Mathematics for Economists Chapter 1: Basic De…nitions 1. Operations on sets Intersection X \ Y; union X [ Y; complement Z ! Z c ; di¤erence X Y = X \ Y c: Distributivity X \ (Y [ Z) = (X \ Y ) [ (X \ Z) X [ (Y \ Z) = (X [ Y ) \ (X [ Z) (X \ Y )c = X c [ Y c ; (X [ Y )c = X c \ Y c union, intersection are commutative and associative Cartesian product: X Y is the set of ordered pairs (x; y) with x 2 X; y 2 Y: The product X X is not the same as the set of subsets of X with two elements. 2. Functions, mappings f : X ! Y has domain X and values/image in Y; it associates with every x 2 X an element y = f (x) in Y: The range of f is the set f (X) = fy 2 Y j 9 x 2 X : y = f (x)g The graph of f is the subset of X Y : graph (f ) = f(x; f (x)) j x 2 Xg It is sometime convenient to identify a function with its graph: then a func tion is de…ned as a subset of X Y such that every xsection contains exactly one y 2 Y: The set of functions from X into Y is denoted Y X : Example: the set of subsets of X is denoted f0; 1gX or 2X : The function f 2 Y X is onto Y if f (X) = Y ; it is onetoone if x 6= x0 ) f (x) 6= f (x0 ); it is a bijection if it is one to one and onto Y: In the later case, we can speak of the inverse of f; the function f 1 from Y into X de…ned by y = f (x) , x = f 1 (y): Note: if f 2 Y X is not a bijection, its inverse as a function from Y to X is not de…ned. However, if f is onetoone, we may identify it with a function g from X into f (X); and that function has an inverse g 1 from f (X) into X: Hence the need to always specify the domain of a function. Composition of f 2 Y X ; g 2 Z Y : g f (x) = g(f (x)): For instance, if idX denotes the identity mapping of X; i.e., idX (x) = x for all x; and f 2 Y X is a bijection, we have f f 1 = idY ; f 1 f = idX : The composition is an associative operation, it is not commutative. If f 2 Y X and g 2 X Y are such that g f = idX ; we cannot deduce g = f 1 ; or f = g 1 : If g f = idX and f g = idY ; we can. Exercise 1 What can we deduce from g f = idX , or f g = idY ? 1 Exercise 2 If f 2 Y X and g 2 Z Y are onto, so is g f: If they are one to one, so is g f: A bijection from X into itself is called a permutation. Their set is denoted A(X): The permutations of X form a group for the composition: every f 2 A(X) has an inverse f 1 2 A(X): Exercise 3 If X contains three or more distinct elements, construct f; g 2 A(X) such that f g 6= g f: Exercise 4 If X is …nite, jXj = n; then A(X) is …nite and A(X) = n! = n(n 1):::2:1: Exercise 5 A permutation h is called elementary if it exchanges two elements x; y of A and leaves every other element …xed: h(x) = y; h(y) = x and h(z) = z for all z 6= x; y: If jXj is …nite, every permutation f 2 A(X) can be written as a product of elementary permutations. Exercise 6 If X = f1; 2; :::; ng and a permutation f of X is written as fi1 ; i2 ; :::; in g; where ik = f (k) for k = 1; :::; n; we can check the parity of f as follows. Denote sign(z) = +1 (resp: 1; resp:0) if z is positive (resp: negative, resp: zero). Then de…ne the signature of a permutation f as (f ) = sign (il ik ) . 1 k;l n k<l Compute the signature of the identity permutation and of an elementary permu tation. Then show (f g) = (f ) (g) for all permutations f,g. Deduce that if a permutation is the product of an even ( resp. odd) number of elementary permutations , then it can not be the product of an odd ( resp. even) number of such e. p.s. We can speak of a permutation being ”odd” or ”even”. Show that A(X) is evenly split between these 2 types of permutations. Correspondences: F; with domain X and values in Y associates with every x 2 X a nonempty subset F (x) of Y: Notation: F: X Y: We use the notation insert to distinguish correspondences from functions. The graph of F is graph(F ) = f(x; y) 2 X Y j y 2 F (x)g Alternative de…nition of a correspondence: a subset of X Y such that every xsection is non empty. The range of F is F (X) = [ F (x) x2X Composition of F : X Y and G : Y Z : G F (x) = [ G(y) y2F (x) The composition is associative. 3. Binary relations A binary relation R on X; Y; is a subset of X Y: The conventional notation is: 2 xRy , (x; y) 2 R def X For instance, a function f 2 Y can be viewed as a binary relation. An equivalence relation on X is a binary relation R on X; X that is re‡exive: xRx for all x 2 X symmetric: xRy , yRx for all x; y 2 X transitive: xRy and yRz , xRz for all x; y; z 2 X A partition of X is a family a subsets Xt ; t 2 T (namely an element in (2X )T ) such that Xt \ Xt0 = ; for all t; t0 2 T and [ Xt = X: t2T An equivalence relation on X determines a partition of X by its equiva lence classes; the set of the equivalence classes is denoted X= : Conversely, a partition of X determines an equivalence relation. So the concepts of equiva lence relations and of partition are interchangeable. Example: aRb i¤ “a lives within 100 miles of b” is not an e.r. aRb i¤ “a and b have the same blood type” is an e.r. What about “a and b sat in the same class at least once”? “a and b have a common friend”? Exercise 7 Counting the number pn of distinct partitions of a set X with n elements is not a simple matter. Show the recursive formula: pn (k) = pn 1 (k) k + pn 1 (k 1) where pn (k) is the number of partitions with k elements in a nset. Deduce pn for small n : p2 = 2; p3 = 5; p4 = 15; p5 =?; p6 =? A linear order is a binary relation R on X that is complete, transitive and antisymmetric: complete: xRy or yRx for all x; y 2 X: antisymmetric: xRy and yRx ) x = y; for all x; y: Conventional notation x y; larger than, preferred to, higher than; or x y; lower than, etc. When X is …nite,the set of linear orders can be identi…ed with A(X) and in particular its cardinality is n! . This is not true when X is in…nite. Preordering/Preference relation/rational preferences: a binary relation R on X that is complete, and transitive. Conventional notation x % y; greater than or equal to, preferred or indi¤er ent to, not lower than. 3 The indi¤erence relation s associated with % is a x s y , fx % y and def y % xg: It is an equivalence relation on X: Every preordering % on X can be decomposed into i) the partition of its indi¤erence classes and ii) a linear ordering on X= s; the set of its indi¤erence classes. Conversely, any pair of an equivalence relation s on X and a linear ordering on X= s de…nes a unique preordering on X: Example: X Rn ; pick a function f 2 RX and de…ne x % y by f (x) % f (y): Then % is a preordering. (If % is a preference relation, f is a utility function representing %): If X is …nite, every preference relation on X can be represented by a ( by many) utility functions. This is not true when X is in…nite. The simplest example is the lexicographic ordering . Example: X = Rn ; x = (x1 ; :::; xn ); the lexicographic ordering of Rn is de…ned by: x y , fx1 > y1 g or fx1 = y1 ; and x2 > y2 g or def ... or fxi = yi for i = 1; :::; n 1 and xn > yn g It is a linear ordering of Rn : Exercise 8 Check that the recursive formula to count the number rn of di¤ erent preorderings over a set X with n elements is : rn (k) = k (rn 1 (k)+rn 1 (k 1)). Compute r2 ; r3; r4 ; r5: 4. Cardinals The binary relation over the set of all sets (X; Y ) 2 R , there exists f 2 Y X and onto Y def , there exists g 2 X Y and onetoone on Y is a preordering. Its indi¤erence classes are written jXj and R is written jXj jY j : We call jXj the cardinal of X: Also jXj > jY j means jXj jY j and Not jY j jXj : The equivalence relation \X and Y have the same cardinality" is character ized as follows: jXj = jY j , there is a onetoone f 2 Y X and a onetoone g 2 X Y , there is an onto f 2 Y X and an onto g 2 X Y , there is a bijection h from X into Y: Note: the latter equivalence requires a non trivial proof, given in the Appen dix. If X Y;then jXj jY j: A set X is …nite i¤ for any proper subset X 0 of X, we have jX 0 j < jXj : It is in…nite i¤ for some proper subset X 0 of X; we have jX 0 j = jXj (note that jX 0 j jXj is clear, if X 0 is a subset of X): 4 The set N of natural integers is the set of …nite cardinals. It is an in…nite set. Conventional notation N = f0; 1; 2; :::; n; :::g: The preordering of …nite cardinals is called the natural order of N: Note that jNj is the smallest in…nite cardinal. A set with cardinality jNj is called enumerable. Examples : Q, the set of rational numbers, is enumerable. R; the set of real numbers, is not: jRj > jNj : (this important fact is an easy consequence of the decimal development of real numbers: see Chapter 2). N =N ::: N is enumerable: Nk = jNj :  {z } k tim es Continuum hypothesis: there is no cardinal between jRj and jNj : Exercise 9 Rk = jRj Exercise 10 If X; Y are …nite: jX [ Y j = jXj + jY j jX \ Y j and Y X = jXj jY j Exercise 11 Show the following properties for intervals of real numbers: j[0; 1]j = j[0; 1[j = j]0; 1[j Exercise 12 For any set, …nite or not, 2X > jXj Exercise 13 The lexicographic ordering of R2 cannot be represented by a utility function. Suppose such a function u(x1 ; x2 ) exists, then we can pick for all x1 2 R a rational number r(x1 ) such that u(x1 ; 2) > r(x1 ) > u(x1 ; 1): Deduce a contradiction. Appendix: Two proofs on Cardinals 1. The relation jXj jY j is complete. Given two sets X; Y; consider the set Z of triples z = (A; B; f ) where A X; B Y; and f is a bijection from A into B. The binary relation % on Z : (A; B; f ) % (A0 ; B 0 ; f 0 ) , A0 A; B 0 B; f = f 0 def on A0 ; is clearly transitive (but not complete). Given any subset Z0 of Z on which % is complete, we construct an upper bound ze of Z0 (i.e., ze % z for all e fe) where A e B; z 2 Z0 ) as follows: ze = (A; e = [B and fe = f on A for e = [A; B all z = (A; B; f ) in Z0 : An important result in set theory, known as Zorn’s lemma, says that the relation % must have a maximal element z in Z; namely such that z z holds for no z 2 Z: Let z = (A ; B ; f ): If X A and Y B are both non empty, we can “augment” z by adding a 2 X A to A ; b 2 Y B to B and setting f (a) = b: This contradicts the maximality of z : Therefore A = X or B = Y or both. The former implies jXj jY j and the latter jY j jXj : 5 2. If jXj = jY j ; there exist a bijection from X into Y: a. We prove an auxiliary result …rst. Let A; A1 ; A2 be 3 nested sets, A2 A1 A; and f a bijection from A into A2 : Then there exists a bijection g from A into A1 : We construct a nested sequence A2 A3 A4 ::: as follows: A3 = f (A1 ) () A3 f (A) = A2 ) A4 = f (A2 ) () A4 f (A1 ) = A3 ) A5 = f (A3 ) () A5 f (A2 ) = A4 ) etc. Note that f (A A1 ) = A2 A3 (because f is a bijection) and similarly f (A2 A3 ) = A4 A5 (for the same reason) f (A4 A5 ) = A6 A7 etc. De…ne g 2 AA g = f on A A1 ; A2 A3 ; A4 A5 ; ::: g = identity on A1 A2 ; A3 A4 ; A5 A6 ; ::: Write B = (A A1 ) [ (A2 A3 ) [ (A4 A5 ) [ ::: C = (A1 A2 ) [ (A3 A4 ) [ (A5 A6 ) [ ::: and note that f (B) B; id(C) = C: Therefore g is onetoone on A: More over f (B) [ C = A1 ;therefore g is a bijection from A into A1 : b. We …x X; Y such that jXj jY j and jY j jXj : Thus there are two one toone mappings f; g; where f is from X into Y; and g from Y into X: De…ne A = X; A1 = g(Y ); A2 = (g f )(X): Note that g f is a bijection from A into A2 ; and moreover A2 A1 A: By the auxiliary result, there is a bijection h from A into A1 : Thus h 1 g is the desired bijection from Y into X: 6 Chapter 2: Real Analysis 1. Sequences Z = f:::; n; :::; 2; 1; 0; 1; 2; :::; n; :::g is the set of relative integers. Q = f pq ; p; q 2 Zg is the set of rational numbers. R =] 1; +1[ is the set of real numbers R+ = [0; +1[ is the set of non negative real numbers R++ =]0; +1[ is the set of positive real numbers Integer part: for all x 2 R; this is the unique relative integer p; p 2 Z; such that p x < p + 1; it is written bxc. R is archimedian: 8b 2 R 9p 2 N p > a Exercise 14 8a > 0; a 2 R; 8b 2 R 9p 2 N p:a > b Exercise 15 8" > 0; " 2 R; 9 2 Q : 0 < <" A sequence of real numbers is an element of RN ; and the conventional no tation is (xn )n2N or (x0 ; x1 ; :::; xn ; :::) or (x1 ; :::; xn ; :::): Note: a sequence in an arbitrary set X is an element of X N : The sequence (xn ) is bounded if there is a 2 N such that jxn j a for all n: The sequence (xn ) of real numbers converges to x if we have 8" > 0 9m 2 N 8n m jxn xj " a property that we write lim xn = x n Note: in the de…nition above, we can limit the choice of " to the numbers 1 p ; p = 1; 2::: Thus, for instance, lim xn = x is equivalent to n!1 1 8p 2 N 9m 2 N 8n m jxn xj p Decimal development of a real number. Given x 2 R; we construct inductively a sequence (an ) of integers such that 0 an 9 for all n b10 xc = 10 bxc + a1 b102 xc = 102 bxc + 10 a1 + a2 ... b10n xc = 10n bxc + 10n 1 a1 + 10n 2 a2 + :::10 an 1 + an i.e., for all n the decimal development of the integer b10n xc 10n bxc is a1 ; a2; :::; an : P n The sequence xn = bxc + 10 k ak converges to x. k=1 Conversely if the sequence (an ) of integers between 0 on 9 is such that: Pn limfbxc + 10 k ak g = x n k=1 then this sequence is the decimal development of x; unless an = 9 for all n large enough. A number is rational if andf only if its decimal development is periodic. Exercise 16 Prove the two statements above. 7 Two important applications of the decimal development : Between any two distinct real numbers, there is an in…nityof rational num bers and an in…nity of irrational numbers. jRj > jNj : Operations on limits: limxn + limyn = lim(xn + yn ) n n n (limxn ) (limyn ) = lim(xn yn ) n n n flimxn 6= 0g ) lim( x1n ) = 1 limxn n n n where we assume that the lefthand limits exist, and deduce that the right hand limit exists, too. f(xn ) bounded and limyn = 0g ) limxn yn = 0 n n Exercise 17 The sum (or product) of two non convergent sequences may be convergent. Exercise 18 The product of a bounded sequence and a convergent (resp. a non convergent) sequence, may or may not be convergent. Cauchy sequences: A Cauchy sequence is a sequence (xn ) such that : 8n xn yn ) limxn limyn n n Cauchy sequences: A Cauchy sequence is a sequence (xn ) such that : 8" > 09m 2 N8n; n0 m jxn xn0 j " Fundamental result: a sequence of real numbers is convergent if and only if it is a Cauchy sequence. Note that the result is in a way tautological, because the set of real numbers is de…ned as the set of Cauchy sequences in Q, up to a quotient operation. In Q, a Cauchy sequence may not be convergent. Example: a non periodic decimal development. Exercise 19 Prove this claim. Upper and lower bounds: the subset X of R is bounded above (resp. below ) if there exists an integer a such that 8x 2 X; x a (resp. x a): In this case a is called an upper bound of X (resp. a lower bound ). A key application of the fundamental result : every set X bounded above has a smallest upperbound denoted sup(X) and called the supremum of X. Every set X bounded below has a largest lowerbound inf(X) called the in…mum of X. 8 Exercise 20 The supremum a = sup(X) is characterized by the two following properties: a is an upper bound of X 8" > 0 9x 2 X : x > a " (a similar statement holds for the in…mum). The sequence (xn ) is called nondecreasing (resp. non increasing) if xn xn+1 for all n (resp. xn xn+1 ): A non decreasing (resp. non increasing) sequence that is bounded above (resp. below) converges to the supremum (resp. in…mum) of its values. If In is a nested sequence of closed intervals In = [an ; bn ]; i.e., In+1 In for all n; then the intersection \In is non empty. If, moreover, lim(bn an ) = n n 0; then \In contains exactly one real number. n The lim inf and lim sup of a bounded sequence (xn ) are de…ned as follows: lim sup(xn ) = inff sup xm j n = 1; 2; :::g = lim( sup xm ) m n n m n lim inf(xn ) = supf inf xm j n = 1; 2; :::g = lim( inf xm ) m n n m n Fact: lim inf(xn ) lim sup(xn ) with equality if and only if (xn ) converges, in which case its limit is the common value. Exercise 21 The lim sup and lim inf operators do not commute with addition, or with multiplication. Exercise 22 Prove : lim sup(xn + yn ) lim sup xn + lim sup yn . If xn ; yn are positive prove: lim sup(xn yn ) (lim sup xn ) (lim sup yn ) Similar properties hold for lim inf : What is the composition of lim sup/lim inf with the inverse operator: xn ! 1 xn ? Limit points of a sequence (xn ) : a number y such that 8" > 0 8m 2 N 9n m jxn yj ": A subsequence of the sequence (xn ) 2 RN is a sequence (x'(n) ) = (x'(0) ; x'(1) ; :::; x'(n) ; :::) where ' is a strictly increasing function of N into itself. Every subsequence of a converging sequence converges to the same limit. The number y is a limit point of the sequence (xn ) if and only if there exists a subsequence (x'(n) ) of (xn ) converging to y: If (xn ) is a bounded sequence, lim sup xn and lim inf xn are, respectively, its n n largest and its smallest limit point. Corollary: a bounded sequence is convergent if and only if it has a unique limit point. Corollary: every bounded sequence has at least one limit point. Note: The set of limit points of a sequence (xn ) can be any enumerable subset of R: Given a …nite set f 0 ; :::; K 1 g the sequence xn = k if k is the remainder of (n : K) has the limit points 0 ; :::; K 1 and only those. 9 Exercise 23 Given an arbitrary sequence ( n ) of real numbers, construct a sequence (xn ) of which the limit points are exactly the values of the sequence ( n ) and its limit points. 2. Continuity Interval: an interval of R is a subset I such that 8x; y = fx 2 I and y 2 Ig ) [x; y] I There are nine types of interval depending upon their endpoints being real numbers or 1; and on whether an endpoint belongs to I or not. Limit of a function over a subset: Let A be a subset of R and a be in the closure A of A (that is, a is the limit of at least one sequence in A), we say that f 2 RA converges to b 2 R when x converges to a in A (or simply, the limit of f in A at a is b) and we write lim f (x) = b; the following property: x!a x2A 8" > 0 9 > 0 8x 2 A : jx aj ) jf (x) bj " We say that the limit of f in A at a is +1 if we have: 8n 2 N 9 > 0 8x 2 A : jx aj ) f (x) n We say that the limit of f in A at +1 is b if we have: 8" > 0 9n 2 N 8x 2 A x n ) jf (x) bj " Exercise 24 De…ne similarly lim f (x) = 1; lim f (x) = +1; etc. x!a x! 1 x2A x2A Equivalent formulation: for any sequence (xn ) in A : limxn = a ) limf (xn ) = b n n Examples: lim Logx = 1; lim bxc = 1; lim bxc = 2 x!0 x!2 x!2 x>0 x<2 x>2 If f is monotone (non increasing or non decreasing) then f has a rightlimit and a leftlimit (in A) everywhere: lim f (x) and lim f (x) exist for all a 2 A x!a x!a x<a x>a x2A x2A (in the above statement, the limit may be 1; and a may be replaced by 1): Continuity of a function f 2 RA where A is a bounded interval: [ ; ]; ] ; ]; [ ; [; ] ; [: If a is interior to A : def f is continuous at a , lim f (x) = f (a) , limf (xn ) = f (a) for every (xn ) x!a n s.t. limxn = a n if a = 2 A : def f is continuous at a , lim f (x) = f (a) , limf (xn ) = f (a) for every (xn ) x! n x> s.t. xn and limxn = a n 10 Exercise 25 If a = 2 A write the similar de…nition of limxn = a: n Right continuity, left continuity Same premises as above. If a is interior to A : def f is rightcontinuous at a , lim f (x) = f (a) x! x> def f is leftcontinuous at a , lim f (x) = f (a) x! x< If a = : f is rightcontinuous at a , f is continuous at a left continuity is not de…ned. Operations on continuous functions: the sum, the product, the quotient respect the property of continuity at a given point, that of right continuity, and that of left continuity. The composition of functions respects the continuity property: assume f 2 RA ; g 2 RB ; f (A) B; then {f continuous at a; and g continuous at b = f (a)g ) g f continuous at a: Exercise 26 The composition of rightcontinuous functions may not be right continuous. The supremum (resp. in…mum) of …nitely many functions respects the prop erties of continuity, rightcontinuity, leftcontinuity. fk 2 RA ; k = 1; :::; K; g(x) = sup fk (x) k=1;:::;K ffor all k; fk is continuous at ag ) fg is continuous at ag Examples: f (x) = bxc; A = R; is rightcontinuous everywhere and continuous for non integer values of x: f (x) = sin x if x 6= 0; f (0) = 0; A = R is neither right nor leftcontinuous at 0; is continuous everywhere else. f (x) = 0 if x 2 Q; f (x) = 1 if x 2 R Q; is neither right nor leftcontinuous anywhere. If f is monotone on R; it is continuous at all points, with the exception of at most an enumerable set of points. Exercise 27 Construct a monotone function on R that is discontinuous at every rational number and continuous at every irrational number. What about a function discontinuous at every irrational number and continuous at every rational number ? A closed bounded interval of R is a set [a; b] = fx j a x bg where a; b 2 R: If f is continuous on A (i.e., continuous at every point of A); and [a; b] A; then f ([a; b]) is a closed bounded interval. This crucial observation has two important corollaries. 11 Intermediate value theorem: if f is continuous on the closed, bounded inter val [a; b]; f takes all the values between f (a) and f (b) : 8x 2]f (a); f (b)[ 9c 2]a; b[ f (c) = x Maximum theorem: if f is continuous on the closed, bounded interval [a; b]; f reaches its maximum and its minimum: 9c; d 2 [a; b] : f (c) = sup f (x); f (d) = inf f (x) a x b a x b Variant of the IVT: if f is continuous on the interval I; then f (I) is an interval of R: Exercise 28 We may have I =]a; b[ and f (I) = [c; d]; or I = [a; +1[ and f (I) =]c; d[; etc. An application of the IVT: a polynomial of odd degree has at least one real root. Homeomorphism. Let I be an interval and f 2 RI : The four following properties are equivalent: i) f is a bijection from I into f (I), f is continuous on I and f 1 is continuous on f (I); ii) f is continuous and onetoone on I; iii) f is continuous and strictly monotone (strictly increasing or strictly de creasing), iv) f is strictly monotone and f (I) is an interval. Uniform Continuity: f 2 RA is uniformly continuous on A if we have: 8" > 0 9 > 0 8x; y 2 A : jx yj ) jf (x) f (y)j " Let I be a closed bounded interval and f 2 RI . Thenff is continuous on Ig , ff is uniformly continuous on Ig: On the other hand, if A is not bounded, uniform continuity is a more de manding property than continuity: f (x) = x2 is c: but not u:c: on R+ f (x) = sin x is u:c: on R Exercise 29 Show that f (x) = x1 is uniformly continuous on [1; +1[: Show that f (x) = sin x1 is not uniformly continuous on ]0,1]. sin(x2 ) Is f (x) = sin(x2 ) u.c. on R? What about f (x) = x for x 6= 0; f (0) = 0? 3. Derivatives Given f 2 RA ; where A is an interval, and a an interior point of A; we say that f is di¤ erentiable (or derivable) at a if lim f (x)x fa (a) exists. This limit is x!a x6=a 0 called the derivative of f at a and denoted f (a): 12 Rightderivative of f at a : it is the limit lim f (x)x f (a) a 0 = f+ (a); when this x!a x>a limit exists Leftderivative of f at a : it is the limit lim f (x)x f (a) a = f 0 (a) x!a x<a If f is di¤erentiable (resp. rightdi¤erentiable; resp. leftdi¤erentiable) at a; then it is continuous as well (resp. rightcontinuous, resp. leftcontinuous). Example of a function continuous everywhere but not di¤erentiable at 0 : 1 1 f (x) = (4p + 1) jxj 2 if 2p+1 jxj 2p 1 1 = (4p + 3) jxj + 2 if 2p+2 jxj 2p+1 Check lim sup f (x) f (x) x = 1; lim inf x = 1 x!0 x!0 x6=0 x6=0 Exercise 30 Example of a function continuous over the interval [0,1], but not di¤ erentiable anywhere on [0,1[: x = 0; a1 a2 :::an :::decimal development f (x) = 0; 0a2 0a4 0a6 0::: Operations on di¤erentiable functions: The following operations respects the property “f is di¤erentiable at a" : the sum of functions, their product, their quotient, their composition. The same applies to rightdi¤erentiability, or leftdi¤erentiability, except for the composition. Exercise 31 What is a correct statement for the composition of right or left di¤ erentiable functions? Computation P Prules0 0 ( f i i ) = i fi where i ; i = 1; :::; n; are constant i i P (f1 f2 ::: fn )0 = fi0 (f1 ::: fi 1 fi+1 ::: fn ) i 0 ( f1 )0 = ff2 (f g)0 (x) = f 0 (g(x)) g 0 (x) chain rule If f is di¤erentiable at x and f 0 (x) 6= 0; then f is a bijection from an interval around x (i.e., an interval containing x as an interior point), to an interval around f (x); and the inverse function f 1 is di¤erentiable at y = f (x); with the following derivative: (f 1 )0 (y) = f 0 (f 11 (y)) = f 01(x) Successive derivatives are denoted f 0 ; f "; f (3) ; :::; f (p) ; ::: f is in the class C p ; or simply “f is C p " on an open interval A if f (p) exists everywhere in A and is continuous on A: If f (p) exists at a; we say that f is p times di¤erentiable at a; in this case f is C p 1 in an interval around a; but not necessarily C p : 13 Exercise 32 Example of a function di¤ erentiable everywhere on R but of which the derivative is not a continuous function: f (x) = x2 sin x1 if x 6= 0 f (0) = 0 The set of C p functions is stable (respected) by the sum, the product, the quotient and the composition of functions. Rolle theorem: If f is continuous on the interval [a; b] in R; and di¤erentiable on ]a; b[, there exists a number c 2]a; b[ such that f (b) f (a) = (b a) f 0 (c) This fundamental result has many applications. Fix any interval I of R and a function f continuous on I and di¤erentiable in all interior points of I : 1. f is constant on I if and only if f 0 is the null function, 2. f is non decreasing if and only if f 0 is non negative on the interior of I; 3. f is (strictly) increasing if and only if f 0 is positive on the interior of I: Local extremum: we say that x is a local extremum of f if there is an interval V I around x such that f reaches at x its maximum on V or its minimum on V (we speak of a local maximum, or a local minimum). Under the same premises as above: 4. x is a local extremum of f only if f 0 (x) = 0: Let f; g be two functions continuous on [a; b] and di¤erentiable on ]a; b[: 5.fjf 0 (x)j g 0 (x) for all x 2]a; b[g ) jf (b) f (a)j g(b) g(a); in particular 6.fjf 0 (x)j K for all x 2]a; b[g ) jf (b) f (a)j K (b a) Taylor development: This is a more sophisticated application of Rolle’s theorem, allowing to ap proximate a C n+1 function by a polynomial of degree n: Global Taylor formula: Let f be of class C n on the (bounded) interval [a; b] and di¤erentiable (n+1) times on ]a; b[; then there exists c 2]a; b[ such that: P n (b a)k (k) a)n+1 f (b) = f (a) + k! f (a) + (b(n+1)! f (n+1) (c) k=1 Corollary: if the (n + 1) th derivative satis…es f (n+1) (c) K for all c 2]a; b[; then we have the following approximation: P n (b a)k (k) a)n+1 f (b) f (a) k! f (a) K (b(n+1)! k=1 Local Taylor formula: Let f be n times di¤erentiable at a; then the function (x) de…ned by: 14 P n (x a)k (k) (0) = 0 and for all x 6= 0 : f (x) = f (a) + k! f (a) + (x a)n (x) k=1 satis…es lim (x) = 0: x!0 Applications: 1. A polynomial of degree n coincides with its Taylor development of order n: 2. If f and g have the same k th derivative at a; for k = 0; 1; 2; :::; n; the di¤erence (f g)(x) = (x a)n (x) is a very ‡at function. 3. The exponential function is C 1 and coincides with its “Taylor series” P 1 n x ex = 1 + n! : n=1 Example of a C 1 function that does not coincide with its “Taylor series”: 1 f (x) = e x2 for x 6= 0; f (0) = 0: Note that f is C 1 and f (k) (0) = 0 for any k = 0; 1; 2::: But f is ‡atter than any polynomial at 0: 15 Chapter 3: Linear Algebra 1. The vector space Rn The elements of Rn are called vectors. A linear combination of the vectors x1 ; :::; xp takes the form 1 x1 + ::: + p xp where 1 ; :::; p are real numbers. The set of vectors fx1 ; :::; xp g is linearly independent if for all 1 ; :::; p ; f 1 x1 + ::: + p xp = 0g ) f 1 = ::: = p = 0g: Exercise 33 This implies xi 6= xj if i 6= j; and xi 6= 0 all i (but the converse implication is false). Example: fx1 ; x2 g is linearly independent i¤ x1 ; x2 are not collinear. A vector space (or subspace, or linear subspace) is any subset X Rn stable by addition of vectors and scalar multiplications: for all x; y 2 Rn fx; y 2 Xg ) fx + y 2 Xg for all x 2 Rn ; all 2 R fx 2 Xg ) f x 2 Xg Equivalently, X is a vector space i¤ it contains all linear combinations of its elements. If S is any subset of Rn ; the set E(S) of all linear combinations of vectors in S is a vector space, called the span of S : E(S) = fx 2 Rn 9p 2 N; 9x1 ; :::; xp 2 S; 9 1 ; :::; p 2 R : x = 1 x1 +::: p xp Examples: The span of a single (non zero) vector is a straight line (containing 0), the span of two linearly independent vectors is a plane. Dimension of a vector space. dim X = r reads: the dimension of the vector space X is r(r is a nonnegative integer), and means: X contains a linearly independent set of r vectors but none of r + 1: Conventionally the null space {0} has dimension 0. Examples: straight lines and planes have respective dimensions 1 and 2: Basis of a vector space. A basis of the vector space X is a linearly inde pendent subset of X spanning X: Equivalently, it is a subset of dim X linearly independent vectors of X: If fx1 ; :::; xr g is a basis of X; each vector x in X is uniquely written as a linear combination of the basis: P r x = 1 x1 + ::: + r xr = i xi ; for a unique set of numbers 1 ; :::; r called i=1 the coordinates of x: Example: The standard basis fe1 ; :::; en g of Rn : ei has all its coordinates 0 except 1 at the ith coordinates. Important fact: If the vector space X is spanned by a set of r vectors, then dim X r: Corollary: dim Rn = n: The incomplete basis theorem: If fx1 ; :::; xr g is a linearly independent subset of the vector space X (hence r r = dim X) we can …nd vectors xr+1 ; :::; xr in X such that fx1 ; :::; xr g is a basis of x: In particular X has a basis. The sum of the subsets S; T Rn is de…ned as: 16 S + T = fx 2 Rn 9y 2 S; z 2 T : x = y + zg Remark: S + S 6= 2S: If X is a vector space, we have X + X = X: A sum of vector spaces is a vector space. Direct sum of subspaces: If X1 ; X2 are two vector spaces such that X1 \ X2 = f0g we say that the sum X1 + X2 is direct. In general the sum of vector spaces X1 + ::: + Xp = Y is direct if and only if (de…nition) we have for all x1 ; :::; xp P p fxi 2 Xi for all i = 1; :::; p and xi = 0g ) fxi = 0 for all i = 1; :::; pg i=1 In this case every vector y 2 Y has a unique decomposition as P p y= xi where xi 2 Xi for all i = 1; :::; p i=1 Exercise 34 If the sums X1 +X2 and X2 +X3 are direct, the sum X1 +X2 +X3 is not necessarily direct. Why? Example: if each Xi is a straight line spanned by xi ; the sum X1 + ::: + Xp is direct i¤ the set fx1 ; :::; xp ) is linearly independent. This generalizes: The sum X1 + ::: + Xp is direct i¤ when we pick a basis fxi1 ; :::; xiri g for Xi ; for i = 1; :::; p; the set fx11 ; :::; x1r1 ; x21 ; :::; xp1 ; :::; xprp g is linearly independent. 2. The Euclidean Space Rn The scalar product of two vectors x; y in Rn is de…ned by: Pn x y= xi yi ; where x = (x1 ; :::; xn ); y = (y1 ; :::; yn ) i=1 The scalar product is commutative, and linear in each variable: for all x; x0 ; y 2 Rn ; all ; 2 R : x y = y x; ( x + x0 ) y = x y + x0 y The euclidean norm of a vector x in Rn is de…ned by: P n 1 1 jxj = ( x2i ) 2 = (x x) 2 i=1 Properties of the norm: for all x; y 2 Rn ; all 2 R: i) jxj 0; jxj = 0 if and only if x = 0 ii) j xj = j j jxj iii) jx yj jxj jyj (Schwartz’s inequality) iv) jx + yj jxj + jyj (triangular inequality) v) In (iii) (resp. iv) equality holds if and only if fx; yg are linearly dependent. Say that x and y are orthogonal if x y = 0: For any x; y; we have 2 2 2 jx + yj = jxj + jyj + 2x y; hence 2 2 2 jx + yj = jxj + jyj if and only if x y = 0 (Pythagora’s formula) 2 2 2 2 jx yj + jx + yj = 2(jxj + jyj ) We say that fx1 ; :::; xp g is an orthogonal set of vectors if xi ; xj are pairwise orthogonal for all i; j: 17 An orthogonal set of vectors where each vector is nonzero is linearly inde pendent. An orthonormal set of vectors is an orthogomal set of vectors fx1 ; :::; xp g such that, in addition, jxi j = 1 for all i = 1; :::; p: In particular, it is linearly independent. An important fact: Every vector space of Rn has an orthonormal basis. For instance, the standard basis in an orthonormal basis of Rn : In an orthonormal Pp basis fx1 ; :::; xp g of a subspace, the coordinates of y are y = (y xi ) xi : i=1 The incomplete basis theorem extends to orthonormal basis: if fx1 ; :::; xr g is an orthonormal set of vectors in the vector space X; we can …nd vectors fxr+1 ; :::; xr g in X such that fx1 ; :::; xr g is an orthonormal basis of X: Corollary: If X is a vector space of Rn with dimension r ; r < n; we can …nd an orthonormal set of n r = p vectors z1 ; :::; zp in Rn such that: x 2 X if and only if x z1 = ::: = x zp = 0 In this way, X is represented by p linear equations. Example: A vector space with dimension n 1 is called a hyperplane. It is described by a single equation X = fx 2 Rn =x z = 0g; for some nonzero vector z 2 Rn : 3. Linear operators, matrices The mapping A from Rn to Rm (where n and m are arbitrary integers) is linear if for all x; y 2 Rn ; all 2 R : A(x + y) = Ax + Ay A( x) = Ax This implies that A commutes with linear combinations: P n P n A( i xi ) = i Axi i=1 i=1 The set of linear operators from Rn into Rm is denoted L (Rn , Rm ): It is a vector space for the addition of operators and multiplication by a real number. In the standard basis of Rn and Rm ; a linear operator A is written as a m n matrix A = [aij ] where i is the index of rows, i = 1; :::; m; and j is that of columns, j = 1; :::; n: The jth column of A is formed by the coordinates of Aej so that: P n P n P n P n P n A( j ej ) = j Aej = ( a1j j ; a2j j ; :::; am j ) i=1 j=1 j=1 j=1 j=1 The composition of linear operators respects linearity. It corresponds to the multiplication of matrices. A 2 L(n; m); B 2 L(m; p) ) C = B A 2 L(n; p) or in matrix form A = [aij ]; B = [bki ] ) BA = [ckj ] P n ckj = bki aij k = 1; :::; p ; i = 1; :::; m ; j = 1; :::; n i=1 18 Matrix notation for linear operators: we identify x 2 Rn with the operator in L(1; n) : t ! t x; of which the matrix is the column vector formed by the coordinates of x: Then the equation y = Ax is simply a product of matrices, (m; n) (n; 1) = (m; 1): For instance P the matrix of a linear form y 2 L(Rn ; R) is a row vector and y x = yi xi is both the scalar product of y and x (vectors in Rn ) and the i value of the linear form y at x: Computational properties the multiplication is associative (AB)C=A(BC) and distributive w.r.t. addition and scalar multiplication: ( A + A0 )B = AB + A0 B A( B + B 0 ) = AB + AB 0 Caution: the multiplication of square matrices (i.e., matrices of dimension n n) is not commutative. Exercise: give an example with square matrices of size 2. Norm of a matrix/linear operator: If jxj stands for the euclidean norm in Rn or Rm ; the norm of A 2 L(n; m) is de…ned as: kAk = sup jAxj where x is any vector such that jxj 1: The norm kAk is positive if A is not the zero operator (Ax = 0 for all x); and it is bounded above as follows: A = [aij ] ) kAk n:sup jaij j i;j The norm meets all the properties listed in Section 4.1. below. The kernel of A 2 L(n; m) is the inverse image of 0 : x 2 ker(A) , Ax = 0: The range of A is just ARn (same de…nition as for any mapping). Both the kernel and range are vector spaces, respectively of Rn and Rm : Important fact for all A 2 L(n; m) : dim(ker A) + dim(range A) = n: The dimension of range (A) is called its rank. Fix a linear operator A 2 L(n; m): Then: i) A has full rank (rank A = m) only if n m ii) A is onetoone (Ax = Ay ) x = y) if and only if its kernel is f0g which is equivalent to rank (A) = n: This is possible only if n m: iii) A is a bijection if and only if fn = m and A has full rankg or equivalently fn = m and A is onetoone}. In the latter case we say that A is invertible. We denote by I(n) the set of invertible operators from Rn into itself. A noninvertible element of L(n; n) is called singular. Thus an invertible operator is also called nonsingular. The set I(n) is open in the space L(n; n): If A 2 I(n) and kBk < kA1 1 k ; then A + B 2 I(n): Thus we can add a small matrix to an invertible operator and still avoid singular operators. 4. The Algebra of Square Matrices 19 A linear operator from Rn into itself is written as a square matrix n n: Their set is denoted M (n): Endowed with addition, scalar multiplication, and multiplication it is an algebra. >From the above discussion we get: A 2 I(n) i¤ for some B 2 M (n) : AB = BA = I In this case the matrix B is unique and denoted B = A 1 : A matrix A 2 M (n) is invertible i¤ it is onto (ARn = Rn ), and i¤ it is onetoone (Ax = 0 ) x = 0): Computation rules on inverse matrices: A 2 I(n) ) A 1 2 I(n) AB 2 I(n) ) A; B 2 I(n) ( :A) 1 = 1 A 1 when 2 R f0g; (A 1 ) 1 = A; (AB) 1 = B 1 A 1 Determinants. The transposed of the m n matrix A = [aij ] is the n m matrix t A = [aji ]: We can write any n n matrix as the list of its column vectors A = [a1 ; ::::an ]; all in Rn : Thus t A = [b1 ; ::; bn ] where bi is the i th row vector of A: Denote Sn = A(f1; :::; ng) the set of permutations of f1; :::; ng and recall from Chapter 1 the de…nition of the sign of a permutation : The determinant of the square matrix A is: X det(A) = sign( ) a1 (1) :::an (n) 2Sn For a 2 2 or 3 3 matrix, this gives: det(A) = a11 a22 a12 a21 det(A) = a11 a22 a33 + a12 a23 a31 + a13 a21 a32 a11 a23 a32 a12 a21 a33 a13 a22 a31 Geometric interpretation. The determinant of A = [a1 ; :::; an ] is, up to a P n i sign, the volume of the polytope P = f ia j 0 i 1g; namely the n i=1 dimensional parallelogram with edges a1 ; :::; an : Two key properties. Given A = [a1 ; :::; an ] det(A) 6= 0 , A is non singular , fa1 ; :::; an g are linearly independent det(t A) = det(A); det(I) = 1; det(AB) = det(A) det(B) Computing a determinant. It is useful to think of det(A) = det[a1 ; :::; an ] = det[b1 ; :::; bn ] as a function of the n column vectors aj or of the n row vectors bi : det[a1 ; :::; an ] is linear in each ai det[a1 ; :::; an ] = sign( ) det[a (1) ; :::; a (n) ] for some i; j det[a1 ; :::; an ] = 0 if ai = aj for some i; j (similar properties hold w.r.t. row vectors). Decomposition along a row i (or column j): 20 Denote A i;;j the (n 1) (n 1) matrix where row i and column j have been deleted: n X det(A) = ( 1)k+i aik det(A i ;k ) k=1 Exercise 35 Gram criterion. The p vectors bi ;...,bp in Rq are linearly inde pendent if and only if the p p matrix G = [gij ]; where gij = bi bj is non singular. Two important applications of the determinants are …nding the rank of a matrix of arbitrary size, and computing the inverse of a matrix. The rank of the m n matrix A = [a1 ; :::an ] is the maximal number of its column vectors forming a linearly independent set. Thus A has full rank m (A 2 L(n; m)) if and only if we can …nd m columns forming a nonzero determinant. The rank of any matrix A is the maximal dimension of a non singular submatrix of A: Cramer’s formula for the inverse matrix. Let A = [a1 ; :::an ] be a non singular matrix, A 2 I(n): For all x; y 2 Rn we have: 1 Ay = x , yi = det(a1 ; :::; ai 1 ; x; ai+1 ; :::; an ) det(A) 1 det(A i;j ) Equivalently, A = [bij ] with bij = ( 1)i+j det(A) Solving a system of linear equations. Consider the system Ax = b with unknown x 2 Rn and parameters b 2 Rn and the m n matrix A: a) Find the rank r of A and a nonsingular submatrix A of A; A = [aij ]; 1 i; j r: b) Solve (e.g. by Cramer’s formula) A x = b where b ; x 2 Rr and b is the projection of b on Rf1;:::;rg : c) Check the system Ax e = b where A = [aij ]; 1 i m; 1 j r : if false: the initial system Ax = b has no solution if true: all solutions of the system are x = x+ker(A); where x is the n vector obtained by completing the vector x with (n r) zeros. d) To …nd (n r) linearly independent vectors in ker(A), the kernel of A; we write aj for the jth column of A and aj for its projection on Rf1;:::;rg ; for r + 1 j n: Solve the system A x = aj and write its solution z (j) 2 Rr : Then the vector z(j) de…ned as follows: zi (j) = zi (j) for 1 i r; zj (j) = 1 : zj 0 (j) = 0 for r + 1 j0 n and j 0 6= j 21 is in ker(A) and the set fz(r + 1); :::; z(n)g is a basis of ker(A): Example: 2 3 32 1 1 1 1 4 1 6 0 2 0 2 2 7 6 0 7 A=6 4 2 0 1 1 7 b=6 7 5 5 4 2 5 1 3 1 3 2 2 3 2 3 2 3 1 1 1 1  31 6 0 7 A =4 0 2 0 5 x =4 0 5 Ax = 6 4 2 5 7 2 0 1  34  53 5 6= 3 ) stop 3 2 2 3 1 2 = 53 : A y = a4 ) y 1 = 4 1 5 ; A y = a5 ) y 2 = 4 1 5 1 1 2 1 3 2 3 2 3 3 1 2 6 0 7 6 1 7 6 1 7 6 4 7 6 7 6 7 )x=6  6 3 7 7 + 1 6 1 7 + 2 6 1 7 for any 1 ; 2 6 7 6 7 4 0 5 4 1 5 4 0 5 0 0 1 5. Diagonalization of square matrices Changing the basis for a linear operator If f"1 ; :::; "n g is a basis of Rn and a certain vector has coordinates x in the standard basis of Rn , its coordinates x in the new basis are x = Q 1 x where Q = ["1 ; :::; "n ]: If a certain operator in L(n; n) has a matrix A in the standard basis of Rn ; its matrix in the new basis is A = Q 1 AQ: Matrices A; A are similar if there exists a nonsingular matrix Q such that A = Q 1 AQ: This de…nes an equivalence relation on M (n); respecting the determinant of matrices: A = Q 1 AQ ) det A = det A Hence the determinant is attached to a linear operator, independently of the basis chosen for its matrix representation. Given a linear operator (or its matrix in the standard basis) of Rn ; we seek a basis of Rn where the new matrix is as simple as possible. Eigenvectors, eigenvalues: An eigenvector of A 2 M (n) is any nonzero vector x such that Ax is collinear to x : Ax = x for some 2 R (possibly zero). The scalar is called an eigenvalue, or a characteristic root, of A. The polynomial P ( ) = det(A I) (of degree n) is called the characteristic polynomial of A: It is invariant when A is replaced by a similar matrix. The eigenvalues of A are the zeros of the polynomial P: They are not a¤ected by changes of the basis. 22 Exercise 36 The eigenvalues of A 1 are the inverse of the eigenvalues of A; its eigenvectors are the same as those of A: Exercise 37 Trace operator: the trace of the matrix A is de…ned as the sum of its diagonal terms. It is written tr(A): In the characteristic polynomial, the constant term is det(A) and the coe¢ cient of n 1 is, up to a sign, tr(A): Check tr(A + B) = tr(A) + tr(B) and tr(AB) = tr(BA): Thus two similar matrices have the same trace. CayleyHamilton theorem: if P is the characteristic polynomial of A; then P (A) = 0: Thus An is a linear combination of fI; A; :::; An 1 g: A diagonal n n matrix ( 1 ; :::; n ) has all its o¤diagonal entries zero, and 1 ; :::; n on the diagonal: ( 1 ; ::: n ) = [ 1 e1 ; 2 e2 ; :::; n en ]: The matrix A is diagonalizable if it is similar to a diagonal matrix. A = Q 1 DQ where D = ( 1 ; :::; n ) In this case 1 ; :::; n are the eigenvalues of A, counted with their multiplicity. A matrix is diagonalizable if and only if we can form a basis of Rn containing eigenvectors of A only (some of these eigenvectors may correspond to the same eigenvalue). If A is diagonalizable, computing its successive powers is easy: At = Q 1 ( t1 ; :::; tn ) Q for t = 1; 2::: If 1 ; :::; k are k distinct eigenvalues of a certain matrix A; and x1 ; :::; xn are corresponding eigenvectors, then fx1 ; :::; xn g is linearly independent. Corol lary: if the characteristic polynomial of A has n distinct real roots, then A is diagonalizable. Exercise 38 Prove the above two statements by …rst computing the determinant of the k k matrix B = [bij ]; bij = ji 1 ; 1 i; j k: ab The case of a 2 2 matrix ; with characteristic polynomial P ( ) = cd 2 (a + d) + (ad bc): 1. (a d)2 + 4bc > 0 : P has two distinct real roots 1 ; 2 ; and each root has a unique eigenvector (up to a multiplicative factor), e.g., xi = ( b; a i) i = 1; 2 (or x0i = (d i ; c)): 2. (a d)2 + 4bc < 0 : P has two distinct complex roots + i ; i : The matrix is not diagonalizable, however it is similar to the matrix . A basis in which the operator takes this form is any f"1 ; "2 g where A "i = "i + "j : If we write + i = p(cos + i sin ); we get the successive powers of as follows: t cos(t ) sin(t ) = pt sin(t ) cos(t ) 23 3. (a d)2 + 4bc = 0 : P has a single root of multiplicity 2. If b = c = 0; A = I is diagonal. Otherwise A is not diagonalizable because there is a single eigenvector (up to a multiplicative factor). The matrix A is then similar 0 1 to ; of which the successive powers are easily computed: why? 0 0 Blockdiagonalization theorem: If the characteristic polynomial of the n n matrix A has only distinct (real or complex) roots 1 ; :::; r in R and 1 i 1 ; :::; s i s in C; n = r + s, then A is similar to a “blockdiagonal” matrix with “blocks” 1 ; :::; r 1  1 s  s and ; :::; : Computing the successive powers of a 1 1 s s block diagonal matrix is easy. Corollary 39 Exercise 40 Nilpotent matrices: these are square matrices A such that At = 0 for some integer t: Such a matrix is diagonalizable only if A = 0: Any matrix A = [aij ]; 1 i; j n; where i < j ) aij = 0; is nilpotent. A nilpotent matrix is similar to a matrix N of the form N = [aij ] where only the (n 1) terms ni;i+1 ; 1 i n 1; may be non zero. 6. Symmetric square matrices and quadratic forms A square matrix A; A 2 M (n); is symmetric if A =t A; or equivalently aij = aji for all i; j: Useful properties of the transposition operator: t (AB) =t B t A; y Ax =t Ay x for all x; y 2 Rn If A is a symmetric square matrix, two eigenvectors corresponding to distinct eigenvalues are orthogonal. Moreover, all eigenvalues of A are real (not necessarily distinct) and there exists an orthonormal basis of Rn made up of eigenvectors of A: Call a matrix Q orthogonal if t QQ = I ,t Q = Q 1 , kQxk = kxk for all x; where the norm is the euclidean one. Spectral theorem: If A is a symmetric matrix, there exists a diagonal matrix and an orthogonal matrix Q such that A =t Q Q: Exercise 41 Corollary of the spectral theorem: the euclidean norm of any square matrix A is kAk2 = max j j where is an eigenvalue of t AA: By euclidean norm, we mean kAk2 = max kAxk over kxk 1 where both norms on the right are euclidean. Quadratic forms If A is a square matrix in M (n); the realvalued function with domain Rn ; x ! x Ax is called a quadratic form. In a polynomial of degree two, the terms 24 of degree two de…ne a quadratic form. We can always write a quadratic form with a symmetric matrix B; because x Ax = x 12 (A +t A)x ) B = 21 (A +t A) The quadratic form is positive (resp. de…nite positive) if x Bx 0 for all x (resp. x Bx > 0 for all x 6= 0): By the spectral theorem, this is equivalent to say that all eigenvalues of B are non negative (resp. all strictly positive). Exercise 42 If A is any n x matrix, then t AA de…nes a positive quadratic form. If A is non singular, then tAA is a de…nite positive quadratic form. If B is a de…nite positive quadratic form, then B 1 is too. Two methods for testing positiveness: The criterion of principal minors: in the n n matrix A = [aij ]1 i;j n the principal minors are the n determinants det[aij ]1 i;j k all k = 1; :::; n: The quadratic form de…ned by the symmetric matrix A is positive de…nite if and only if its principal minors are all strictly positive. Gauss’Reduction Algorithm Developing x Ax in the standard basis, one eliminates in sequence all vari ables as in the reduction of the usual degree 2 polynomial in one variable. For instance: x21 + 2x22 + 4x23 6x1 x2 + 2x1 x3 = [x1 + (x3 = 3x2 )]2 7x22 + 3x23 + 6x2 x3 = [x1 + x3 3x2 ]2 7(x2 + 37 x3 )2 + 30 2 7 x3 which is neither positive nor negative. 7. Positive matrices A square matrix A is positive if all its entries aij are positive (aij > 0): PerronFroebemius theorem: a positive matrix A has a unique eigenvalue (A) which has the greatest absolute value. This eigenvalue is positive and simple, and the corresponding eigenvectors are on a positive line (i.e., in Rn++ and Rn ): Asymptotic behavior: for any vector x in Rn+ ; the sequence of vectors 1t (Atx ) converges toward a positive eigenvector associated with (A) lim 1 (A)t Atx = z where Az = (A) z; z 2 Rn++ t This means that the system xt+1 = Axt converges from any initial position in Rn+ toward the canonical direction of the eigenvectors. Example: positive Markov matrix A Markov matrix is a square matrix A = [aij ] such that: n X aij 0 for all i; j; aij = 1 for all j i=1 25 Interpretation: aij is the probability that the system moves from state j (at time t) to state i (at time t + 1) The eigenvalue of largest absolute value of a Markov matrix is 1. Any as sociated eigenvector is non negative. If A is positive, the eigenvectors are on a positive line. Asymptotic behavior: if A is a positive Markov matrix, there is a unique probability vector x that is a …xed point of A : X n fAx = x and xi = 1g ) fx = x g i=1 For any probability vector x0 ; the sequence xt+1 = Axt starting at x0 con verges to x : 26 8. Exercises on Chapter 3 1. The binary relation on Rn Rn “is orthogonal to”is not transitive; is the relation “is not orthogonal to” transitive? 2. Suppose fx1 ; x2 g are linearly independent in Rn and so are fx3 ; x4 g: Suppose also xi xj = 0 for i = 1; 2 and j = 3; 4: Is the set fx1 ; x2 ; x3 ; x4 g linearly independent? 3. In R6 consider 4 vectors: x1 = (a1 ; 0; 0; 0; 0; 0) x2 = (a2 ; b2 ; 0; 0; 0; 0) x3 = (a3 ; b3 ; c3 ; 0; 0; 0) x4 = (a4 ; b4 ; c4 ; d4 ; 0; 0) when is this set orthogonal? linearly independent? 4. Say that U; V are 2 vector spaces in Rn such that u v = 0 for all u 2 U; v 2 V: Prove that the sum U + V is direct. In Exercises 5 to 7, a vector space of Rn is given by a basis, or a spanning subset, and you must …nd an orthogonal basis of the space a system of linear equations representing the subspace an orthogonal system of linear equations representing the subspace 5. (2; 1; 2; 3); (3; 1; 2; 2); ( 1; 2; 4; 5) 6. (1; 1; 0; 2; 1); (0; 2; 3; 1; 1); 1; 0; 2; 1 1) 7. (a; b; a; b) (b; c; b; c) (c; a; c; a) In Exercises 8 to 11, a vector space is given by a system of linear equations, and you must …nd: an orthogonal system of equations for this subspace a basis an orthogonal basis 8. x1 + x2 + x3 = 0; 2x2 x3 = 0 9. x1 + 2x2 + 3x3 + 4x4 + 5x5 = 0 10. 2x2 + x3 x4 = 0 x1 x2 + 3x4 = 0 x2 + x3 = 0 11. a x1 x5 = 0 27 b x2 x1 = 0 c x3 x4 = 0 12. Find a matrix B such that for any 3 3 matrix A; the matrix AB obtains from A by switching columns 1 and 2. Find a matrix C such that for any 3 3 matrix A; the matrix CA obtains from A by switching columns 1 and 2. 13. For any two n n matrices, is the following equation true: (A + B)(A B) = A2 B 2 ? Give the development of (A + B)3 : 14. A square matrix is triangular if all entries below the diagonal are zero. When is a triangular matrix invertible? What are its eigenvalues? 15. Suppose a square matrix A is diagonal, and its diagonal entries are all nonzero and distinct. Characterize the matrices B such that AB = BA: *Generalize to the case where the diagonal entries of A are nonzero but not necessarily distinct.* 16. Say A; B are two n n matrices, B is not the null matrix and AB = 0: Show that A cannot be invertible. What about if B is non null and BA = 0? 17. If A is an invertible n n matrix and B is any n m matrix, show rank (AB) = rank (B): If C is any p n matrix, show rank (CA) = rank (C): 18. If A is n m and B is m p; show rank (AB) min {rank (A); rank (B)g: 19. Show by an example that it is possible for two invertible matrices A; B to satisfy the equation: (A + B) 1 = A 1 + B 1 In exercises 20 to 26, a matrix is given and you must …nd: its rank a basis of its kernel a system of equations describing its range. Then you must give the general form of the solutions of Ax = b : existence and parametric representation. 2 3 0 20. 2 1 1 2 3 2 7 6 0 1 7 21. 6 4 10 2 5 7 3 1 2 3 1 1 1 1 6 1 1 1 1 7 22. 6 4 1 1 1 1 5 7 1 1 1 1 2 3 a b b b 6 b a b b 7 23. 6 4 b b a b 5 7 b b b a 28 2 3 1 1 1 1 24. 4 1 1 1 1 5 1 1 1 1 2 3 4 6 2 6 1 4 3 7 25. 6 4 2 3 1 5 7 1 1 2 2 3 0 3 2 1 6 1 0 3 2 7 26. 6 4 2 1 0 3 5 7 3 2 1 0 27. In this exercise, all matrices are 2 2 a) The sum of two diagonalizable matrices may not be diagonalizable. b) The product of two diagonalizable matrices may not be diagonalizable. c) If A is diagonal and B is diagonalizable, the sum A + B may not be diagonalizable, and the same holds for AB: In exercises 28 to 33 you must diagonalize, or block diagonalize the given matrix A: This means you must compute Q; the change of basis matrix, and 1 the diagonal 2 or block 3 diagonal matrix, such that A = Q Q : 2 0 0 29. 4 0 1 0 5 2 0 2 3 3 1 0 0 30.4 1 2 0 5 2 1 0 1 3 1 0 1 31. 4 0 0 2 5 2 0 1 0 3 1 3 2 32. 4 2 1 3 5 3 2 1 2 3 5 3 3 3 6 3 5 3 3 7 33. 6 4 3 3 5 3 5 7 3 3 3 5 In exercise 2 34 and 3 35 you must also …nd an orthogonal basis of eigenvectors. 1 4 0 34. 4 4 2 1 5 2 0 1 1 3 0 1 0 1 6 1 1 1 0 7 35. 6 4 0 1 0 1 5 7 1 0 1 1 29 Chapter 4: Topology in metric spaces In this chapter, all examples of metric spaces (E; d) involve a subset E of some vector space of …nite dimension. Yet the theory of metric spaces and their topology, applies to arbitrary sets E: 1. Metric spaces, Banach spaces, Hilbert spaces For any set E; a distance on E is a function d from E E into R+ such that for all x; y; z 2 E: d(x; y) = 0 , x = y d(x; y) = d(y; x) d(x; y) d(x; z) + d(z; y) The latter is called the triangular inequality. the pair (E; d) is called a metric space. Two simple properties of distances: d(x1 ; xn ) d(x1 ; x2 ) + d(x2 ; x3 ) + ::: + d(xn 1 ; xn ) jd(x; z) d(y; z)j d(x; y) “A sequence (xn ) in E converges to x 2 E"; is the following property: 8" > 0 9N 2 N 8n N : d(xn ; x) " Exercise 43 A converging sequence has a unique limit. A Cauchy sequence in E is a sequence (xn ) such that 8" > 09N 2 N8n; m>N : d(xn ; xm ) " Exercise 44 A converging sequence is a Cauchy sequence, but the converse is not true in all metric spaces. Why? A metric space (E; d) is complete if every Cauchy sequence is convergent. For instance, with E = Qn and d is the euclidean distance, the metric space (E; d) is not complete. Let E be a vector space. In this chapter, E is a subspace of a vector space Rn ; but the theory of normed spaces is widely used for vector spaces of in…nite dimension. A norm on E is a mapping kk from E into R+ such that: kxk = 0 , x = 0 for all x 2 E kx + yk = kxk + kyk for all x; y 2 E k xk = k k kxk for all x 2 E; 2R 30 The pair (E; kk) is called a normed space. On a normed space, d(x; y) = kx yk is a distance, invariant by translation. A Banach space is a normed space (E; kk); complete for the associated dis tance. Important norms in Rn : P n kxk1 = kxi k for x = (x1 ; :::; xn ) i=1 kxk1 = sup kxi k i=1;:::;n P n kxkp = ( kxpi k where 1 < p < +1 i=1 Note: the triangular inequality for kxkp is often called the Minkowsky in equality. The euclidian norm is the norm kk2 : Exercise 45 Draw the unit balls of kk1 ; kk1 and kkp in R2 All norms of Rn are equivalent in the following sense: if kk is a norm, and p any number, 1 p 1; there exists two positive numbers a; A such that: for all x 2 Rn a kxkp kxk A kxkp This implies that the notions of converging sequence, closed, open and com pact sets, and continuous functions coincide for all norms. In particular, Rn is a Banach space for any norm. We can speak of the topology of Rn : Example: The space S(n; m) of linear operators is a vector space of dimension n m; hence its topology is that of Rn m : Rather than the familiar norms kxkp ;it is convenient to use norms related to the structure of linear operators: f or all A 2 S(n; m) : kAk = sup kAk kxk 1 n m where the two norms in R and R are arbitrary. Recall from Chapter 3 that if n = m and the norm on Rn is euclidean, kAk is the absolute value of the largest eigenvalue of A: A scalar product on the vector space E is a symmetric, bilinear form (x; y) !< x; y >; namely a mapping E E into R, linear in each variable and such that: < x; x >0 and < x; x >= 0 , x = 0 p To a scalar product <,>, we associate the norm kxk = < x; x >: Simple properties of a scalar product: j< x; y >j kxk kyk Schwartz inequality 31 2 2 2 2 kxk + kyk = 12 (kx + yk + kx yk ) 2 2 2 < x; y >= 0 ) kx + yk = kxk + kyk In Rn all scalar products takeP the form < x; y >= (Ax) y; where is the canonical scalar product x y = xi yi ; and A is symmetric positive de…nite i (see Section 3.6). A Hilbert space is a pair (E; <; >) where E is a vector space, and <,> is a scalar product such that E is complete for the associated norm. The space Rn is a Hilbert space for any scalar product. Holder and Minkowsky’s inequalities: If p > 1 is a real number, denote p the number de…ned by p1 + p1 = 1; so P n 1 that p > 1 as well. De…ne kxkp = ( xpi ) p for all p 1: For all x; y 2 Rn and i=1 p > 1; we have: jx yj kxkp kykp (Holder) kx + ykp kxkp + kykp (Minkowsky) 2. Open and closed sets We …x a metric space (E; d) throughout this section. Open ball center x;radius " : 0 B(x; ") = fy 2 E j d(x; y) < "g Closed ball center x;radius " : B(x; ") = fy 2 E j d(x; y) "g Diameter of a subset A E: (A) = sup d(x; y) x;y2A The set A is said to be bounded i¤ its diameter is …nite. The boundedness property is stable: by …nite union by any intersection Exercise 46 Prove (V B(x; r)) 2r in any metric space .Equality holds in any normed vector space, but not in a general metric space. Give a counterexample. 32 The point x 2 E is called a limit point of A E i¤ it satis…es one of the two equivalent properties below: there exists a sequence (yn ) s.t. yn 2 A; for all n and limyn = x; n 8n 2 N B(x; n1 ) \ A 6= ;: The closure A of A is the set of limit points of A: Note that A A. The set A is called closed i¤ A = A: Examples. In any metric space (E; d) a closed ball is a closed set the set E is closed a …nite subset of E is closed a complete set is closed Exercise 47 Prove these claims. If the metric space (E; d) is complete, and A is a subset of E; the metric space (A; d) is complete if and only if A is closed in (E; d) Thus, when A is a subset of Rn and d is a distance on Rn compatible with its canonical topology (e.g., d is a norm) closedness of A and completeness of (A; d) are equivalent properties. Properties of closed sets: A B)A B A[B =A[B A\B A\B and strict inclusion is possible. Any intersection, …nite or not, of closed sets is closed. A …nite union of closed sets is closed. Exercise 48 In Rn ; the sum of two closed sets is not necessarily closed If (E; d) is complete, and (An ) is a decreasing sequence of nonempty closed sets (i.e., An An+1 for all n); then lim (An ) = 0 ) \An = fxg (i.e., the n n intersection of the sets An is a single point). The points x 2 E is called an interior point of A E i¤ it satis…es one of the two equivalent properties below: 9" > 0 : B(x; ") A c = A where Ac = E A is the complement of A x2 The interior A of A is the set of interior points of A: The set A is called open i¤ A = A: Examples: 33 an open ball is an open set, the set E is open. The set A is open if and only if Ac is closed. Properties of open sets: A B)A B A\B =A\B Ac = (A)c ; (Ac ) = (A)c any union, …nite or not, of open sets is open a …nite intersection of open sets is open Exercise 49 In Rn the sum of an open set A and any non empty set B is open. 3. Compact sets We …x a metric space (E; d): A point 2 E is a limit point of the sequence (xn ); xn 2 E; if x is the limit of at least one subsequence of (xn ): Two equivalent de…nitions are: 8" > 0 8n 9m n : d(xn ; x) " x 2 \ fxm j m ng n2N The set of limit points of any sequence is closed. Exercise 50 The closure of the range of the sequence (xn ) contains all limit points of (xn ) but the inclusion may be strict A convergent sequence has a unique limit point (but the converse is not true!). A Cauchy sequence has at most one limit point. A non convergent sequence may have no limit point (example: E = R; xn = n) or the set of its limit points can be the entire set E (example: E = R; xn = the n th rational number, in any enumeration of Q): A subset A E; is called compact, in the metric space (E; d) , if any sequence in A has at least one limit point in A: Properties of compact sets: If A is a compact subset of an arbitrary metric space: A is closed 34 A is bounded A is complete A has a dense sequence (an ); namely a sequence of which every point in A is a limit point. If A is a subset of Rn ; A is compact for the canonical topology if and only if A is closed and bounded. Compactness of subsets (like closedness) is preserved: by any intersection, …nite or not by …nite union The intersection A \ B is compact if A is compact and B is closed. 4. Continuous functions Let f be a mapping from the metric space (E; d) into the metric space (F; ): We say that f is continuous at x 2 E; i¤ it satis…es (one of) the two equivalent properties: i)8(xn ) 2 E N flimxn = xg ) flimf (xn ) = f (x)g n n ii) 8" > 0 9 > 0 8y 2 B(x; ) : f (y) 2 B(f (x); ") A mapping f from (E; d) into (F; ) is continuous if it is continuous at every point x 2 E: This property has three additional equivalent formulations: i) B open in F ) f 1 (B) open in E ii) B closed in F ) f 1 (B) closed in E iii) 8A E : f (A) f (A) The image of an open (resp. closed) subset of (E; d) by a continuous mapping f is not necessarily open (resp. closed) [this fact is already true when E = F = R; see Chapter 2]. Uniform continuity: f mapping (E; d) into (F; ) is said to be uniformly continuous i¤ 8" > 0 9 > 0 8x; y 2 E : d(x; y) ) (f (x); f (y)) " A uniformly continuous function transforms a Cauchy sequence into a Cauchy sequence; this property characterizes uniform continuity. Properties of continuous (resp. uniformly continuous) functions: the composition of functions respects continuity, and uniform continuity linear combinations respect c: and u:c: (when F is a normed vector space) 35 the product of functions E ! R respects c: and u:c: the quotient of functions E ! R respects c: but not u:c: the supremum (or in…mum) of a …nite number of functions E ! R respects c: and u:c: Exercise 51 The supremum of an in…nite family of continuous functions may not be continuous, example: E = [0; 1]; fk (x) = xk ; k = 1; 2... The pointlimit of a sequence of continuous functions may not be continuous. Two important relations between continuity and compactness: If f is a continuous mapping from (E; d) into (F; ) and A is a compact subset of E; then f (A) is a compact subset of F and f is uniformly continuous on A: If f is a bijection from E into F; and is continuous from (E; d) into (F; ); and if (E; d) is compact, then f 1 is continuous from (F; ) into (E; d): 5. Product spaces The product of the metric spaces (E1 ; d1 ); :::; (Ep ; dp ) is (E1 ::: Ep ; d) where d(x; y) = max di (xi ; yi ) for all x; y 2 E i=1;:::;p It is a metric space. Alternative de…nition: we could use any norm kk of Rn and de…ne: d (x; y) = k(di (xi ; yi )k for all x; y 2 E This could give a di¤erent distance but the same topology on E1 ::: Ep : same open and closed sets, same continuous functions, etc. Properties of the product: a sequence (xn ) in E1 ::: Ep converges if and only if each sequence (xin ) converges a subset A1 ::: Ap is bounded/closed/open/compact if and only if each subset Ai is bounded/etc. the projection i from E1 ::: Ep onto Ei is uniformly continuous. if A is bounded/open/compact in E; so is its projection i (A): Exercise 52 The projection of a closed set in E may not be closed in Ei : Ex ample: E = R R A = f(x1 ; x2 ) j x1 x2 = 1g: A is closed in R2 , i (A) is not closed in R: 36 A function f from (F; ) into (E1 ::: Ep ; d1 ::: dp ) is continuous if and only if each function fi = i f is continuous from F into Ei : If the function f from (E1 ::: Ep ; d1 ::: dp ) into (F; ) is continuous, each function f a i ; f a i (xi ) = f (xi ; a i ); from Ei into F , is continuous for all i; all a i 2 E i : Exercise 53 The converse of the above property is false. Example: f (x1 ; x2 ) = xx21+x x2 2 if (x1 ; x2 ) 6= (0; 0) 1 2 f (0; 0) = 0 The graph of a continuous function from (E; d) into (F; ) is closed in (E F; d ): Conversely: if (F; ) is compact, and if the graph of f is closed, then f is continuous. Exercise 54 Prove the two claims above 6. Continuity of correspondences The correspondence f from (E; d) into (F; ) is called upperhemicontinuous at x 2 E i¤ 8V F fV open and f (x) V g )9" > 0 f (B(x; ")) V It is called lowerhemicontinuous at x 2 E i¤ 8V F fV open and f (x) \ V 6= ;g )9" > 0 8y 2 B(x; ") : f (y) \ V 6= ; If f is a function, both de…nitions coincide with the continuity at x: A useful criterion to check u:h:c:: if (F; ) is compact and f (x) is closed in F for all x the correspondence f is upperhemicontinuous on E if and only if its graph is closed. Examples of correspondences uhc and not lhc at 0 f (x) = f0g if x < 0; f (0) = [0; 1]; f (x) = f1g if x > 0 lhc and not uhc at 0 f (x) = [0; 1] if x < 0; f (x) = 1 if x 0 neither lhc nor uhc at 0 f (x) = [0; 1] if x < 0; f (0) = [1; 2]; f (x) = f2g if x > 0 7. Connected sets The metric space (E; d) is said to be connected i¤ E is not the union of two disjoint open sets. 37 Thus (E; d) is connected i¤ the only subsets open and closed are E and ;: The subset A of (E; d) is said to be connected i¤ the metric space (A; d) is connected. The intermediate value theorem: the image of a connected set by a continuous mapping is connected. Example: the connected subsets of R are the intervals, namely the convex subsets of I (x; y 2 I ) [x; y] I): Properties of connected sets: A connected and A not connected is possible A connected and A B A ) B connected (in particular A connected) A connected) If Ak is any family of connected sets such that \ Ak 6= ;; then [ Ak is k2K k2K connected. A convex set in a normed vector space is connected(this follows from the property above). Application: construction of a continuous utility function (Debreu). Let % be a rational preference (complete, transitive) on Rn+ such that: 8x 2 Rn+ x%0 9e 2 Rn+ : 8 1 ; 2 > 0 : 1 e 2 e 8x 2 Rn+ 9 >0:x e 8x 2 Rn+ fy j y % xg and fy j y % xg are closed. Then there exists a continuous utility function un ; from Rn+ into R such that 8x; y 2 Rn+ x  y , u(x) 6 u(y): 38 8. Exercises on Chapter 4 1) A vector space in Rn is open? closed? bounded? connected? Answer the same questions for the subset Nn : 2) If f : (E; d) ! (F; ); maps every converging sequence of E into a con verging sequence of F; then f is continuous. 3) Proof of the equivalence of norms in Rn : Let kk be any norm, we prove below that kk is equivalent to kk 1: Check …rst that this implies the desired conclusion. Show next: supfkxk j x 2 Rn ; kxk1 6 1g is …nite. Deduce that x ! kxk is continuous on (Rn ; kxk1 ): Show next: supfkxk1 j x 2 Rn ; kxk 6 1g is …nite as well. By contradiction: if not, there is a sequence xn such that andkxk1 = 1; lim kxn k = 0: n Conclude. 4) Proof of Hölder’s inequality For any x; y > 0 and any p > 1 show the inequality: Rx Ry 1 x y 6 0 tp 1 dt + 0 t p 1 dt (Hint: draw the graph of t ! tp 1 and distinguish the cases where (x; y) is above or below this graph). Check that we have equality if and only if y = xp 1 : ai bi Applying the above result to x = kak and y = kbk p p deduce Hölder’s inequality for a; b 2 Rn : 5)Proof of Minkowsky’s inequality Fix x; y 2 Rn and p > 1: We have n X n X n X (xi + yi )p = xi (xi + yi )p 1 + yi (xi + yi )p 1 i=1 i=1 i=1 Apply Hölder’s inequality twice to deduce Minkowsky’s. 6) In the metric space (E; d), de…ne the pseudodistance of two subsets A; B as d (A; B) = inf d(x; y) x2A;y2B a) Check the triangular inequality for d : b) Show: A \ B 6= ; ) d (A; B) = 0: Check that the converse implication does not hold, even if A; B are closed sets in Rn : Check that it holds, however, if A; B are compact. = B(x; r) ) d (y; B(x; r)) > d(x; y) r: c) Show for all x; y 2 E; r > 0 : y 2 Show that equality holds if (E; d) is a normed vector space, but not for any metric space. d) Show that x ! d (x; A) is uniformly continuous on (E; d); for any A E: 7) a) Show that d(x; y) = x3 y 3 de…nes a distance on R: b) Show this distance de…nes the same open subsets and the same Cauchy sequences as the usual distance d0 : 39 c) Show that f (x) = x; from (R; do ) into (R; d) is not uniformly continuous, and neither is its inverse function. 8) Let (xn ) be a sequence in the metric space (E; d); such that the three subsequences:(x2n ),(x2n+1 ),(x3n ) are convergent. Show that (xn ) is convergent. Show that if we only assume two of the above subsequences to converge, it does not follow that (xn ) converges. 9) Give an example of a sequence (xn ) in R such that: (xn ) is not convergent for all integer k > 2; the subsequence (xkn ) is convergent. 10) Let (xn ) be a sequence in the metric space (E; d): True, false, explain: a) if limxn = x; then fxg [ fxn j n 2 Ng is a compact/subset of E: n b) if (xn ) has a unique limit point x then limxn = x: n 11) If (E; d) is a metric space, the mapping (x; y) ! d(x; y) from (E 2 ; d d) into R is uniformly continuous. 12) Let f be a continuous function from [0; 1]2 into R: De…ne the function g as follows: g(x) = sup(f (x; y) j 0 6 y 6 xg for all x 2 [0; 1]: Show that oj is uniformly continuous on [0,1]. 13) Show an example of two sets A; B in R such that A \ B is non empty and closed, and the inclusion A \ B A \ B is strict. Show A \ B A \ B; and that the inclusion may be strict. 14) If A is a compact subset of Rn and B a closed subset, show that A + B is closed. If A and B are compact, show that A + B is compact. 15) If A and B are closed subsets of Rn+ ; show that A + B is closed (compare with Exercise 5!). 16) If A; B are subsets of the metric spaces (E; d) and (F; ) respectively, show A B = A B: 17) If (E; d) is a metric space, the diagonal of E E is closed in the product space: = f(x; y) 2 E E j x = yg 18) If Ai Ei is a connected set in the metric space (Ei ; di ); for i = 1; :::; n; show that A1 ::: An is connected in the product space (E1 ::: En ; d): 19) Let Ak be a connected set in (E; d) for k = 1; :::; K: Suppose Ak \Ak+1 6= ; for k = 1; :::; K 1: Show that [ Ak is connected as well. k=1;:::;K 20) If x 2 R2 ; show that R2 fxg is connected. Deduce that there is no continuous bijection from R2 into R: 21) Let (E; d) be a metric space, A an arbitrary subset of E and B a con nected subset of E: Show that if B \ A and B \ (A)c A is non empty. Note: c A A is called the frontier of A and A is the exterior of A: Thus the property reads: fB meets the interior and the exterior of Ag ) fB meets its frontierg. 40
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