Lecture 1 Mathematical Preliminaries
A. Banerji July 26, 2016
(Delhi School of Economics)
Introductory Math Econ
- July 26, 2016
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Outline
1
Preliminaries
Sets Logic Sets and Functions Linear Spaces
(Delhi School of Economics)
Introductory Math Econ
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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
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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
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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.
5 / 27
(Delhi School of Economics)
Introductory Math Econ
July 26, 2016
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. y2 = 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, y2 = 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
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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
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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
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Preliminaries
Sets and Functions
Sets
Arbitrary Unions and Intersections
Let A be a collection of sets. Then
S
= {x|x ∈ A for at least one A ∈ A}
A∈A A∈A
T
= {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
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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
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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
2. Using the notation < for the Cartesian product < × < (representing
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the plane), let f : < → < be defined by f(x1, x2) = x1x2, ∀(x1, x2) ∈ < .
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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 (x1, ..., xn). The order counts. We write Xn for the set of all n-tuples of
n
elements of X. The leading example is < , n-dimensional Euclidean space.
(Delhi School of Economics)
Introductory Math Econ
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Preliminaries
Sets and Functions
Functions
Definition
S
Let A1, ...An be a collection of sets, and let X = n Ai. The cartesian
1
product of this collection of sets, written as A1 × ... × An or Πni=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
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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}.
Let B0 ⊆ B. The preimage of B0 under f, f−1(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
For the function f(x) = x2, let A0 = [−2, 2]. Then, f(A0) = [0, 4]. Let B0 = [−2, 9]. Then, f−1(B0) = [−3, 3].
(Delhi School of Economics)
Introductory Math Econ
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Preliminaries
Sets and Functions
Functions - Injective, Surjective
Fact. Let f : A → B, A0, A1 ⊆ A, B0, B1 ⊆ B. Then (i)
B0 ⊆ B1 ⇒ f−1(B0) ⊆ f−1(B1).
(ii) f−1(B0 ∗ B1) = f−1(B0) ∗ f−1(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
f : A → B is injective (one-to-one) if [f(a) = f(a0 )] ⇒ [a = a0 ]. 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
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Preliminaries
Sets and Functions
Functions, Inverse
Fact. Let f : A → B, A0 ⊆ A, B0 ⊆ B. Then (i)f−1(f(A0)) ⊇ A0; equality holds if f is injective. (ii) f(f−1(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 also bijective. Indeed, suppose b = b0 and f−1(b) = f−1(b0 ) = a. Then f(a) = b and f(a) = b0 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
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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.
f injective. Suppose a = a0 and f(a) = f(a0 ) = b. Then g(f(a)) = g(b) = g(f(a0 )). So g(b) cannot be equal to both a and a0 . 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
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Preliminaries
Sets and Functions
Simultaneous Equations
- n
- m
- m
Let f : < → < . Fix y ∈ < . Then the equation f(x) = y represents a system of m simultaneous equations in n variables. This is clear since the equation can be rewritten as
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
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whether f−1({y}) is a nonempty set. Exercise. f : < → < ,
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
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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
n
Let g : < → < and let y ∈ <. The contour set of g at y,
n
Cg(y) = {x ∈ < |g(x) = y}. The upper contour set Ug(y) = {x|g(x) ≥ y}. The lower contour set Lg(y) = {x|g(x) ≤ y}.
Observe that Cg(y) = Ug(y) ∩ Lg(y).
(Delhi School of Economics)
Introductory Math Econ
- July 26, 2016
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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
- n
- n
- n
to a system of equations F(x) = 0, where F : < → < and 0 ∈ < . General equilbrium is sometimes described as a fixed point of a
- n
- n
function. (i.e., if say f : < → < , x is a fixed point if it satisfies f(x) = x). But x is a fixed point of f if and only if it is a zero of F(x) ≡ f(x) − x, so the the question is really of finding the zeros of F i.e. solving F(x) = 0. Continuity of F is an important player in the existence of a solution. (Just as in 1-dimensional case: if f : < → <, is continuous, and f(x1) > 0 > f(x2), then the intermediate value theorem assures a solution x ∈ (x1, x2).) For the more general higher dimensional case, in computational economics methods like Gauss-Jacobi make use of 1-dimensional solutions to the n different equations in an iterative way to converge to a solution. (see Judd - Numerical Methods in Economics).
(Delhi School of Economics)
Introductory Math Econ
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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 X2, 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".