Lecture 1 Mathematical Preliminaries
A. Banerji
July 26, 2016
(Delhi School of Economics) Introductory Math Econ July 26, 2016 1 / 27 Outline
1 Preliminaries Sets Logic Sets and Functions Linear Spaces
(Delhi School of Economics) Introductory Math Econ July 26, 2016 2 / 27 Preliminaries Sets Basic Concepts
Take as understood the notion of a set. Usually use upper case letters for sets and lower case ones for their elements. Notation. a ∈ A, b ∈/ A. A ⊆ B if every element of A also belongs to B. A = B if A ⊆ B and B ⊆ A are both true. A ∪ B = {x|x ∈ A or x ∈ B}. The ‘or’ is inclusive of ‘both’. A ∩ B = {x|x ∈ A and x ∈ B}. What if A and B have no common elements? To retain meaning, we invent the concept of an empty set, ∅. A ∪ ∅ = A, A ∩ ∅ = ∅. For the latter, note that there’s no element in common between the sets A and ∅ because the latter does not have any elements. ∅ ⊆ A, for all A. Every element of ∅ belongs to A is vacuously true since ∅ has no elements. This brings us to some logic.
(Delhi School of Economics) Introductory Math Econ July 26, 2016 3 / 27 Preliminaries Logic Logic
Statements or propositions must be either true or false. P ⇒ Q. If the statement P is true, then the statement Q is true. But if P is false, Q may be true or false. For example, on real numbers, let P = x > 0 and Q = x2 > 0. If P is true, so is Q, but Q may be true even if P is not, e.g. x = −2. We club these together and say P ⇒ Q. Let P = x2 < 0 and Q = x = 5. Then P ⇒ Q is vacuously true. ∼ Q ⇒∼ P is the contrapositive of P ⇒ Q. For example: If x3 ≤ 0, then x ≤ 0 is the contrapositive of “if x > 0, then x3 > 0 ".
(Delhi School of Economics) Introductory Math Econ July 26, 2016 4 / 27 Preliminaries Logic Logic
Claim. A statement and its contrapositive are equivalent. Proof. Suppose P ⇒ Q is true. Suppose Q is false. Then P must be false, for if not, then via P ⇒ Q, Q would be true, contradicting our assumption that Q is false. On the other hand, suppose P ⇒ Q is false. It follows that P is true and Q is false (for if P is false, then P ⇒ Q is vacuously true). But then, ∼ Q ⇒∼ P cannot be true.
Definition Q ⇒ P is called the converse of the statement P ⇒ Q. No relationship between a statement and its converse. e.g., if x > 0 then x2 > 0 is true, but its converse is not. On the other hand, if some statement and its converse are both true, we say P if and only if Q or P ⇔ Q. (Delhi School of Economics) Introductory Math Econ July 26, 2016 5 / 27 Preliminaries Logic Quantifiers and Negation
2 logical quantifiers: ‘for all’ and ‘there exists’. P : For all a ∈ A, property Π(a) holds. The negation of P (i.e. ∼ P) is the statement: There exists at least one a ∈ A s.t. property Π(a) does not hold. e.g. P : For every x ∈ <, x2 > 0. The negation of P : There exists x ∈ < s.t. x2 ≤ 0.
Q : There exists b ∈ B s.t. property Θ(b) holds. ∼ Q : For all b ∈ B, property Θ(b) does not hold. Note. Order of quantifiers matters. e.g. ‘for all x > 0, there exists y > 0 s.t. y 2 = x’, says that every positive real has a positive square root. This is not the same as ‘there exists y > 0 s.t. for all x > 0, y 2 = x’, which says some number y is the common square root of every positive real.
(Delhi School of Economics) Introductory Math Econ July 26, 2016 6 / 27 Preliminaries Logic Logic
Let Π(a, b) be a property defined on elements a and b in sets A and B respectively. Let P : For every a ∈ A there exists b ∈ B s.t. Π(a, b) holds. Then, ∼ P : There exists a ∈ A s.t. for all b ∈ B, Π(a, b) does not hold.
Necessary and Sufficient Conditions Let P ⇒ Q be true. We say Q is necessary for P. Or P is sufficient for Q. So, if P ⇔ Q, P is necessary and sufficient for Q.
(Delhi School of Economics) Introductory Math Econ July 26, 2016 7 / 27 Preliminaries Sets and Functions Sets
Set Difference: A − B = {x|x ∈ A and x ∈/ B}. Also called the complement of B relative to A. More familiar is the notion of the universal set X, and X − B = Bc. Some set-theoretic ‘laws’ Distributive Laws (i) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) (ii) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) Proof. (i) Suppose x ∈ LHS. So, x ∈ A and x ∈ (B ∪ C). So x ∈ A and either x ∈ B or x ∈ C or both. So, either x ∈ (A ∩ B) or x ∈ (A ∩ C) (or both). Converse? DeMorgan’s Laws (i) A − (B ∪ C) = (A − B) ∩ (A − C) or more familiarly, (B ∪ C)c = Bc ∩ Cc (ii) A − (B ∩ C) = (A − B) ∪ (A − C) or (B ∩ C)c = Bc ∪ Cc. (Delhi School of Economics) Introductory Math Econ July 26, 2016 8 / 27 Preliminaries Sets and Functions Sets
Arbitrary Unions and Intersections Let A be a collection of sets. Then S A∈A = {x|x ∈ A for at least one A ∈ A} T A∈A = {x|x ∈ A for every A ∈ A} Cartesian Products We’ll say (a, b) is an ordered pair if the order of writing these 2 objects matters: i.e. if (a, b) and (b, a) are not the same thing. (Think of points on the plane). Alternatively, we can derive the notion of ordered pair from the more primitive notion of a set as follows. Define (a, b) = {{a}, {a, b}}. On the right is a set of 2 sets; the first of these is the singleton that we want to be ‘first’ in the ordered pair. The 2nd is the set of both objects (obviously, {a, b} = {b, a}, so that alone cannot be sufficient to define an ordered pair). Thus (b, a) is defined to be the set {{b}, {a, b}}. Let A and B be sets. We then have
(Delhi School of Economics) Introductory Math Econ July 26, 2016 9 / 27 Preliminaries Sets and Functions Functions
Definition The Cartesian product A × B = {(x, y)|x ∈ A and y ∈ B}.
We can formally define a function using the notion of Cartesian product, as follows. Let C, D be 2 sets. A rule of assignment r is a subset of C × D s.t. elements of C appear as first coordinates of ordered pairs belonging to r at most once. The set A of elements of C appearing as first coordinates in r is called the domain of r. The set of elements of D comprising 2nd coordinates of r is called the image set of r. Then Definition A function f is a rule of assignment r along with a set B that contains the image set of r.
A is called the domain of f and the image set of r is called the image set or range of f . (Delhi School of Economics) Introductory Math Econ July 26, 2016 10 / 27 Preliminaries Sets and Functions Functions
We write f : A → B and think of f as a rule carrying every element a ∈ A to exactly one element b ∈ B. Examples. 1. f : < → < defined by f (x) = x2, ∀x ∈ <. 2. Using the notation <2 for the Cartesian product < × < (representing 2 2 the plane), let f : < → < be defined by f (x1, x2) = x1x2, ∀(x1, x2) ∈ < . 2 2 2 3. g : < → < defined by g(x1, x2) = (x1x2, x1 + x2), ∀(x1, x2) ∈ < . 4. Arbitrary Cartesian Products. We first formally define n-tuples in terms of functions. Let X be a set and define the function x : {1, ..., n} → X. This function is called an n-tuple of elements of X. It’s image at i ∈ {1, ..., n}, x(i), is written as xi . The n-tuple is written as n (x1, ..., xn). The order counts. We write X for the set of all n-tuples of elements of X. The leading example is (Delhi School of Economics) Introductory Math Econ July 26, 2016 11 / 27 Preliminaries Sets and Functions Functions Definition Sn Let A1, ...An be a collection of sets, and let X = 1 Ai . The cartesian n product of this collection of sets, written as A1 × ... × An or Πi=1Ai , is the set of all n-tuples (x1, ..., xn) such that xi ∈ Ai , ∀i ∈ {1, ..., n}. 5. Let X be a set. A sequence or infinite sequence of elements of X is a function x : Z++ → X.(Z++ is the set of positive integers). Sequences are written by collecting images in order. We write x = (x1, x2, ...... ). This definition generalizes the notion of infinite sequences of real numbers. (Delhi School of Economics) Introductory Math Econ July 26, 2016 12 / 27 Preliminaries Sets and Functions Functions - Images, Preimages Let f : A → B. Definition Let A0 ⊆ A. The image of A0 under f , denoted f (A0), is the set {b|b = f (a), for some a ∈ A0}. −1 Let B0 ⊆ B. The preimage of B0 under f , f (B0) = {a|f (a) ∈ B0}. So the image of a set is the collection of images of all its elements, and the preimage or inverse image of a set is the collection of all elements in the domain that map into this set. Example 2 For the function f (x) = x , let A0 = [−2, 2]. Then, f (A0) = [0, 4]. Let −1 B0 = [−2, 9]. Then, f (B0) = [−3, 3]. (Delhi School of Economics) Introductory Math Econ July 26, 2016 13 / 27 Preliminaries Sets and Functions Functions - Injective, Surjective Fact. Let f : A → B, A0, A1 ⊆ A, B0, B1 ⊆ B. Then (i) −1 −1 B0 ⊆ B1 ⇒ f (B0) ⊆ f (B1). −1 −1 −1 (ii) f (B0 ∗ B1) = f (B0) ∗ f (B1), where ∗ can be ∪, ∩, −. i.e., f −1 preserves set inclusion, union, intersection and difference. f only preserves the first two of these. The 3rd holds with ⊆ and the 4th with ⊇. To find counterexamples, many-to-one functions (to which we now move) are helpful. Proofs of the above claims are homework. Definition 0 0 f : A → B is injective (one-to-one) if [f (a) = f (a )] ⇒ [a = a ]. It is surjective (onto) if for every b ∈ B, there exists a ∈ A s.t. b = f (a).A function that is both of these is called bijective. For example, f : < → < defined by f (x) = x2 is many-to-one and not surjective. If the domain is <+ instead, then f is injective, and further if the codomain is <+, then it is bijective. (Delhi School of Economics) Introductory Math Econ July 26, 2016 14 / 27 Preliminaries Sets and Functions Functions, Inverse −1 Fact. Let f : A → B, A0 ⊆ A, B0 ⊆ B. Then (i)f (f (A0)) ⊇ A0; equality holds if f is injective. −1 (ii) f (f (B0)) ⊆ B0; equality holds if f is surjective. You should also show by examples that equality does not in general hold. Now we use the f −1 to mean something different, namely the inverse function. If f is bijective, then define a function f −1 : B → A by f −1(b) = a if a is the unique element of A s.t. f (a) = b. Note that f −1 is 0 0 also bijective. Indeed, suppose b 6= b and f −1(b) = f −1(b ) = a. Then 0 f (a) = b and f (a) = b which is not possible. So f −1 is injective. Moreover, for every a ∈ A, there is a b ∈ B s.t. b = f (a), since f is a function. So, there is a b ∈ B s.t. f −1(b) = a. So f −1 is also surjective; hence it is bijective. (Delhi School of Economics) Introductory Math Econ July 26, 2016 15 / 27 Preliminaries Sets and Functions Functions One way to check whether f is bijective uses the following Lemma Let f : A → B. If there are 2 functions g, h from B to A s.t. g(f (a)) = a, ∀a ∈ A and f (h(b)) = b∀b ∈ B, then f is bijective and g = h = f −1. Proof. 0 0 f injective. Suppose a 6= a and f (a) = f (a ) = b. Then 0 0 g(f (a)) = g(b) = g(f (a )). So g(b) cannot be equal to both a and a . Contradiction. f is also surjective. Indeed, suppose there is a b ∈ B with no preimage under f . However, we require f (h(b)) = b. This implies h(b) is a preimage. Contradiction. (Delhi School of Economics) Introductory Math Econ July 26, 2016 16 / 27 Preliminaries Sets and Functions Simultaneous Equations Let f : f1(x1, ..., xn) = y1 ...... fm(x1, ..., xn) = ym where y = (y1, ..., ym), x = (x1, ..., xn), n f (x) = (f1(x), ..., fm(x)), ∀x ∈ < , where for each i ∈ {1, ..., m}, the n component function fi : < → <. An x which satisfies f (x) = y is called a solution to the equation or system of equations. Note that whether the system of equations has a solution is the same question as whether f −1({y}) is a nonempty set. Exercise. f : <2 → <2, f (x1, x2) = (x1x2, x1 + x2). For what values of y in the codomain does the equation f (x) = y have a solution? (Delhi School of Economics) Introductory Math Econ July 26, 2016 17 / 27 Preliminaries Sets and Functions Simultaneous Equations - contour surfaces One way to look at a solution: the intersection of (hyper)-surfaces. For example, in the exercise above, given y = (y1, y2), the equations 2 x1x2 = y1 and x1 + x2 = y2. These are 1-dimensional curves in < , and their intersection is the set of solutions. For the more general n-variable case, fi (x1, ..., xn) = yi describes an (n − 1)-dimensional surface, and the solution set is the intersection of the m such surfaces. Definition Let g : Observe that Cg(y) = Ug(y) ∩ Lg(y). (Delhi School of Economics) Introductory Math Econ July 26, 2016 18 / 27 Preliminaries Sets and Functions Simultaneous Equations in Economics First order conditions in optimization problems; general equilibrium; Nash equilibrium in some problems. All these can be cast as solutions to a system of equations F(x) = 0, where F : (Delhi School of Economics) Introductory Math Econ July 26, 2016 19 / 27 Preliminaries Sets and Functions Relations Recall that we can define a function as a subset of a Cartesian product C × D such that elements of C appear as first coordinates of ordered pairs at most once. Relations are more general in a way. Definition A relation on a set X is a subset of X 2, i.e. ⊆ X × X. Conventionally, if (x, y) ∈, we write x y. As in the case of defining functions, the idea of the definition is to not take anything more than the meaning of a set to be understood, and to successively define things in terms of it. (Set -> Cartesian Product -> Relation). But as in the case of a function, we think of relations in specific ways not directly related to the definition. For example, in consumer theory, if X is the consumption set and x, y ∈ X, we think of x y directly as “x is preferred to y". (Delhi School of Economics) Introductory Math Econ July 26, 2016 20 / 27 Preliminaries Linear Spaces Vector Spaces I want to develop the very small amount of material on vector spaces that we will need. We wish to add vectors and scale them up and down, for which we need scalars. Scalars are drawn from a field.A field is a set K of objects on which we can define two operations (addition and multiplication) that follow commutative and associative laws, and additive and multiplicative identities (called 0 and 1 must exist), as must additive and multiplicative inverses for all objects in K (with the exception of multiplicative inverse of 0). For instance, < is a field. Definition A linear space or vector space V over a field K of scalars is a set of elements (called vectors), along with two binary operations (+ and ) (called vector addition and scalar multiplication) that satisfies the following properties. (Delhi School of Economics) Introductory Math Econ July 26, 2016 21 / 27 Preliminaries Linear Spaces Vector Spaces contd Definition For every v, w ∈ V , v + w ∈ V (vector addition is closed); v + w = w + v (commutes); for every v, w, z ∈ V , (v + w) + z = v + (w + z) (associates); there exists a vector (called 0) s.t. v + 0 = v, ∀v ∈ V (vector-additive identity); and for all v ∈ V , there exists w s.t. v + w = 0 (additive inverse). For every v ∈ V and c ∈ K , cv ∈ V ; for every v ∈ V and c, d ∈ K , c(dv) = d(cv) = (cd)v; the field multiplicative identity 1 satisfies 1v = v ∀v ∈ V . Finally, for v, w ∈ V and c, d ∈ K , c(v + w) = cv + cw, and (c + d)v = cv + dv. (Delhi School of Economics) Introductory Math Econ July 26, 2016 22 / 27 Preliminaries Linear Spaces Vector Spaces 3 The most salient vector space for us is v ∈ V is a linear combination of a set of vectors {v1,..., vn} if there Pn are scalars c1,..., cn s.t. v = i=1 ci vi . Definition Span({v1,..., vn}) is the set of all linear combinations of the set of vectors {v1,..., vn}. Definition {v1,..., vn} is a set of linearly dependent vectors if there exist scalars Pn c1,..., cn, not all zero, s.t. i=1 ci vi = 0. . (Delhi School of Economics) Introductory Math Econ July 26, 2016 23 / 27 Preliminaries Linear Spaces Vector Spaces 4 Pn So if {v1,..., vn} is a linearly independent set, then i=1 ci vi = 0 implies ci = 0, ∀i = 1,..., n. Definition {v1,..., vn} is a basis in a vector space V if (i) {v1,..., vn} are linearly independent. (ii) Span({v1,..., vn}) = V . For example, check that (i) {(1, 0), (0, 1)} and {(1, 1), (1, 2)} are two different bases in <2. (Delhi School of Economics) Introductory Math Econ July 26, 2016 24 / 27 Preliminaries Linear Spaces Unique Representation Theorem If {v1,..., vn} is a basis in a vector space V , then for every vector v ∈ V , there exists a unique corresponding set of scalars {c1,..., cn} Pn s.t. v = i=1 ci vi . Proof. Let v ∈ V . Note that since {v1,..., vn} is a basis, v can be expressed as some linear combination of these vectors. Suppose there are two such representations, so there are {c1,..., cn} and {d1,..., dn} s.t. Pn Pn v = i=1 ci vi = i=1 di vi Pn Pn Pn So 0 = i=1 ci vi − i=1 di vi , i.e. 0 = i=1(ci − di )vi . But since {v1,..., vn} are linearly independent, ci − di = 0, ∀i, or ci = di ∀i. (Delhi School of Economics) Introductory Math Econ July 26, 2016 25 / 27 Preliminaries Linear Spaces Application: Linear system of equations Theorem Suppose A is an n × n matrix of full rank. Then given any vector b ∈ Proof. Let A = (a1 ... an) in terms of the n columns or column vectors. Since A has full rank, these columns are linearly independent, and therefore they form a basis in (Delhi School of Economics) Introductory Math Econ July 26, 2016 26 / 27 Preliminaries Linear Spaces Linear Transformations We used in the proof above the fact that every set of n linearly independent vectors in Definition Let V and W be vector spaces. A function T : V → W is a linear transformation if for vectors v1, v2 ∈ V and all scalars c1, c2, T (c1v1 + c2v2) = c1T (v1) + c2T (v2). So an m × n matrix A is a linear transformation from (Delhi School of Economics) Introductory Math Econ July 26, 2016 27 / 27