inf{bn − an : n ∈ N} = 0 then the number ξ contained in In for all n is unique. 4
1.3.3
The Uncountability of R
Theorem 6. The set R is uncountable. Proof. It is sufficient to show that the unit interval I := [0, 1] is an uncountable set as this would imply that R is an uncountable set3 . Suppose, to the contrary, that I is countable. Then we can enumerate the set as I = {x1 , x2 , . . . , xn , . . .}. Select first a closed subinterval I1 ⊆ I such that x1 ∈ / I1 . Then select a closed subinterval I2 ⊆ I1 such that x2 ∈ / I2 , and so on until we obtain the closed intervals I1 ⊆ I2 ⊆ · · · ⊆ In ⊆ · · · such that In ⊆ I and xn ∈ / In for all n ∈ N. But the Nested Intervals Property implies that there exists ξ ∈ I such that ξ ∈ In for all n, and since xn ∈ / In for all n, ξ 6= xn for all n ∈ N. The enumeration is not a complete listing of the elements of I as claimed. Hence, I is an uncountable set.
3 If
R were countable, then the subset I would be countable
5
6
Chapter 2
Sequences and Series 2.1 2.1.1
Sequences and Their Limits Sequences
Definition 3. A sequence of real numbers (a sequence in R) is a function whose domain is defined on the set N = {1, 2, . . .}, and whose range is contained in the set R. In other words, a sequence X in R is a function X : N → R that assigns to every natural number n = 1, 2, . . . a uniquely determined real number xn . The values of xn are the terms (or elements) of the sequence. A sequence may be denoted by any one of the following notations: X,
(xn ),
(xn : n ∈ N)
It should be noted that the notation (xn : n ∈ N) denotes an ordered set, while the notation {xn : n ∈ N} denotes an unordered set in the range of the sequence. Sequences may be listed, but are more frequently defined by specifying a formula for the general nth term xn .
2.1.2
The Limit of a Sequence
The limit of a sequence is the most basic of limit concepts in real analysis. Definition 4. A sequence X = (xn ) in R is said to converge to a limit x ∈ R if for every ε > 0 there exists some K(ε) ∈ N such that for all n ≥ K(ε) |xn − x| < ε If a sequence X = (xn ) has a limit x, then we denote this by lim X = x,
or
lim(xn ) = x,
or xn → x
If a sequence has a limit, then the sequence is said to be convergent. Otherwise, if the limit of the sequence does not exist, then the sequence is said to be divergent. Now that we have established the existence of a limit for a convergent sequence, the next question to ask is whether or not the limit of a particular convergent sequence is unique; that is, whether the sequence has only one limit or possibly many limits. Theorem 7 (Uniqueness of Limits). A sequence in R has at most one limit. 7
Proof. Suppose that x0 and x00 are both limits the the sequence (xn ) in R. It is sufficient to show that x0 = x00 . It follows from the definition that for every ε > 0 there exists a K 0 such that |xn − x0 | < ε/2 for all n ≥ K 0 , and there exists a K 00 such that |xn − x00 | < ε/2 for all n ≥ K 00 . Let K > K 0 , K 00 so that for all n ≥ K we have |xn − x0 | < ε/2 and |xn − x00 | < ε/2. Applying the Triangle Inequality |x0 − x00 | = |x0 − xn + xn − x00 | ≤ |x0 − xn | + |xn − x00 |
0 is arbitrary, we conclude that x0 − x00 = 0, and hence x0 = x00 . There are several ways of saying that a sequence (xn ) converges to x: Theorem 8. Let X = (xn ) be a sequence in R, and let x ∈ R. Then the following statements are equivalent: 1. X converges to x. 2. For every ε > 0, there exists a K ∈ N such that for all n ≥ K, the terms xn satisfy |xn − x| < ε. 3. For every ε > 0 there exists a K ∈ N such that for all n ≥ K, the terms xn satisfy x − ε < xn < x + ε. 4. For every ε-neighbourhood Vε (x) := {u ∈ R : |u − x| < ε} of x, there exists a K ∈ K such that for all n ≥ K, the terms xn belong to Vε (x). The final statement of the previous theorem is equivalent to saying that for every ε-neighbourhood Vε (x) of x, all but a finite number of terms x1 , x2 , . . . , xK−1 of X belong to Vε (x). Those terms belonging to Vε (x) are xK , xK+1 , xK+2 , . . .. Clearly, there are infinitely many of them. To show that that a sequence X = (xn ) does not converge to a limit x ∈ R, it is sufficient to show that there is at least one ε0 > 0 such that irrespective of how large K is chosen, there exists an nK satisfying nK ≥ K such that |xNK − x| ≥ ε0 .
2.1.3
Tails of a Sequence
If the first m ∈ N terms of a sequence X are removed, then the resulting sequence Xm will converge if and only if the original sequence X converges. This idea will be expressed in a theorem, but we shall first define the m-tail Xm of a sequence X: Definition 5. If X = (x1 , x2 , . . . , xn , . . .) is a sequence in R, and if m ∈ N is given, then the m-tail of X is the sequence Xm := {xm+n : n ∈ N} = (xm+1 , xm+2 , . . .) Now we show that the m-tail Xm of a sequence X converges if and only if the original sequence X converges: Theorem 9. Let X = (xn : n ∈ N) be a sequence in R, and let m ∈ N. Then the m-tail Xm = (xm+n : n ∈ N) of X converges if and only if X converges, and converges to the limit lim Xm = lim X.
2.1.4
Further Examples
Theorem 10. Let (xn ) be a sequence in R, and let x ∈ R. If (an ) is a sequence of positive reals with lim(an ) = 0, and if there exists some constant C > 0 such that for some m ∈ N |xn − x| ≤ Can , then lim(xn ) = x. 8
for all n ≥ m
Proof. Let ε > 0. Since lim(an ) = 0 there exists K ∈ N such that for all n ≥ K an = |an − 0|
0 is arbitrary, it follows that lim(xn ) = x.
2.2
Limit Theorems
Definition 6. A sequence X = (xn ) in R is bounded if there exists a real number M > 0 such that |xn | ≤ M for all n ∈ N. In other words, a sequence (xn ) is bounded if and only if the set of terms {xn : n ∈ N} is a bounded subset of R. It should be evident that a convergent sequence is bounded. Theorem 11. A convergent seqience of real numbers is bounded. Proof. Suppose that lim(xn ) = x. Let ε := 1. Then there exists K = K(1) such that |xn − x| < 1 for all n ≥ K. Then, if n ≥ K, and using the Triangular Ineqality |xn | = |xn − x + x| ≤ |xn − x| + |x| < 1 + |x| Thus, we can set M := sup{|x1 |, |x2 |, . . . , |xK−1 |, 1 + |x|} and it follows that |xn | ≤ M for all n ∈ N. We may apply the operations of addition, subtraction, multiplication, and division to sequences of real numbers. Definition 7. Let X = (xn ) and Y = (yn ) be sequences of real numbers. Then 1. X ± Y := (xn ± yn ) 2. XY := (xn yn ) 3. X/Y := (xn /yn ) 4. cX := (cxn ) for c ∈ R Applying these arithmetic operations to convergent sequences result in new sequences whose limits can be determined. Theorem 12. 1. Let X = (xn ) and Y = (yn ) be sequences in R that converge to limits lim X = x and lim Y = y respectively, and let c ∈ R. Then the sequences X ± Y , XY , and cX converge to the limits x ± y, xy, and cx respectively. 2. If X = (xn ) converges to x and Z = (zn ) is a sequence of nonzero real numbers that converges to z 6= 0, then the quotient sequence X/Z converges to x/z. We can extend these results by Mathematical Induction to a finite number n of convergent sequences A = (zn ), B = (bn ), . . . , Z = (zn ). Thus, the sum A + B + · · · + Z = (an + bn + · · · + zn ) is also a convergent sequence with limit lim(an + bn + · · · + zn ) = lim(an ) + lim(bn ) + · · · + lim(zn ) Similar results may be obtained for the product of a finite number n of convergent sequences. A convergent sequence, each of whose terms is positive, will converge to a limit which is also positive. 9
Theorem 13. If X = (xn ) is a convergent sequence of real numbers, and if xn ≥ 0 for all n ∈ N, then lim(xn ) ≥ 0. Proof. Suppose, to the contrary, that the sequence (xn ) with xn converges with limit x = lim(xn ) < 0. Then ε := −x is positive, and since X converges to x, there exists a K ∈ K such that x − ε < xn < x + ε for all
n≥K
In particular, for n = K we have xK < x + ε = x + (−x) = 0. But this contradicts the hypothesis that xn ≥ 0 for all n ∈ N. This contradiction implies that the assumption x < 0 is incorrect, and we conclude that x ≥ 0. Theorem 14. If X = (xn ) and Y = (yn ) are convergent sequences in R with xn ≤ yn for all n ∈ N, then lim(xn ) ≤ lim(yn ). Theorem 15. If X = (xn ) is a convergent sequence and a ≤ xn ≤ b for all n ∈ N, then a ≤ lim(xn ) ≤ b. The next result, known as the Squeeze Theorem is an important result. Essentially, it says that that if a sequence is squeezed between two sequences that converge to the same limit, then it must also converge to the same limit. Theorem 16 (Squeeze Theorem). Suppose that X = (xn ), Y = (yn ), and Z = (zn ) are sequences in R such that xn ≤ yn ≤ zn for all n ∈ N and that X, Y converge with limits lim(xn ) = lim(zn ). Then, the sequence Y is convergent and lim(xn ) = lim(yn ) = lim(zn ) Proof. TODO... Theorem 17. Let the sequence X = (xn ) in R converge to lim(xn ) = x ∈ R. Then the sequence (|xn |) of absolute values converges to lim(|xn |) = |x|. Proof. As a consequence of the Triangular Inequality, for all n ∈ N ||xn | − |x|| ≤ |xn − x| Then, since (xn ) converges to x, (|xn |) must converge to |x|. Theorem 18. Let the sequence X = (xn ) in R converge to lim(xn ) √ = x, and suppose that xn ≥ 0. Then √ √ the sequence ( xn ) of positive square roots converges to lim( xn ) = x. Proof. TODO... Theorem 19 (The Ratio Test). Let (xn ) be a sequence of positive reals such that the limit L := lim(xn+1 /xn ) exists. If L < 1, then (xn ) converges to the limit lim(xn ) = 0. Proof. TODO...
2.3
Monotone Sequences
Thus far we have employed several techniques to show that a sequence X := (xn ) in R is convergent. However, most of these techniques require that we already know the value of the limit. There may, however, be instances where we do not have a candidate for the limit of a sequence. In the following sections we establish several results that may be employed to show that a sequence is convergent even though the value of the limit is not known. 10
Definition 8. Let X := (xn ) be a sequence in R. Then, 1. X is increasing if its terms satisfy x1 ≤ x2 ≤ · · · ≤ xn ≤ xn+1 ≤ · · · 2. X is decreasing if its terms satisfy x1 ≥ x2 ≥ · · · ≥ xn ≥ xn+1 ≥ · · · Definition 9 (Monotone). A sequence in R is monotone if it is neither increasing nor decreasing. The following theorem establishes the existence of the limit of a bounded monotone sequence, since if a sequence is convergent, then a limit must exist. It also provides a simple means for determining the value of the limit of the sequence provided we can evaluate the supremum or infimum. Theorem 20 (Monotone Convergence Theorem). A monotone sequence X = (xn ) in R is convergent if an only if it is bounded. Furthermore, 1. If X is increasing, then lim(xn ) = sup{xn : n ∈ N} 2. If X is decreasing, then lim(xn ) = inf{xn : n ∈ N} Proof. For the forward implication, suppose that the sequence X is convergent. But, by Theorem 3.2.2 in Bartle, a convergent sequence is bounded. Conversely, for the reverse implication, suppose that the sequence X is a bounded monotone sequence. Since the sequence is monotone, the sequence must be either increasing or decreasing. We consider both cases: 1. If X is bounded increasing, then there exists some M ∈ R such that xn ≤ M for all n ∈ N. Since the sequence is bounded above, the Completeness Property of R ensures that a supremum exists, say x∗ = sup{xn : n ∈ N}. We must show that x∗ = lim X. Since x∗ is the supremum, there must exist some xK ∈ R such that for any ε > 0 given x∗ − ε < xK ≤ xn < x∗ < x∗ + ε for n ≥ K Thus |xn − x∗ |varepsilon for all n ≥ K, and since ε > 0 is arbitrary, we conclude that (xn ) converges to x∗ . 2. If Y = (yn is bounded decreasing, then X = −Y = (−yn ) is a bounded increasing sequence. Then lim X = − lim Y = sup{−yn : n ∈ N}. But, from an earlier result sup{−yn : n ∈ N} = − inf{yn : n ∈ N}, and so lim Y = − lim X = inf{yn : n ∈ N}.
Again, the Monotone Convergence Theorem states that a bounded monotone sequence is convergent, so a limit must exist. It further states that every bounded increasing sequence converges to the supremum of the underlying set, and every bounded decreasing sequence converges to the infimum of the underlying set. 11
2.4
Subsequences and the Balzano-Weierstrass Theorem
A subsequence is a selection of terms from a sequence such that the selected terms form a new sequence. Definition 10. Let X = (xn ) be a sequence in R and let n1 < n2 < · · · < nk < · · · be a strictly increasing sequence of natural numbers. Then the sequence X 0 = (xnk ) given by (xn1 , xn2 , . . . , xnk , . . .) is a subsequence of X. It is natural to ask what properties of a sequence a subsequence of that sequence may inherit. Primarily, we are interested in the convergence and limit of a sequence, in which case, is there a relationship between the convergence and limit of a sequence and any of its subsequences? Simply, subsequences of the same convergent sequence are themselves convergent, and converge to the same limit as the sequence. This is stated in the following theorem: Theorem 21. If a sequence X = (xn ) in R converges to x ∈ R, then any subsequence X 0 = (xnk ) of X also converges to x. Proof. Let ε > 0 be given, and let K(ε) ∈ N be such that n ≥ K(ε). Then |xn − x| < ε. Since n1 < n2 < · · · < nk < · · · is an increasing sequence in N, by Induction, nk ≥ k. Thus, if k ≥ K(ε), then nk > k > K(ε), and consequently |xnk − x| < ε. Hence, the sequence (xnk ) converges to x. Theorem 22. Let X = (xn ) be a sequence in R. The the following statements are equivalent: 1. The sequence X does not converge to x ∈ R. 2. There exists some ε0 > 0 such that for any k ∈ N there exists some nk ∈ N such that nk ≥ k and |xnk − x| ≥ ε0 . 3. There exists some ε0 > 0 and a subsequence X 0 = (xn ) of X such that |xnk − x| ≥ ε0 for all k ∈ N. We know that subsequences of a convergent sequence must converge to the same limit. We also know that convergent sequences are bounded. The following theorem shows how we can use these facts as criteria for divergence. Theorem 23 (Divergence Criteria). If a sequence X = (xn ) in R. Then X is divergent if either 1. X has two convergent subsequences X 0 and X 00 whose limits are not equal, or 2. X is unbounded.
2.4.1
The Existence of Monotone Subsequences
While not every sequence is monotone, it can be shown that every sequence has at least one subsequence that is monotone. Theorem 24 (Monotone Subsequence Theorem). Every sequence in R has a subsequence of X that is monotone. Proof. TODO... 12
2.4.2
The Bolzano-Weierstrass Theorem
Theorem 25 (The Bolzano-Weierstrass Theorem). A bounded sequence of real numbers has a convergent subsequence. Proof. Let X = (xn ) be a bounded sequence in R. By the Monotone Subsequence Theorem there exists a subsequence X 0 = (xnk ) of X that is monotone. But the subsequence X 0 is also bounded, so that by the Monotone Convergence Theorem, the subsequence X 0 is convergent. It should be noted that a bounded sequence may have various subsequences which converge to different limits or even diverge. A simple example is the sequence ((−1)n ), whose subsequence (−1) converges to −1, while (1) converges to 1. Subsequences of subsequences are also subsequences of the original sequence. Theorem 26. Let X = (xn ) be a bounded sequence in R. If every convergent subsequence of X converges to x ∈ R, then X converges to x.
2.5
The Cauchy Criterion
The Monotone Convergence Theorem allowed us to determine the convergence of a sequence without first knowing the value of the limit of the sequence. However, we are still incumbered by the fact that the Monotone Convergence Theorem requires us to know whether the sequence is monotone or not. In this section we introduce the so-called Cauchy Criterion, which will provide us with a condition which implies convergence without requiring that we know the value of the limit or whether the sequence is monotone in advance. The main result of this section will be a theorem which will assert that a sequence in R is convergent if an only if it is a Cauchy sequence. The definition of a Cauchy sequence follows: Definition 11. A sequence X = (xn ) of real numbers is said to be a Cauchy sequence if for every ε > 0 there exists a natural number H(ε) such that for all n, m ∈ N with the condition n, m ≥ H(ε), the terms xn , xm satisfy |xn − xm | < ε. Before presenting the Cauchy Convergence Criterion, the following two lemmas will be useful in proving, firstly, that a convergent sequence is a Cauchy sequence, and secondly, that a Cauchy sequence is bounded: Lemma 3. If X = (xn ) is a convergent sequence of real numbers, then X is a Cauchy sequence. Proof. By definition, if x := lim X, then given any ε > 0 there exists a K(ε/2) ∈ N such that if n ≥ K(ε/2), then |xn − x| < ε. If we let H(ε) := K(ε/2) and n, m ≥ H(ε), then by the Triangular Inequality |xn − xm | = |(xn − x) + (x − xm )| ≤ |xn − x| + |xm − x| < ε/2 + ε/2 = ε Since ε > 0 is arbitrary, it follwos that the sequence (xn ) is a Cauchy sequence. Lemma 4. A Cauchy sequence of real numbers is bounded. Proof. Let X := (xn ) be a Cauchy sequence, and let ε := 1. If n ≥ H := H(1), then |xn − xH | < 1, and by the Triangular Inequality |xn − xH | ≤ |xn | − |xH | < 1 ⇒ |xn | ≤ |xH | + 1 for all n ≥ H. If we set M := sup{|x1 , x2 , . . . , |xH−1 |, |xH | + 1} Hence, |xn ≤ M | for all n ∈ N; that is, the Cauchy sequence (xn ) is bounded. We now make use of these two results in proving the Cauchy Convergence Criterion: Theorem 27 (Cauchy Convergence Criterion). A sequence of real numbers is convergent if and only if it is a Cauchy sequence. Proof. TODO...
13
14
Chapter 3
Limits 3.1
Limits of Functions
Informally, a function f has a limit L at a point c if the values of f (x) are close to L when x is close to c. By close we mean infinitesimally close but not equal to c. A formal ε-δ definition of the limit of a function follows, but it must be kept in mind that we require the function f to be defined at points near to c even if f is not defined at c. Before formally defining the limit of a function, it may be necessary to explore the meaning of close to a point c, as we have used it in our informal discussion. Definition 12. A point c ∈ R is called a cluster point of the a set A ⊆ R if for every δ > 0 there exists at least one point x ∈ A, x = 6 c such that |x − c| < δ. We could similarly say that c ∈ R is a cluster point of A ⊆ R if every δ-neighbourhood Vδ (c) = (c−δ, c+δ) of the point c contains at least one point of A distinct from c. Note that c need not belong to A. Theorem 28. A number c ∈ R is a cluster point of A ⊆ R if and only if there exists a sequence (an ) in A such that lim(an ) = c, and an 6= c for all n ∈ N. Proof. For the forward implication, let c be a cluster point of A. Then for any n ∈ N we may choose δ = 1/n such that every 1/n-neighbourhood V1/n (c) of c contains at least one point an of A distinct from c. Since an ∈ A, an 6= c, and |an − c| < 1/n, this implies that lim(an ) = c. Conversely, if there exists a sequence (an ) in the set A {c} with lim(an ) = c, then for every δ > 0 there must exist some K ∈ N such that if n ≥ K then |an − c| < δ; that is, an ∈ Vδ (c), an 6= c. Since the δ-neighbourhood Vδ (c) of the point c contains the points an for n ≥ K belonging to A but distinct from c, c is a cluster point of A.
3.1.1
The Definition of the Limit
In the following definition of the limit of a function f at a point c it is important, again, to remember that the function f need not be defined at the point c. Definition 13. Let A ⊆ R, and let c be a cluster point of A. Then, a function f : A → R has a limit L ∈ R at the point c, written L = lim f (x) x→c
if for any ε > 0 there exists a δ(ε) > 0 such that if x ∈ A and 0 < |x − c| < δ, then |f (x) − L| < ε. If the limit of f at c does not exist, then f diverges at c. In reading the above definition, it is useful to visualise 0 < |x − c| < δ as the set of all x ∈ A within the neighbourhood (c − δ, c + δ) but not including c itself (as implied by the inequality 0 < |x − c|). The following theorem expresses our definition in the language of neighbourhoods: 15
Theorem 29. Let f : A → R, and let c be a cluster point of A. Then the following statements are equivalent: 1. limx→c f (x) = L 2. Given any ε-neighbourhood Vε (L) of L, there exists a δ-neighbourhood Vδ (c) of c such that if x 6= c is any point in Vδ (c) ∩ A, then f (x) belongs to Vε (L). Now that we have formally defined the limit L of a function f at a point c, it is natural to ask whether the limit L is unique. Theorem 30. Let f : A → R and c be a cluster point of A. Then f can have at most one limit at c.
3.1.2
Sequential Criterion for Limits
Theorem 31 (Sequential Criterion). Let f : A → R, and let c be a cluster point of A. Then the following statements are equivalent: 1. limx→c f = L 2. For every sequence (xn ) in A that converges to c such that xn 6= c for all n ∈ N, the sequence (f (xn )) converges to L. Proof. TODO...
3.1.3
Divergence Criteria
Theorem 32 (Divergence Criteria). Let A ⊆ R, and let f : A → R and c ∈ R be a cluster point of A. 1. If L ∈ R, then f does not have a limit L at c if and only if there exists a sequence (xn ) in A with xn 6= c for all n ∈ N such that the sequence (xn ) converges to c but the sequence (f (xn )) does not converge to L. 2. The function f does not have a limit at c if and only if there exists a sequence (xn ) in A with xn 6= c for all n ∈ N such that the sequence (xn ) converges to c but the sequence (f (xn )) does not converge in R.
3.2
Limit Theorems
The theorems in this section are useful in calculating the limits of functions. Definition 14. Let A ⊆ R and c ∈ R be a cluster point of A. If f : A → R, then f is bounded on a neighbourhood of c if there exists a δ-neighbourhood Vδ (c) of c and a constant M > 0 such that |f (x)| ≤ M for all x ∈ A ∩ Vδ (c) Theorem 33. If A ⊆ R, and f : A → R has a limit at c ∈ R, then f is bounded on some neighbourhood of c. Theorem 34. Let A ⊆ R, and let f, g be functions defined on A to R. Then for all x ∈ A (f ± g)(x) := f (x) ± g(x),
(f g)(x) := f (x)g(x)
Furthermore, if b ∈ R, then for all x ∈ A (bf )(x) := bf (x) Finally, if h(x) 6= 0, then for all x ∈ A f f (x) (x) := g g(x) 16
Theorem 35. Let A ⊆ R, and let c ∈ R be a cluster point of A. If f, g are functions on A to R, and b ∈ R, then 1. If limx→c f = L and limx→c g = M lim (f ± g) = L ± M,
x→c
lim (f g) = LM,
x→c
lim (bf ) = bL
x→c
2. If h : A → R with h(x) 6= 0 for all x ∈ A, and if limx→c h = H 6= 0, then f L lim = x→c h H Theorem 36. Let A ⊆ R, and let c ∈ R be a cluster point of A. If f : A → R and a ≤ f (x) ≤ b for all x ∈ A, x 6= c, and if limx→c f exists, then a ≤ limx→c f ≤ b. Theorem 37 (Squeeze Theorem). Let A ⊆ R, and let c be a cluster point on A. If f, g, h : A → R and f (x) ≤ g(x) ≤ h(x) for all x ∈ A, x 6= c, and if limx→c f = L = limx→c h, then limx→c g = L. Theorem 38. TODO...
17
18
Chapter 4
Continuous Functions 4.1
Continuous Functions
Definition 15. Let A ⊆ R, and let c ∈ R. If f : A → R, then f is continuous at c if, given any number ε > 0, there exists a δ > 0 such that if x is any point of A satisfying |x − c| < δ, then |f (x) − f (c)| < ε. If f is not continuous at the point c, then it is discontinuous at c. We can similarly formulate a definition of continuity at a point in terms of neighbourhoods: Theorem 39. A function f : A → R is continuous at a point c ∈ A if and only if given any ε-neighbourhood Vε (f (c)) there exists a δ-neighbourhood Vδ (c) such that if x is any point of A ∩ Vδ (c), then f (x) belongs to Vε (f (c)); that is, f (A ∩ Vδ (c)) ⊆ Vε (f (c)) TODO: Read remarks to theorem carefully... Theorem 40 (Sequential Criterion for Continuity). A function f : A → R is continuous at the point c ∈ A if and only if for every sequence (xn ) in A that converges to c, the sequence (f (xn )) converges to f (c). Theorem 41 (Discontinuity Criterion). Let A ⊆ R, and let c ∈ A. If f : A → R, then f is discontinuous at c if and only if there exists a sequence (xn ) in A such that (xn ) converges to c, but the sequence (f (xn )) does not converge to f (c). Thus far we have discussed continuity at a point. What about continuity on a set? A function is continuous on a set if it is continuous at every point on the set. Definition 16. Let A ⊆ R and let f : A → R. If B is a subset of A, then f is continuous on the set B if f is continuous at every point of B.
4.2 4.2.1
Combination of Continuous Functions Continuity
The following theorem applies to the continuity of combinations of functions at a point in R: Theorem 42. Let A ⊆ R and b ∈ R. If f, g : A → R are continuous at c ∈ A, then 1. f ± g, f g, and bf are continuous at c 2. If h : A → R is continuous at c and h(x) 6= 0 for all x ∈ A, then f /h is continuous at c. The following theorem applies to the continuity of combinations of functions on a subset of R: 19
Theorem 43. Let A ⊆ R and b ∈ R. If f, g : A → R are continuous on A, then 1. f ± g, f g, and bf are continuous on A 2. If h : A → R are continuous at A and h(x) 6= 0 for all x ∈ A, then f /h is continuous on A.
4.2.2
Composition of Functions
Theorem 44. Let A, B ⊆ R and let f : A → R and g : B → R be functions such that f (A) ⊆ B. If f is continuous at c ∈ A, and g is continuous at b = f (c) ⊆ B, then the composition g ◦ f : A → R is continuous at c. Theorem 45. Let A, B ⊆ R, and let f : A → R be continuous on A and g : B → R be continuous on B. If f (A) ⊆ B, then the composite function g ◦ f : A → R is continuous.
20
