Advanced Honors Calculus, I and II (Fall 2017 and Winter 2018) Volker Runde

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1 Advanced Honors Calculus, I and II (Fall 27 and Winter 28) Volker Runde April 23, 28

2 Contents Introduction 3 The Real Number System and Finite-Dimensional Euclidean Space 4. The Real Line Functions The Euclidean Space R N Topology Limits and Continuity Limits of Sequences Limits of Functions Global Properties of Continuous Functions Uniform Continuity Differentiation in R N 6 3. Differentiation in One Variable: A Review Partial Derivatives Vector Fields Total Differentiability Taylor s Theorem Classification of Stationary Points Integration in R N 9 4. Content in R N The Riemann Integral in R N Evaluation of Integrals in One Variable: A Review Fubini s Theorem Integration in Polar, Spherical, and Cylindrical Coordinates

3 5 The Implicit Function Theorem and Applications Local Properties of C -Functions The Implicit Function Theorem Local Extrema with Constraints Change of Variables and the Integral Theorems by Green, Gauß, and Stokes Change of Variables Curves in R N Curve Integrals Green s Theorem Surfaces in R Surface Integrals and Stokes Theorem Gauß Theorem Infinite Series and Improper Integrals Infinite Series Improper Riemann Integrals Sequences and Series of Functions Uniform Convergence Power Series Fourier Series A Linear Algebra 255 A. Linear Maps and Matrices A.2 Determinants A.3 Eigenvalues B Stokes Theorem for Differential Forms 265 B. Alternating Multilinear Forms B.2 Integration of Differential Forms B.3 Stokes Theorem C Limit Superior and Limit Inferior 272 C. The Limit Superior C.2 The Limit Inferior

4 Introduction The present notes are based on the courses MATH 27 and 37 as I taught them in the academic year 24/25 and later, again, in 26/7 and then in 27/8. It is an updated (and debugged) version of previous incarnations of these notes. The most distinctive notion of this version is that it includes exercises. Also, some new material has been added to Sections 6.3 (on conservative vector fields) and 8.3 (Weierstraß Approximation Theorem). The notes are not intended replace any of the many textbooks on the subject, but rather to supplement them by relieving the students from the necessity of taking notes and thus allowing them to devote their full attention to the lecture. Of course, the degree of originality conveyed in these notes is (very) limited. In putting them together, I mostly relied on the following sources:. James S. Muldowney, Advanced Calculus Lecture Notes for Mathematics Third Edition. (available online); 2. Robert G. Bartle, The Elements of Real Analysis. Second Edition. Jossey-Bass, 976; 3. Otto Forster, Analysis 2. Vieweg, 984; 4. Harro Heuser, Lehrbuch der Analysis, Teil 2. Teubner, 983. It ought to be clear that these notes may only be used for educational, non-profit purposes. Volker Runde, Edmonton April 23, 28 3

5 Chapter The Real Number System and Finite-Dimensional Euclidean Space. The Real Line What is R? Intuitively, one can think of R as of a line stretching from to. Intuitition, however, can be deceptive in mathematics. In order to lay solid foundations for calculus, we introduce R from an entirely formalistic point of view: we demand from a certain set that it satisfies the properties that we intuitively expect R to have, and then just define R to be this set! What are the properties of R we need to do mathematics? First of all,we should be able to do arithmetic. Definition... A field is a set F together with two binary operations + and satisfying the following: (F) for all x, y F, we have x + y F and x y F as well; (F2) for all x, y F, we have x + y = y + x and x y = y x (commutativity); (F3) for all x, y, z F, we have x + (y + z) = (x + y) + z and x (y z) = (x y) z (associativity); (F4) for all x, y, z F, we have x (y + z) = x y + x z (distributivity); (F5) there are, F with such that for all x F, we have x + = x and x = x (existence of neutral elements); 4

6 (F6) for each x F, there is x F such that x + ( x) =, and for each x F \ {}, there is x F such that x x = (existence of inverse elements). Items (F) to (F6) in Definition.. are called the field axioms. For the sake of simplicity, we use the following shorthand notation: xy := x y; x + y + z := x + (y + z); xyz := x(yz); x y := x + ( y); x y := xy (where y ); x n := x x }{{} n times x :=. (where n N); Examples.. Q, R, and C are fields. 2. Let F be any field then { } p F(X) := : p and q are polynomials in X with coeffients in F and q q is a field. 3. Define + and on {A, B} through the following tables: + A B A A B B B A and A B A A A B A B This turns {A, B} into a field as is easily verified. 4. Define + and on {,, }: + and This turns {,, } into a field as is also routinely verified. 5

7 5. Let F[X] := {p : p is a polynomial in X with coefficients in F}. Then F[X] is not a field because, for instance, X has no multiplicative inverse. 6. Both Z and N are not fields. There are several properties of a field that are not part of the field axioms, but which, nevertheless, can easily be deduced from them:. The neutral elements and are unique: Suppose that both and 2 are neutral elements for +. Then we have = + 2, by (F5), = 2 +, by (F2), = 2, again by (F5). A similar argument works for. 2. The inverses x and x are uniquely determined by x: Let x, and let y, z F be such that xy = xz =. Then we have y = y(xz), by (F5) and (F6), = (yx)z, by (F3), = (xy)z, by (F2), = z(xy), again by (F2), = z, again by (F5) and (F6). A similar argument works for x. 3. x = for all x F. Proof. We have x = x( + ), by (F5), = x + x, by (F4). This implies = x x, by (F6), = (x + x) x, = x + (x x), by (F3), = x, which proves the claim. 6

8 4. ( x)y = xy holds for all x, y F. Proof. We have xy + ( x)y = (x x)y =. Uniqueness of xy then yields that ( x)y = xy. 5. For any x, y F, the identity ( x)( y) = (x( y)) = ( xy) = xy holds. 6. If xy =, then x = or y =. Proof. Suppose that x, so that x exists. Then we have which proves the claim. y = y(xx ) = (yx)x =, Of course, Definition.. is not enough to fully describe R. Hence, we need to take properties of R into account that are not merely arithmetic anymore: Definition..2. An ordered field is a field O together with a subset P with the following properties: (O) for x, y P, we have x + y P as well; (O2) for x, y P, we have xy P, as well; (O3) for each x O, exactly one of the following holds: (i) x P ; (ii) x = ; (iii) x P. Again, we introduce shorthand notation: x < y : y x P ; x > y : y < x; x y : x < y or x = y; x y : x > y or x = y. As for the field axioms, there are several properties of odered fields that are not part of the order axioms (Definition..2(O) to (O3)), but follow from them without too much trouble: 7

9 . x < y and y < z implies x < z. Proof. If y x P and z y P, then (O), implies that z x = (z y)+(y x) P as well. 2. If x < y, then x + z < y + z for any z O. Proof. This holds because (y + z) (x + z) = y x P. 3. x < y and z < u implies that x + z < y + u. 4. x < y and t > implies tx < ty. Proof. We have ty tx = t(y x) P by (O2). 5. x < y and t < s implies tx < sy. 6. x < y and t < implies tx > ty. Proof. We have tx ty = t(x y) = t(y x) P because t P by (O3). 7. x 2 > holds for any x. Proof. If x >, then x 2 > by (O2). Otherwise, x > must hold by (O3), so that x 2 = ( x) 2 > as well. In particular = 2 >. 8. x > for each x >. Proof. This is true because holds. x = x x x = (x ) 2 x >. 9. < x < y implies y < x. 8

10 Proof. The fact that xy > implies that x y = (xy) >. It follows that holds as claimed. y = x(x y ) < y(x y ) = x Examples.. Q and R are ordered. 2. C cannot be ordered. Proof. Assume that P C as in Definition..2 does exist. We know that P. On the other hand, we have = i 2 P, which contradicts (O3). 3. {, } cannot be ordered. Proof. Assume that there is a set P as required by Definition..2. Since P and / P, it follows that P = {}. But this implies = + P contradicting (O). Similarly, it can be shown that {,, 2} cannot be ordered. The last two of these examples are just instances of a more general phenomenon: Proposition..3. Let O be an ordered field. Then we can identify the subset {, +, + +,...} of O with N. Proof. Let n, m N be such that } + {{ + } = } + {{ + }. n times m times Without loss of generality, let n m. Assume that n > m. Then = } + {{ + } n times + + }{{} m times must hold, which is impossible. Hence, we have n = m. = } + {{ + } > n m times Hence, if O is an ordered field, it contains a copy of the infinite set N and thus has to be infinite itself. This means that no finite field can be ordered. Both R and Q satisfy (O), (O2), and (O3). Hence, (F) to (F6) combined with (O), (O2), and (O3) still do not fully characterize R. Definition..4. Let O be an ordered field, and let S O. Then C O is called: (a) an upper bound for S if x C for all x S (in this case S is called bounded above); 9

11 (b) a lower bound for S if x C for all x S (in this case S is called bounded below). If S is both bounded above and below, simply call it bounded. Example. The set {q Q : q and q 2 2} is bounded below (by ) and above by 28. Definition..5. Let O be an ordered field, and let S O. Then: (a) an upper bound for S is called the supremum of S (in short: sup S) if sup S C for every upper bound C for S; (b) a lower bound for S is called the infimum of S (in short: inf S) if inf S C for every lower bound C for S. Remark. It is easy to see that, whenever a set has a supremum or an infimum, then they are unique. Example. The set S := {q Q : 2 q < 3} is bounded such that inf S = 2 and sup S = 3. Clearly, 2 is a lower bound for S and since 2 S, it must be inf S. Cleary, 3 is an upper bound for S; if r Q were an upper bound of S with r < 3, then can not be in S anymore whereas 2 (r + 3) > (r + r) = r 2 2 (r + 3) < (3 + 3) = 3 2 implies the opposite. Hence, 3 is the supremum of S. Do infima and suprema always exist in ordered fields? We shall soon see that this is not the case in Q. Definition..6. An ordered field O is called complete if sup S exists for every S O which is bounded above. We shall use completeness to define R: Definition..7. R is a complete ordered field. It can be shown that R is the only complete ordered field (see Exercise 8 below) even though this is of little relevance for us: the only properties of R we are interested in are those of a complete ordered field. From now on, we shall therefore rely on Definition..7 alone when dealing with R. Here are a few consequences of completeness:

12 Theorem..8. R is Archimedean, i.e., N is not bounded above. Proof. Assume otherwise. Then C := sup N exists. Since C < C, it is impossible that C is an upper bound for N. Hence, there is n N such that C < n. This, in turn, implies that C < n +, which is impossible. Corollary..9. Let ɛ >. Then there is n N such that < n < ɛ. Proof. By Theorem..8, there is n N such that n > ɛ. This yields n < ɛ. Example. Let S := { n } : n N R Then S is bounded below by and above by. Since S, we have inf S =. Assume that sup S <. Let ɛ := sup S. By Corollary..9, there is n N with < n < ɛ. But this, in turn, implies that n > ɛ = sup S, which is a contradiction. Hence, sup S = holds. Corollary... Let x, y R be such that x < y. x < q < y. Then there is q Q such that Proof. By Corollary..9, there is n N such that n < y x. Let m Z be the smallest integer such that m > nx, so that m nx. This implies Division by n yields x < m n < y. nx < m nx + < nx + n(y x) = ny. Theorem... There is a unique x R \ Q with x such that x 2 = 2. Proof. Let S := {y R : y and y 2 2}. Then S is non-empty and bounded above, so that x := sup S exists. Clearly, x holds. We first show that x 2 = 2. Assume that x 2 < 2. Choose n N such that n < 5 (2 x2 ). Then ( x + ) 2 = x 2 + 2x n n + n 2 x 2 + (2x + ) n x n < x x 2 < 2

13 holds, so that x cannot be an upper bound for S. Hence, we have a contradiction, so that x 2 2 must hold. Assume now that x 2 > 2. Choose n N such that n < 2x (x2 2), and note that ( x ) 2 = x 2 2x n n + n 2 x 2 2x n x 2 (x 2 2) = 2 y 2 for all y S. This, in turn, implies that n y for all y S. Hence, x n upper bound for S, which contradicts the definition of sup S. All in all, x 2 = 2 must hold. To prove the uniqueness of x, let z be such that z 2 = 2. It follows that < x is an = 2 2 = x 2 z 2 = (x z)(x + z), so that x + z = or x z =. Since x, z, x + z = would imply that x = z =, which is impossible. Hence, x z = must hold, i.e., x = z. We finally prove that x / Q. Assume that x = m n with n, m N. Without loss of generality, suppose that m and n have no common divisor except. We clearly have 2n 2 = m 2, so that m 2 must be even. Therefore, m is even, i.e. there is p N such that m = 2p. Thus, we obtain 2n 2 = 4p 2 and consequently n 2 = 2p 2. Hence, n 2 is even and so is n. But if m and n are both even, they have the divisor 2 in common. This is a contradiction. The proof of this theorem shows that Q is not complete: if the set {q Q : q and q 2 2} had a supremum in Q, this this supremum would be a rational number x with x 2 =. But the theorem asserts that no such rational number can exist. For a, b R with a < b, we introduce the following notation: [a, b] := {x R : a x b} (a, b) := {x R : a < x < b} (a, b] := {x R : a < x b}; [a, b) := {x R : a x < b}. (closed interval); (open interval); Theorem..2 (Nested Interval Property). Let I, I 2, I 3,... be a decreasing sequence of closed intervalls, i.e., I n = [a n, b n ] such that I n+ I n for all n N. Then n= I n. 2

14 Proof. For all n N, we have a a n a n+ < b n+ b n b. Hence, each b m is an upper bound for {a n : n N} for any m N. Let x := sup{a n : n N}. Hence, a n x b m holds for all n N, i.e., x I n for all n N and thus x n= I n. a a n a n + b n + b n b Figure.: Nested Interval Property The theorem becomes false if we no longer require the intervals to be closed: Example. For n N, let I n := (, ] n, so that In+ I n. Assume that there is ɛ n= I n, so that ɛ >. By Corollary..9, there is n N with < n < ɛ, so that ɛ / I n. This is a contradiction. Definition..3. For x R, let x := { x, if x, x, if x. Proposition..4. Let x, y R, and let t. Then the following hold: (i) x = x = ; (ii) x = x ; (iii) xy = x y ; (iv) x t t x t; (v) x + y x + y (triangle inequality); (vi) x y x y. Proof. (i), (ii), and (iii) are routinely checked. (iv): Suppose that x t. If x, we have t x = x t; for x, we have x and thus t x t. This implies t x t. Hence, t x t holds for any x with x t. Conversely, suppose that t x t. For x, this means x = x t. For x, the inequality t x implies that x = x t. 3

15 (v): By (iv), we have x x x and y y y. Adding these two inequalities yields ( x + y ) x + y x + y. Again by (iv), we obtain x + y x + y as claimed. (vi): By (v), we have x = x y + y x y + y and hence x y x y. Exchanging the rôles of x and y yields ( x y ) = y x y x = x y, so that x y x y holds by (iv). Exercises. Let + and be defined on {,,, A} through: + A A A A A A A A A A. Do these turn {,,, A} into a field? 2. Show that [ ] { Q 2 := p + q } 2 : p, q Q, with + and inherited from R, is a field. (Hint: Many of the field axioms are true for Q [ 2 ] simply because they are true for R; in this case, just point it out and don t verify the axiom in detail.) 3. Let O be an ordered field, and let x, y, z, u O: 4

16 (a) suppose that x < y and z < u, and show that x + z < y + u; (b) suppose that x < y and z < u, and show that xz < yu. You may use the axioms of an ordered field and all the properties that were derived from them in class. 4. Let S R be bounded below, and let S := { x : x S}. Show that: (a) S is bounded above. (b) S has an infimum, namely inf S = sup( S). 5. Find sup S and inf S in R for S := ( {( ) n ) } : n N. n Justify, i.e., prove, your findings. 6. Let S, T R be non-empty and bounded above. Show that S + T := {x + y : x S, y T } is also bounded above with sup(s + T ) = sup S + sup T. 7. An ordered field O is said to have the nested interval property if n= I n for each decreasing sequence I I 2 I 3 of closed intervals in O. Show that an Archimedean ordered field with the nested interval property is complete. 8. Let R be a complete ordered field, and let ι : Q R be the canonical embedding. Show that defines a bijective map satisfying: ι: R R, x sup{ι (q) : q Q, q x} ι(x + y) = ι(x) + ι(y) for x, y R; ι(xy) = ι(x)ι(y) for x, y R; ι(x) > if x >. 9. Let x, y R with x < y. Show that there is z R \ Q such that x < z < y. 5

17 .2 Functions In this section, we give a somewhat formal introduction to functions and introduce the notions of injectivity, surjectivity, and bijectivity. We use bijective maps to define what it means for two (possibly infinite) sets to be of the same size and show that N and Q have the same size whereas R is larger than Q. Definition.2.. Let A and B be non-empty sets. A subset f of A B is called a function or map if, for each x A, there is a unique y B such that (x, y) f. For a function f A B, we write f : A B and y = f(x) : (x, y) f. We then often write f : A B, x f(x). The set A is called the domain of f, and B is called its co-domain. Definition.2.2. Let A and B be non-empty sets, let f : A B be a function, and let X A and Y B. Then f(x) := {f(x) : x X} B is the image of X (under f), and f (Y ) := {x A : f(x) Y } A is the inverse image of Y (under f). Example. Consider sin: R R, i.e., {(x, sin(x)) : x R} R R. Then we have: sin(r) = [, ]; sin([, π]) = [, ]; sin ({}) = {nπ : n Z}; sin ({x R : x 7}) =. Definition.2.3. Let A and B be non-empty sets, and let f : A B be a function. Then f is called: (a) injective if f(x ) f(x 2 ) whenever x x 2 for x, x 2 A; (b) surjective if f(a) = B; (c) bijective if it is both injective and surjective. 6

18 Examples.. The function f : R R, x x 2 is neither injective nor surjective, whereas is injective, but not surjective, and is bijective. f 2 : [, ) }{{} R, x x 2 :={x R:x } f 3 : [, ) [, ), x x 2 2. The function sin: [, 2π] [, ], x sin(x) is surjective, but not injective. For finite sets, it is obvious what it means for two sets to have the same size or for one of them to be smaller or larger than the other one. For infinite sets, matters are more complicated: Example. Let N := N {}. Then N is a proper subset of N, so that N should be smaller than N. On the other hand, N N, n n + is bijective, i.e., there is a one-to-one correspondence between the elements of N and N. Hence, N and N should have the same size. We use the second idea from the previous example to define what it means for two sets to have the same size : Definition.2.4. Two sets A and B are said to have the same cardinality (in symbols: A = B ) if there is a bijective map f : A B. Examples.. If A and B are finite, then A = B holds if and ony if A and B have the same number of elements. 2. By the previous example, we have N = N even though N is a proper subset of N. 3. The function f : N Z, n ( ) n n 2 is bijective, so that we can enumerate Z as {,,, 2, 2,...}. As a consequence, N = Z holds even though N Z. 7

19 4. Let a, a 2, a 3,... be an enumeration of Z. We can then write Q as a rectangular scheme that allows us to enumerate Q, so that Q = N. a a 2 a 3 a 4 a 5 a 2 a 2 2 a 3 2 a 4 2 a 5 2 a 3 a 2 3 a 3 3 a 4 a a 4 a 2 4 a 3 a 4 a a 5 a 2 a 3 a 4 a Figure.2: Enumeration of Q 5. Let a < b. The function f : [a, b] [, ], x x a b a is bijective, so that [a, b] = [, ]. Definition.2.5. A set A is called countable if it is finite or if A = N. A set A is countable, if and only if we can enumerate it, i.e., A = {a, a 2, a 3,...} where the sequence a, a 2, a 3,... may break off after a finite number of terms. As we have already seen, the sets N, N, Z, and Q are all countable. But not all sets are: Theorem.2.6. The sets [, ] and R are not countable. 8

20 Proof. We only consider [, ] (this is enough because it is easy to see that a an infinite subsets of a countable set must again be countable). Each x [, ] has a decimal expansion x =.ɛ ɛ 2 ɛ 3 (.) with ɛ, ɛ 2, ɛ 3,... {,, 2,..., 9}. Assume that there is an enumeration [, ] = {a, a 2, a 3,...}. Define x [, ] using (.) by letting, for n N, { 6, if the n-th digit of a n is 7, ɛ n := 7, if the n-th digit of a n is not 7 Let n N be such that x = a n. Case : The n-th digit of a n is 7. Then the n-th digit of x is 6, so that a n x. Case 2: The n-th digit of a n is not 7. Then the n-th digit of x is 7, so that a n x, too. Hence, x / {a, a 2, a 3,...}, which contradicts [, ] = {a, a 2, a 3,...}. The argument used in the proof of Theorem.2.6 is called Cantor s Diagonal argument. Exercises. For any set S, its power set P(S) is defined to be the set consisting of all subsets of S. Show that there is no surjective map from S to P(S). (Hint: Assume that there is a surjective map f : S P(S) and consider the set {x S : x / f(x)}.).3 The Euclidean Space R N Recall that, for any sets S,..., S N, their (N-fold) Cartesian product is defined as S S N := {(s,..., s N ) : s j S j for j =,..., N}. The N-dimensional Euclidean space is defined as R N := R } {{ R } = {(x,..., x N ) : x,..., x N R}. N times An element x := (x,..., x N ) R N is called a point or vector in R N ; the real numbers x,..., x N R are the coordinates of x. The vector := (,..., ) is the origin or zero vector of R N. (For N = 2 and N = 3, the space R N can be identified with the plane and three-dimensional space of geometric intuition.) 9

21 We can add vectors in R N and multiply them with real numbers: For two vectors x = (x,..., x N ), y := (y,..., y N ) R N and a scalar λ R define: x + y := (x + y,..., x N + y N ) (addition); λx := (λx,..., λx N ) (scalar multiplication). The following rules for addition and scalar multiplication in R N are easily verified: x + y = y + x; (x + y) + z = x + (y + z); + x = x; x + ( )x = ; x = x; x = ; λ(µx) = (λµ)x; λ(x + y) = λx + λy; (λ + µ)x = λx + µx. This means that R N is a vector space. Definition.3.. The inner product on R N is defined by N x y := x j y j for x = (x,..., x N ), y := (y,..., y N ) R N. Proposition.3.2. The following hold for all x, y, z R N and λ R: (i) x x ; (ii) x x = x = ; (iii) x y = y x; (iv) x (y + z) = x y + x z; (v) (λx) y = λ(x y) = x λy. Definition.3.3. The (Euclidean) norm on R N is defined by x := x x = N for x = (x,..., x N ). j= j= x 2 j 2

22 For N = 2, 3, the norm x of a vector x R N can be interpreted as its length. The Euclidean norm on R N thus extends the notion of length in 2- and 3-dimensional space, respectively, to arbitrary dimensions. Lemma.3.4 (Geometric versus Arithmetic Mean). For x, y, the inequality holds with equality if and only if x = y. xy (x + y) 2 Proof. We have x 2 2xy + y 2 = (x y) 2 (.2) with equality if and only if x = y. This yields xy xy + 4 (x2 2xy + y 2 ) = xy + 4 x2 2 xy + 4 y2 = 4 x2 + 2 xy + 4 y2 = 4 (x2 + 2xy + y 2 ) = 4 (x + y)2. Taking roots yields the desired inequality. It is clear that we have equality if and only if the second summand in (.3) vanishes; by (.2) this is possible only if x = y. Theorem.3.5 (Cauchy Schwarz Inequality). We have x y N x j y j x y j= for x = (x,..., x N ), y := (y,..., y N ) R N. Proof. The first inequality is clear due to the triangle inequality in R. If x =, then x = = x N =, so that N j= x jy j = ; a similar argument 2

23 applies if y =. We may therefore suppose that x y. We then obtain N x j y j N x y = ( ) 2 ( ) 2 xj yj x x j= j= [ N ( ) 2 ( ) ] 2 xj yj +, by Lemma.3.4, 2 x x j= = N 2 x 2 x 2 j + N y 2 y 2 j j= j= = [ ] x 2 2 x 2 + y 2 y 2 =. Multiplication by x y yields the claim. Proposition.3.6 (Properties of ). For x, y R N and λ R, we have: (i) x ; (ii) x = x = ; (iii) λx = λ x ; (iv) x + y x + y ( Triangle Inequality); (v) x y x y. Proof. (i), (ii), and (iii) are easily verified. For (iv), note that x + y 2 = (x + y) (x + y) = x y + x y + y x + y y = x 2 + 2x y + y 2 x x y + y 2, by Theorem.3.5, = ( x + y ) 2. Taking roots yields the claim. For (v), note that by (iv) with x and y replaced by x y and y x = (x y) + y x y + y, holds, so that x y x y. 22

24 Interchanging x and y yields y x y x = x y, so that x y x y x y. This proves (v). We now use the norm on R N to define two important types of subsets of R N : Definition.3.7. Let x R N and let r >. Then: (a) the open ball in R N centered at x with radius r is the set B r (x ) := {x R N : x x < r}. (b) the closed ball in R N centered at x with radius r is the set B r (x ) := {x R N : x x r}. For N =, B r (x ) and B r [x ] are nothing but open and closed intervals, respectively, namely Moreover, if a < b, then B r (x ) = (x r, x + r) and B r [x ] = [x r, x + r]. (a, b) = (x r, x + r) and [a, b] = [x r, x + r] holds, with x := 2 (a + b) and r := 2 (b a). For N =, B r (x ) and B r [x ] are just disks with center x and radius r, where the circle is not include in the case of B r (x ), but is included for B r [x ]. Finally, if N = 3, then B r (x ) and B r [x ] are balls in the sense of geometric intuation. In the open case, the surface of the ball is not included, but it is include in the closed ball. Definition.3.8. A set C R N is called convex if tx + ( t)y C for all x, y C and t [, ]. In plain language, a set is convex if, for any two points x and y in the C, the whole line segment joining x and y is also in C. 23

25 y x Figure.3: A convex subset of R 2 24

26 x y Figure.4: Not a convex subset of R 2 Proposition.3.9. Let x R N. Then B r (x ) and B r [x ] are convex. Proof. We only prove the claim for B r (x ) in detail. Let x, y B r (x ) and t [, ]. Then we have tx + ( t)y x = t(x x ) + ( t)(y x ) t x x + ( t) y x < tr + ( t)r = r, so that tx + ( t)y B r (x ). The claim for B r [x ] is proved similarly, but with instead of < in (.3). Let I,..., I N R be closed intervals, i.e. I j = [a j, b j ] where a j < b j for j =,..., N. 25

27 Then I := I I N is called a closed interval in R N. We have I = {(x,..., x N ) R N : a j x j b j for j =,..., N}. For N = 2, a closed interval in R N, i.e., in the plane, is just a rectangle. For N = 3, a closed interval in R 3 is a rectangular box. Theorem.3. (Nested Interval Property in R N ). Let I, I 2, I 3,... be a decreasing sequence of closed intervals in R N. Then n= I n holds. Proof. Each interval I n is of the form I n = I n, I n,n with closed intervals I n,,..., I n,n in R. For each j =,..., N, we have I,j I 2,j I 3,j, i.e., the sequence I,j, I 2,j, I 3,j,... is a decreasing sequence of closed intervals in R. By Theorem..2, this means that n= I n,j, i.e., there is x j I n,j for all n N. Let x := (x,..., x N ). Then x I n, I n,n holds for all n N, which means that x n= I n. Exercises. For x = (x,..., x N ) R N, set x := x + + x N and x := max{ x,..., x N }. (a) Show that the following are true for j =,, x, y R N and λ R: (i) x j and x j = if and only if x = ; (ii) λx j = λ x j ; (iii) x + y j x j + y j. (b) For N = 2, sketch the sets of those x for which x x. (c) Show that, x, and x N x N x for all x R N. 2. Let x, y R N. Show that x y = x y holds if and only if x and y are linearly dependent. 26

28 3. Show that x + y 2 = x 2 + y 2 x y = for any x, y R N. 4. Let C be a family of convex sets in R N. Show that C C C is again convex. Is C C C necessarily convex?.4 Topology The word topology derives from the Greek and literally means study of places. mathematics, topology is the discipline that provides the conceptual framework for the study of continuous functions: Definition.4.. Let x R N. A set U R N is called a neighborhood of x if there is ɛ > such that B ɛ (x ) U. In B ε ( x ) U x x ~ Figure.5: A neighborhood of x, but not of x Examples.. If x R N is arbitrary, and r >, then both B r (x ) and B r [x ] are neighborhoods of x. 27

29 2. The interval [a, b] is not a neighborhood of a: To see this assume that is is a neighborhood of a. Then there is ɛ > such that B ɛ (a) = (a ɛ, a + ɛ) [a, b], which would mean that a ɛ a. This is a contradiction. Similarly, [a, b] is not a neighborhood of b, [a, b) is not a neighborhood of a, and (a, b] is not a neigborhood of b. Definition.4.2. A set U R N is open if it is a neighborhood of each of its points. Examples.. and R N are trivially open. 2. Let x R N, and let r >. We claim that B r (x ) is open. Let x B r (x ). Choose ɛ r x x, and let y B ɛ (x). It follows that hence, B ɛ (x) B r (x ) holds. y x y x + x x }{{} <ɛ < r x x + x x = r; 28

30 ε x r x x x B ε (x ) B r ( x ) Figure.6: Open balls are open In particular, (a, b) is open for all a, b R such that a < b. On the other hand, [a, b], (a, b], and [a, b) are not open. 3. The set S := {(x, y, z) R 3 : y 2 + z 2 =, x > } is not open. Proof. Clearly, x := (,, ) S. Assume that there is ɛ > such that B ɛ (x ) S. It follows that (,, + ɛ ) B ɛ (x ) S. 2 On the other hand, however, we have ( + 2) ɛ 2 >, so that (,, + ɛ 2) cannot belong to S. 29

31 To determine whether or not a given set is open is often difficult if one has nothing more but the definition at one s disposal. The following two hereditary properties are often useful: Proposition.4.3. The following are true: (i) if U, V R N are open, then U V is open; (ii) if I is any index set and {U i : i I} is a collection of open sets, then i I U i is open. Proof. (i): Let x U V. Since U is open, there is ɛ > such that B ɛ (x ) U, and since V is open, there is ɛ 2 > such that B ɛ2 (x ) V. Let ɛ := min{ɛ, ɛ 2 }. Then B ɛ (x ) B ɛ (x ) B ɛ2 (x ) U V holds, so that U V is open. (ii): Let x U := i I U i. Then there is i I such that x U i. Since U i there is ɛ > such that B ɛ (x ) U i U. Hence, U is open. is open, Example. The subset n= B n 2 ((n, )) of R2 is open because it is the union of a sequence of open sets. Definition.4.4. A set F R N is called closed if F c := R N \ F := {x R N : x / F } is open. Examples.. and R N are (trivially) closed. 2. Let x R N, and let r >. We claim that B r [x ] is closed. To see this, let x B r [x ] c, i.e., x x > r. Choose ɛ x x r, and let y B ɛ (x). Then we have y x y x x x x x y x > x x x x + r = r, so that B ɛ (x) B r [x ] c. It follows that B r [x ] c is open, i.e., B r [x ] is closed. 3

32 x ε x x Bε(x ) x r B r [ x ] Figure.7: Closed balls are closed In particular, [a, b] is closed for all a, b R with a < b. 3. For a, b R with a < b, the interval (a, b] is not open because (b ɛ, b + ɛ) (a, b] for all ɛ >. But (a, b] is not open either because (a ɛ, a + ɛ) R \ (a, b]. Proposition.4.5. The following are true: (i) if F, G R N are closed, then F G is closed; (ii) if I is any index set and {F i : i I} be a collection of closed sets, then i I F i is closed. Proof. (i): Since F c and G c are open, so is F c G c = (F G) c by Proposition.4.3(i). Hence, F G is closed. (ii): Since F c i is open for each i I, Proposition.4.3(ii) hields the openness of ( ) c Fi c = F i, i I 3 i I

33 which, in turn, means that i I F i is closed. Example. Let x R N. Since {x} = r> B r[x], it follows that {x} is closed. Consequently, if x,..., x n R N, then is closed. {x,..., x n } = {x } {x N } Arbitrary unions of closed sets are, in general, not closed again. Definition.4.6. A point x R N is called a cluster point of S R N if each neighborhood of x contains a point y S \ {x}. Example. Let n S := { } n : n N. Then is a cluster point of S. Let x R be any cluster point of S, and assume that x. If x S, it is of the form x = n for some n N. Let ɛ := n n+, so that B ɛ (x) S = {x}. Hence, x cannot be a cluster point. If x / S, choose n N such that < x 2. This implies that n < x 2 for all n n. Let It follows that (because x { } x ɛ := min, x,..., 2 n x >., 2,..., n / B ɛ(x) k ɛ for k =,..., n. For n n, we have n x x 2 all, we have n / B ɛ(x) for all n N. Hence, is the only accumulation point of S. Definition.4.7. A set S R N is bounded if S B r [] for some r >. ɛ. All in Theorem.4.8 (Bolzano Weierstraß Theorem). Every bounded, infinite subset S R N has a cluster point. Proof. Let r > such that S B r []. It follows that S [ r, r] [ r, r] =: I }{{}. N times We can find 2 N closed intervals I (),..., I(2N ) such that I = 2 N j= I(j), where I (j) = I (j), I(j),N for j =,..., 2 N such that each interval I (j),k has length r. 32

34 Since S is infinite, there must be j {,..., 2 N } such that S I (j ) is infinite. Let I 2 := I (j ). Inductively, we obtain a decreasing sequence I, I 2, I 3,... of closed intervals with the following properites: (a) S I n is infinite for all n N; (b) for I n = I n, I n,n and l(i n ) = max{length of I n,j : j =,..., N}, we have l(i n+ ) = 2 l(i n) = 4 l(i n ) = = 2 n l(i ) = r 2 n. r I (2) = I 2 I I 2 () I 2 (2) r () I (3) I S 2 I 2 (4) r I (3) I (4) r Figure.8: Proof of the Bolzano Weierstraß Theorem From Theorem.3., we know that there is x n= I n. We claim that x is a cluster point of S. Let ɛ >. For y I n note that max{ x j y j : j =,..., N} l(i n ) = 33 r 2 n 2

35 and thus N x y = x j y j 2 j= N max{ x j y j : j =,..., N} N r = 2 n 2. 2 Choose n N so large that N r < ɛ. It follows that I 2 n 2 n B ɛ (x). Since S I n is infinite, B ɛ (x) S must be infinite as well; in particular, B ɛ (x) contains at least one point from S \ {x}. Theorem.4.9. A set F R N is closed if and only if it contains all of its cluster points. Proof. Suppose that F is closed. Let x R N be a cluster point of F and assume that x / F. Since F c is open, it is a neighborhood of x. But F c F = holds by definition. Suppose conversely that F contains its cluster points, and let x R N \ F. Then x is not a cluster point of F. Hence, there is ɛ > such that B ɛ (x) F {x}. Since x / F, this means in fact that B ɛ (x) F =, i.e. B ɛ (x) F c. For our next definition, we first give an example as motivation: Example. Let x R N and let r >. Then S r [x ] := {x R N : x x = r} is the the sphere centered at x with radius r. We can think of S r [x ] as the surface of B r [x ]. Suppose that x S r [x ], and let ɛ >. We claim that both B ɛ (x) B r [x ] and B ɛ (x) B r [x ] c are not empty. For B ɛ (x) B r [x ], this is trivial because S r [x ] B r [x ], so that x B ɛ (x) B r [x ]. Assume that B ɛ (x) B r [x ] c =, i.e., B ɛ (x) B r [x ]. Let t >, and set y t := t(x x ) + x. Note that y t x = t(x x ) + x x = (t )(x x ) = (t )r. Choose t < + ɛ r, then y t B ɛ (x). On the other hand, we have y t x = t x x > r, so that y t / B r [x ]. Hence, B ɛ (x) B r [x ] c is empty. Define the boundary of B r [x ] as B r [x ] := {x R N : B ɛ (x) B r [x ] and B ɛ (x) B r [x ] c are not empty for each ɛ > }. 34

36 By what we have just seen, S r [x ] B r [x ] holds. Conversely, suppose that x / S r [x ]. Then there are two possibilities, namely x B r (x ) or x B r [x ] c. In the first case, we find ɛ > such that B ɛ (x) B r (x ), so that B ɛ (x) B r [x ] c =, and in the second case, we obtain ɛ > with B ɛ (x) B r [x ] c, so that B ɛ (x) B r [x ] =. It follows that x / B r [x ]. All in all, B r [x ] is S r [x ]. This example motivates the following definition: Definition.4.. Let S R N. A point x R N is called a boundary point of S if B ɛ (x) S and B ɛ (x) S c for each ɛ >. We let S := {x R N : x is a boundary point of S} denote the boundary of S. Examples.. Let x R N, and let r >. As for B r [x ], one sees that B r (x ) = S r [x ]. 2. Let x R, and let ɛ >. Then the interval (x ɛ, x + ɛ) contains both rational and irrational numbers. Hence, x is a boundary point of Q. Since x was arbitrary, we conclude that Q = R. Proposition.4.. Let S R N be any set. Then the following are true: (i) S = (S c ); (ii) S S = if and only if S is open; (iii) S S if and only if S is closed. Proof. (i): Since S cc = S, this is immediate from the definition. (ii): Let S be open, and let x S. Then there is ɛ > such that B ɛ (x) S, i.e., B ɛ (x) S c =. Hence, x is not a boundary point or S. Conversely, suppose that S S =, and let x S. Since B r (x) S for each r > (it contains x), and since x is not a boundary point, there must be ɛ > such that B ɛ (x) S c =, i.e., B ɛ (x) S. (iii): Let S be closed. Then S c is open, and by (iii), S c S c =, i.e., S c S. With (ii), we conclude that S S. Suppose that S S, i.e., S S c =. With (ii) and (iii), this implies that S c is open. Hence, S is closed. Definition.4.2. Let S R N. Then S, the closure of S, is defined as S := S {x R N : x is a cluster point of S}. 35

37 Theorem.4.3. Let S R N be any set. Then: (i) S is closed; (ii) S is the intersection of all closed sets containing S; (iii) S = S S. Proof. (i): Let x R N \ S. Then, in particular, x is not a cluster point of S. Hence, there is ɛ > such that B ɛ (x) S {x}; since x / S, we then have automatically that B ɛ (x) S =. Since B ɛ (x) is a neighborhood of each of its points, it follows that no point of B ɛ (x) can be a cluster point of S. Hence, B ɛ (x) lies in the complement of S. Consequently, S is closed. (ii): Let F R N be closed with S F. Clearly, each cluster point of S is a cluster point of F, so that S F {x R N : x is a cluster point of F } = F. This proves that S is contained in every closed set containing S. Since S itself is closed, it equals the intersection of all closed set scontaining S. (iii): By definition, every point in S not belonging to S must be a cluster point of S, so that S S S. Conversely, let x S and suppose that x / S, i.e., x S c. Then, for each ɛ >, we trivially have B ɛ (x) S c, and since x must be a cluster point, we have B ɛ (x) S as well. Hence, x must be a boundary point of S. Examples.. For x R and r >, we have B r (x ) = B r (x ) B r (x ) = B r (x ) S r [x ] = B r [x ]. 2. Since Q = R, we also have Q = R. Definition.4.4. A point x S R N is called an interior point of S if there is ɛ > such that B ɛ (x) S. We let int S := {x S : x is an interior point of S} denote the interior of S. Theorem.4.5. Let S R N be any set. Then: (i) int S is open and equals the union of all open subsets of S; (ii) int S = S \ S. 36

38 Proof. For each x int S, there is ɛ x > such that B ɛx (x) S, so that int S B ɛx (x). (.3) x int S Let y R N be such that there is x int S such that y B ɛx (x). Since B ɛx (x) is open, there is δ y > such that B δy B ɛx (x) S. It follows that y int S, so that the inclusion (.3) is, in fact, an equality. Since the right hand side of (.3) is open, this proves the first part of (i). Let U S be open, and let x U. Then there is ɛ > such that B ɛ (x) U S, so that x int S. Hence, U int S holds. For (ii), let x int S. Then there is ɛ > such that B ɛ (x) S and thus B ɛ (x) S c =. It follos that x S \ S. Conversely, let x S such that x / S. Then there is ɛ > such that B ɛ (x) S = or B ɛ (x) S c =. Since x B ɛ (x) S, the first situation cannot occur, so that B ɛ (x) S c =, i.e., B ɛ (x) S. It follows that x is an interior point of S. Example. Let x R N, and let r >. Then int B r [x ] = B r [x ] \ S r [x ] = B r (x ) holds. Definition.4.6. An open cover of S R N is a family {U i : i I} of open sets in R N such that S i I U i. Example. The family {B r () : r > } is an open cover for R N. Definition.4.7. A set K R N is called compact if every open cover {U i : i I} of K has a finite subcover, i.e., there are i,..., i n I such that K U i U in. Examples.. Every finite set is compact. Proof. Let S = {x,..., x n } R N, and let {U i : i I} be an open cover for S, i.e., x,..., x n i I U i. For j =,..., n, there is thus i j I such that x j U ij. Hence, we have S U i U in. Hence, {U i,, U in } is a finite subcover of {U i : i I}. 2. The open unit interval (, ) is not compact. 37

39 Proof. For n N, let U n := ( n, ). Then {U n : n N} is an open cover for (, ). Assume that (, ) is compact. Then there are n,..., n k N such that (, ) = U n U nk. Without loss of generality, let n < < n k, so that which is nonsense. (, ) = U n U nk = U nk = ( ),, n k 3. Every compact set K R N is bounded. Proof. Clearly, {B r () : r > } is an open cover for K. Since K is compact, there are < r < < r n such that which is possible only if K is bounded. K B r () B rn () = B rn (), Lemma.4.8. Every compact set K R N is closed. Proof. Let x K c. For n N, let U n := B [x] c, so that n K R N \ {x} U n. n= Since K is compact, there are n < < n k in N such that It follows that Hence, K c is a neighborhood of x. K U n U nk = U nk. B n k (x) B n k [x] = U c n k K c. Lemma.4.9. Let K R N be compact, and let F K be closed. Then F is compact. Proof. Let {U i : i I} be an open cover for F. Then {U i : i I} {R N \ F } is an open cover for K. Compactness of K yields i,..., i n I such that Since F (R N \ F ) =, it follows that K U i U in R N \ F. F U i U in. Since {U i : i I} is an arbitrary open cover for F, this entails the compactness of F. 38

40 Theorem.4.2 (Heine Borel Theorem). The following are equivalent for K R N : (i) K is compact; (ii) K is closed and bounded. Proof. (i) = (ii) is clear (no unbounded set is compact, as seen in the examples, and every compact set is closed by Lemma.4.8). (ii) = (i): By Lemma.4.9, we may suppose that K is a closed interval I in R N. Let {U i : i I} be an open cover for I, and suppose that it does not have a finite subcover. As in the proof of ( the) Bolzano Weierstraß Theorem, we may find closed intervals I (),..., ) I(2N with l I (j) = 2 l(i ) for j =,..., 2 N such that I = 2 N j= I(j). Since {U i : i I} has no finite subcover for I, there is j {,..., 2 N } such that {U i : i I} has no finite subcover for I (j ). Let I 2 := I (j ). Inductively, we thus obtain closed intervals I I 2 I 3 such that: (a) l(i n+ ) = 2 l(i n) = = 2 n l(i ) for all n N; (b) {U i : i I} does not have a finite subcover for I n for each n N. Let x n= I n, and let i I be such that x U i. Since U i is open, there is ɛ > such that B ɛ (x) U i. Let y I n. It follows that Choose n N so large that y x N max y j x j j=,...,n N 2 n l(i ) < ɛ. It follows that I n B ɛ (x) U i, so that {U i : i I} has a finite subcover for I n. N 2 n l(i ). Definition.4.2. A disconnection for S R N is a pair {U, V } of open sets such that: (a) U S V S; (b) (U S) (V S) = ; (c) (U S) (V S) = S. If a disconnection for S exists, S is called disconnected; otherwise, we say that S is connected. 39

41 Note that we do not require that U V =. U V S S Figure.9: A set with disconnection Examples.. Z is disconnected: Choose ( U :=, ) 2 and V := ( ) 2, ; the {U, V } is a disconnection for Z. 2. Q is disconnected: A disconnection {U, V } is given by U := (, 2) and V := ( 2, ). 3. The closed unit interval [, ] is connected. Proof. We assume that there is a disconnection {U, V } for [, ]; without loss of generality, suppose that U. Since U is open, there is ɛ >, which we can suppose without loss of generality to be from (, ), such that ( ɛ, ɛ ) U and thus [, ɛ ) U S. Let t := sup{ɛ > : [, ɛ) U [, ]}, so that < ɛ t. Assume that t U. Since U is open, there is δ > such that (t δ, t + δ) U. Since t δ < t, there is ɛ > t δ such that [, ɛ) with [, ɛ) U, so that [, t + δ) [, ] U [, ]. 4

42 If t <, we can choose δ > so small that t + δ <, so that [, t + δ) U [, ], which contradicts the definition of t. If t =, this means that U [, ] = [, ], which is also impossible because it would imply that V [, ] =. We conclude that t / U. It follows that t V. Since V is open, there is θ > such that (t θ, t + θ) V. Since t θ < t, there is ɛ > t θ such that [, ɛ) U [, ]. Pick t (t θ, ɛ). It follows that t (U [, ]) (V [, ]), which is a contradiction. All in all, there is no disconnection for [, ], and [, ] is connected. Theorem Let C R N be convex. Then C is connected. Proof. Assume that there is a disconnection {U, V } for C. Let x U C and let y V C. Let Ũ := {t R : tx + ( t)y U} and Ṽ := {t R : tx + ( t)y V }. We claim that Ũ is open. To see this, let t Ũ. It follows that x := t x + ( t )y U. Since U is open, there is ɛ > such that B ɛ (x ) U. For t R with t t < we thus have (tx + ( t)y) x = (tx + ( t)y) (t x + ( t )y) t t ( x + y ) < ɛ ɛ x + y, and therefore tx + ( t)y B ɛ (x ) U. It follows that t Ũ. Analoguously, one sees that Ṽ is open. The following hold for {Ũ, Ṽ }: (a) Ũ [, ] Ṽ [, ]: since x = x+( ) y U and y = x+( ) y V, we have U and V ; (b) (Ũ [, ]) (Ṽ [, ]) = : if t (Ũ [, ]) (Ṽ [, ]), then tx + ( t)yin(u C) (V C), which is impossible; (c) (Ũ [, ]) (Ṽ [, ]) = [, ]: for t [, ], we have tx + ( t)y C = (U C) (V C) due to the convexity of C, so that t (Ũ [, ]) (Ṽ [, ]). Hence, {Ũ, Ṽ } is a disconnection for [, ], which is impossible. Example., R N, and all closed and open balls and intervals in R N are connected. 4

43 Corollary The only subsets of R N R N. which are both open and closed are and Proof. Let U R N be both open and closed, and assume that U R N. Then {U, U c } would be a disconnection for R N. Exercises. Let S R N. Show that x R N is a cluster point of S if and only if each neighbourhood of x contains an infinite number of points in S. 2. Let S R N be any set. Show that S is closed. 3. For j =,..., N, let I j = [a j, b j ] with a j < b j, and let I := I I N. Determine I. (Hint: Draw a sketch for N = 2 or N = 3.) 4. Which of the following sets are compact: (a) {x R N : r x R} with < r < R; (b) {(x, y) R 2 : x y [, ]}; (c) {(t cos t, t sin t) : t (, )}. Justify your answers. 5. Show that: (a) if U R N and U 2 R M are open, then so is U U 2 R N+M ; (b) if F R N and F 2 R M are closed, then so is F F 2 R N+M ; (c) if K R N and K 2 R M are compact, then so is K K 2 R N+M. 6. Show that a subset K of R N is compact if and only if it has the finite intersection property, i.e., if {F i : i I} is a family of closed sets in R N such that K i I F i =, then there are i,..., i n I such that K F i F in =. 7. Show that the subset { (x, y) R 2 : y x } of R 2 is not convex, but nevertheless connected. 8. Let C R N be connected. Show that C is also connected. 9. Determine whether or not the set { (x, y, z) R 3 : x 2 + y 2 4, z {, } } is (a) open, (b) closed, (c) compact, or (d) connected. 42

44 . Let S R N be arbitrary, and let U R N be open. Show that is open. S + U := {x + y : x S, y U} 43

45 Chapter 2 Limits and Continuity 2. Limits of Sequences Definition 2... A sequence in a set S is a function s: N S. When dealing with a sequence s : N S, we prefer to write s n instead of s(n) and denote the whole sequence s by (s n ) n=. We shall also consider, when the occasion arises, sequences indexed over subsets of Z other than N, e.g., {n Z : n 333}. Definition A sequence (x n ) n= in RN converges or is convergent to x R N if, for each neighborhood U of x, there is n U N such that x n U for all n n U. The vector x is called the limit of (x n ) n=. A sequence that does not converge is said to diverge or to be divergent. Equivalently, the sequence (x n ) n= converges to x RN if, for each ɛ >, there is n ɛ N such that x n x < ɛ for all n n ɛ. If a sequence (x n ) n= in RN converges to x R N n, we write x = lim n x n or x n x or simply x n x. Proposition Every sequence in R N has at most one limit. Proof. Let (x n ) N n= be a sequence in RN with limits x, y R N. Assume that x y, and set ɛ := x y 2. Since x = lim n x n, there is n x N such that x n x < ɛ for n n x, and since also y = lim n x n, there is n y N such that x n y < ɛ for n n y. For n max{n x, n y }, we then have which is impossible. x y x x n + x n y < 2ɛ = x y, 44

46 y x n 2 x x n Figure 2.: Uniqueness of the limit Proposition Every convergent sequence in R N is bounded. We omit the proof which is almost verbatim like in the one-dimensional case. (( )) Theorem Let (x n ) n= = x () n,..., x (N) n be a sequence in n= RN. Then the following are equivalent for x = ( x (),..., x (N)) : (i) lim n x n = x. (ii) lim n x (j) n = x (j) for j =,..., N. Proof. (i) = (ii): Let ɛ >. Then there is n ɛ N such that x n x < ɛ for all n n ɛ, so that x (j) n x (j) xn x < ɛ holds for all n n ɛ and for all j =,..., N. This proves (ii). (ii) = (i): Let ɛ >. For each j =,..., N, there is n (j) ɛ N such that x (j) n x (j) ɛ < N 45

47 { holds for all j =,..., N and for all n n (j) ɛ. Let n ɛ := max that and thus for all n n ɛ. Examples.. The sequence max j=,...,n x n x N x (j) n x (j) < ɛ N max j=,...,n x (j) n x (j) < ɛ ( ) n, 3, 3n2 4 n 2 + 2n n= n () ɛ converges to (, 3, 3), because n, 3 3 and 3n2 4 3 in R. n 2 +2n 2. The sequence ( n 3 + 3n, ( )n ) n= diverges because (( ) n ) n= does not converge in R. },..., n (N) ɛ. It follows Since convergence in R N is nothing but coordinatewise convergence, the following is a straightforward consequence of the limit rules in R: Proposition 2..6 (Limit Rules). Let (x n ) n=, (y n) n= be convergent sequences in RN, and let (λ n ) n= be a convergent sequence in R. Then the sequences (x n + y n ) n=, (λ n x n ) n=, and (x n y n ) n= are also convergent such that: and lim (x n + y n ) = lim x n + lim y n, n n n lim λ nx n = ( lim λ n)( lim x n) n n n lim (x n y n ) = ( lim x n) ( lim y n). n n n Definition Let (s n ) n= be a sequence in a set S, and let n < n 2 <. Then (s nk ) k= is called a subsequence of (x n) n=. As in R, we have: Theorem Every bounded sequence in R N has a convergent subsequence. Proof. Let (x n ) n= be a bounded sequence in RN, and let S := {x n : n N}. If S is finite, (x n ) n= obviously has a constant and thus convergent subsequence. 46