Lecture 1 Mathematical Preliminariesecondse.org/wp-content/uploads/2012/02/math_prelim.pdf · Math Econ Preliminaries Sets Logic Sets and Functions Linear Spaces Logic Deals with

IntroductoryMath Econ

PreliminariesSets

Logic

Sets and Functions

Linear Spaces

Lecture 1 Mathematical Preliminaries

A. Banerji

July 25, 2017


PreliminariesSets

Logic

Sets and Functions

Linear Spaces

Outline

1 PreliminariesSetsLogicSets and FunctionsLinear Spaces


PreliminariesSets

Logic

Sets and Functions

Linear Spaces

Basic Concepts• Take as understood the notion of a set. Usually use

upper case letters for sets and lower case ones for theirelements. Notation. a ∈ A, b /∈ A.

• A ⊆ B if every element of A also belongs to B. A = B ifA ⊆ 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 nocommon elements? To retain meaning, we invent theconcept of an empty set, ∅.

• A ∪ ∅ = A, A ∩ ∅ = ∅. For the latter, note that there’s noelement in common between the sets A and ∅ becausethe latter does not have any elements.

• ∅ ⊆ A, for all A. Every element of ∅ belongs to A isvacuously true since ∅ has no elements. This bringsus to some logic.


PreliminariesSets

Logic

Sets and Functions

Linear Spaces

Logic

• Deals with propositions.• A proposition can be either true or false. So, it can have

one of two truth values.• Propositions are denoted by upper case letters.

P = It is raining outside. Q = There is a cloud above.∼ P is the negation of P. (It is not raining outside).

• For any 2 propositions P and Q, there are 4 possibleassignments of truth values.1. P is true and Q is true.2. P is true and Q is false.3. P is false and Q is true.4. P is false and Q is false.


PreliminariesSets

Logic

Sets and Functions

Linear Spaces

Logic• We can combine 2 propositions.• (P & Q).

e.g. “It is raining outside" and "There is a cloud above".The (combined) proposition (P & Q) is true in case 1above: when both P and Q are true. It is false in cases2 to 4.

• (P or Q).e.g. “It is raining outside" or “There is a cloud above"(or both).This combined proposition means P or Q or both. So itis true in cases 1, 2, 3, and false in case 4.

• P ⇒ Q. If the statement P is true, then the statement Qis true. But if P is false, Q may be true or false.For example, on real numbers, let P = x > 0 andQ = x2 > 0. If P is true, so is Q, but Q may be trueeven if P is not, e.g. x = −2. We club these togetherand say P ⇒ Q.


PreliminariesSets

Logic

Sets and Functions

Linear Spaces

Logic

• So, P ⇒ Q is true in cases 1 (both P and Q are true), 3(P false, Q true) and 4 (P false, Q true).And P ⇒ Q is false in case 2 (P true and Q false).

• Another e.g. Let P = x2 < 0 and Q = x = 5. ThenP ⇒ Q is vacuously true. These are cases 3 and 4.

• ∼ 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 ".


PreliminariesSets

Logic

Sets and Functions

Linear Spaces

Logic

Claim. A statement and its contrapositive are equivalent.

Proof.∼ Q ⇒∼ P is true in case 4 (Q false, P false), and cases 1and 3 (Q true, P true or false).∼ Q ⇒∼ P is false in case 2 (Q is true, P is false).So, ∼ Q ⇒∼ P is true or false in exactly those cases forwhich P ⇒ Q is true of false.

DefinitionQ ⇒ 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 theother hand, if some statement and its converse are bothtrue, we say P if and only if Q or P ⇔ Q.


PreliminariesSets

Logic

Sets and Functions

Linear Spaces

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 existsat 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 realhas a positive square root. This is not the same as ‘thereexists y > 0 s.t. for all x > 0, y2 = x ’, which says somenumber y is the common square root of every positive real.


PreliminariesSets

Logic

Sets and Functions

Linear Spaces

Logic

Let Π(a,b) be a property defined on elements a and b insets A and B respectively. LetP : 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 nothold.

Necessary and Sufficient Conditions

Let P ⇒ Q be true. We say Q is necessary for P. Or P issufficient for Q. So, if P ⇔ Q, P is necessary and sufficientfor Q.


PreliminariesSets

Logic

Sets and Functions

Linear Spaces

Sets

Set Difference: A− B = {x |x ∈ A and x /∈ B}.Also called the complement of B relative to A. More familiaris 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 ∈ Aand either x ∈ B or x ∈ C or both. So, either x ∈ (A ∩ B) orx ∈ (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 .


PreliminariesSets

Logic

Sets and Functions

Linear Spaces

SetsArbitrary Unions and IntersectionsLet A be a collection of sets. Then⋃

A∈A = {x |x ∈ A for at least one A ∈ A}⋂A∈A = {x |x ∈ A for every A ∈ A}

Cartesian ProductsWe’ll say (a,b) is an ordered pair if the order of writingthese 2 objects matters: i.e. if (a,b) and (b,a) are not thesame thing. (Think of points on the plane). Alternatively, wecan derive the notion of ordered pair from the moreprimitive notion of a set as follows. Define(a,b) = {{a}, {a,b}}. On the right is a set of 2 sets; the firstof these is the singleton that we want to be ‘first’ in theordered pair. The 2nd is the set of both objects (obviously,{a,b} = {b,a}, so that alone cannot be sufficient to definean ordered pair). Thus (b,a) is defined to be the set{{b}, {a,b}}. Let A and B be sets. We then have


PreliminariesSets

Logic

Sets and Functions

Linear Spaces

Functions

DefinitionThe Cartesian product A× B = {(x , y)|x ∈ A and y ∈ B}.We can formally define a function using the notion ofCartesian product, as follows. Let C,D be 2 sets. A rule ofassignment r is a subset of C ×D s.t. elements of C appearas first coordinates of ordered pairs belonging to r at mostonce. The set A of elements of C appearing as firstcoordinates in r is called the domain of r . The set ofelements of D comprising 2nd coordinates of r is called theimage set of r . Then

DefinitionA function f is a rule of assignment r along with a set B thatcontains the image set of r .A is called the domain of f and the image set of r is calledthe image set or range of f .


PreliminariesSets

Logic

Sets and Functions

Linear Spaces

Functions

We write f : A→ B and think of f as a rule carrying everyelement 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 the plane), let f : <2 → < be defined byf (x1, x2) = x1x2,∀(x1, x2) ∈ <2.3. g : <2 → <2 defined byg(x1, x2) = (x1x2, x1 + x2),∀(x1, x2) ∈ <2.4. Arbitrary Cartesian Products. We first formally definen-tuples in terms of functions. Let X be a set and define thefunction x : {1, ...,n} → X . This function is called an n-tupleof elements of X . It’s image at i ∈ {1, ...,n}, x(i), is writtenas xi . The n-tuple is written as (x1, ..., xn). The order counts.We write X n for the set of all n-tuples of elements of X . Theleading example is <n, n-dimensional Euclidean space.


PreliminariesSets

Logic

Sets and Functions

Linear Spaces

Functions

DefinitionLet A1, ...An be a collection of sets, and let X =

⋃n1 Ai . The

cartesian product of this collection of sets, written asA1 × ...× An or Πn

i=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 ofelements of X is a function x : Z++ → X . (Z++ is the set ofpositive integers). Sequences are written by collectingimages in order. We write x = (x1, x2, ......). This definitiongeneralizes the notion of infinite sequences of real numbers.


PreliminariesSets

Logic

Sets and Functions

Linear Spaces

Functions - Images, Preimages

Let f : A→ B.

DefinitionLet A0 ⊆ A. The image of A0 under f , denoted f (A0), is theset {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 itselements, and the preimage or inverse image of a set is thecollection of all elements in the domain that map into thisset.

ExampleFor the function f (x) = x2, let A0 = [−2,2]. Then,f (A0) = [0,4]. Let B0 = [−2,9]. Then, f−1(B0) = [−3,3].


PreliminariesSets

Logic

Sets and Functions

Linear Spaces

Functions - Injective, SurjectiveFact. 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 anddifference. f only preserves the first two of these. The 3rdholds 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.

Definitionf : 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 isboth of these is called bijective.For example, f : < → < defined by f (x) = x2 is many-to-oneand not surjective. If the domain is <+ instead, then f isinjective, and further if the codomain is <+, then it isbijective.


PreliminariesSets

Logic

Sets and Functions

Linear Spaces

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 ingeneral hold.Now we use the f−1 to mean something different, namelythe inverse function. If f is bijective, then define a functionf−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, supposeb 6= b

′and f−1(b) = f−1(b

′) = a. Then 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. Sof−1 is also surjective; hence it is bijective.


PreliminariesSets

Logic

Sets and Functions

Linear Spaces

Functions

One way to check whether f is bijective uses the following

LemmaLet 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 isbijective and g = h = f−1.

Proof.f injective. Suppose a 6= a

′and f (a) = f (a

′) = b. Then

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 apreimage. Contradiction.


PreliminariesSets

Logic

Sets and Functions

Linear Spaces

Simultaneous EquationsLet f : <n → <m. Fix y ∈ <m. Then the equation f (x) = yrepresents a system of m simultaneous equations in nvariables. This is clear since the equation can be rewrittenas

f1(x1, ..., xn) = y1. . . . . . . . . . . . . . . . . .fm(x1, ..., xn) = ym

where y = (y1, ..., ym), x = (x1, ..., xn),f (x) = (f1(x), ..., fm(x)), ∀x ∈ <n, where for eachi ∈ {1, ...,m}, the component function fi : <n → <. An xwhich satisfies f (x) = y is called a solution to the equationor system of equations. Note that whether the system ofequations has a solution is the same question aswhether f−1({y}) is a nonempty set. Exercise.f : <2 → <2, f (x1, x2) = (x1x2, x1 + x2). For what values of yin the codomain does the equation f (x) = y have asolution?


PreliminariesSets

Logic

Sets and Functions

Linear Spaces

Simultaneous Equations -contour surfaces

One way to look at a solution: the intersection of(hyper)-surfaces. For example, in the exercise above, giveny = (y1, y2), the equations x1x2 = y1 and x1 + x2 = y2.These are 1-dimensional curves in <2, and their intersectionis the set of solutions. For the more general n-variablecase, fi(x1, ..., xn) = yi describes an (n − 1)-dimensionalsurface, and the solution set is the intersection of the msuch surfaces.

DefinitionLet g : <n → < and let y ∈ <. The contour set of g at y ,Cg(y) = {x ∈ <n|g(x) = y}. The upper contour setUg(y) = {x |g(x) ≥ y}. The lower contour setLg(y) = {x |g(x) ≤ y}.Observe that Cg(y) = Ug(y) ∩ Lg(y).


PreliminariesSets

Logic

Sets and Functions

Linear Spaces

Simultaneous Equations inEconomics

First order conditions in optimization problems; generalequilibrium; Nash equilibrium in some problems. All thesecan be cast as solutions to a system of equations F (x) = 0,where F : <n → <n and 0 ∈ <n. General equilbrium issometimes described as a fixed point of a function. (i.e., ifsay f : <n → <n, 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 ofF (x) ≡ f (x)− x , so the the question is really of finding thezeros of F i.e. solving F (x) = 0.Continuity of F is an important player in the existence of asolution. (Just as in 1-dimensional case: if f : < → <, iscontinuous, and f (x1) > 0 > f (x2), then the intermediatevalue theorem assures a solution x ∈ (x1, x2).) For the moregeneral higher dimensional case, in computationaleconomics methods like Gauss-Jacobi make use of1-dimensional solutions to the n different equations in aniterative way to converge to a solution. (see Judd -Numerical Methods in Economics).


PreliminariesSets

Logic

Sets and Functions

Linear Spaces

Relations

Recall that we can define a function as a subset of aCartesian product C × D such that elements of C appear asfirst coordinates of ordered pairs at most once. Relationsare more general in a way.

DefinitionA 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 caseof defining functions, the idea of the definition is to not takeanything 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 afunction, we think of relations in specific ways not directlyrelated to the definition. For example, in consumer theory, ifX is the consumption set and x , y ∈ X , we think of x � ydirectly as “x is preferred to y".


PreliminariesSets

Logic

Sets and Functions

Linear Spaces

Vector Spaces

I want to develop the very small amount of material onvector spaces that we will need. We wish to add vectors andscale them up and down, for which we need scalars.Scalars are drawn from a field. A field is a set K of objectson which we can define two operations (addition andmultiplication) that follow commutative and associative laws,and additive and multiplicative identities (called 0 and 1must exist), as must additive and multiplicative inverses forall objects in K (with the exception of multiplicative inverseof 0). For instance, < is a field.

DefinitionA linear space or vector space V over a field K of scalars isa set of elements (called vectors), along with two binaryoperations (+ and ) (called vector addition and scalarmultiplication) that satisfies the following properties.


PreliminariesSets

Logic

Sets and Functions

Linear Spaces

Vector Spaces contd

DefinitionFor 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 (additiveinverse).

For every v ∈ V and c ∈ K , cv ∈ V ; for every v ∈ V andc,d ∈ K , c(dv) = d(cv) = (cd)v ; the field multiplicativeidentity 1 satisfies 1v = v ∀v ∈ V . Finally, for v ,w ∈ V andc,d ∈ K , c(v + w) = cv + cw , and (c + d)v = cv + dv .


PreliminariesSets

Logic

Sets and Functions

Linear Spaces

Vector Spaces 3

The most salient vector space for us is <n over the field <.Verify that it satisfies all the requirements of a vector space.Let V be a vector space.

Definitionv ∈ V is a linear combination of a set of vectors {v1, . . . , vn}if there are scalars c1, . . . , cn s.t. v =

∑ni=1 civi .

DefinitionSpan({v1, . . . , vn}) is the set of all linear combinations of theset of vectors {v1, . . . , vn}.

Definition{v1, . . . , vn} is a set of linearly dependent vectors if thereexist scalars c1, . . . , cn, not all zero, s.t.

∑ni=1 civi = 0.

.


PreliminariesSets

Logic

Sets and Functions

Linear Spaces

Vector Spaces 4

So if {v1, . . . , vn} is a linearly independent set, then∑ni=1 civi = 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.


PreliminariesSets

Logic

Sets and Functions

Linear Spaces

Unique Representation

TheoremIf {v1, . . . , vn} is a basis in a vector space V , then for everyvector v ∈ V, there exists a unique corresponding set ofscalars {c1, . . . , cn} s.t. v =

∑ni=1 civi .

Proof.Let v ∈ V . Note that since {v1, . . . , vn} is a basis, v can beexpressed as some linear combination of these vectors.Suppose there are two such representations, so there are{c1, . . . , cn} and {d1, . . . ,dn} s.t.v =

∑ni=1 civi =

∑ni=1 divi

So 0 =∑n

i=1 civi −∑n

i=1 divi , i.e. 0 =∑n

i=1(ci − di)vi .But since {v1, . . . , vn} are linearly independent,ci − di = 0, ∀i , or ci = di ∀i .


PreliminariesSets

Logic

Sets and Functions

Linear Spaces

Application: Linear system ofequations

TheoremSuppose A is an n × n matrix of full rank. Then given anyvector b ∈ <n, there exists a unique vector x ∈ <n thatsolves the system of equations Ax = b.

Proof.Let A = (a1 . . . an) in terms of the n columns or columnvectors. Since A has full rank, these columns are linearlyindependent, and therefore they form a basis in <n. So, bythe previous theorem, there is a unique set of scalars{x1, . . . , xn} s.t.

∑ni=1 xiai = b, i.e. Ax = b.


PreliminariesSets

Logic

Sets and Functions

Linear Spaces

Linear Transformations

We used in the proof above the fact that every set of nlinearly independent vectors in <n will in fact span <n, andwe take it as given here.

DefinitionLet V and W be vector spaces. A function T : V →W is alinear transformation if for vectors v1, v2 ∈ V and all scalarsc1, c2, T (c1v1 + c2v2) = c1T (v1) + c2T (v2).So an m × n matrix A is a linear transformation from <n to<m.

Documents

Lecture 1 Mathematical Preliminariesecondse.org/wp-content/uploads/2012/02/math_prelim.pdf · Math Econ Preliminaries Sets Logic Sets and Functions Linear Spaces Logic Deals with